Forces - Nelson Phys12 Textbook Solutions

Unit 1 Forces and Motion: DynamicsARE YOU READY? (Pages 2–3) Knowledge and Understanding Scalar quantities include distance (metre, 5.0 m), time (second, 15 s), mass (kilogram, 65 kg), and frequency (hertz, 60 Hz). Vector quantities include velocity (metres per second, 15 m/s [E]), displacement (metre, 6.5 m [S]), acceleration (metres per second squared, 9.8 m/s2 [down]), and force (newton, 25 N [forward]). 2. (a) Both masses will hit the floor at the same time since the speed at which an object falls is independent of mass, and is related only to acceleration due to gravity (neglecting air resistance). (b) 1. (c) m = 20 g = 0.02 kg Fg = ? Fg = mg = (0.02 kg)(9.8 N/kg [down]) Fg = 0.2 N [down] The weight of the 20-g mass is 0.2 N [down]. (d) One example is the force of Earth pulling downward on the 20-g mass and the force of the 20-g mass pulling upward on Earth. GM E mMoon The magnitude of the force of gravity between Earth and the Moon depends linearly on the masses of 3. FG = r2 Earth and the Moon, and depends inversely as the square of the distance between the centres of Earth and the Moon. 4. (a) Kinematics is the study of motion (e.g., analyzing motion with constant acceleration). Dynamics is the study of the causes of motion (e.g., analyzing forces by applying Newton’s three laws of motion). total distance change of position . Average velocity is a vector quantity, vav = . (b) Average speed is a scalar quantity, v = av total time time interval (c) Static friction is a force that acts to prevent a stationary object from starting to move. Kinetic friction is a force that acts against a moving object. For a given situation, kinetic friction tends to be less than maximum static friction. (d) Helpful friction is needed in many cases (e.g., turning a doorknob, walking, and travelling around a corner on a highway). Unwanted friction usually increases the production of waste heat (e.g., friction in the moving parts of an engine). (e) Frequency is the number of cycles of a vibration per unit time; it is measured in hertz (Hz) or s−1. Period is the time for one complete cycle of a vibration; it is measured in seconds (s). (f) Rotation is the spinning of an object on its own axis (e.g., Earth rotates daily). Revolution is the motion of one body around another (e.g., Earth revolves around the Sun once per year). Inquiry and Communication 5. (a) The units of acceleration are m/s2, so the inspector could use a metre stick to measure the distance in metres (m), and a stopwatch to measure the time in seconds (s). (b) The acceleration is the dependent variable, and the distance and time are the independent variables. (Students will discover in Chapter 3 that the distance is actually the radius of the circle and the time is the period of rotation of the ride.) Copyright © 2003 Nelson Unit 1 Are You Ready? 1 6. Error analysis can be reviewed by referring to page 755 of the text. Note that possible error is also called uncertainty. (a) The possible error is ± half of the smallest division of the measurement, or ±0.05 m/s2. possible error × 100% (b) % possible error = measurement = ±0.05 m/s 9.4 m/s 2 2 × 100% % possible error = ±0.5% The percent possible error is ±0.5%. measured value − accepted value × 100% (c) % error = accepted value = 9.4 m/s − 9.8 m/s 9.8 m/s 2 2 2 × 100% % error = − 4.0% The percentage error is –4.0%. difference in values (d) % difference = × 100% average of values = 9.7 m/s − 9.4 m/s 1 2 2 2 (9.7 m/s 2 + 9.4 m/s 2 ) × 100% 7. % difference = 3.0% The percentage difference is 3.0%. A prediction is a stated outcome expected from an experiment. A hypotenuse is a tentative explanation of what is expected in an experiment. Making Connections 8. The rapid spinning in the centrifuge would cause the liquids to separate according to their densities, with the liquid of greatest density moving farthest away, that is, toward the base of the tube. 9. (a) Let Fg = force of gravity on the truck FN = normal force of the road on the truck Fs = force of static friction on the truck θ = angle of the banked curve in (a) and (c) (a) (b) (c) (b) Choice (c) would be the best because the force of the road on the truck (the normal force) will help a lot in forcing the truck to the right. 2 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson Math Skills 10. (a) The equation is rearranged as follows. 1 ∆d = vi ∆t + a ∆t 2 2 1 ∆d − vi ∆t = a ∆t 2 2  ∆d vi  a = 2 2 −  ∆t   ∆t (b) The quadratic formula is used to solve an equation in the form ax 2 + bx + c = 0 As shown on page 750 of the text, the quadratic equation is x= Solving for ∆t, we have: −b ± b 2 − 4ac 2a 1 ∆d = vi ∆t + a ∆t 2 2 a 2   ∆t = + (vi ) ∆t − ∆d 2 ∆t = −v ± i (v ) i 2 − 2 ( a ) ∆d a ( ) 11. (a) Using the scale indicated in the question: A = 29.0 m/s [35° N of E] The north and east components of this vector are, respectively: A (sin 35°) = 29.0 m/s (sin 35°) = 17 m/s A (cos 35°) = 29.0 m/s (cos 35°) = 24 m/s Notice that the answers are written to two significant digits because the angle is stated to two significant digits. (b) The vectors can be added by using a vector scale diagram (adding the vectors head-to-tail), by using components, or by applying trigonometry (the laws of sines and cosines). (c) Scale: 1.0 cm = 5.0 cm B + A = 3.9 cm × 5.0 cm = 20 cm [65° N of E] A − B = 3.3 cm × 5.0 cm = 17 cm [2° S of E] Copyright © 2003 Nelson Unit 1 Are You Ready? 3 Technical Skills and Safety 12. (a) The total time is 6(0.10 s) = 0.60 s. (b) As shown in the illustration, the x-component of each displacement vector is about 1.0 cm in the diagram, or 5.0 cm using the scale indicated. We can conclude that the motion in the x direction is constant-speed motion. (c) The y-component of the displacements constantly changes, becoming smaller as the puck rises, and then becoming larger as the puck descends. (d) The displacement (or change of position) from the initial position to the final position is: ∆d = 6.15 cm × 5.0 cm/cm = 31 cm [right] ∆t = 0.60 s vav = ? ∆d vav = ∆t 31 cm [right] = 0.60 s vav = 52 cm/s [right] The average velocity of the puck is 52 cm/s [right]. 13. (a) Using a stopwatch, determine the total time for a certain number of complete revolutions of the stopper (e.g., 20 cycles). Then apply the following relationships: number of cycles frequency: f = total time period: T = total time number of cycles (b) The string should be strong, the stopper should be securely attached to the string, and the lab partner should hold the string securely while twirling the stopper a safe distance away from objects or people. (c) Typical sources of error are: • starting and stopping the stopwatch at the precise instant required (also called human reaction time error.) • choosing the same position to indicate both the starting and finishing locations of the motion • keeping the stopper moving at a constant speed and/or in a horizontal circle • measuring the radius of the circle of motion 4 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson CHAPTER 1 KINEMATICS Reflect on Your Learning (Page 4) 1. (a) (b) Ball B will land first because it has an initial downward component of velocity. Balls A and C will land at the same instant because, as students will discover in the Try This Activity, page 5, the horizontal component of the velocity of ball C does not affect its downward acceleration. Ball D will land last because its initial velocity has an upward vertical component. 2. (a) (b) The canoeists arrive at the north shore at the same time. The motion of the canoeist in the river is perpendicular to the flow of the river, so the motion is not affected by the flow of the river. (c) The figure below shows that the canoeist must aim upstream in order to arrive directly north of the starting position. This trip will take longer because the component of the velocity perpendicular to the shore is less than it was previously. Try This Activity: Choose the Winner It is important in setting up this demonstration that the device be fixed horizontally. For an alternative suggestion to the apparatus shown in the text, refer to Section 1.4 Questions, page 51, question 10, and the corresponding solution. (a) Predictions may vary. Refer to the Reflect on your learning questions 1(b) above. (b) Balls A and B will land simultaneously. The horizontal motion of the ball projected horizontally is independent of its downward acceleration. Copyright © 2003 Nelson Chapter 1 Kinematics 5 76 m/s. 2. 5.04 × 10 2 km = 2.42 × 102 km/h.76 m/s The swimmer’s average speed is 0. (a) d = 16 m ∆t = 21 s vav = ? d vav = ∆t 16 m = 21s vav = 0.04 × 102 km vav = ? d vav = ∆t 8.1.69 h vav = ? d vav = ∆t 8.32 h vav = ? d vav = ∆t 8. (c) ∆t = 2. It does not indicate any direction. (a) ∆t = 6.20 × 10 2 km/h The average speed in 1911 was 1. 6 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . (a) The force exerted by an elevator cable is a vector measurement.69 h d = 4. 4. (c) The motion of a basketball moving through the air toward the hoop is in two dimensions. (b) The motion of a tennis ball that falls vertically downward and then bounces is in one dimension.32 h vav = 2.99 × 102 km/h. (c) The gravitational force of Earth on you is a vector measurement. A car’s speedometer indicates instantaneous speed. (b) The reading on a car’s odometer is a scalar measurement.69 h vav = 1. (a) The motion of a tennis ball that falls vertically downward is in one dimension. (d) The motion of a curve ball is in three dimensions.20 × 102 km/h. (d) The number of physics students in your class is a scalar measurement. (e) Your age is a scalar measurement.04 × 10 2 km = 6. (f) The motion of a roller coaster is in three dimensions.42 × 10 2 km/h The average speed in 1965 was 2.04 × 10 2 km = 3.1 SPEED AND VELOCITY IN ONE AND TWO DIMENSIONS PRACTICE (Pages 7–8) Understanding Concepts 1. (e) The motion of a passenger seat of a Ferris wheel is in two dimensions.69 h vav = 2. a scalar quantity.02 km × 200 laps = 8. 3.99 × 10 2 km/h The average speed in 1990 was 2. (b) ∆t = 3. 74 km/h ∆t = 60. For example.54 × 10–2 s d=? d = vav ∆t = (342 m/s) (3. or 1. PRACTICE (Page 10) Understanding Concepts Typical examples of different types of vector quantities include a displacement while walking at a constant velocity.74 km/h) (60. Copyright © 2003 Nelson Chapter 1 Kinematics 7 . the distance travelled is 70 km. and the force of gravity.04 × 105 m.0 h) d = 104 km The distance travelled is 104 km. or if the motion is in two or three dimensions. (b) vav = 1. 7. 6. 10. if a car travels 50 km [N] and then 20 km [S]. (a) vav = 342 m/s ∆t = 3.1 m. 9. (a) d = 12 km + 12 km = 24 km  1h  ∆t = 24 min + 24 min = (48 min)   = 0. Thus.26 m ∆t = ? d vav = ∆t d ∆t = vav = 50. Yes.0 h d=? d = vav ∆t = (1. (a) It is possible for the total distance travelled to be equal to the magnitude of the displacement if the motion is all in one direction.80 h  60 min  vav = ? d vav = ∆t 24 km = 0.0 × 101 km/h The average speed of the bus for the entire route is 3. an acceleration while on a school bus. 8.54 × 10−2 s) d = 12.80 h vav = 3. the total distance is always equal to or greater than the magnitude of the final displacement vector.0 × 101 km/h.26 m 0. (c) It is not possible for the magnitude of the displacement to exceed the total distance travelled.1m The distance travelled is 12. but the displacement is 30 km [N]. the average speed can equal the magnitude of the average velocity if the motion is all in one direction. The total distance is the sum of the magnitudes of all of the vectors.76 m/s ∆t = 66s It would take the swimmer 66 s to swim around the edge of the pool.(b) circumference = πD = π (16 m) = 50. (b) It is possible for the total distance travelled to exceed the magnitude of the displacement if the motion is in one dimension along a path that turns back on itself. 0 km Since the total displacement is 0. so its displacement is zero and its average velocity is zero. (b) Calibrating the windsock involves finding how the angle of the sock above the vertical depends on the speed of the wind. the average velocity of the bus for the entire route is 0.0 × 101 km/h [E].6 ×10 4 km[S] 21km/h  1d  ∆t = 7. the sock hangs vertically downward.40 h vav = 3.6 × 104 km vav =21 km/h [S] ∆t = ? ∆d ∆t = vav = 1.) Thus. The easiest way to do this would be to hold the sock securely out of an open window of a car as the car travels at different speeds on a calm day. an experiment must be devised to measure the angle at various increasing speeds (e.g.6 × 102 h or 32 days. (a) The windsock indicates both the approximate speed and the approximate direction of the wind. Speed with a direction is velocity. etc. the bus returns to its starting position. which is a vector.40 h  60 min  vav = ? ∆d vav = ∆t 12 km[E] = 0.. at 10 km/h. (At a wind speed of zero. 20 km/h.0 km. In (b) the bus has reached its maximum displacement.6 m [fwd] during the time it takes for the driver to react.).0 × 101 km/h [E] The average velocity of the bus is 3. (b) ∆d = 12 km [E]  1h  ∆t = (24 min)   = 0.32 s vav = 27 m/s [fwd] ∆d = ? ∆d = vav ∆t = (27 m/s [fwd])(0. 12. a vector quantity. ∆t = 0.6 m [fwd] The truck is displaced 8.32 s) ∆d = 8.00 km/hr. (Students are not expected to attempt such an experiment. ∆d = 1. (d) The answers in (b) and (c) are different because the average velocity is determined by the displacement.) 8 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . 11. (c) vav = ? First we must calculate the total displacement: ∆d = 12 km [E] + 12 km [W] = 12 km [E] + ( − 12 km [E]) ∆d = 0. Applying Inquiry Skills 13. however in (c).6 × 102 h    24 h  ∆t = 32 d The tern’s journey takes 7. Try This Activity: Graphing Linear Motion (Page 12) The required graphs are drawn with the assumption that the zero displacement is the location of the motion sensor. PRACTICE (Pages 13–14) Understanding Concepts 14. (c) The motion begins west of the zero displacement position and is eastward but slowing down (westward acceleration). and the positive displacement direction is away from the sensor. (b) The motion begins at the zero displacement position and is upward but slowing down (downward acceleration). 15. beginning north of the zero displacement position. (a) The motion. (a) Copyright © 2003 Nelson Chapter 1 Kinematics 9 . is southward with constant velocity to a position south of the zero displacement position. (b) (c) 16. 10 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .5 m [N]. the initial velocity is equal to the slope of the tangent at the time indicated. and 7 m/s [W].20s) 2 = 3. ∆t 13 m/s [W].5 m [N] Thus. In each case.5 m [N] Area = 4.20s) + (15 m/s [N])(0.0 m [N] + 1. the area is 4. 17. students should calculate approximate answers of 7 m/s [E]. The diagram with typical tangents is shown below: ∆d Using the equation v = slope = m = . 7 m/s [W]. Assume two significant digits.40s − 0. 1 Area = (15 m/s [N])(0. The area under the velocity-time graph represents the displacement. 0 m/s. x = 22 m (cos 33°) ∆d1. Using a vector scale diagram to solve this problem.PRACTICE (Page 16) Understanding Concepts 18.0 m Copyright © 2003 Nelson Chapter 1 Kinematics 11 .0 m ∆d y = 7. y = 11m (sin 28°) ∆d 2. x = 18 m ∆d1. y − ∆d 2. Using components would yield the same result. y = 5 m ∆d y = ∆d1. with +x eastward and +y northward: ∆d1 = 22 m ∆d 2 = 11m ∆d1. x + ∆d 2.6 m [24° E of S]. 19. y = 12 m − 5. the vector sum of the displacements is found to be 5. x = 10 m ∆d 2. x = 11m (cos 28°) ∆d 2. (a) (b) Solving Sample Problem 5 (c) using components. y = 22 m (sin 33°) ∆d1. x = 18 m + 10 m ∆d x = 28 m ∆d 2. y = 12 m ∆d x = ∆d1. 6 × 102 m)(cos136°) ∆d = 1.0 m  θ = tan −1    28 m  θ = 14° The total displacement is 29 m [14° N of E].0 m)2 ∆d = 29 m tan θ = ∆d y ∆d x  7.3 × 103 m [42° N of E]. 12 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .6 × 102 m [21° E of N]  60 s  ∆t = 4.5 × 102 m [25° N of E] ∆d 2 = 5.6 × 102 m)2 − 2(8.3 × 103 m   θ = 17° The angle of the displacement is 17° + 25° = 42° The skater’s displacement is 1.6 × 10 m 1.6 × 102 m) sin136°  θ = sin −1     1.3 × 103 m Use the sine law to solve for the angle of the displacement’s direction: sin θ sin φ = ∆d 2 ∆d sin θ sin136° = 2 5.To calculate the total displacement: ∆d = ∆d x 2 + ∆d y 2 = (28 m) 2 + (7.3 × 103 m  (5. ∆d1 = 8. 20.5 × 102 m) 2 + (5.5 ×102 m)(5.2 min   = 252 s  1 min  (a) Add the vectors together: Use the cosine law to find the magnitude of the resultant displacement: 2 2 2 ∆d = ∆d1 + ∆d 2 − 2 ∆d1 ∆d2 cos φ 2 ∆d = (8. Copyright © 2003 Nelson Chapter 1 Kinematics 13 . (b) d = 3. (a) vav = 3. 3.56 s The laser light takes 2.97 × 102 s to travel from the Sun to Earth. A car travelling at a constant speed around a circular track has a velocity that is constantly changing.2 m/s [42° N of E] The skater’s average speed is 5. velocity (m/s) × time (s) = distance (m). 4. In the example in (c) above.84 × 105 km = 3. if the bus returns to the station its average speed is greater than zero. but its average velocity is zero. (e) A car travelling around a circular track has an average speed greater than zero.3 × 103 m [42° N of E] = 252 s vav = 5.1 Questions (Page 17) Understanding Concepts scalar scalar scalar vector vector (equal to the displacement) A car travelling at constant speed in one direction is at constant velocity.6 m/s and average velocity is 5.(b) vav = ? vav = d ∆t (8.84 ×108 m 3.00 × 108 m/s d = 1. Measurements with different dimensions cannot be added.49 × 1011 m 3. for example.56 s to travel to the Moon and back to Earth.2 m/s [42° N of E].84 × 108 m ∆t = ? d ∆t = vav = 2 3.00 × 10 m/s 8 ( ) ∆t = 2. but when it reaches its starting position. velocity cannot be added to time.5 × 102 m) + (5. (a) (b) (c) (d) (e) 2.97 × 102 s Light takes 4.6 ×102 m) = 252 s vav = 5. Measurements with different dimensions can be multiplied.00 × 108 m/s 1.6 m/s To calculate the average velocity: vav = ? ∆d vav = ∆t 1. A bus travelling from a station to a bus stop and then travelling back along the same route.49 × 1011 m ∆t = ? d ∆t = vav = 1. its average velocity is zero. for example. (a) (b) (c) (d) ∆t = 4. Section 1. The average speed between 0.0 9. The data to draw the position-time graph are found by calculating the area under the line at several instances and adding 8.0 s ∆t = 4.0 s ∆d = 40 m vav = ? vav = ∆d ∆t 40 m = 8.0 s and 8.0 m/s The instantaneous speed at 6.0 s is 2.31 m/s [E].0 5.0 s 4.0 s is 2.5 m[E] 45 m[W] 5.0 m/s and at 4.0 m = = 4.5 m/s v = 0. (a) ∆t = 4. The slope of the line can be calculated from a position-time graph to indicate velocity.5 m/s [E].5 3.0 m[E] = = = 1. the average velocity between 12 s and 16 s is 11 m/s [W].0 s and 16 s is 0.0 s 16s vav = 2.0 m [E] to the area in each case.0 m/s vav = 5.0 s and 9.0 s v = 2.5 m/s.0 s is the slope of the line at that instant.31m/s[E] The average velocity between 8.0 s and t = 9. Thus. ∆d v = slope = m = ∆t 45 m [W] = 4.5 m [E] ∆d =45 m [W] ∆d = 5. (c) The instantaneous speed at t = 6.0 s is 5.5 14 18 22 The required graph: 14 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .0 d (m [E]) 8.0 1.0 m/s The average speed between 4. 6.0 s is 0. t (s) 0.0 s 4. and the average velocity between 0. If the graph is curved. The table shows the results.5.0 s ∆d = 0 m vav = ? vav = ∆d ∆t 0m = 4.0 m/s.0 m/s.0 [E] vav = ? vav = ? vav = ? ∆d ∆d ∆d vav = vav = vav = ∆t ∆t ∆t 2. the slope of the tangent to the curve indicates the instantaneous velocity.0 s ∆t = 16 s ∆d = 2.0 4. (d) The instantaneous velocity at t = 14 s is the slope of the line at that instant. 7.0 s v = 11 m/s [W] The instantaneous velocity at 14 s is 11 m/s [W]. Thus.5 m/s[E] vav = 11m/s[W] vav = 0.0s ∆t = 8. (b) ∆t = 1.0 s is 0.0 s and 8. ∆d ∆d v=m= v = slope = m = ∆t ∆t 10 m 0.0s vav = 0. 73 m/s The average speed is 0.73 m/s. the total distance travelled is 87 m. (b) vav = ? d vav = ∆t 87 m = 120 s vav = 0. the total displacement is 65 m [43° S of E]. Thus. (c) Add the vectors together as shown in the illustration: Use the cosine law to find the magnitude of the resultant displacement: 2 2 2 ∆d = ∆d1 + ∆d 2 − 2 ∆d1 ∆d 2 cos φ 2 ∆d = (22 m) 2 + (65 m) 2 − 2(22 m)(65 m)(cos 79°) ∆d = 65 m Use the sine law to solve for the angle of the displacement’s direction: sin θ sin φ = ∆d ∆d 2 ∆d sin φ sin θ = ∆d 2  (65 m) sin 79°  θ = sin −1   65 m   θ = 79° The angle of the displacement is 79° − 36° = 43°.0 min = 120 s d1 = 22 m [36° N of E] d2 = 65 m [25° E of S] (a) d = ? d = d1 + d 2 = 22 m + 65 m d = 87 m Thus. ∆t = 2.8. Copyright © 2003 Nelson Chapter 1 Kinematics 15 . 4 m/s2 [fwd].5 s and t = 4. 3. (b) When the flock’s acceleration is negative. (c) When the flock’s acceleration is zero. draw the tangent parallel to the line joining the points at t = 2. (a) Students can refer to the Learning Tip titled “The Image of a Tangent Line” on page 13 of the text to understand how to use a plane mirror to check the accuracy of their tangents. Use the equation d = vav∆t to complete the table. (b) It is possible to have acceleration when the velocity is zero. 2. Making Connections 10. at the instant that a ball tossed vertically upward comes to a stop. the flock is moving south with decreasing velocity. its acceleration is still downward.54 m/s [43° S of E]. (b) One way to draw tangents accurately is to use the plane mirror method. (a) It is possible to have an eastward velocity with a westward acceleration. All five examples could be units of acceleration. (a) When the flock’s acceleration is positive.5 s.g.3 m/s [fwd] ∆t = 3.3 m/s [fwd] − 0 = 3.2 ACCELERATION IN ONE AND TWO DIMENSIONS PRACTICE (Page 20) Understanding Concepts 1. For example..9 s aav = ? vf − vi aav = ∆t 9. Another way that is useful for a displacement-time graph of uniform acceleration is to draw the tangent parallel to an imaginary line joining two points that are equal times away from the tangent time (e.54 m/s [43° S of E] The average velocity is 0. (d) vav = ? ∆d vav = ∆t 65 m [43° S of E] = 120 s vav = 0. Speed 17 m/s (60 km/h) 25 m/s (90 km/h) 33 m/s (120 km/h) no alcohol 14 m 20 m 26 m Reaction Distance 4 bottles 34 m 50 m 66 m 5 bottles 51 m 75 m 99 m 1.9 s aav = 2.5 s). as described in the Learning Tip. For example. Applying Inquiry Skills 9. 16 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . a truck moving eastward whle slowing down has a westward acceleration. the flock is moving south with increasing velocity.4 m/s 2 [fwd] The runner’s average acceleration is 2. the flock is moving south with constant velocity. 4. at t = 3. vi = 0 vf = 9. (d) vav = ? ∆d vav = ∆t 65 m [43° S of E] = 120 s vav = 0.54 m/s [43° S of E] The average velocity is 0.54 m/s [43° S of E]. Applying Inquiry Skills 9. (a) Students can refer to the Learning Tip titled “The Image of a Tangent Line” on page 13 of the text to understand how to use a plane mirror to check the accuracy of their tangents. (b) One way to draw tangents accurately is to use the plane mirror method, as described in the Learning Tip. Another way that is useful for a displacement-time graph of uniform acceleration is to draw the tangent parallel to an imaginary line joining two points that are equal times away from the tangent time (e.g., at t = 3.5 s, draw the tangent parallel to the line joining the points at t = 2.5 s and t = 4.5 s). Making Connections 10. Use the equation d = vav∆t to complete the table. Speed 17 m/s (60 km/h) 25 m/s (90 km/h) 33 m/s (120 km/h) no alcohol 14 m 20 m 26 m Reaction Distance 4 bottles 34 m 50 m 66 m 5 bottles 51 m 75 m 99 m 1.2 ACCELERATION IN ONE AND TWO DIMENSIONS PRACTICE (Page 20) Understanding Concepts 1. All five examples could be units of acceleration. 2. (a) It is possible to have an eastward velocity with a westward acceleration. For example, a truck moving eastward whle slowing down has a westward acceleration. (b) It is possible to have acceleration when the velocity is zero. For example, at the instant that a ball tossed vertically upward comes to a stop, its acceleration is still downward. 3. (a) When the flock’s acceleration is positive, the flock is moving south with increasing velocity. (b) When the flock’s acceleration is negative, the flock is moving south with decreasing velocity. (c) When the flock’s acceleration is zero, the flock is moving south with constant velocity. 4. vi = 0 vf = 9.3 m/s [fwd] ∆t = 3.9 s aav = ? vf − vi aav = ∆t 9.3 m/s [fwd] − 0 = 3.9 s aav = 2.4 m/s 2 [fwd] The runner’s average acceleration is 2.4 m/s2 [fwd]. 16 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson 5. vi = 0 vf = 26.7 m/s [fwd] a = 9.52 m/s2 (a) ∆t = ? ∆t = = vf − vi aav 26.7 m/s − 0 9.52 m/s 2 ∆t = 2.80 s The Espace takes 2.80 s to achieve a speed of 26.7 m/s. (b) v = ? m 1 km 3600 s v = 26.7 × × s 1000 m 1h v = 96.1 km/h The speed is 96.1 km/h. ? v −v (c) ∆t = f i aav L L − T T T=  L   2 T  ? ? L  T2    T =    T  L  6. 7. = 42 km/h [fwd ] + (14 (km/h)/s [fwd ]) ( 4.7 s ) vf =108 km/h [fwd ] Thus, the final velocity is 108 km/h [fwd]. aav = 1.37 × 103 m/s2 ∆t = 3.12 × 10–3 s vf =0 m/s vi = ? v − vi aav = f ∆t vi = vf − aav ∆t vi = 42.8 m/s [E ] = 0 m/s − 1.37 ×103 m/s 2 [W ] 3.12 × 10−2 s T=T Thus, the equation is dimensionally correct. aav =14 (km/h)/s [fwd] ∆t = 4.7 s vi = 42 km/h [fwd] vf = ? v − vi aav = f ∆t vf = vi + aav ∆t ? ( )( ) The velocity of the arrow as it hits the target is 42.8 m/s [E]. Copyright © 2003 Nelson Chapter 1 Kinematics 17 Try This Activity: Graphing Motion with Acceleration (Page 23) The required graphs are shown below, in which the position of zero displacement is located where the cart is near the bottom of the ramp but is not experiencing a push. PRACTICE (Pages 23–24) Understanding Concepts 8. (a) To determine the average acceleration from a velocity-time graph, determine the slope of the line if the acceleration is constant. (b) To determine the change in velocity from an acceleration-time graph, determine the area under the line. 9. (a) The motion starts with a westward velocity, but constant eastward acceleration. The motion then slows down to zero velocity, then accelerates westward with increasing westward velocity. The magnitude of the westward acceleration is somewhat less than the magnitude of the eastward acceleration. (b) The motion is southward with northward acceleration slowing down to zero velocity. (c) The motion is forward with constant acceleration forward. After a period of time, the motion increases to a higher constant acceleration forward. (d) The motion starts with northward acceleration then increases its northward acceleration. It starts to slow down (southward acceleration), and then decreases its southward acceleration to zero. 10. (a) 18 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson (b) 11. 12. The car’s displacement is the area under the velocity-time graph. It is determined by adding the areas of rectangles and triangles contained in each time segment. Referring to the figure in the text: displacement = total area = A4 (0 to 3 s) + A5 (3 s to 5 s) +A6 (5 s to 9 s) 1 (12 m/s [S])(3.0s) = 18 m [S] 2 1 A5 = (12 m/s [S])(2.0s) + (18 m/s [S] − 12 m/s [S])(2.0s) = 30 m [S] 2 1 A6 = (18 m/s [S])(4.0s) + (24 m/s [S] − 18 m/s [S])(4.0s) = 84 m [S] 2 A4 = displacement = A4 + A5 + A6 = 132 m [S] The car’s displacement is 132 m [S]. Making Connections 13. (a) The word “idealized” means that the acceleration changes instantaneously from one value to another. In real situations, changes from one acceleration value to another occur over a finite time interval. (b) Calculations are much easier if idealized examples are used. For example, to find the change in velocity for an idealized acceleration graph, we can find the area of a rectangle on the graph. That is much easier than finding the area under a curved line. (c) 14. The solution to this question depends on the software, calculator, or planimeter available. Each device is accompanied by a set of instructions that students can follow to analyze graphs. Copyright © 2003 Nelson Chapter 1 Kinematics 19 PRACTICE (Page 27) 2 a ( ∆t ) 15. (a) ∆d = vi ∆t + . 2 1 (b) ∆d = (vi + vf ) ∆t . 2 2 2 16. vf = vi + 2a∆d 2 2 Understanding Concepts L ?L  L   T  =  T  + 2  2  (L )     T  L ?L L  T  = T  + 2 T        Since the dimensions of each term are the same, the equation is dimensionally correct. 1 17. (a) ∆d = (vi + vf ) ∆t 2 2 ∆d ∆t = vi + vf 1 ∆d = (vi + vf ) ∆t (b) 2 2 ∆d vi + vf = ∆t  2 ∆d  vf =  −v  ∆t   i   18. Start with the defining equation for constant acceleration and the equation for displacement in terms of average velocity: ∆v a= ∆d = vav ∆t ∆t (vi + vf ) (vf − vi ) ∆d = ∆t a= 2 ∆t (a) To derive the constant acceleration equation in which the final velocity has been eliminated: first solve acceleration equation for vf , then substitute into the equation for displacement. (vi + vf ) ∆d = ∆t 2 (vi + vi + a ∆t ) vf − vi ∆ = ∆t d a= 2 ∆t 2vi ∆t + a ∆t 2 vf = vi + a ∆t = 2 1 ∆d = vi ∆t + a ∆t 2 2 2 2 2 20 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson 9 × 10–3 s a =? v −v a= f i ∆t 47 m/s[N] − 41m/s[S] = 1. a =18 m/s [E] ∆t = 1.6 s) = 0.6s) vf = 44 m/s[W] The velocity of the birdie is 44 m/s [W].9 × 10−3 s a = 4.6 s vi =73 m/s [W] vf = ? v − vi a= f ∆t vf = vi + a ∆t = 73m/s[W] + 18 m/s 2 [E](1. then substitute into the equation for displacement. 21.6 × 104 m/s2 [N].0 m + 2 2 ∆d = 15 m [fwd] The sprinter’s displacement is 15 m [fwd].6 s (a) ∆d = ? vi = 0 1 2 ∆d = vi ∆t + a ( ∆t ) 2 (2.(b) To derive the constant acceleration equation in which the initial velocity has been eliminated: first solve the acceleration equation for vi . 20. we have kept the vector notation in order to show what the final direction is. we can omit the vector notation when vf 2 or vi 2 terms are involved. page 26.3 m/s2 [fwd] ∆t = 3. pages 24 and 25. in the solution shown here as well as the solution for question 24.6 × 104 m/s 2 [N] The acceleration is 4. (vi + vf ) ∆d = ∆t 2 (vf − a ∆t + vf ) ∆d = ∆t 2 vi = vf − a ∆t 2vf ∆t − a ∆t 2 = 2 1 ∆d = vf ∆t − a ∆t 2 2 2 19. vi = 41 m/s [S] vf = 47 m/s [N] ∆t = 1. Note: As was described in the text. However.3m/s [fwd ]) (3. and applied in Sample Problem 6. 2 Copyright © 2003 Nelson Chapter 1 Kinematics 21 . a = 2.9 ms = 1. slide the book at a constant speed of about 0. Remove the pushing force from the book and determine the displacement from that instant to the stopping position.3 m/s [fwd].50 m/s for 2. 22.3 m/s 2 [fwd] (3..72 ×10 m/s [E]) + (2.478 m [E]) a = −5.0 s (e. where vf = 0. Applying Inquiry Skills 23.72 ×10 m/s [E]) = 2 7 2 (7.72 × 107 m/s [E] vf = 2.3 m/s [fwd] The sprinter’s final velocity is 8.60 × 1015 m/s2 [W]. Making Connections  1000 m  1 h  24.0 km/h [N])    = 20. (b) vf = ? vf − vi a= ∆t vf = vi + a ∆t vf = 8. The acceleration can be found by applying the equation vf 2 = vi 2 + 2a∆d .0 s).46 × 107 m/s [E] ∆d = 0. vi = 75. vi is the measured value of the speed while the book is being pushed.39 × 10–9 s.80 m/s2 [S] reaction time = ? 22 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .60 × 1015 m/s 2 [W] The electron’s acceleration is 5. and ∆d is the distance the book slides after it is no longer pushed. (b) ∆t = ? 1 ∆d = (vi + vf ) ∆t 2 2∆d t ∆ = vi + vf = 2 ( 0. The experiment can be simple.39 × 10−9 s The acceleration occurs over 9. vi = 7.6 s) ( ) 2(0.60 × 1015 m/s 2 [E] a = 5. the only equipment required is a metre stick and a stopwatch.478 m [E] (a) a = ? vf 2 = vi 2 + 2a ∆d v 2 −v2 a = f i 2 ∆d 7 = 0 m/s + 2.g.478 m [E]) 7 (2.46 ×10 m/s [E]) − (7. Besides the book and the desk.46 ×10 m/s [E]) 7 ∆t = 9. Slide the book along the desk at a constant speed for a predetermined distance that is long enough that the time interval to cover the distance is at least 2.0 km/h [N] = (75.8 m/s [N]  km  3600 s  a = 4. 2 m = 2.0 m – 45. Therefore.2 m[N] Calculate reaction distance (distance before stopping) = 48.4 m/s [45° S of W ] Copyright © 2003 Nelson Chapter 1 Kinematics 23 .8 m reaction distance reaction time = vi = 2.80 m/s 2 [S]) ∆d = 45.4 m/s2 [45° S of W].8 m/s = (0 m/s )2 − ( 20.3 m/s tanθ = ∆v y ∆v x = ( 25 m/s )2 + ( 25 m/s )2 Thus. with +x eastward and +y southward: ∆v x = vfx + ( − vix ) ∆v y = vfy + ( − viy ) ∆v x = −25 m/s ∆v y = 25 m/s ∆v = ∆v x 2 + ∆v y 2 ∆v = 35.First we must calculate the change in displacement: 2 2 vf = vi + 2a ∆d 2 2 v −v ∆d = f i 2a 2(4.8 m 20. the reaction time is 0.13 s.13s Thus. ∆v = 35m/s [45° S of W ]. the car’s average acceleration is 2.  25 m/s  θ = tan −1    25 m/s  θ = 45° So. PRACTICE (Page 29) Understanding Concepts 25.8 m/s[N])2 reaction time = 0. vi = 25 m/s [E] vf =25 m/s [S] ∆t = 15 s aav = ? Using components. ∆v aav = ∆t 35 m/s [45° S of W ] = 15 s 2 aav = 2. y = aav.4 m/s [E] + 2.x = ∆v x ∆t −4. 26.5 × 10−3 s = −1. vi = 26 m/s [22° S of E] vf = 21 m/s [22° N of E] aav = ? Using +x eastward and +y northward: ∆vx = ( 21m/s ) cos 22° − ( 26 m/s ) cos 22° ∆vx = −4.4 m/s [E] aav = 2.9 ×10 m/s ) + (7.5s ) ( ) (6. 27.0 ×10 m/s ) 3 2 2 3 2 2 tan θ =  7.0 m/s )2  5.9 ×103 m/s 2 aav = aavx 2 + aavy 2 = ∆v y = ( 21m/s ) sin 22° − ( −26 m/s ) sin 22° ∆v y = 17.9 × 10 m/s  θ = 75° Thus.6 m/s = 2.0 × 103 m/s 2  θ = tan −1  3 2   1. the average acceleration of the puck is 7.0 ×103 m/s 2 aav = 7.0 m/s [S] vf = vfx 2 + vfy 2 vf = 8.6 m/s = 2.4 m/s [E] + 5.1 m/s [38° S of E].6 m/s ∆v y aav.4 m/s )2 + (5.3 × 103 m/s2 [75° N of W].4 m/s  θ = 38° The final velocity of the watercraft is 8.y ∆t 17.6 m/s aav. vi = 6.0 m/s2 [S] ∆t = 2.5 s vf = ? v − vi aav = f ∆t vf = vi + aav ∆ t vf = 6.1m/s tan θ = vfy vfx = = 6.0 m/s 2 [S] ( 2.x aav.5 × 10−3 s = 7. 24 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .3 × 103 m/s 2 aavy aavx (−1.0 m/s  θ = tan −1    6. Instantaneous acceleration equals average acceleration during motion of constant acceleration.x = 9. +x east and +y north. Copyright © 2003 Nelson Chapter 1 Kinematics 25 . the change in velocity and.8 m/s (sin12. thus.0s ) ( ) (17 m/s )2 + ( 2.8 m/s (cos12.  82. It is possible to have a northward velocity with westward acceleration if there is a change in direction. ∆v aav.7° ) − 22.5 × 10–2 m/s2. 2.0 km   1000 m   1 h  29. aav =9.0 × 10–3 m/s2 and the y-component is –2. 28. vi = 82.2° ) 9.00 × 103 s aav.0 s vf =24 m/s [45° below horizontal] vi = ? v − vi aav = f ∆t vi = vf − aav ∆t vi = 24 m/s [45° below the horizontal] − 19.7° ) − 22. Section 1.0 km/h =     = 22.0 × 10 −3 m/s 2 aav.2 Questions (Pages 30–31) Understanding Concepts 1.8 m/s (sin 38.8 m/s  60 s  3 ∆t = 15 min = 15 min   = 9.00 × 10 s  min  As stated in the question. if a truck is initially moving northward at 50 km/h and changes direction to obtain a final velocity of 50 km/h [45° W of N].8 m/s2 [down] ∆t = 2.8 m/s  1 h   1 km   3600 s  vi = vf = 22.6 m/s [down ] vix = ( 24 m/s ) cos 45° vix = 17 m/s vi = vix 2 + viy 2 vi = 17 m/s tanθ = viy vix = viy = ( −24 m/s ) sin 45° − ( −20 m/s ) viy = 2.6 m/s = 24 m/s [45° below the horizontal] − 9.y = = ∆v y ∆t −22.00 × 103 s aav.8 m/s 2 [down ] ( 2. the acceleration just at the instant of the initial velocity.8 m/s (cos 38.x = x ∆t 22.2° ) = 9. is westward. For example. the x-component of the average acceleration is 9.5 × 10 −2 m/s2 Thus.6 m/s )2  2.6 m/s  θ = tan −1    17 m/s  θ = 10° The ball’s initial velocity is 17 m/s [10° above the horizontal].y = −2. 54 (km/h)/s     km  3600 s  a = 0. (a) Students can determine the data for the velocity-time graph by using the constant acceleration equation ∆d = vav ∆t (applied at the times indicated). or by drawing the position-time graph and finding the tangents to the curve.427 m/s 2 [E] The average acceleration of the aircraft is 0.0 s.54 (km/h)/s [E].54 (km/h)/s [E ] The average acceleration of the aircraft is 1.427 m/s2 [E]. The velocity-time graph is a straight line.4 5.12 × 103 km/h [W] ∆t = 345 s (a) a = ? vf − vi a= ∆t 1.4 v (m/s [W]) 26 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . The table below gives the data.  1000 m  1 h  (b) a = 1. 5. (a) (b) To determine the instantaneous acceleration at t = 2.3. 4.6 0.8 10.6 7.2 2.12 × 103 km/h [W] − 1. and its slope indicates the acceleration.2 0. t (s) 0 0 0. vi = 1.8 0.65 × 103 km/h [W] vf = 1. calculate the slope of the tangent to the curve indicated on the graph.65 × 103 km/h [W ] = 345 s a = 1. 5 m/s 2 [E])(2.0 3.5 6.7 m/s2 [fwd])(2. The data points for the position-time graph can be found by finding the total area on the velocity-time graph up the each second. The results are shown in the table.8 8.7 m/s2 [fwd] ∆t = 2. The motion then undergoes increasing velocity in the opposite direction until reaching the initial position. = 0 − ( −9. (a) The motion starts at a specific location at a high velocity with a negative acceleration.0 82.80 s ) 2 8.5 4. eventually reaching zero velocity.0 2.5 11. (b) This motion is increasing velocity in the negative direction. 9.5 5. (c) The motion in this graph starts with a constant velocity.5 9. 7.0 67.0 67.0 56. (b) a = ? 2 a ( ∆t ) ∆d = vi ∆t + 2 2 ∆d a= 2 ( ∆t ) = 2(4.8 10.0 52.0 78.5 m/s2 [E] ∆t = 2.0 10.0 78. followed by a decreasing velocity in the same direction. reaching zero velocity midway through the motion.5 d (m [W]) Copyright © 2003 Nelson Chapter 1 Kinematics 27 .2 12.5 7. vi =26 m/s [E] a = 5.5 2. then accelerates at a high rate for a short time before slowing down with negative acceleration at a lower rate to zero velocity.5 m/s2 [W] = −5. a = −9.16 m [W]) (where vi = 0) a = 13 m/s 2 6.0 22. This is followed by increasing velocity in the positive direction. The magnitudes of the negative and positive accelerations are equal.6 s vf = ? vf − vi a= ∆t vf = vi + a ∆t vf = 12 m/s[E] The car’s velocity is 12 m/s [E].9 s) vi = 28 m/s[fwd] The car’s initial speed is 28 m/s.0 52. t (s) 0 0 1.9 s vi = ? vf − vi a= ∆t vi = vf − a ∆t = 26 m/s[E] + ( −5.6s) (0.0 37. 0 × 1015 m/s2 [E]. 11.0 × 107 m/s [E] ∆d = 0.0 × 107 m/s[E])2 = 2(0.0 × 1015 m/s 2 [E] = 0 + 4.10 m [E]) a = 2.4 s) ( ) ( ) The acceleration of the electron is 2.The acceleration-time graph is generated by calculating the slopes of the line segments on the velocity-time graph.4 m/s2 [fwd] ∆t = 3. (b) ∆d = ? 1 2 ∆d = vi ∆t + a ( ∆t ) 2 1 = 0 + 4.4 m/s 2 [fwd ] (3. 28 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . a = 4.4 m/s 2 [fwd] (3.10 m [E] (a) a = ? vf 2 = vi 2 + 2a ∆d v2 a= f 2 ∆d (2. 10. vi = 0 vf = 2.4 s)2 2 ∆d = 25 m [fwd] The jumper’s displacement is 25 m [fwd].4 s vi = 0 (a) vf = ? v − vi a= f ∆t vf = vi + a ∆t vf = 15 m/s [fwd] The jumper’s final velocity is 15 m/s [fwd]. 1 ×10 m/s [fwd] The average velocity of the bullet is 2.2 ×10 2 m/s [fwd] = 2 2 vav = 2.0 ×1015 m/s 2 [E] ∆t = 1. vi = 204 m/s [fwd] vf = 508 m/s [fwd] ∆t = 29.V ∆d C = Atriangle.V + Arectangle.56 m [fwd] (a) vav = ? vi + vf vav = 2 0 + 4. 13.56 m [fwd] 2.05 × 104 m [fwd]. the car and the van have the same velocity (from the graph).0 × 10−8 s The electron takes 1.1× 10 2 m/s [fwd] ∆t = 2.0 × 107 m/s [E] − 0 = 2. 12. The displacements of the two vehicles are equal at some time t.C = (vV1 )av ∆tV1 + vV2 ∆tV2 = (vC1 )av ∆tC1 + vC2 ∆tC2  20 m/s + 0 m/s  =  60s + ( 20 m/s )(t − 60s ) 2    15 m/s + 0 m/s  =  30s + (15 m/s )(t − 30s ) 2   ∆d V = 600 m+ ( 20 m/s )(t − 60s ) ∆d C = 225 m+ (15 m/s )(t − 30s ) Copyright © 2003 Nelson Chapter 1 Kinematics 29 . (a) After 45 s.05 × 104 m [fwd] The displacement of the rocket is 1.0 × 10–8 s to reach its final velocity. 14.4 s ) 2 ∆d = 1.(b) ∆t = ? v − vi a= f ∆t vf − vi ∆t = a 2.7 × 10−3 s The uniform acceleration occurs over 2. vi = 0 vf = 4.1 × 102 m/s [fwd]. ∆d V = Atriangle.2 × 102 m/s [fwd] ∆d = 0.4 s ∆d = ? 1 ∆d = ( vi + vf ) ∆t 2 ( 204 m/s [fwd] + 508 m/s [fwd]) = (29.C + Arectangle.7 × 10–3 s. and can be found by determining the areas under the lines on the graphs. (b) ∆t = ? ∆d = vav ∆t ∆d ∆t = vav = 0. (b) Let the subscript V represent the van and the subscript C represent the car. 66 m/s 2 a = ax 2 + a y 2 = a = 0. ∆d V = (vV1 )av ∆tV1 + vV2 ∆tV2  20 m/s + 0 m/s  =  60s + ( 20 m/s )( 75s − 60s ) 2   = 600 m+ 300 m ∆d V = 900 m The displacement from the intersection when V overtakes C is 900 m.5s a x = 0. substitute into the equation for ∆d V or ∆d C . vAx = 4. 15.5s 2 vBx = 7.3 m/s ax = vBx − vAx ∆t 7. vA = 4.3m/s ) 8.4 m/s (sin 31° ) vAy = −2.66 m/s ) 2 2 2 2  0.4 m/s (cos 31° ) vBx = 7.8 m/s [25° N of E] ∆t = 8. the van V overtakes the car C at 75 s. 30 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .8 m/s = 8.39 m/s a y = 0.76 m/s 2 tan θ = ay ax (0.1m/s − 3.39 m/s  θ = 31° The bird’s average acceleration is 0.3 m/s − ( −2.66 m/s 2  θ = tan −1  2   0.1m/s vBy = 7.5 s a =? Using +x east and +y north.4 m/s [31° S of E] vB = 7.76 m/s2 [31° E of N]. Set ∆d V = ∆dC and solve for t: 600 m+ ( 20 m/s )(t − 60s ) = 225 m+ (15 m/s )(t − 30s ) 375 m 5 m/s (20 m/s ) t − (15 m/s ) t = 225 m − (15 m/s )(30 s ) + ( 20 m/s )(60 s ) − 600 m t= t = 75s Thus.8 m/s vAy = −4.8 m/s (cos 25° ) vAx = 3.3m/s ay = = vBy − vAy ∆t 3.39 m/s ) + (0. we find the components of the velocities and then the accelerations. (c) Using t = 75 s.8 m/s (sin 25° ) vBy = 3. 7° Thus.8 × 10 m/s [fwd] The average acceleration of the bullet is –6. (b) The stopping distance can be measured by adding the total crunch distance of the car plus any change of position of the barrels. Then the variables can be used in the equation vf 2 = vi 2 + 2a∆d . the acceleration is: 2 2 vf − vi aav = 2 ∆d 2(1.15 m [fwd]) 5 2 aav = −6.09 (km/h)/s  θ = 52. Estimates of the acceleration will vary if they are just guesses.5s a x = −2. 16.5 s a =? Using components with +x east and +y south: vfy − viy v −v ay = a x = fx ix ∆t ∆t 0 km/h − 155 km/h 118 km/h − 0 km/h = = 56.74 (km/h)/s )2 + ( 2. Assuming the value is 15 cm [fwd].5s 56.5 m [fwd]) 2 2 aav = −1.09 (km/h)/s ( −2. In both cases.74 (km/h)/s aav = a x 2 + a y 2 aav = 3.9 102 m/s2 [fwd].45 (km/h)/s tan θ = ay ax = a y = 2.45 (km/h)/s [52. = 0 − (24 m/s) [fwd]) 2 = 0 − (175 m/s [fwd]) 2 Copyright © 2003 Nelson Chapter 1 Kinematics 31 .7° W of S].09 (km/h)/s )2  2.8 105 m/s2 [fwd].5 m [fwd]. the variable most difficult to measure is the time interval for the acceleration. but should be fairly close if they are determined by a quick calculation with estimated quantities. the acceleration is: 2 2 v −v aav = f i 2 ∆d 2(0. it remains to find the displacement the object undergoes during the acceleration. Since the final speed is zero and the initial speed is given. Applying Inquiry Skills 17. (a) The displacement of the bullet during stopping can be found by measuring the penetration of the bullet plus the distance the wood moves. vi = 155 km/h [E] vf = 118 km/h [S] ∆t = 56. the helicopter’s average acceleration is 3. Assuming the value is 1.9 × 10 m/s [fwd] The average acceleration of the test car is –1.74 (km/h)/s  θ = tan −1    2. and changing the initial speed to 24 m/s. (b) Students have to combine the colours on the map with the colours in the legend. the time is shorter than if you include acceleration time for all runners. and they have to compare the contours and locations on the map to a conventional atlas of the same region. stretch inland somewhat in British Columbia.3 ACCELERATION DUE TO GRAVITY PRACTICE (Pages 32–33) Understanding Concepts 1. There are no areas unaffected. and Oregon. 1. 2. (This device is available commercially from scientific supply companies. as depicted by blue-greens. (about 117° west longitude). and from the middle of Vancouver Island (125° west longitude) to just east of Trail. the eastern parts of Washington and Oregon. Areas affected slightly. as depicted by yellows. and fourth runners have already accelerated to vf before they handoff to the next runner. Since there is no atmosphere on the Moon. Washington. especially in Northern California and Southern Oregon.Making Connections 18. In the relay. as depicted by white. lie near the west coast of the continent. as depicted by deep reds.C. The regions affected most severely. The experimental setup would require a vacuum chamber in which the coin and the feather are released simultaneously.) Making Connections 4. falling objects do not experience air resistance. the second. Areas affected moderately.80 m/s2 [down] (a) ∆d = 5. Air resistance increases with velocity and cannot be neglected for the skydiver.00 m vf = ? 32 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . The skydiver’s velocity is much greater than the diver’s velocity. Aristotle’s reasoning that heavy objects fell faster than lighter ones provides an example of the disadvantage of not using experimentation to determine the dependency of one variable on another. third. The disadvantage is that what might seem logical or reasonable does not agree with what actually happens. B. and a long way in California. Case Study: Predicting Earthquake Accelerations (Page 34) (a) The map stretches from Northern California (40° north latitude) to a few kilometres north of Vancouver (50° north latitude). PRACTICE (Page 35) Understanding Concepts 5. are in the northern and eastern parts of British Columbia. Thus. vi = 0 a = 9. One reason for this is some events occur too rapidly for our senses to be able to observe slight differences. Applying Inquiry Skills 3. in this case the dependency of the acceleration of a falling body on the mass of the body. and the state of Idaho. as well as areas fairly close to Vancouver and Victoria. the time is shorter than if you include acceleration time for all runners. especially in Northern California and Southern Oregon. as depicted by blue-greens. and fourth runners have already accelerated to vf before they handoff to the next runner.C. in this case the dependency of the acceleration of a falling body on the mass of the body. and Oregon. Aristotle’s reasoning that heavy objects fell faster than lighter ones provides an example of the disadvantage of not using experimentation to determine the dependency of one variable on another.) Making Connections 4. (This device is available commercially from scientific supply companies. and a long way in California. (about 117° west longitude). Areas affected slightly. The skydiver’s velocity is much greater than the diver’s velocity. In the relay. lie near the west coast of the continent. There are no areas unaffected. the second. The experimental setup would require a vacuum chamber in which the coin and the feather are released simultaneously. stretch inland somewhat in British Columbia. Case Study: Predicting Earthquake Accelerations (Page 34) (a) The map stretches from Northern California (40° north latitude) to a few kilometres north of Vancouver (50° north latitude). Washington. vi = 0 a = 9. Applying Inquiry Skills 3. One reason for this is some events occur too rapidly for our senses to be able to observe slight differences. PRACTICE (Page 35) Understanding Concepts 5. B. are in the northern and eastern parts of British Columbia.Making Connections 18. (b) Students have to combine the colours on the map with the colours in the legend. Air resistance increases with velocity and cannot be neglected for the skydiver.00 m vf = ? 32 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . and the state of Idaho.3 ACCELERATION DUE TO GRAVITY PRACTICE (Pages 32–33) Understanding Concepts 1. 1. as depicted by deep reds. as well as areas fairly close to Vancouver and Victoria. 2. as depicted by yellows. as depicted by white. The regions affected most severely. Since there is no atmosphere on the Moon. and they have to compare the contours and locations on the map to a conventional atlas of the same region. Areas affected moderately. Thus. The disadvantage is that what might seem logical or reasonable does not agree with what actually happens. falling objects do not experience air resistance. the eastern parts of Washington and Oregon. third.80 m/s2 [down] (a) ∆d = 5. and from the middle of Vancouver Island (125° west longitude) to just east of Trail. The actual answer depends on the student’s hand span.80 m/s 2 )(10.80 m/s 2 )(5.20 s For a hand span of 0. (b) v yi = 0 a y = 9.0 m[down]) vf = 2(9.20 m) 9.90 m/s × 1km × 3600 s/h 1000 m vf = 35.90 m/s[down] = 9.0 m/s × 1km × 3600 s/h 1000 m vf = 50.80 m/s 2 [down])(10.8 m/s 2 2 2 ∆y = 0.20 s. Students can improve their skills by trying to estimate answers to problems before performing calculations. or 200 ms.00 m[down]) vf = 2(9. (b) ∆d = 10. (a) vi = 0 ∆d = 1.600 s a =? 1 2 Rearranging the equation ∆d = vi ∆t + a ( ∆t ) : 2 Copyright © 2003 Nelson Chapter 1 Kinematics 33 .6 km/h [down].20 m (a typical hand span) 1 2 a y ∆t 2 = ∆t = 0.4 km/h [down] The diver’s velocity is 14.0 m/s [down] or 50.00 m) vf = 9.20 m. and by practising estimating quantities in everyday transactions and activities.4 km/h [down]. Applying Inquiry Skills 6.0 m) vf = 14.80 m/s 2 [down])(5.0 m/s [down] = 14. the time interval is 0. (c) Comparisons will vary.90 m/s [down] or 35.55 m [down] ∆t = 0. vf 2 = vi 2 + 2a ∆d vf 2 = 0 + 2(9. 7.6 km/h [down] The diver’s velocity is 9.8 m/s ∆t = ? ∆y = v yi ∆t + ∆y = ∆t = 1 2 a y ∆t 2 ∆y ay 2(0.00 m vf = ? vf 2 = vi 2 + 2a ∆d vf 2 = 0 + 2(9. (a) A good guess would be between 150 ms and 250 ms. 34 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . Making Connections 8.8 m/s2 [down]).61 m/s 2 − 9. (c) a = 9. so they would not likely be made for changes in altitude. 2∆d a= 2 ( ∆t ) = 2 1.61 m/s2 [down]. Answers may vary.81 m/s 2 9. The change in the gravitational field strength at various altitudes is quite small.600 s ) a = 8. find the slopes of the tangents at three or more times.) Another important source of error is measuring the time interval of the fall. and deciding how much adjustment would be required might be controversial. (b) Plot the position-time graph. even when it is at the top of the flight. Furthermore.81 m/s2 [down] measured value − accepted value % error = × 100% accepted value = 8. and calculate the slope of the line on that graph to find the acceleration. if records were adjusted downward for events that are easier at higher altitudes. (d) The high percent error results from a variety of possible errors. plot the corresponding velocity-time graph. PRACTICE (Pages 37–38) Understanding Concepts 9.61m/s 2 [down ] Thus.2% The percent error is 12. The ball’s velocity at the top of the flight is zero. (Since the ball is very light. The ball’s acceleration during the entire motion is the acceleration due to gravity (9. The initial and final velocities would be equal in magnitude but opposite in direction. (a) (b) (c) (d) (e) The time the ball takes to rise equals the time the ball takes to fall since air resistance is negligible. with air resistance likely the most crucial influence. air resistance has a greater effect than if the ball were more massive but the same size.55 m [down ] 2 (0. the acceleration of the ball is 8. then calls would be made for records to be adjusted upward for events that are more difficult in the rarified atmosphere at higher altitudes. Adjustments are not made for the effects of low winds.2 %.81 m/s 2 ×100% ( ) % error = −12. 0 m Using the equation ∆y = viy ∆t + equation.80 m/s 2 )(3. (a) viy = 0 a y = 9.6 m/s at impact.5 m) vfy = 245 m 2 /s2 vfy = 15. rearrange as  a y  ( ∆t )2 + viy ∆t − ∆y = 0 .33s ∆t = 3. ∆t = ? vfy = ? ∆t = 15.10. which is a form of the quadratic 2  ay = vfy − viy 2 −b ± b 2 − 4ac 2a  ay  −viy ± viy 2 − 4   ( −∆y )  2  = a  y 2   2  ∆t = −viy ± viy 2 + 2a y ∆y ay (a) The initial velocity is up. 12.80 m/s2 ∆y = 12.6 m/s The shellfish has a speed of 15. (b) viy = 0 a y = 9.80 m/s2 ∆y = 15.0 m/s )2 + 2(9. We could solve Sample Problem 1(b) using the following equations: (viy + vfy ) 1 1 ∆y = viy ∆t + a y ( ∆t )2 ∆y = vfy ∆t − a y ( ∆t )2 ∆y = ∆t 2 2 2 11.80 m/s 2 (only the positive root is applicable) = 1.80 m/s 2 )(12.37 s vfy = ? ∆t vfy = viy + a y ∆t = 0 + (9.5 m vfy = ? vfy 2 = viy 2 + 2 a y ∆y vfy = 0 + 2(9.80 m/s2 )(15. Thus. Define +y as down.0 m/s at impact.37 s) vfy = 33.86 s Copyright © 2003 Nelson Chapter 1 Kinematics 35 .0 m/s ± ( −15.0 m/s a y = 9.0 m) 9.0 m/s The steel ball has a speed of 33. viy = −15. ∆t = a y ( ∆t ) 2 2 1  .80 m/s2 ∆t = 3.53s ± 2. 8 m/s 1 ∆y2 = viy .794 s ) Thus.80 m/s2 vfy = 22. Rules for rounding off have not been followed exactly here.2 = 9.80 m/s2 36 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .0 s − 0.0 m) = −1.0 s) + 9.80 m/s 2 (only the positive root is applicable) = 15.8 m/s 2 (1. Let ∆y1 represent the distance the ball travels from t = 0.86 s and the speed of impact is 22.0 s − 0. ∆t = ? vfy = ? ∆t = −15.2 = viy .9 m ) Therefore.0s) + 9.0 s. the initial velocity viy.0 s to 1.0 s) 2 2 ∆y1 = 4. a ball travels three times as far from 1.7 m ) = = 3. the initial velocity viy.0s − 1. the total flight time is 3.86s ) vfy = 22.vfy = viy + a y ∆t = −15.2 is the velocity at t = 1.0s) 2 2 ∆y2 = 14.0 s to t = 1.0 m/s ± ( ) (15.) (c) The final velocity is independent of whether the ball is thrown up or down (at the same speed).0 s as from 0.0 m/s ) 2 + 2(9.9 m For ∆y2.1 = 0 1 Using the equation ∆y = viy ∆t + a y ( ∆t )2 : 2 1 ∆y1 = viy .0 ∆y1 (4. Define +y as up.1 s 2 ay = –9.794s vfy = viy + a y ∆t 9. (Notice that the answers are written to three significant digits in order to compare them to the answers in (a). the total flight time is 0.53s ± 2. Define +y as down.80 m/s2 (3.0 m/s + 9.0 s and let ∆y2 represent the distance the ball travels from t = 1.0 s to t = 2. Thus.8 m/s2 (1. 14.1 + a y ( ∆t1 ) ( ) = 0 m/s + 9.8 m/s Thus. (b) The initial velocity is down.2s ∆t = = 2. 13.0 s.0 s) viy .0 m/s + 9.33s ∆t = 0.0s − 1.794 s and the speed of impact is 22.80 m/s 2 )(15.8 m/s.0 s.0 s to 2.0 s − 0.1∆t1 + a y ( ∆t1 )2 2 1 = 0 m/s (1. For ∆y1.8 m/s ( ) (0.8 m/s if the initial velocity is down.2 ∆t2 + a y ( ∆t2 )2 2 1 = 9. vfy = 0 m/s 4.8 m/s 2 (2.7 m ∆y2 (14. viy .8 m/s (2. 15.80 m/s 2 )(3.3 m) (1. viy = −2.3 m ∆t = 1.8 s ay = 9.7 s viy = 0 (a) g = ? ∆y = viy ∆t + ∆y = g= = 1 g ( ∆t ) 2 2 1 g ( ∆t ) 2 2 2∆y ( ∆t ) 2 2(2. Define +y as down. the pitcher threw the ball with a velocity of 21 m/s [up]. ∆d = 2.80 m/s 2 )(2. (b) ∆y = ? 1 ∆y = viy ∆t + a y (∆t ) 2 2 1 = 21 m/s (2.80 m/s2 )(2.8s) + 2 ∆y = 63m The balloon was 63 m when the ballast was released.8s)2 = (−2.80 m/s2 (a) ∆y = ? a y (∆t ) 2 ∆y = viy ∆t + 2 (9.7 s)2 g = 1.1 m/s + (9. (b) vfy = ? vfy − viy ay = ∆t vfy = viy + a y ∆t = −2.(a) viy = ? ay = vfy − viy ∆t viy = vfy − a y ∆t = 0 m/s − ( −9. Copyright © 2003 Nelson Chapter 1 Kinematics 37 .1m/s)(3. 16.1s) viy = 21m/s Thus.1 m/s ∆t = 3.80 m/s 2 )(3.1 s) 2 2 ∆y = 22 m The ball rises 22 m.6 m/s 2 [down] The acceleration of gravity on the Moon is 1.1 s) + (−9.6 m/s2 [down]. Define +y as down.8 s) vfy = 35 m/s The velocity of the ballast at impact was 35 m/s [down]. (a) Students can use graphing techniques or uniform acceleration equations to determine the acceleration.40 s )2 2 a y1 = 9.40 s a y1 = ? ∆y = viy ∆t + ∆y = a y1 = = 1 2 a y 1 ∆t 1 2 2 a y 1 ∆t 2 2 ∆y ∆t 2 2(0.2 m/s 2 + 9.6 m/s 2 Thus.2 m/s − 9.736 m) ( 0.7 m/s2.40 s a y2 = ? ∆y = viy ∆t + ∆y = ay 2 = = 1 2 1 2 2 a y 2 ∆t 2 a y 2 ∆t 2 ∆y ∆t 2 2(0.8 m/s 2 = = 6. For mass 1: viy = 0 ∆y = 0.7 m/s 2 ) × 100% % difference = 5.736 m ∆t = 0. the ratio of g Earth to g Moon is 6.776 m) ( 0.7 m/s 1 2 2 2 (9. Sample calculations using the final data points are shown below.(b) g Earth 9.1 g Moon 1. (b) To find the percent difference: difference in values × 100% % difference = average of values = 9. respectively. Let +y be down.2 m/s For mass 2: viy = 0 ∆y = 0.40 s )2 2 a y 2 = 9.3% 38 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .7 m/s The accelerations of the two masses are 9. Applying Inquiry Skills 17.1:1.2 m/s2 and 9.776 m ∆t = 0. Making Connections 18. create a prototype. (b) Since there are so many factors that could be tested. The distance can be increased if the landing occurs on the down slope of a hill or if the fall is broken by trees. Increasing the stopping or deceleration distance reduces the chances of death or serious injury. There is no “terminal speed” related to air resistance on the Moon because the Moon does not have an atmosphere. running. Air resistance increases dramatically with increased speed. bushes. Section 1. Each element moved in a straight line to its proper place where it could be at rest. and water. The graph is shown below. For objects with the same surface area. Another advantage is that objects subject to toppling over might be more stable. air resistance is negligible for all but the lowest density objects that can fall in a classroom setting.(c) Likely. and contents of the package. a high speed followed by a much lower terminal speed when the parachute is opened. and the ease with which the package can be opened when it is found. the material of the package itself. There are two main factors to consider. Thus. Walking. similar objects were attracted to each other. so air resistance is negligible for most common objects that fall for a short distance. The terminal speed values are found on page 39 of the text (Table 5). Aristotle believed that the central region of Earth was made up of four elements: earth. 9. The two main factors that affect terminal speed are mass and surface area. for example for a construction worker on a slanted roof. the method of securing the outer layer of the package. the materials of the packets within the package. 20. Applying Inquiry Skills 22. (For solid spherical objects. the greater the mass the greater is the terminal speed. air. PRACTICE (Page 39) Understanding Concepts 19. Once the terminal speed of a body is reached. One advantage is that friction would be increased when it might be helpful. Making Connections 23. Each element had its proper place. fire. dimensions. the height from which the package will be dropped. or water. a layer of snow. and then perform a controlled experiment by dropping the sample of the design from various heights onto various surfaces to determine how well the package can survive the fall. objects made from the Copyright © 2003 Nelson Chapter 1 Kinematics 39 . 21. a person falling from a greater height can reduce the terminal speed by using the “spread-eagle” orientation. The slope of the line near the beginning of the motion is equal to the magnitude of the acceleration due to gravity at that location. Modifications and further testing are part of the process. or it may even drop as the air becomes more dense nearer the ground. For example. it would be wise to predict what choices would be best for the design. Answers will vary. the speed remains constant. For example.) For objects of the same mass. the type of terrain on which the package will land. Furthermore. (a) Examples of factors to consider in designing the package are the mass. and going up stairways are just a few examples of the activities that would be more difficult with a higher acceleration due to gravity. an increased surface area contacting the air causes the terminal speed to become lower. is attributable to the ticker-tape timer because friction between the paper strip and the timer would reduce the acceleration. 2. which was determined by its relative heaviness.2 m/s2. There are two terminal speeds in this case. the lower experimental acceleration. terminal speed and stopping distance. the forward speed of the aircraft that drops the package. this is equivalent to saying that the terminal speed depends on the object’s density.3 Questions (Page 40) Understanding Concepts 1. 2 s vfy = ? vfy − viy ay = ∆t vfy = viy + a y ∆t = 0 + (9.8 m/s2 ∆y = 36 m vfy = ? vfy 2 = viy 2 + 2a y ∆y vfy 2 = 0 + 2(9. Galileo believed that all objects fall toward Earth at the same acceleration. the jumper achieved a height of 1.112 m/s)2 2( −9.336 m. the landing speed of the diver is 27 m/s or 97 km/h.112 m/s v fy = 0 ∆y = ? In Java.8 m/s 2 )(3.1× 102 km/h 4. In each case. viy = 5. g = −9.330 m. size.2 s)  1 km   3600 s  = (31m/s)     1000 m   1 h  vfy = 1.8 m/s2 ∆t = 3. g = −9.1 × 102 km/h. Thus. when gravity is the only force acting on them.823 m/s2: ∆y = = −viy 2 2a y −(5. we define +y as down. the landing speed of the stone is 31 m/s or 1. 3. the jumper achieved a height of 1.330 m In Java.823 m/s 2 ) ∆y = 1.336 m In London. Evidently. (a) viy = 0 ay = 9. he proved his theory by dropping two objects of different mass from the top floor of the Leaning Tower of Pisa. (b) viy = 0 ay = 9.8 m/s 2 )(36 m)  27 m   1 km   3600 s  vfy =      s   1000 m   1 h  vfy = 97 km/h Thus. in London.782 m/s2: vfy 2 = viy 2 + 2a y ∆y 2a y ∆y = vfy − viy ∆y = = −viy 2 2 2 2a y −(5. Define +y as up. Aristotle also believed that heavier objects fell faster than less massive objects of the same shape. Fire tended to rise from Earth. regardless of their mass. or shape.782 m/s 2 ) ∆y = 1.earth were attracted down to Earth.112 m/s)2 2( −9. 40 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . ∆t = 3.4 s The total time in flight would be 2(3.5. the time for the ball to rise is 1. viy = 13 m/s ay = g = −3. 6.80 m ay = 9.3s) viy = 13m/s The velocity is 13 m/s [up] when the golf ball leaves the person’s hand.4 s) = 6. +y is up.3 s. Let +y be down.1 × 104 km/h.9 ×103 m/s)     1000 m   1 h  vfy = 1.3 s viy = ? vfy − viy ay = ∆t viy = vfy − a y ∆t = 0 − ( −9.0 min = 60 s (two significant digits) v fy = ? vfy − viy ay = ∆t vfy = a y ∆t = 5(9.3 s Therefore. Let +y be the forward direction of the shuttle.9 m/s or 1.80 m/s 2 )(60 s)  1 km   3600 s  = (2.6 s ∆t = 2 ∆t = 1. ∆t = 0.8 m/s2 Copyright © 2003 Nelson Chapter 1 Kinematics 41 .8 m/s 2 7.087 s ∆y = 0. assumed to be a straight line.8 m/s2 )(1. Let +y be up. 2.80 m/s2) ∆t = 1.8 m/s2 ∆t = ? vfy − viy ay = ∆t vfy − viy ∆t = ay = 0 − 13m/s −3.80 m/s2 (a) ∆t = ? The time for the ball to rise will be half of the total time.8 s. ∆t = 2.6 s ay = g = −9.1× 10 4 km/h The shuttle’s speed is 2. (b) vfy = 0 ∆t = 1. viy = 0 ay = 5(9. (c) Again. 00 s after the flowerpot is released. But with uniform acceleration. . Therefore.8 m/s 2 )(21m) 9.087 s) 0. after some time ∆tF = ∆tB − 1. the flowerpot will have travelled 28. and let B represent the ball.1 ∆y = vfy ∆t − a y (∆t ) 2 2 ∆y 1 vfy = + a y ( ∆t ) ∆t 2 0.1s or − 3.9s The stone will take 1. we have: 42 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson 1 2 ay∆t 2 . let F represent the flowerpot. viy = 14 m/s ay = 9.e. 10.1 s to reach the water below. units are omitted until the final answer is written. Let +y be down.0 m = 2.5 m − 26. 9. (b) The positive root is the actual answer when the stone is thrown vertically downward.00. let us assume the ball will pass the flowerpot. Thus.087 s 2 ( ) vfy = 9.8 m/s2 (a) ∆y = 21 m ∆t = ? ay 1 Using the equation ∆y = viy ∆t + a y ( ∆t )2 . if vi = −14 m/s in this case).8 m/s 2 ∆t = 1. Let +y be down. The velocity of the ball as it hits the floor is 9. The negative root is the time that the stone would have travelled had the initial velocity been upward rather than downward (i. ∆yF = ∆yB + 2.5 m farther than the ball.5. rearrange as ( ∆t )2 + viy ∆t − ∆y = 0 .6 m/s [down].6 m/s 8. ∆t = −b ± b 2 − 4ac 2a  ay  −viy ± viy 2 − 4   ( −∆y )  2  =  ay  2   2  = = −viy ± viy 2 + 2a y ∆y ay −14 m/s ± (14 m/s) 2 + 2(9. which is a form of the quadratic 2 2 equation. ∆y = viy ∆t + Combining these relationships. Since the ball is thrown 1.8 m/s 2 (0.. Note: To make the solution more compact. At this instant.80 m 1 = + 9. 5%. ∆y.9 ∆t F 2 − 9.∆yF = ∆yB + 2.809 060 m/s 2 (b) 9.8 × 10–6 m/s2 has 2 significant digits.0005 m/s2.801 m/s2 has 4 significant digits.9 ∆t F 2 = 12 ∆t F − 12 + 4.8 ×10−6 m/s 2 13.005 m/s 2 × 100% = ±0.9 ∆t F 2 = 12 ( ∆t F − 1) + 4.05%. and percent possible error of ±0.9 ∆t F 2 − 2 ∆t F + 1 + 2.9∆t F 2 = 12 ( ∆t F − 1) + 4.80 m/s2 has 3 significant digits.2 ∆tF = 2.8 m/s2 has 2 significant digits.5 4. and a skydiver in spread-eagle orientation.000 005%.80 m/s 2 (d) 9.6 2.9 ( ∆t F − 1) + 2. a skydiver diving headfirst.5 2 0 + 4.5 m − 21. 9.9 + 2.005 m/s2. 9.5%. as shown below. (a) 9.05× 10–6 m/s2.6 ( ) ( ) The flowerpot is 28.000 000 5 m/s 2 × 100% = ±0. a possible error of ±0. and percent possible error of ±0. Without warning.8∆t F + 4.05 m/s 2 × 100% = ±0. 11.09 s ) 2 ∆yF = 21.10 m above the ground when the ball passes it. the flowerpot has fallen: 1 2 ∆y F = a y ∆t F 2 1 2 2 = 9.9 + 2. such as the 50-cm mark.05 m/s2.0005 m/s 2 × 100% = ±0.18 m ay = 9. The ranking from highest terminal speed to lowest is: pollen.5 viyF ∆tF + 1 2 a y ∆tF 2 = viyB ∆tB + 1 2 a y ∆tB 2 + 2. Assume ∆y = 18 cm = 0. Hold a metre stick vertically while your partner holds his or her separated index finger and thumb ready to catch the stick at a specific mark.09 s After this time interval. (a) Students may recall performing this activity in a previous grade.8 m/s 2 (c) 9.8∆t F + 4.5 4. drop the stick and determine how far the stick falls before the partner catches it. Let +y be down.2∆t F − 4.4 m ∆t F = 4. 9. Use this value in the appropriate equation. and percent possible error of ±0.5 0 = 12 ∆t F − 12 − 9.5 0 = 2. Repeat several times and take an average of the distances.005%.4 m = 7. and percent possible error of ±0. 9.8 m/s2 viy = 0 ∆t = ? Applying Inquiry Skills Copyright © 2003 Nelson Chapter 1 Kinematics 43 . 9.05 ×10 −6 m/s 2 ×100% = ±0. a possible error of ±0. a possible error of ±0. 12.801 m/s 2 (e) 9. a basketball.809 060 m/s2 has 7 significant digits. a possible error of ±0.000 000 5.80 m/s ( 2. and percent possible error of ±0. a possible error of ±0. a ping-pong ball. he formed the hypothesis that the speed of the object was directly proportional to the distance the object fell. 2.8 m/s 2 ∆t = 0. Thus.18 m) 9. The initial position is the position where the marble leaves the table. in entering the words Luis Alvarez. particular results are inferred from a general law. an advanced word search on the Internet.ceemast. (a) Aristotle and other ancient scientists used deductive reasoning. (b) Galileo used the process of inductive reasoning. 15. Inductive reasoning involves making and collecting observations. (a) Horizontally (constant vix ): vix = 1. and then developing general theories or hypotheses to account for the observations. 3. Students can simulate this situation by engaging in distracting conversation with the lab partner whose reaction time is being tested. conclusions follow from premises. Stated another way for deduction. and dinosaurs. a general law is inferred from particular results. In mathematics. (c) Various ways are used to contrast these types of reasoning. Through experiments. Aristotle and Galileo influenced the philosophy and scientific thought of their respective eras.csupomona. When he observed that heavy objects fall with increasing speed. that is the reasoning goes from the general to the specific. found more than 20 thousand hits. Students can find information about these science “giants” in books and encyclopedias.93 m/s ∆x = ? ∆t = ? 44 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . In induction.html 1. For example. For example. induction involves proving a theorem using a process in which the theorem is verified for a small value of an integer. and in both cases their influence lasted long after they died. Let +x be to the right and +y be downward. premises lead to the conclusions. many of which featured discussions of the same topics featured in the text. or on the Internet. and then extending the verification to greater values of the integer.edu/nova/alvarez2. An airplane has a propulsion system and does not follow a trajectory. more than 400 sites were found. he hypothesized that the speed of a falling object is directly proportional to the time. or the reasoning goes from the specific to the general. whereas in inductive reasoning. not the distance. in deductive reasoning. When this hypothesis proved false.com/scienceastronomy/planetearth/deep_impact_991228. Two examples are: www. entering only the words Aristotle and Galileo.html www.∆y = viy ∆t + ∆y = ∆t = 1 2 a y ∆t 2 ∆y ay 1 2 2 a y ∆t 2 = 2(0. an airplane is not a projectile. Deductive reasoning involves using theories to account for specific experimental results. Thus. For example.4 PROJECTILE MOTION PRACTICE (Page 46) Understanding Concepts 1. Yucatan. he was able to verify his hypothesis. Making Connections 14. 16.19 s (b) Talking on a cell phone would likely increase reaction time. many of them highly credible. Many sites can be found by doing an advanced word search on the Internet. The projectile experiences constant downward acceleration due to gravity (vertical acceleration) and the horizontal component of acceleration is zero.space. deductive reasoning uses ideas to explain observed phenomena. A projectile is an object that moves through the air without a propulsion system and follows a curved path. many of them highly credible. and in both cases their influence lasted long after they died. Deductive reasoning involves using theories to account for specific experimental results. premises lead to the conclusions.93 m/s ∆x = ? ∆t = ? 44 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . found more than 20 thousand hits. Thus. Many sites can be found by doing an advanced word search on the Internet. Through experiments. 3. Let +x be to the right and +y be downward. 16. For example. 15. Yucatan. not the distance.19 s (b) Talking on a cell phone would likely increase reaction time. more than 400 sites were found. particular results are inferred from a general law. An airplane has a propulsion system and does not follow a trajectory. When he observed that heavy objects fall with increasing speed. in entering the words Luis Alvarez. Inductive reasoning involves making and collecting observations. that is the reasoning goes from the general to the specific. In induction. Students can find information about these science “giants” in books and encyclopedias. or on the Internet. For example. a general law is inferred from particular results. he hypothesized that the speed of a falling object is directly proportional to the time. For example. Two examples are: www. conclusions follow from premises. and dinosaurs. whereas in inductive reasoning. or the reasoning goes from the specific to the general. When this hypothesis proved false. Making Connections 14. (a) Horizontally (constant vix ): vix = 1. In mathematics. (b) Galileo used the process of inductive reasoning. and then developing general theories or hypotheses to account for the observations.∆y = viy ∆t + ∆y = ∆t = 1 2 a y ∆t 2 ∆y ay 1 2 2 a y ∆t 2 = 2(0. and then extending the verification to greater values of the integer. an advanced word search on the Internet. The projectile experiences constant downward acceleration due to gravity (vertical acceleration) and the horizontal component of acceleration is zero.csupomona. The initial position is the position where the marble leaves the table. many of which featured discussions of the same topics featured in the text. (c) Various ways are used to contrast these types of reasoning. deductive reasoning uses ideas to explain observed phenomena.ceemast. in deductive reasoning. he formed the hypothesis that the speed of the object was directly proportional to the distance the object fell. Stated another way for deduction.18 m) 9.4 PROJECTILE MOTION PRACTICE (Page 46) Understanding Concepts 1. A projectile is an object that moves through the air without a propulsion system and follows a curved path. induction involves proving a theorem using a process in which the theorem is verified for a small value of an integer.space.html 1.html www.com/scienceastronomy/planetearth/deep_impact_991228.8 m/s 2 ∆t = 0. (a) Aristotle and other ancient scientists used deductive reasoning. he was able to verify his hypothesis. an airplane is not a projectile. Thus. Aristotle and Galileo influenced the philosophy and scientific thought of their respective eras. 2. Students can simulate this situation by engaging in distracting conversation with the lab partner whose reaction time is being tested. entering only the words Aristotle and Galileo.edu/nova/alvarez2. 87 m/s 1. (c) To determine the marble’s final velocity.8 m/s 2 )(0. (a) Let the +x direction be to the right and the +y direction be downward.93 m/s. and 3. determine its horizontal and vertical components.80 m/s2 ∆t = ±0. 4.395 s) ∆x = 0. 2.33 m/s [63. Horizontally (constant vix = 8.0 m The ∆x values at t = 0.395 s to hit the floor.0 s) ∆x = 8.0 s ∆x = ? vix ): ∆x = vix ∆t = (8.5 m 1 The vertical motion. (b) ∆x = ? =± Using the equation for horizontal motion: ∆x = vix ∆t = (1.93 m/s)2 + (3. 1. the final velocity of the marble just prior to landing is 4.33 m/s θ = tan −1 = tan −1 vfy vfx 3. The x-component is constant at 1.87 m/s Using the law of Pythagoras and trigonometry: vf = vfx + vfy 2 2 2 vf = (1.93 m/s θ = 63. Copyright © 2003 Nelson Chapter 1 Kinematics 45 .93 m/s)(0.763m or 76.765 m) 9.5° Thus.0 s.0 m/s ∆t = 1. The y-component is: vfy = viy + a y ∆t = 0 m/s + (9.3 cm The horizontal range is 76.80 m/s 2 ∆y = 76.0 s are shown in Table 1.5° below the horizontal]. ∆y = viy ∆t + a y (∆t ) 2 can be simplified to: 2 2 ∆y ∆t = ± ay 2(0.0 s.395 s) vfy = 3.0 s.Vertically (constant ay): viy = 0 a y = + g = 9.0 m/s)(1. the marble takes 0.395 s Since only the positive root applies.3 cm.87 m/s)2 vf = 4. The y-component at 1. vf = 13 m/s [51° below the horizontal]. The vf values for ∆t = 0.0 s. Determine the final velocity using its horizontal and vertical components.9 13 [51° below horizontal] 2.8 m/s2 ∆t = 1.8° Therefore.0 16 20 21 [68° below horizontal] 3.0 m/s.8 m/s 2 )(1.8 m/s 2 )(1.0 s.0 s are shown in Table 1.0 s is: vfy = viy + a y ∆t = 0 m/s + (9.0 s) 2 = 2 ∆y = +4.0 1. Table 1 Calculated Horizontal and Vertical Displacements and Instantaneous Velocity at Select Times t (s) ∆x (m) ∆y (m) 0.8 m/s 8.0 s ∆y = ? a y ): 1 ∆y = viy ∆t + a y (∆t ) 2 2 1 = a y ( ∆t ) 2 2 (9.65 m/s θ = tan −1 = tan −1 vfy vfx 9. 1.0 0. 1.0 0. 2. and 3.0 4.0 s.0 s.9 m The ∆y values at t = 0.8 m/s Using the law of Pythagoras and trigonometry: 2 2 2 vf = vfx + vfy vf = (8.0 24 44 30 [75° below horizontal] vf (m/s) 46 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .0 m/s) 2 + (9. The x-component is constant at 8.0 s.0 s are shown in Table 1.0 s.0 s) vfy = 9.0 0.0 8.8 m/s) 2 vf = 12. 2.Vertically (constant viy = 0 m/s ay = +g = 9. and 3.0 m/s θ = 50. 8 m/s 1.6 m/s v3y = 29.0 s and 3.4 m/s − 19. The horizontal component of the acceleration is zero because vx = constant = 8.4 m/s ∆t = 1.0 s: ay = = v3 y − v2 y ∆t 29.0 s and 2.0 s ay = ? For the vertical acceleration between 1.0 s: ay = = v2 y − v1 y ∆t 19. (c) (d) We can use components to find the average acceleration.8 m/s v2y = 19.6 m/s 1. we can conclude that the vertical acceleration is constant. and it equals the acceleration due to gravity.0 m/s. Copyright © 2003 Nelson Chapter 1 Kinematics 47 . v1y = 9.0 s 2 a y = 9.8 m/s Thus.6 m/s − 9.8 m/s2 For the vertical acceleration between 2.(b).0 s a y = 9. After the trials are complete. The apparatus shown in the text. Show that the vertical component of the motion is a vertical acceleration of magnitude 9.8 m/s 2 ∆t = ±0. ∆y = 83 cm = 0. Show that the horizontal component of the motion is a constant speed (i. Making Connections 7. Repeat this procedure with the target plate at increasing distances from the ramp (e.41 s Now consider the horizontal component of the motion: ∆x vix = ∆t 18. etc. Allow the steel ball.8 m/s2.4 m ay = 9.83 m ∆x = 18. page 46.. it is more fun to have the characters defy physics by appearing to be suspended in midair before plummeting downward at a great velocity. page 41. (b) The diagram looks like the projectile part shown in the text.g. Of course.800 m/s 2 ) The values for the maximum height are given in Table 2. Try This Activity: Comparing Horizontal Range (Page 49) A spreadsheet program using the correct equations can be used to create the table.0 cm.00 m/s (sin θ )) 2 2(−9.e.. (a) Start with the vertical target plate close to the bottom of the launching ramp. Solve for maximum height using vfy = 0 m/s and +y downward: vfy 2 = viy 2 + 2a y ∆y ∆y = ∆y = viy 2 2a y − ( 25. compare all the horizontal displacements).5.0 cm. 6. join the dots with a smooth curve to observe the path of the projectile.). Let +x be forward and +y be downward.0 cm. they would follow a projectile path after running off the edge of a cliff. while always releasing the ball from the same height. 4. is available commercially and is recommended as a relatively inexpensive way of performing projectile motion experiments. to roll down from the top of the ramp.41 s vix = 45 m/s The ball’s initial horizontal speed is 45 m/s.4 m = 0. Mark the exact location where the ball strikes the target plate. although the initial velocity is to the left rather than to the right. If cartoon characters obeyed the laws of physics. 48 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . Applying Inquiry Skills 6. initially at rest. say at a separation of 2.83m) 9.8 m/s2 vix = ? Determine ∆t: ∆t = ± =± 2 ∆y ay 2(0. Figure 3. 779 3.061 1.045 4.61 19.89 25.928 4.67 47.26 60.418 4.211 3.38 63.39 42.666 The following conclusions can be made for a projectile with an initial velocity at some angle above the horizontal.074 5.67 37.80 Horizontal Range (m) 6.54 31.26 20.71 25.2670 0.26 19.63 14.60 55.039 5. Solve for the horizontal range using the total time: ∆x = vix ∆t ∆x = 25 m/s (cos θ )∆t The values for the horizontal range are given in Table 2.128 4.378 2.26 55.60 47.94 19.275 6.94 31.38 60.87 22.65 58.075 2. the greater the maximum height of the projectile and the greater the time of flight.3484 0.61 27. Copyright © 2003 Nelson Chapter 1 Kinematics 49 .84 29.991 5. • The maximum horizontal range occurs when the angle of the initial velocity is 45° above the horizontal.792 3.11 31.92 25. • The greater the angle above the horizontal becomes.551 2.26 6.23 51.661 4.546 4.000 3.321 1.572 7.49 31.965 4.972 9.75 30.08734 0.43 63.095 Maximum Height (m) 0.39 51.414 3.459 11.5333 0.89 37.279 4.136 3.852 4.94 17.49 42. assuming that air resistance can be neglected.608 3.828 2.65 62.095 5.666 13. Table 2 Launch Angle 3° 6° 9° 12° 15° 18° 21° 24° 27° 30° 33° 36° 39° 42° 45° 48° 51° 54° 57° 60° 63° 66° 69° 72° 75° 78° 81° 84° 87° Time of Flight (s) 0.02 12.763 4.43 62.23 15.51 31.7981 1.316 2.7803 1.23 58.71 13.78 63.79 28.577 1. Values of the horizontal range are identical for angles equidistant on either side of 45°.43 23.32 26.Solve for the total time of flight (twice the time taken to get to maximum height): 1 ∆y = vfy ∆t − a y (∆t ) 2 2 −2∆y ∆t = ay total time = 2∆t The values for the time of flight are given in Table 2. Let +y be up.PRACTICE (Page 50) Understanding Concepts 8. 50 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . (a) vi = 2.6 × 102 m/s 4. (a) The vertical component of the ball’s velocity at the top of the flight is zero (vy = 0 m/s).8 m/s2 [down].6 × 10 2 m/s ∆t − 4.2 × 102 m/s)(cos 45°) vix = 1.8 m/s 2 ) ∆y = 1. (b) ∆t = ? 1 ∆y = viy ∆t + a y (∆t ) 2 2 0 =1.6 ×102 m/s At the highest position. (c) The rise time is equal to the fall time when the ball lands at the same level from which it was struck. (b) The ball’s acceleration at the top of the flight is 9. the y-component of the instantaneous velocity is zero (vfy = 0 m/s).9 m/s 2 ( ∆t ) 0 = ∆t 1. 9. Thus.2 × 103 m The maximum height of the cannonball is 1.9 m/s 2 ∆t ∆t = 0 or 2 ( ) (1.6 ×102 m/s)2 −2(−9.8 m/s2 ∆y = ? Find the horizontal and vertical components of the initial velocity: Horizontally (constant vix ): vix = vi cos θ = (2. 2 2 vfy = viy + 2a y ∆y 0 = viy + 2a y ∆y ∆y = = viy 2 2 −2 a y (1.2 × 102 m/s [45°above the horizontal] ay = –9.2 ×102 m/s)(sin 45°) viy = 1.6 × 10 2 m/s − 4.9 m/s 2 ∆t = 32s Therefore.2 × 103 m.9 m/s 2 ∆t ) = 0 ∆t = 1. the cannonball was fired at ∆t = 0 and the cannonball lands at ∆t = 32 s.6 × 102 m/s Vertically (constant a y ): viy = vi sin θ = (2.6 × 102 m/s − 4. 0 m/s 1 ∆y = viy ∆t + a y ( ∆t )2 2 −9. ∆t = = −b ± b 2 − 4ac 2a (where a = 4. but is at an angle of 45° below the horizontal.0 m/s)2 + 2( −9. and c = − 9.9 m/s 2 ( ∆t )2 − 8.5 m) vfy = 16 m/s Copyright © 2003 Nelson Chapter 1 Kinematics 51 .9 m/s)(2.6 ×10 2 m/s)(32 s) ∆x = 4.5 m Solve using the quadratic formula.4 s) ∆x = 22 m The width of the moat is 22 m.8 m/s2 ∆y = –9.9 m/s 2 )(−9.9 × 103 m.9 m/s vf = ? vfy 2 = viy 2 + 2 a y ∆y vfy = (8.4 s Thus.4 s.(c) ∆x = ? ∆x = vix ∆t = (1.2 × 102 m/s [45° below the horizontal].5 m ∆t = ? Find the horizontal and vertical components of the initial velocity: Horizontally (constant vix ): vix = vi cos θ = (12 m/s)(cos 42°) vix = 8.9 m/s Vertically (constant a y ): viy = vi sin θ = (12 m/s)(sin 42°) viy = 8. (d) The final velocity has the same magnitude as the initial velocity.9 m/s2 ( ∆t )2 0 = 4.9 m/s 2 ) ∆t = 2.9 × 103 m The horizontal range is 4. (b) ∆x = ? ∆x = vix ∆t = (8. b = − 8. (a) vi = 12 m/s [42°above the horizontal] ay = –9. the time of flight is 2. 10.0 m/s ∆t − 4.0 m/s)2 − 4(4.0 m/s ∆t − 9.9 m/s 2 . the final velocity is 2.8 m/s 2 )( − 9.0 m/s.5 m = 8.5 m) 2(4.0 m/s) ± (−8.5 m) −(−8. Let +y be up. Thus. (c) vfx = vfy = 8. (a) ∆y = –2. (a) For a projectile with the launch point lower than the landing point. Let +y be up. The initial velocity of the projectile is 29 m/s [horizontally].8 m/s2 viy = 0 m/s vi = ? First we must solve for the change in time: a y ( ∆t ) 2 ∆y = viy ∆t + 2 2∆y ∆t = ay = 2(−1. Section 1. ∆x = 16 m ∆y = –1. 3. The vertical acceleration of a projectile is the same throughout its trajectory and is equal to the acceleration due to gravity. the magnitude of the velocity is at a maximum just before landing (final velocity) and is at a minimum at the top of its trajectory. Let +y be up.9 m/s  θ = 60° The final velocity just before landing is 18 m/s [60° below the horizontal].8 m/s 2 ∆t = 0.9 m/s) 2 + (16 m/s)2 vf = 18 m/s tan θ = vfy vfx  16 m/s  θ = tan −1    8.5 m) −9.vf = vfx 2 + vfy 2 = (8.55s = vi = 29 m/s 4.4 Questions (Pages 50–51) Understanding Concepts 1. (b) For a projectile with the launch point higher than the landing point.55 s To calculate the initial velocity: vi = vix ∆x ∆t 16 m = 0.8 m/s2 viy = 0 m/s ∆t = ? Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson 52 . the magnitude of the velocity is at a maximum at the initial part of the flight (initial velocity) and is at a minimum at the top of its trajectory. 2.5 m ay = –9.5 m ay = –9. 71s) ∆x = 17 m The horizontal displacement of the ball is 17 m [fwd].∆y = viy ∆t + ∆t = = 2 ∆y ay 1 a y ( ∆t ) 2 2 2( −2.0 m/s)2 vf = 25 m/s tan θ = vfy vfx vfx = vix = 24 m/s  7.71s The tennis ball is in the air for 0. (c) vf = ? vfy 2 = viy 2 + 2a y ∆y vfy = (0 m/s) 2 + 2(−9.8 m/s 2 )( − 2. and then subtract the height of the net. (d) To calculate the distance d that the ball clears the net.5 m + ∆y = ? d = h – 0. (b) vix = 24 m/s ∆x = ? ∆x = vix ∆t = (24 m/s)(0.8 m/s2 ∆t = 0.0 m/s  θ = tan −1    24 m/s  θ = 16° The ball’s maximum velocity just prior to landing on the court surface is 25 m/s [16° below the horizontal]. first calculate the height of the ball above the ground at the net h. ∆x = 12 m h = 2.5 m) vfy = 7.9 m = ? First we must determine the change in time: ∆x ∆t = vix = 12 m 24 m/s ∆t = 0.71 s.5 m) −9.0 m/s vf = vfx 2 + vfy 2 = (24 m/s) 2 + (7.50s Copyright © 2003 Nelson Chapter 1 Kinematics 53 . b = 1. and c = −5.7 m/s) 2 + 2(9.7 m/s 54 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .7 m/s. (c) vf = ? vfy 2 = viy 2 + 2a y ∆y vfy = (1. ∆y = 6.5 m − 1.7 m/s (a) ay = 9. we can determine h and d: h = 2.0 m = 5.2 m) −(1.87 s) ∆x = 2.7 m/s Vertically (constant ay): viy = vi sin θ = (3.8 m/s 2 )(∆t )2 0= + 1.22 m = 1.2 m) 2(4. Let +y be down. we determine the change in the vertical component: a y ( ∆t ) 2 ∆y = viy ∆t + 2 2 ( −9. the ball clears the net by 0.87 s The ball is airborne for 0.38 m.8 m/s 2 )(5.3 m The child should hold the glove 2.8 m/s )(0.7 m/s)2 − 4(4. (b) ∆x = ? ∆x = vix ∆t = (2.9 m/s 2 .7 m/s) ± (1.22 m Now.2 m ∆y = viy ∆t + 1 a y ( ∆t ) 2 2 (9.2 m − 1. Horizontally (constant vix ): vix = vi cos θ = (3.9 m/s 2 ) ∆t = 0.Next.38 m Thus.28 m d = 1.2 m/s vfx = vix = 2.2 m/s)(cos 33°) vix = 2.9 m/s 2 )(−5.7 m/s ∆t − 5. Determine the horizontal and vertical components of the initial velocity.28 m − 0.2 m 2 Solve using the solution to the quadratic equation: ∆t = = −b ± b 2 − 4ac 2a (where a = 4.2 m/s)(sin 33°) viy = 1.7 m/s)(0.2 m) vfy = 10.3 m from the edge of the roof.87 s.50s)2 = 2 ∆y = −1.90 m = 0.8 m/s2 ∆t = ? 5. 6 × 102 s Therefore.7 m/s)2 + (10.8 × 102 m/s (a) Solve for the time of flight using the equation: 1 ∆y = viy ∆t + a y ( ∆t ) 2 2 1 0 = viy ∆t + a y (∆t )2 2 1 −viy ∆t = a y ( ∆t )2 2 −2viy ∆t = a y ( ∆t ) 2 ∆t = = −2viy ay −2 7. Copyright © 2003 Nelson Chapter 1 Kinematics 55 . the launch angles are 54° (same range as 36°).8 × 102 m/s)(1.2 × 105 m The horizontal range is 1.1 × 103 m/s)(cos 45°) vix = 7.2 m/s)2 6.4° (same range as 45.  10. Thus.8 × 102 m/s vix ): Vertically (constant a y ): viy = vi sin θ = (1. Let +y be up.6°). 7.6 × 102 s. Find the horizontal and vertical components of the initial velocity: Horizontally (constant vix = vi cos θ = (1.2 × 105 m.6 × 102 s) ∆x =1.vf = vfx + vfy vf = 11m/s tan θ = vfy vfx 2 2 = (2.8 m/s ∆t = 1.7 m/s  θ = 75° The ball’s final velocity just before it lands in the glove is 11 m/s [75° below the horizontal]. 74° (same range as 16°). and 44.8 × 102 m/s 2 ( ) −9. each shell was airborne for 1. As determined in the Try This Activity on page 49.2 m/s  θ = tan −1    2. (b) To determine the horizontal range: ∆x = vix ∆t = (7.1 × 103 m/s)(sin 45°) viy = 7. the horizontal range in this type of situation is identical for angles “equidistant” on either side of 45°. (b) ∆t = ? Solve for the time of flight using the equation: 1 ∆y = viy ∆t + a y ( ∆t )2 2 −15 m = 18 m/s ∆t − 0.6 m/s2 vi = 32 m/s [35° above the Moon’s horizontal] ∆yf = –15 m (a) Determine the horizontal and vertical components of the initial velocity.80 m/s 2 ( ∆t ) 2 2 0 = 0. vfy 2 = viy 2 + 2a y ∆y 0 = viy + 2a y ∆y ∆y = = −viy 2 2 2a y −(18 m/s)2 2( −1.8 m/s2 ) 8. the y-component of the instantaneous velocity is zero (vfy = 0 m/s). g = 1.6 m/s2 ) ∆y = 1. Horizontally (constant vix ): vix = vi cos θ = (32 m/s)(cos 35°) vix = 26 m/s Vertically (constant g): viy = vi sin θ = (32 m/s)(sin 35°) viy = 18 m/s Let +y be down.(c) At the highest position. 2 a y = − g = −1.6 m/s At the highest position.1 × 102 m. ∆y = 3. the y-component of the instantaneous velocity is zero (vfy = 0 m/s).80 m/s2 ( ∆t ) − 18 m/s ∆t − 15 m 56 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .1 × 104 m The maximum height of each shell is 3. Thus.1 × 104 m or 31 km.1 × 102 m The maximum height of the golf ball is 1.8 × 102 m/s)2 −2( −9. Thus. vfy 2 = viy 2 + 2 a y ∆y 0 = viy + 2a y ∆y ∆y = = viy 2 2 −2 a y (7. (c) ∆x = ? ∆x = vix ∆t = (26 m/s)(24 s) ∆x = 6. (In that text.2 ×10 2 m The horizontal range is 6. Applying Inquiry Skills 9. 10. vix = ? Using the vertical motion: ∆y = viy ∆t + ∆y = ∆t = 1 2 1 2 a y ∆t 2 a y ∆t 2 2 ∆y ay Using the horizontal motion: vix = = ∆x ∆t ∆x 2 ∆y ay vix = ∆x ay 2∆y The initial speed can be found by substituting the appropriate values into this equation. b = −18 m/s. ay = 9. refer to page 505. Let +x be the horizontal direction of the initial velocity and +y be down.8.2 × 102 m.80 m/s2 )( −15 m) 2(0. Part 3 of Investigation 16.8 m/s2 ∆x = ? Copyright © 2003 Nelson Chapter 1 Kinematics 57 . while the other coin will drop vertically downward.80 m/s2 . Figure 1 on that page shows that the device is simple to make and very inexpensive.2 m vi = 14 m/s [42° above the horizontal] g = 9.8 m/s 2 ) ∆t = 24 s The time of flight is 24 s.) Fold a piece of thin cardboard lengthwise so a vertical component facing upward separates two horizontal components.8 m/s2 viy = 0 ∆x and ∆y can both be measured with the metre stick.Solve using the solution to the quadratic equation: ∆t = = −b ± b2 − 4 ac where a = 0. If students used the Nelson Science 9 text. (a) ∆y = 2. One coin will project outward. and c = −15 m 2a −( −18 m/s) ± ( −18 m/s)2 − 4(0. Making Connections 11. they may recall designing and testing the type of device needed for this question. Place a coin on either side of the vertical component and rotate the device quickly on the horizontal plane.  −v sin θ + v 2 sin 2 θ + 2 g ∆y 2 2vi sin θ cos θ i i ∆x = 0. (b) vTS = 1.7 m/s [fwd] When walking toward the stern.9 m/s [fwd] When walking toward the bow. which is positive. Perhaps the world record holder has longer arms than the equation considers and was able to throw the shot with a slightly higher initial velocity than the value stated in the question. (c) Let the +x direction be forward and the +y direction be to the right.8 m/s2 9. Use the subscripts S for the ship. and T for the tourist group.30 m +      −(14 m/s) sin 42° + (14 m/s) 2 sin 2 42° + 2(9.1 m/s [fwd] vTW = ? vTW = vTS + vSW = 1.8 m/s2   = 0.3 m (b) (c) ∆x = 22 m The range of the shot is 22 m.7 m/s [fwd]. W for the water.1 m/s [backward] = –1.8 m/s2 )(2.8 m/s [fwd] vTW = 1. rather than using ay.30 m + 19.6 m + 2.30 m + + vi cos θ   g g  = 0. 1. vTS = 1.1 m/s [right] vTW = ? 58 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .8 m/s [fwd] vTW = 3.1 m/s [fwd] vTW = ? vTW = vTS + vSW = −1.5 FRAMES OF REFERENCE AND RELATIVE VELOCITY PRACTICE (Page 56) Understanding Concepts 1. The first term in the equation takes into consideration the fact that the thrower’s hand goes about 30 cm beyond the line from which the horizontal range is measured. The equation is set up using g. the group’s velocity relative to the water is 3. the group’s velocity relative to the water is 1. (a) vSW = 2.1m/s [fwd] + 2.2 m)  2(14 m/s)2 sin 42° cos 42°   + ° 14 m/s(cos42 )   9.1m/s [fwd] + 2. depending on which direction is defined as +y. The calculated value using the equation is slightly less than the world record. which can be positive or negative.9 m/s [fwd].8 m/s [fwd] vTS = 1. The second term is the same as the equation derived on page 49 for the situation in which the projectile lands at the same level from which it began. vLP = vLM + vMN + vNO + vOP 2. (a) vLE = vLD + vDE (b) vAC = vAB + vBC (c) vMN = vMT + vTN vNM = vNT + vTM (d) Replace vML with vLM . The third term takes into consideration the fact than the horizontal range increases when the projectile lands at a level lower than its original level. 8 m/s [fwd] vTW = 1.5 FRAMES OF REFERENCE AND RELATIVE VELOCITY PRACTICE (Page 56) Understanding Concepts 1. which is positive. the group’s velocity relative to the water is 3. Perhaps the world record holder has longer arms than the equation considers and was able to throw the shot with a slightly higher initial velocity than the value stated in the question. The calculated value using the equation is slightly less than the world record.1m/s [fwd] + 2. The third term takes into consideration the fact than the horizontal range increases when the projectile lands at a level lower than its original level.1 m/s [fwd] vTW = ? vTW = vTS + vSW = 1.7 m/s [fwd].6 m + 2.1m/s [fwd] + 2. The equation is set up using g.7 m/s [fwd] When walking toward the stern.8 m/s2 )(2. vLP = vLM + vMN + vNO + vOP 2.1 m/s [fwd] vTW = ? vTW = vTS + vSW = −1. 1.1 m/s [backward] = –1. vTS = 1. The second term is the same as the equation derived on page 49 for the situation in which the projectile lands at the same level from which it began. and T for the tourist group.8 m/s2 9.9 m/s [fwd] When walking toward the bow. (b) vTS = 1. depending on which direction is defined as +y. −v sin θ + v 2 sin 2 θ + 2 g ∆y 2 2vi sin θ cos θ i i ∆x = 0. which can be positive or negative. (c) Let the +x direction be forward and the +y direction be to the right.1 m/s [right] vTW = ? 58 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . (a) vLE = vLD + vDE (b) vAC = vAB + vBC (c) vMN = vMT + vTN vNM = vNT + vTM (d) Replace vML with vLM .2 m)  2(14 m/s)2 sin 42° cos 42°   + ° 14 m/s(cos42 )   9. (a) vSW = 2.30 m + 19.9 m/s [fwd]. The first term in the equation takes into consideration the fact that the thrower’s hand goes about 30 cm beyond the line from which the horizontal range is measured.30 m +      −(14 m/s) sin 42° + (14 m/s) 2 sin 2 42° + 2(9.8 m/s [fwd] vTW = 3. W for the water. rather than using ay. the group’s velocity relative to the water is 1.30 m + + vi cos θ   g g  = 0.3 m (b) (c) ∆x = 22 m The range of the shot is 22 m.8 m/s2   = 0.8 m/s [fwd] vTS = 1. Use the subscripts S for the ship. W for the water.2 m/s vTW.y tan θ =  v  TW.8 m/s + 2.y = (320 km/h)sin28° + 72 km/h vPG.1 m/s [E] vSW = 2.1m/s) 2 vTW = 3.x 2 + vTW.  1.y = 1.8 m/s [N] vWC = 2.y = 1. and G for the ground.y tan θ =  v  TW.4 m/s [N] vTW = ? vTC = vTS + vSW + vWC vTW.1m/s) 2 vTW = 5.y = 222 km/h Copyright © 2003 Nelson Chapter 1 Kinematics 59 . T for the tourist. Use the subscripts S for the ship.3 m/s [12° E of N].8 m/s vTW.x = 2. the group’s velocity relative to the water is 3.  1. vTS = 1.x = (320 km/h)cos28° vPG. Let the +x direction be north and the +y direction be east.2 m/s)2 + (1.3 m/s  vTW.0 m/s  vTW. Let the +x direction be west and the +y direction be south.x = 2. vPA = 320 km/h [28° S of W] vAG = 72 km/h [S] ∆t = 2.1m/s vTW = vTW.y 2 = (5.0 m/s [21° right of fwd]. and C for the coast.1m/s  θ = tan −1    2.x     3.x 2 + vTW.1m/s  θ = tan −1    5. A for the wind (air).2 m/s  θ = 12° The velocity of the tourist group relative to the coast is 5. Use the subscripts P for the plane.8 m/s  θ = 21° When walking toward the starboard.y 2 = (2.x     4.1m/s vTW = vTW.4 m/s = 5. vTW = vTS + vSW vTW.8 m/s) 2 + (1.0 h d PG = ? vPG = vPA + vAG vPG.x = 283 km/h vPG. Making Connections 5.x = 24 m/s vHG. In travelling westward.and low-pressure weather systems.x = (55 m/s)cos 35° − 21 m/s vHG. Let vKW be the velocity of either kayaker relative to the water and ∆d the magnitude of the displacement across the river. vPG = 360 km/h [38°S of W] .y = 32 m/s 60 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .x      222 km/h  θ = tan −1    283 km/h  θ = 38° Therefore. (a) vHA = 55 m/s [35° N of W] vAG = 21 m/s [E] vHG = ? vHG = vHA + vAG vHG. from Tokyo to Toronto). Section 1. they generally move from west to east and influence the motion of most of the major high. The jet streams are high-speed winds in the high regions of the troposphere.2 × 102 km [38°S of W] The plane’s displacement from Winnipeg is 7. perpendicular to the shores. or north or south of it. the kayaker who aims straight across the river reaches the far side first. A for the wind (air). For the kayaker that aims straight across the river: vKW = ∆t1 = ∆d ∆t1 ∆d vKW ∆d ∆t 2 For the kayaker who aims upstream at an angle θ to the near shoreline: vKW sin θ = ∆t 2 = ∆d vKW sin θ 2. and G for the ground. Pilots can take advantage of a jet stream only when travelling from west to east (e.5 Questions (Page 57) Understanding Concepts 1.. Since ∆t2 > ∆t1.y tan θ =  v  PG. d PG = vPG ∆t = (360 km/h [38°S of W] )(2 h) d PG =7.2 × 102 km [38° S of W].g. Let the +x direction be west and the +y direction be north.y 2 = (283 km/h)2 + (222 km/h)2 vPG = 360 km/h  vPG.y = (55 m/s)sin 35° vHG. pilots try to avoid the jet stream by flying below it. vPG = vPG.x 2 + vPG. Like other prevailing winds. Use the subscripts H for the helicopter. y tan θ =  v  HG.y = 72 m 0.75 m/s [across the river] ∆y = 72 m ∆x = 54 m (a) vWG = ? ∆y ∆t = vSW.y = (55 m/s)sin 35° + (21 m/s)cos 22° vHG.x = (55 m/s)cos 35° + (21 m/s)sin 22° vHG.75 m/s ∆t = 96s Copyright © 2003 Nelson Chapter 1 Kinematics 61 .x = (53m/s)2 + (51m/s)2 vHG.y tan θ =  v  HG. W for the water. and G for the ground.x = (24 m/s) 2 + (32 m/s) 2      32 m/s  θ = tan −1    24 m/s  θ = 53° The velocity of the helicopter relative to the ground is 4. vHG = vHG.x 2 + vHG. vSW = 0. Let the +x direction be downstream from the initial shore and the +y direction be across the river. (b) vHA = 55 m/s [35° N of W] vAG = 21 m/s [22° W of N] vHG = ? vHG = vHA + vAG vHG.x = 53m/s vHG = vHG. Use the subscripts S for the swimmer.y 2 vHG = 40 m/s  vHG. 3.y 2 vHG = 74 m/s  vHG.0 × 101 m/s [53° N of W].x 2 + vHG.y = 51m/s      51m/s  θ = tan −1    53m/s  θ = 44° The velocity of the helicopter relative to the ground is 74 m/s [44° N of W]. Let the +y direction be south and the +x direction be east.56 m/s)2 + (0.56 m/s The speed of the river current is 0. 42° from the shore. ∆x vWG = ∆t 54 m = 96s vWG = 0. and G for the ground.y = vSW = 0. A for the wind (air).75 m/s) 2  0.5 h vAG = 75 km/h [E] vPA = ? 62 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .4 × 103 km [43° E of S] ∆t = 3.56 m/s.94 m/s  vSG.y tanθ =  v  SG.56 m/s  θ = 53° The swimmer’s velocity relative to the shore is 0.50 m/s  vSG tan θ =  v  WG     4. ∆ d = 1. Use the subscripts P for the plane.56 m/s  θ = 42° The direction in which she should have aimed to land directly across from her starting position is upstream.94 m/s [downstream.75 m/s vSG = ? vSG = vHG. 53° from the initial shore]. (b) vSG = vSW + vWG vSG.75 m/s)2 − (0.y 2 vSG = 0.75 m/s  θ = tan −1    0.x     = (0.50 m/s  θ = tan −1    0.x 2 + vHG.56 m/s)2 vSG = 0.x = vWG = 0.56 m/s vSG.  0. (c) vSG = ? vSG = vSW + vWG 2 2 2 vSG = vSW + vWG 2 2 2 vSG = vSW − vWG vSG = (0. d vPG = PG ∆t 1.4 × 103 km [43° E of S] = 3.9 × 102 km/h)2 vPA.y      = (2.5 × 102 km/h [34° E of S] Applying Inquiry Skills 5.5 h 2 vPG = 4.y = (4.0 × 102 km/h)2 + (2. we see that the drops appear to be at an angle θ below the horizontal such that: 2 hand spans tan θ = 1 hand span tan θ = 2 1 Copyright © 2003 Nelson Chapter 1 Kinematics 63 .and y-components: vPA.x = (4.0 × 102 km/h  θ = tan −1   2  2.0 ×10 2 km/h) sin 43° − 75 km/h vPA.y 2 vPA = 3.5 × 102 km/h  vPA.x = 2.x = 2.0 × 102 km/h To calculate the velocity of the plane: vPA = vPA.x 2 + vPA.9 × 102 km/h  2.9 × 10 km/h  θ = 34° The velocity of the plane relative to the air is 3.0 ×10 km/h [43° E of S] vPG = vPA + vAG vPA = vPG − vAG To determine the x.0 × 102 km/h) cos 43° vPA.x tan θ =   vPA. (a) vtrain = 64 km/h vrain = ? Referring to Figure 9 on page 57 of the text. (c) A typical graph of the predicted motions: 64 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . and report on a relatively easy experiment. As the mass increases the downward force of gravity increases. it is reasonable to say that the terminal speed increases as the number of filters added increases. Thus. From the video of the weather report. but the upward force of air resistance may not increase in proportion to the increase in gravity. and then apply the equation: vwind tan θ = vrain vwind = vrain tan θ vwind = (32 km/h)( tan θ ) CHAPTER 1 LAB ACTIVITIES Investigation 1.3. Question (a) A typical question is “How does the terminal speed of a falling coffee filter depend on the mass of the filter?” Hypothesis/Prediction (b) Since the surface area of the flat part of the falling filter in contact with the air is constant for all three trials. (b) Estimating an angle by using hand spans is a major source of error. is that different drops move at different angles. Another source of error.As shown in the diagram: vtrain tan θ = vrain vtrain vrain = = tan θ 64 km/h vrain = 32 km/h The speed of the raindrops is 32 km/h. 2   1 Making Connections 6. estimate the angle θ. (This corresponds to the fact that lighter water droplets have a lower terminal speed than heavier ones. not shown in the text.1: Comparing Terminal Speeds (Page 58) This investigation is one of the few in this text in which the students are required to display all of the Inquiry Skills listed at the top of the page.) Another source of error is knowing or estimating the speed of the train. it provides a good opportunity for assessing students’ abilities to design. carry out. Refer to the solution to question 5(a) above. not shown in the text. carry out.As shown in the diagram: vtrain tan θ = vrain vtrain vrain = = tan θ 64 km/h vrain = 32 km/h The speed of the raindrops is 32 km/h. (c) A typical graph of the predicted motions: 64 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . is that different drops move at different angles.1: Comparing Terminal Speeds (Page 58) This investigation is one of the few in this text in which the students are required to display all of the Inquiry Skills listed at the top of the page. Refer to the solution to question 5(a) above. and report on a relatively easy experiment. it is reasonable to say that the terminal speed increases as the number of filters added increases. 2   1 Making Connections 6. Question (a) A typical question is “How does the terminal speed of a falling coffee filter depend on the mass of the filter?” Hypothesis/Prediction (b) Since the surface area of the flat part of the falling filter in contact with the air is constant for all three trials. As the mass increases the downward force of gravity increases.3. it provides a good opportunity for assessing students’ abilities to design. (b) Estimating an angle by using hand spans is a major source of error.) Another source of error is knowing or estimating the speed of the train. estimate the angle θ. Thus. (This corresponds to the fact that lighter water droplets have a lower terminal speed than heavier ones. and then apply the equation: vwind tan θ = vrain vwind = vrain tan θ vwind = (32 km/h)( tan θ ) CHAPTER 1 LAB ACTIVITIES Investigation 1. Another source of error. but the upward force of air resistance may not increase in proportion to the increase in gravity. From the video of the weather report. it moves away from the vertical and the motion might not even be picked up by the sensor. Question See page 59. Such collisions must be avoided. With good organization and student cooperation. (ii) Add a second filter inside the first one. and then repeat step (i). This error can be minimized by activating the sensor just prior to letting the filter(s) drop. one of the biggest sources of error in this investigation is trying to get the filters to drop straight downward. if the filter is not horizontal. even if there is only one air table. an entire class can perform the observations in a single 70-min period. say between 5 cm and 8 cm higher than the front. (i) The direction of the acceleration of the puck is down the inclined plane. or they can have the computer generate the graph. Hypothesis (a) The “correct” answers to the questions are given here.1: Investigating Projectile Motion (Pages 58–60) The air table shown on page 59. Repeat for accuracy. The table should be elevated a small amount at the back of the table. This is true as long as the horizontal component of the motion is at a constant speed (or zero). and so on. Figure 1 of the text is highly recommended for studying projectile motion. or falling through a single liquid at different. The second part of the experiment could involve a single sphere falling through liquids of different density but the same temperature. (j) One part of the experiment would involve spheres of equal diameter but different masses all falling through the same liquid. the puck will accelerate too quickly and may crash into the lower edge. Materials (f) The materials and apparatus needed per group are: • 3 flat-bottom coffee filters • motion sensor • computer with software to operate the motion sensor Analysis (g) Students can generate their own graph based on their observations. Another source of error is moving the filter vertically while holding it as the sensor is activated.Experimental Design (d) The main safety concern is to be sure that electric wires that may be along the floor are concealed so nobody trips over them. If that acceleration is straight down the inclined plane and is constant in magnitude for all motions on the Copyright © 2003 Nelson Chapter 1 Kinematics 65 . students could use a metre stick and a stopwatch to collect data to determine the approximate terminal speed. Evaluation (h) Each student’s evaluation depends on his or her hypothesis and prediction. they can begin analyzing the data as other groups use the air table. (ii) One way to show the independence is to find the magnitude and direction of the acceleration of the puck on the inclined plane. say water at room temperature. Hold the first filter directly above the sensor. If the table is set at too high an angle. It takes practice to be able to allow the filter to fall horizontally. Furthermore. (e) Refer to the text. The air resistance on a filter is maximum when the bottom surface remains horizontal during the fall. The instructions have been written so that after students gather data for one motion. Investigation 1. Repeating trials helps to minimize the effects of this type of error. and then repeat step (i). Rather than using a motion sensor. which in turn is true in the absence of any friction. (iii) Add a third filter inside the first one. controlled temperatures.4. (i) As students soon discover. Connect the sensor to the computer and set up the program so it can record the motion (speed-time data and graph) of the filter dropped from a preset position. and then let it drop flat side down toward the sensor. (i) Position the motion sensor on the floor so it faces upward. The graph above in part (c) shows the shapes of the curves on a typical graph of the results. although students may not be expected to know the answers before the investigation is complete. Connect the sensor to the computer and set up the program so it can record the motion (speed-time data and graph) of the filter dropped from a preset position.4. which in turn is true in the absence of any friction. and then repeat step (i). if the filter is not horizontal. (j) One part of the experiment would involve spheres of equal diameter but different masses all falling through the same liquid. Such collisions must be avoided. Evaluation (h) Each student’s evaluation depends on his or her hypothesis and prediction. Hold the first filter directly above the sensor. Repeating trials helps to minimize the effects of this type of error. (i) As students soon discover. (iii) Add a third filter inside the first one. It takes practice to be able to allow the filter to fall horizontally. say between 5 cm and 8 cm higher than the front. (ii) Add a second filter inside the first one. and then repeat step (i). If the table is set at too high an angle. or they can have the computer generate the graph. This is true as long as the horizontal component of the motion is at a constant speed (or zero). one of the biggest sources of error in this investigation is trying to get the filters to drop straight downward. Figure 1 of the text is highly recommended for studying projectile motion. or falling through a single liquid at different. although students may not be expected to know the answers before the investigation is complete. they can begin analyzing the data as other groups use the air table.Experimental Design (d) The main safety concern is to be sure that electric wires that may be along the floor are concealed so nobody trips over them. (ii) One way to show the independence is to find the magnitude and direction of the acceleration of the puck on the inclined plane. Question See page 59.1: Investigating Projectile Motion (Pages 58–60) The air table shown on page 59. Repeat for accuracy. controlled temperatures. and so on. say water at room temperature. even if there is only one air table. The instructions have been written so that after students gather data for one motion. the puck will accelerate too quickly and may crash into the lower edge. The table should be elevated a small amount at the back of the table. and then let it drop flat side down toward the sensor. If that acceleration is straight down the inclined plane and is constant in magnitude for all motions on the Copyright © 2003 Nelson Chapter 1 Kinematics 65 . (i) Position the motion sensor on the floor so it faces upward. students could use a metre stick and a stopwatch to collect data to determine the approximate terminal speed. With good organization and student cooperation. The air resistance on a filter is maximum when the bottom surface remains horizontal during the fall. Rather than using a motion sensor. Hypothesis (a) The “correct” answers to the questions are given here. (i) The direction of the acceleration of the puck is down the inclined plane. Another source of error is moving the filter vertically while holding it as the sensor is activated. an entire class can perform the observations in a single 70-min period. it moves away from the vertical and the motion might not even be picked up by the sensor. This error can be minimized by activating the sensor just prior to letting the filter(s) drop. Materials (f) The materials and apparatus needed per group are: • 3 flat-bottom coffee filters • motion sensor • computer with software to operate the motion sensor Analysis (g) Students can generate their own graph based on their observations. Investigation 1. Furthermore. The second part of the experiment could involve a single sphere falling through liquids of different density but the same temperature. (e) Refer to the text. The graph above in part (c) shows the shapes of the curves on a typical graph of the results. (For motion 2. down the inclined plane and equal in magnitude). (c) A sample calculation is shown here using the sample shown in Procedure 1.7°) 2 a y = 0. it is important to move the vectors exactly parallel to themselves. Figure 4.plane. When students perform vector subtractions by completing vector scale diagrams.1 cm = sin −1 61 cm θ = 5. that is fairly easy to do because the vectors are close together. but if supplies are limited..8 cm change in elevation.8 cm − 3.97 m/s2. (ii) Since the acceleration in all three cases is the same (i. Materials The instructions suggest three sheets of construction paper per student.7 cm height of rear of table = 9. or directed at an angle above the horizontal. height of front of table = 3. then the horizontal motion must not be affected by the vertical motion.e.97 m/s 2 ) × 100% . However.1 cm length of table (front to rear). Procedure 1.05 m/s 2 + 0. only one sheet per student would be needed if the first two motions were placed on one side of the sheet and the third motion were placed on the reverse side. the acceleration down the table is 0. this analysis is similar to what is shown in the text. A typical calculation of the angle of the table is given below. the projectile has acceleration in the vertical plane that is independent of whether the initial velocity was zero. students may benefit from the Learning Tip about transferring vectors on pages 264–265 of the text (in Investigation 5.7 cm = 6. ∆y sin θ = L ∆y θ = sin −1 L 6. L = 61 cm Let the angle of the table above the horizontal be θ. in the +x direction.05 m/s2 [down the inclined plane].3. 66 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson ( 1.97 m/s 2 1 2 % difference = 8% (e) (i) The direction of the acceleration of the projectile is straight down the inclined plane.1 where the vectors are relatively far apart). Assume that the average of the accelerations for motion 3 is 1.05 m/s 2 − 0.) The horizontal displacements are constant. a y = g sin θ = (9. Another way is to compare the individual horizontal displacements covered in equal time intervals. difference in values % difference = × 100% average of values = 1. while the vertical displacements are constantly changing.97m/s In this example.8 m/s 2 )(sin 5. Analysis (b) Within experimental error. (d) A sample calculation is shown here for any one of the motions. and then compare the observations with the individual vertical components of the displacements covered in equal time intervals. Another way of showing the independence of the vertical and horizontal components is to compare the horizontal displacements between the dots for motions 2 and 3 with the vertical displacements between the dots.7° Steps 2–6: The three different motions and mathematical manipulations of the vectors are shown in the text. page 42. In this investigation. the magnitude and direction of the accelerations should be the same for all three motions. ∆y = 9. therefore θ1 = θ2 = θ. (j) The diagram below shows the situation.) • transferring the vectors parallel to themselves to perform vector subtractions (Use one or two rules as accurately as possible.) • twisting effects of the cord (Be sure that the cord connected to the air puck is not twisted. Copyright © 2003 Nelson Chapter 1 Kinematics 67 . ay perpendicular to the inclined plane. if the ∆t values become too small.4. the percent uncertainty in measuring the lengths of the vectors.) • measuring the angle of the air table (Be sure the ruler is vertical and reduce parallax error in measuring the elevation of the front and rear edges of the table above the bench top. Brancazio. and measure the vectors as accurately as possible without parallax error by viewing the ruler straight on. (i) The question asks for percent difference rather than percent error because both values result from experimental measurements. So the best value of ∆t is small but not too small. “Physics buys a ticket to the Super Bowl” by Steven Strauss. Although this article is relatively old. which has two components. “The Physics of Kicking a Football” by Peter J. (g) The main random sources of error and how to minimize them are: • levelling the table horizontally (Use a level to be sure the left-to-right orientation of the table is level. and ax parallel to the inclined plane. Science and Physics section. The puck is a projectile only after that force has been removed. both the magnitude and direction of the acceleration will be wrong. If the dots are included in the vector analysis. 403–407. However.) The main source of systematic error is in using dots on the construction paper created when the puck was still being pushed by hand. exercise care in drawing the tangents to the curve. In practice. January 28. β1 = β2. (a) Answers will vary greatly and are not expected to be accurate. Synthesis (h) Theoretically. ) • drawing and measuring the vectors (Use a sharp pencil to draw the vectors. The Physics Teacher. The Globe and Mail. it is better to use smaller values of ∆t because we are trying to find the instantaneous acceleration of the puck at various instants. the advantage of using relatively small ∆t values is that vector subtractions can be carried out more accurately. Now the component of g parallel to the inclined plane is g sinθ. Saturday. October 1985.Evaluation (f) Answers will depend on the initial hypothesis. Here is a potential response: The range of angles will go from about 45° to 60° in order to maximize both the hang time and the horizontal range. pp. 1995.) • rounding errors (Interim answers should be kept in the calculator and answers should be rounded off at the end of the calculations. From the diagram: θ1 + β1 = 90° θ2 + β2 = 90° However. it still acts as a standard for this topic. do not use the first 3 or 4 dots of the motion.1: Hang Time in Football (Pages 60–61) The following references are highly recommended for teachers and motivated students. To minimize this error. The acceleration due to gravity has a magnitude of g. becomes quite high. especially the ∆v vectors. Lab Exercise 1. it still acts as a standard for this topic. if the ∆t values become too small. 1995. (j) The diagram below shows the situation. So the best value of ∆t is small but not too small. the advantage of using relatively small ∆t values is that vector subtractions can be carried out more accurately. β1 = β2. In practice.Evaluation (f) Answers will depend on the initial hypothesis. The acceleration due to gravity has a magnitude of g. Synthesis (h) Theoretically. 403–407.1: Hang Time in Football (Pages 60–61) The following references are highly recommended for teachers and motivated students.) • measuring the angle of the air table (Be sure the ruler is vertical and reduce parallax error in measuring the elevation of the front and rear edges of the table above the bench top. Brancazio.) • transferring the vectors parallel to themselves to perform vector subtractions (Use one or two rules as accurately as possible.) The main source of systematic error is in using dots on the construction paper created when the puck was still being pushed by hand. pp. therefore θ1 = θ2 = θ. ) • drawing and measuring the vectors (Use a sharp pencil to draw the vectors. Lab Exercise 1. both the magnitude and direction of the acceleration will be wrong.4. January 28. If the dots are included in the vector analysis. the percent uncertainty in measuring the lengths of the vectors. (i) The question asks for percent difference rather than percent error because both values result from experimental measurements. (g) The main random sources of error and how to minimize them are: • levelling the table horizontally (Use a level to be sure the left-to-right orientation of the table is level. do not use the first 3 or 4 dots of the motion. “Physics buys a ticket to the Super Bowl” by Steven Strauss. Now the component of g parallel to the inclined plane is g sinθ. However. Although this article is relatively old.) • twisting effects of the cord (Be sure that the cord connected to the air puck is not twisted. (a) Answers will vary greatly and are not expected to be accurate. October 1985. exercise care in drawing the tangents to the curve. which has two components. The puck is a projectile only after that force has been removed. The Globe and Mail. ay perpendicular to the inclined plane.) • rounding errors (Interim answers should be kept in the calculator and answers should be rounded off at the end of the calculations. and ax parallel to the inclined plane. “The Physics of Kicking a Football” by Peter J. especially the ∆v vectors. From the diagram: θ1 + β1 = 90° θ2 + β2 = 90° However. To minimize this error. The Physics Teacher. Saturday. becomes quite high. it is better to use smaller values of ∆t because we are trying to find the instantaneous acceleration of the puck at various instants. Here is a potential response: The range of angles will go from about 45° to 60° in order to maximize both the hang time and the horizontal range. Science and Physics section. and measure the vectors as accurately as possible without parallax error by viewing the ruler straight on. Copyright © 2003 Nelson Chapter 1 Kinematics 67 . while hang time continues to rise. some students can be responsible for determining the horizontal range while other students are responsible for obtaining a video to determine the launch angle and measuring hang time) Synthesis (i) The closed stadium would prevent the source due to error of any wind blowing outdoors. (j) Analysis helps to determine how to maximize certain variables.g. Thus. punters can practise punting so that the launch angle of the football is between 55° and 60°. (Refer to the results of the Try This Activity on page 49 of the text. parallel to the sidelines (g) Sources of random error are: • estimating the horizontal range of the ball in each trial • estimating the angle of each launch • measuring the time the ball was in the air in each trial • any angle between the camera’s line of sight and the kick • any lateral motion of the ball relative to the sidelines The main source of systematic error occurs when estimating the angle of launch based on the view of a camera that is at a level well above the playing field. whereas with punts the players are blocked by the opposing team and need more time to get downfield. (In actual football games. the best launch angle is likely close to that of an “ideal” projectile. Accept any reasonable explanation. and the maximum hang time occurs as the launch angle approaches 90°. A good value of both range and hang time occurs at launch angles around 60°. the maximum horizontal range is reached at launch angles just below and up to 45°. (d) For an “ideal” projectile.6 s. hang time is more important for punts than kickoffs because in kickoffs all the players on the kicking team are running at nearly top speed when the ball is kicked. the maximum horizontal range occurs for a launch angle of 45°. The horizontal range drops substantially at launch angles higher than 60°. 68 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . the best launch angles for kickoffs are around 45° and the best angles for punts are around 60°. (f) The main assumptions are: • the data were analyzed for games in which the wind was negligible • the TV camera was located so its line of sight was along a field line so the launch angle could be estimated from a straight-line view • the height above the field that the ball was kicked and caught was approximately constant • the ball was kicked straight downfield. Thus. For a football punt. (h) Some ways of obtaining accurate data are: • use an actual activity that can be analyzed using a combination of direct measurements and measurements from a video or a digital camera • have several students aid in gathering data (e. Evaluation (e) The answer depends on the prediction. and then the athletes can practise until they approach the ideal.8 s to 4.. and hang time increases as the launch angle increases.) Between 45° and 90° there is a compromise between these two variables.Analysis (b) (c) The angles between 45° and 60° yield horizontal ranges from 49 m to 60 m and hang times from 3. For example. in the region of 55° to 60°. 5 m/s is not an example of motion with constant velocity because the direction is always changing even though the magnitude is constant. 5. Copyright © 2003 Nelson Chapter 1 Kinematics 69 . Florida. 3. the magnitude of the acceleration due to gravity at Miami. Many of the concepts and equations in Chapter 1 are included in the diagram. John’s. F This motion provides an example of one-dimensional motion (not two-dimensional motion) because the ball has travelled along a single line even though it has reversed directions F The magnitude of the velocity just before landing is equal to (not greater than) the magnitude of the initial velocity as the ball leaves your hand.CHAPTER 1 SUMMARY (Page 62) Make a Summary This illustration shows the start of a projectile motion diagram with some of the labels suggested. F The acceleration at the top of the flight is equal to the acceleration due to gravity and is constant throughout the ball’s motion. T F A jogger running four laps around a circular track at 4. 2. CHAPTER 1 SELF QUIZ (Page 63) True/False 1. Newfoundland. because Miami is closer to the equator where the distance to Earth’s centre is greater. 6. 7. is less than that at St. 4. T T F At sea level. 8. 4. 6. Newfoundland. Many of the concepts and equations in Chapter 1 are included in the diagram. 3. is less than that at St. F This motion provides an example of one-dimensional motion (not two-dimensional motion) because the ball has travelled along a single line even though it has reversed directions F The magnitude of the velocity just before landing is equal to (not greater than) the magnitude of the initial velocity as the ball leaves your hand. 7. 8. the magnitude of the acceleration due to gravity at Miami. T T F At sea level. John’s. 2. Florida. F The acceleration at the top of the flight is equal to the acceleration due to gravity and is constant throughout the ball’s motion. CHAPTER 1 SELF QUIZ (Page 63) True/False 1. Copyright © 2003 Nelson Chapter 1 Kinematics 69 . T F A jogger running four laps around a circular track at 4.CHAPTER 1 SUMMARY (Page 62) Make a Summary This illustration shows the start of a projectile motion diagram with some of the labels suggested.5 m/s is not an example of motion with constant velocity because the direction is always changing even though the magnitude is constant. because Miami is closer to the equator where the distance to Earth’s centre is greater. 5. (b) A ball is tossed vertically upward from your hand: the initial position is your hand. then vBA = 8. F If vAB = 8.5 m/s [W] .2 m/s [N] vf = 7. (e) We can use components to solve the problem.0 × 101 s vf = ? vf − vi aav = ∆t vf = vi + aav ∆t = 25 m/s[E] + (−2. vi = 7. 2a 10.5 m/s 2 [E])(2. +y is upward. (e) ax = 2. Multiple Choice 12.2 m/s)2  7.2 m/s 2  θ = tan −1  2   2.2 m/s [45° W of N].5 m/s2 ay = 6. +y is downward. which has the solution ∆t = −b ± b 2 − 4ac . F The quadratic formula must be used to solve problems involving the quadratic equation of the form a ∆t 2 + b∆t + c = 0 . ∆v = 10. ∆v y = vfy − viy ∆v x = vfx − vix ∆v x = 7. 70 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . 14. (c) vi =25 m/s [E] a = 2.2 m/s ∆v y = −7.2 m/s  θ = 45° Thus.2 m/s)2 + (−7. +y is up the ramp.0 × 101 s) vf = −25 m/s [E]or 25 m/s [W] 17. T 11.0 s Let +x be west and +y be north.5 m/s2 [W] = –2.2 m/s ∆v = ∆v x 2 + ∆v y 2 ∆v = 10. 15.2 m/s  θ = tan −1    7.9. 18.5 m/s2 [E] ∆t = 2.2 m/s [W] ∆t = 2.2 m/s2 θ=? tan θ = ay ax  6. +y is upward.5 m/s [E]. (d) A cart is released from rest at the top of a ramp: the initial position is at the top of the ramp. (a) A ball is tossed vertically upward from your hand: the initial position is your hand. 13.5 m/s  θ = 68° The direction of the acceleration is 68° N of E. (d) A rubber stopper is dropped from your raised hand: the initial position is your hand. 16.2 m/s tan θ = ∆v y ∆vx = (7. (b) No. To convert m/s to km/h.5 × 10 km/h  3  1h × 1 10 m    The peregrine falcon can dive at speeds of 3. ? 4. ∆v aav = ∆t 10. (c) Yes. for example.2 m/s [45° W of S] = 2. (a) L × T–1 results in speed. No. 7. two vectors having different magnitudes cannot be combined to give a zero resultant vector. A vector can be broken into components using sine and cosine. (a) Yes.6 10 s     Thus. (b) The instantaneous acceleration is the slope of the line (for constant acceleration) or the slope of the tangent to the curve (for nonconstant acceleration) on a velocity-time graph.6. for example. (c) To convert km/h to m/s. for example. (c) constant vix = 5. 3  1km   3. 2. which have a maximum of one. T  3. In (a). 5. multiply by 3. 5 m [S] + 5 m [N] = 0.6. can total zero. the equation could be one of the following: 1 ∆d = v ∆t + a ∆t 2 i 2 1 2 or ∆d = a ∆t 2 (a) The instantaneous speed is equal to the average speed throughout the motion.0s 2 aav = 5. two vectors having the same magnitude can be combined to give a zero resultant vector by adding two vectors that have opposite directions. (a) ∆d = a (∆t ) 2 Therefore: ?  L  L =  2  × T2 T  L=L The equation is dimensionally correct because both sides of the equation have the same dimensions. Even in one dimension three vectors can add to zero. 2 m [S] + 5 m [N] + 3 m [S] = 0.6 × 10 s  2 (b) 97 m/s    = 3. a component of a vector cannot have a magnitude greater than the vector’s magnitude. (b) The instantaneous velocity is equal to the average velocity throughout the motion. three vectors having different magnitudes can be combined to give zero. (a) The displacement is equal to the area under a velocity-time graph. (c) The instantaneous speed is equal to the magnitude of the average velocity throughout the motion. L  L  (b)  3  T = 2 results in acceleration. Copyright © 2003 Nelson Chapter 1 Kinematics 71 . 6. T T    L  (c)  2  T T = L results in length. Adding three vectors in two dimensions head-to-tail.8 m/s.1 m/s [45° W of S] 19. (a) 100 km/h  3   1km   × 3.5 × 102 km/h. (b) An equation may be dimensionally correct yet still be wrong.8 m/s 1. divide by 3. 100 km/h is 27.5 m/s CHAPTER 1 REVIEW (Pages 64–67) Understanding Concepts  1× 103 m   1h  = 27. 1 m/s [45° W of S] 19. (c) constant vix = 5. 7. (a) 100 km/h  3   1km   × 3. (b) The instantaneous acceleration is the slope of the line (for constant acceleration) or the slope of the tangent to the curve (for nonconstant acceleration) on a velocity-time graph.6 × 10 s  2 (b) 97 m/s    = 3. (c) The instantaneous speed is equal to the magnitude of the average velocity throughout the motion. (b) No.8 m/s 1. (c) Yes. T T    L  (c)  2  T T = L results in length.8 m/s. In (a). for example.0s 2 aav = 5. a component of a vector cannot have a magnitude greater than the vector’s magnitude. multiply by 3. A vector can be broken into components using sine and cosine.5 × 102 km/h. (b) An equation may be dimensionally correct yet still be wrong. 6. (a) The displacement is equal to the area under a velocity-time graph. (c) To convert km/h to m/s. (a) ∆d = a (∆t ) 2 Therefore: ?  L  L =  2  × T2 T  L=L The equation is dimensionally correct because both sides of the equation have the same dimensions. 2. the equation could be one of the following: 1 ∆d = v ∆t + a ∆t 2 i 2 1 2 or ∆d = a ∆t 2 (a) The instantaneous speed is equal to the average speed throughout the motion. No.5 × 10 km/h  3  1h × 1 10 m    The peregrine falcon can dive at speeds of 3. L  L  (b)  3  T = 2 results in acceleration.2 m/s [45° W of S] = 2.5 m/s CHAPTER 1 REVIEW (Pages 64–67) Understanding Concepts  1× 103 m   1h  = 27. which have a maximum of one. 5 m [S] + 5 m [N] = 0. two vectors having different magnitudes cannot be combined to give a zero resultant vector. 2 m [S] + 5 m [N] + 3 m [S] = 0.6 10 s     Thus. can total zero. three vectors having different magnitudes can be combined to give zero. for example. two vectors having the same magnitude can be combined to give a zero resultant vector by adding two vectors that have opposite directions. T  3. Adding three vectors in two dimensions head-to-tail. Even in one dimension three vectors can add to zero. (a) L × T–1 results in speed. Copyright © 2003 Nelson Chapter 1 Kinematics 71 . ? 4. for example. (a) Yes. 100 km/h is 27. ∆v aav = ∆t 10. (b) The instantaneous velocity is equal to the average velocity throughout the motion. To convert m/s to km/h. divide by 3.6. 3  1km   3.6. 5. x = 214 m + (96 m) cos 28° + (12 m) cos 25° ∆d x = 310 m ∆d = ∆d x 2 + ∆d y 2 = (310 m/s)2 + (40 m/s) 2 ∆d = 312 m/s  ∆d y  tan θ =    ∆d x   40 m/s  θ = tan −1    310 m/s  θ = 7.8 km) cos19° ∆d y = ∆d1.1 km) cos 38° + (6. Let C be the addition of the two vectors. d1 = 214 m [E] d 2 = 96 m [28° N of E] d3 = 12 m [25° S of E] (a) (b) ∆d = ? Let the +x direction be to the east and the +y direction be to the north. Let the +x direction be to the east and the +y direction be north. x + ∆d 3.1 × 102 m [7. C y = −3. y + ∆d 2.3 km (a) The vector that would add to the vector C to give a resultant displacement of zero would be equal in magnitude.2 km C y = Ay + By = (5.4° The displacement from the tee needed to get a hole-in-one using components is 312 m/s. Thus. y + ∆d 3.4° N of E]. C = A+ B C x = Ax + Bx = (5.1 km) sin 38° − (6. ∆d = ∆d1 + ∆d 2 + ∆d 3 ∆d x = ∆d1. or 3. but opposite in direction to the vector C . y = (96 m) sin 28° − (12 m) sin 25° ∆d y = 40 m 9. vector C is 72 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . x + ∆d 2.8 km) sin 19° C x = 6.8. 0 km − 6.118 s vav = ? d vav = ∆t 4. for a town of 2000.426 m/s The average speed for the lap is 52.3 km  Dy   tan θ =   Dx     3.118 s vav = 52. If we assume an average nose-to-toes ∆d value of 1 m [down] (when person is sitting).M. 304. assume that everyone is sleeping lying down.29 km 11. (a) At 5 P.0 km  Cy   tan θ =   Cx     3. Thus the resultant displacement averages to 0 m. the vector that must be added is 7.3 km)2 Dy = −C y Dy = −3. (b) Let the vector D represent the desired vector.2 km) 2 + (−3. the resultant displacement vector of the sum of all the nose-to-toes vectors is about 2000 m or about 2 × 103 m [down].2 km)2 + (−3. (b) At 5 A.0 km [28° N of W].0 km [W] − C Dx = −4. C = Cx 2 + C y 2 = (6. Copyright © 2003 Nelson Chapter 1 Kinematics 73 .426 m/s.3 km  θ = tan −1    6. we can assume that the vectors point in random directions and average to 0 m. Because the beds face different directions.2 km D = Dx 2 + Dy 2 D = 11 km = (−10.0 km [W] = A + B + D 4.2 km  θ = 28° Thus.M.2 km Dx = −10. d = = 4.4100 km 69 laps ∆t = 84.2 km  θ = 18° The vector is 11 km [18° N of W].3km) 2 C = 7... 4.3km  θ = tan −1    10. 10.4100 × 103 m = 84.0 km [W] = C + D D = 4. assume that people are sitting down to dinner. (b) vav = ? d vav = ∆t 1.8 m ∆dy = 4.2 × 102 s.2 × 102 s The time interval is 1. ∆dx = 6. (b) Assuming that the average length of a car is 5.2 × 102 s vav = 21 m/s The eagle’s average speed during this motion is 21 m/s.2 × 103 m + 24 m/s 18 m/s ∆t = 1.0 s (a) vav = ? d + d2 vav = 1 ∆t 4. 15. The motion now is a high velocity followed by decreasing velocity back to the initial position. The last part of the motion takes about half as long as the first part.0 s vav = 2.8 m ∆t = 5. (a) Starting from a positive position and a low velocity toward the zero position.12. (b) Starting from rest. (b) Let the +x direction be horizontal and the +y direction be down.5 m + 6. 14.2 × 103 m + 1.2 × 103 m v2 = 18 m/s ∆d2 = 1.5 m d2 = 6.9 m/s)(2.2 ×103 m 1.3 m/s The firefighter’s average speed is 2.2 × 103 m = 1.0 s vav = 115 km/h = 31. (a) v1 = 24 m/s ∆d1 = 1.5 m ∆d = ? 74 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . (a) ∆t = 2. d1 = 4.8 m = 5.0 m. 5.0 s) d = 64 m The distance in metres if the speed of your car is 115 km/h is 64 m. the velocity changes direction.0 m ∆t = ∆t1 + ∆t2 = = ∆d1 ∆d 2 + v1 v2 1.3 m/s. Upon reaching the zero position. followed by a higher uniform acceleration in the opposite direction for a shorter period of time until coming to a stop. the speed increases gradually at first and then dramatically. the number of car lengths is 13. the object undergoes uniform acceleration in one direction. Thus.9 m/s d=? vav = d ∆t d = vav ∆t = (31.2 × 103 m ∆t = ? 64 m = 13 car lengths. The instantaneous velocity at position D is 100 km/h [10° S of E]. vav = 100 km/h The directions are found by estimating the directions of the tangents at the positions D.1m  θ = 61° ∆ d = 17 m [61° S of E] or 17 m [29° E of S] . the player’s average velocity is 2.1 m ∆d = ∆d x 2 + ∆d y 2 ∆d = 17 m = (8.8 m  θ = 33° The firefighter’s average velocity is 1.6 m/s [33° below the horizontal]. 16. Copyright © 2003 Nelson Chapter 1 Kinematics 75 .7 m/s [29° E of S] Thus.7 m/s [29° E of S]. The instantaneous velocity at position E is 100 km/h [10° N of E].1m)2 + (15 m)2 ∆d y = ∆d1 y + ∆d 2 y = (16 m) sin 35° + (22 m) sin 15° ∆d y = 15 m  ∆d y  tan θ =    ∆d x   15 m  θ = tan −1    8. The instantaneous velocity at position F is 100 km/h [75° N of E].4 s (a) ∆d = ? Solve using components. 17. and F. ∆d1 = 16 m [35° S of W] ∆d 2 = 22 m [15° S of E] ∆t = 6. Let the +x direction be to the east and the +y direction be south. the player’s resultant displacement is 17 m [29° E of S]. E. (b) vav = ? ∆d vav = ∆t 17 m [29° E of S] = 6.5 m  θ = tan −1    6.6 m  ∆d y  tan θ =    ∆d x   4.8 m)2 + (4.4 s vav = 2. thus. ∆d = ∆d x 2 + ∆d y 2 = (6. ∆d = ∆d1 + ∆d 2 ∆d x = ∆d1 x + ∆d 2 x = ( −16 m) cos 35° + (22 m) cos 15° ∆d x = 8.5 m)2 ∆d = 1. 8 m/s 2 a = 2.4 (km/h)/s.4 (km/h)/s [fwd] The magnitude of the car’s average acceleration is 2.1 g 20. 19.53km The car travels 0.8 m/s 2 = g 9.2 s (a) a = ? 1 2 ∆d = vi ∆t + a ( ∆t ) 2 1 2 ∆d = 0 + a ( ∆t ) 2 2∆d a= 2 ( ∆t ) = = 2(15 m [fwd]) (1.53 km in the time interval. (a) vi = 42 km/h vf = 105 km/h ∆t = 26 s = 7. vf = 4.0 × 102 m/s [fwd] vi = 0 ∆d = 0.2 s) vf = 25 m/s [fwd] The velocity of the cars at 1.2 s is 25 m/s [fwd]. (c) The magnitude of the acceleration in terms of g can be determined as: a 20. vi = 0 ∆d = 15 m [fwd] ∆t = 1.80 m [fwd] a=? 76 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .2 × 10–3 h ∆d = ? ∆d = vav ( ∆t ) 1 (105 km/h + 42 km/h)(7.8 m/s 2 [fwd])(1. (b) vf = ? vf = vi + a ∆t = 0 + (20.18. (b) aav = ? vf − vi aav = ∆t 105 km/h [fwd] − 42 km/h [fwd] = 26 s aav = 2.2 × 10−3 h) 2 ∆d = 0.2 s) 2 a = 21m/s 2 [fwd] The average acceleration of the cars is 21 m/s2 [fwd]. 23.8 m[fwd]) a = 1.28 × 102 m/s [fwd] a = 6.6 m/s2 [fwd] (a) vi = 0.86 × 103 m [fwd] vf = ? 2 2 vf = vi + 2a ∆d vf = (2. ∆d = 2.8 m/s2 [down] ∆d =1. vf = 0 a = 9.28 × 10 2 m/s[fwd]) 2 + 2(6.9 m [up] vi = ? 2 2 vf = vi + 2a ∆d 0 = vi 2 + 2 a ∆ d vi = −2a ∆d = −2( −9. vi = 2.25 × 101 m/s 2 [fwd])(1.6 m/s 2 [fwd] ∆t = 16s The motion takes 16 s.0 × 105 m/s 2 [fwd] The constant acceleration needed by the bullet is 1.25 × 101 m/s2 [fwd] ∆d = 1.0 × 102 m[fwd]) 1.0 m/s ∆t = ? 1 2 ∆d = vi ∆ t + a ( ∆ t ) 2 1 2 ∆d = 0 + a ( ∆t ) 2 2 ∆d ∆t = a = 2(2. 2 2 vf = vi + 2a ∆d vf 2 = 0 + 2a ∆d v2 a= f 2 ∆d (4.0 × 102 m [fwd] a =1. 22.8 m/s 2 [up])(1. 21.1 m/s [up] The minimum vertical velocity needed by the salmon to jump to the top of the waterfall is 6.9 m [up] ) vi = 6.0 × 102 m/s[fwd])2 = 2(0.33 × 102 m/s[fwd] Thus.0 × 105 m/s2 [fwd]. the rocket’s velocity is 533 m/s [fwd].1 m/s [up]. Copyright © 2003 Nelson Chapter 1 Kinematics 77 .86 × 103 m[fwd]) vf = 5. 0 m/s[fwd] ± (8.0 × 102 m [fwd]) 1.6 m/s 2 [fwd])(2. (b) vi = 8.0 m/s [fwd] ∆t = ? 1 2 ∆d = vi ∆t + a ( ∆t ) 2 1 2 0 = vi ∆t + a ( ∆t ) − ∆d 2 1 2 0 = a ( ∆ t ) + vi ∆ t − ∆ d 2 This is a quadratic equation with solution: ∆t = −b ± b 2 − 4ac 2a a −vi ± vi 2 − 4   ( −∆d ) 2 = a   2  2 = −8. 24.9 m/s 2 tan θ = vx vy = (−109 m/s)2 + (82 m/s) 2 ∆v y = vfy − viy = (220 m/s) cos 28° − (240 m/s) sin 28° ∆v y = 82 m/s  109 m/s  θ = tan −1    82 m/s  θ = 53° The airplane’s average acceleration during this time interval is 3. 78 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . vi = 240 m/s [28° S of E] vf = 220 m/s [28° E of S] ∆t = 35 s aav = ? ∆vx = vfx − vix = (220 m/s) sin 28° − (240 m/s) cos 28° ∆vx = −109 m/s 2 2 ∆v = ∆v x + ∆v y ∆v = 136 m/s ∆v a= ∆t 136 m/s = 35 s a = 3. Let the +x direction be east and the +y direction be south.6 m/s 2 [fwd] ∆t = 12 s The motion takes 12 s.9 m/s2 [53° W of S].0 m/s [fwd]) 2 + 2(1. 4 s ∆t = 0.34 s on the way up and at 3. 26. Copyright © 2003 Nelson Chapter 1 Kinematics 79 .0 s vi = ? To find the initial velocity: vf = vi + aav ∆t vi = vf − aav ∆t = vf + ( − aav ∆t ) = 54 m/s [N] + −19 m/s 2 [45° W of N](4.0 s = 54 m/s [N] + ( −76 m/s [45° W of N] ) vi = 54 m/s [N] + 76 m/s [45° E of S] Using components with +x east and +y north: vix = 0 + 76 m/s ( sin 45° ) vix = 54 m/s viy = 54 m/s − 76 m/s ( cos 45° ) = 54 m/s − 54 m/s viy = 0 m/s vi = 54 m/s [E] The car should enter the curve with a velocity of 54 m/s [E].1 s on the way down.34 s or 3.8 m/s2 [down] ∆t = ? 1 2 ∆d = vi ∆t + a ( ∆t ) 2 1 2 0 = vi ∆t + a ( ∆t ) − ∆d 2 1 2 0 = a ( ∆ t ) + vi ∆ t − ∆ d 2 This is a quadratic equation with solution: ∆t = −b ± b 2 − 4ac 2a ( ) a −vi ± vi 2 − 4   ( −∆d ) 2 = a 2  2 = −17 m/s[up] ± (17 m/s[up]) 2 + 2( −9.8 m/s 2 [up])(5. vf = 54 m/s [N] aav =19 m/s 2 [45° W of N] ∆t = 4. 25.2 m[up]) −9.8 m/s 2 [up] = 1.7 s ± 1.1 s The ball passes the camera at 0. vi = 17 m/s [up] ∆d = 5.2 m a = 9. 08 × 1011 m) = 1.26 ×105 km/h Venus has an average speed of 3.23 × 104 m/s.94 × 107 s (a) v = ? 2π r v= T 2π (1.50 × 104 m/s. (b) The area between the line and the x-axis up to the specific times indicate the total displacements up to those times. 28. rVenus = 1. Thus. (a) The slopes of the line segments on the velocity-time graph indicate the accelerations. (b) vav = ? Determine the average velocity after half a revolution using ∆d = 2r.26 × 105 km/h.94 × 107 s v = 3.50 × 10 4 m/s     1000 m   1 h  v = 1.23 × 104 m/s The magnitude of the average velocity of Venus after it has completed half a revolution around the Sun is 2. or 1.08 ×1011 m) = 1.08 × 1011 m TVenus = 1. ∆d vav = ∆t 2r = T  2   4r = T 4(1.50 × 10 4 m/s  1 km   3600 s  = 3. 80 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .94 × 107 s vav = 2.27. 8 m/s 2 )(0.94 × 107 s a =7.0 m ∆t = ? Find the horizontal component of the initial velocity: Horizontally (constant vix ): vix = 18.02 × 10−2 m/s 2 = 2(7.3 m.22 × 10−3 m/s 2 ) 2 The magnitude of its average acceleration during a quarter of a revolution around the Sun is 1. the horizontal acceleration would be negative (assuming the direction of the horizontal component of the velocity is defined as positive).8 m/s2.0 m/s ∆t = = ∆x vix 9.8 m/s2 ∆x = 9.2 m = 0.0 m 18. The vertical component of the acceleration of a projectile is 9.8 m/s2 toward Earth. (b) Find the vertical component of the initial velocity: Vertically (constant a y ): viy = 0 m/s 1 ∆y = viy ∆t + a y (∆t ) 2 2 1 = a y ( ∆t ) 2 2 (9. 30.5 m above ground. vi = 18 m/s [horizontal] ay = 9.50 s. a =? v − vi a= f ∆t vf − vi = T  4   = 4(3. (b) With air resistance affecting both components.50 × 10 4 m/s[E] − 3.0 m/s ∆t = 0. Choose directions such as east and south that are perpendicular to define the directions at these two positions.50 s) 2 = 2 ∆y = 1.22 ×10−3 m/s 2 [S] a = ax 2 + a y 2 a = 1.50 s The snowball will hit the tree after 0.22 × 10 −3 m/s 2 [E] − 7. the height that the snowball will hit the tree is 1.5 m – 1. Copyright © 2003 Nelson Chapter 1 Kinematics 81 . 29. and the vertical component of the acceleration would be less than 9. (a) Let +y be down.50 × 104 m/s[S]) 1.(c) Use the initial and final velocities to determine the average acceleration after a quarter revolution.02 × 10–2 m/s2. (a) The horizontal component of the acceleration of a projectile is zero.2 m Since the snowball’s original location was 1. 8 m/s 18 m/s θ = 15° The snowball’s velocity as it strikes the tree is 19 m/s [15° below the horizontal]. (b) In the other frame of reference (actually Earth’s frame). ∆y = 1. (c) vf = ? vfy = viy + 2a y ∆y vfy 2 = 0 + 2a y ∆y vfy = 2a y ∆y = 2(9.8 m/s 2 ∆t = 0. (a) Let +y be down. the initial velocity of a ball projected horizontally is 29 m/s [horizontally]. 31.2 m) vfy = 4.8 m/s2 ∆x = 16 m viy = 0 m/s vix = ? 1 ∆y = viy ∆t + a y (∆t ) 2 2 1 ∆y = a y ( ∆t ) 2 2 2 ∆y ∆t = ay 2(1.5 may = 9. the path of the ball is straight downward from the release position. 2 2 2 vf = vfx + vfy vf = (18 m/s) 2 + (4. the path of the ball is parabolic and is shaped just like any projectile whose initial velocity is horizontal.55 s vix = 29 m/s Thus.8 m/s 2 )(1. (a) In the train’s frame of reference.8 m/s Therefore.5 m) 9. 32. (The magnitude of the initial velocity relative to Earth is equal in magnitude to the magnitude of the train’s velocity relative to Earth.) vix = 82 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .55 s = ∆x ∆t 16 m = 0.8 m/s) 2 vf = 19 m/s 2 2 θ = tan −1 = tan −1 vfy vfx 4. x = (285 km/h)cos 45° + (75 km/h)sin 22° vPG. vSW = 0. vPA = 285 km/s [45° S of E] vAG = 75 km/h [22° E of N] vPG = ? vPG = vPA + vAG vPG.y = vSW = 0.33. 34.y vPG = 2.x + vPG. Let the +x direction be downstream from the initial shore and the +y direction be across the river. Let the +x direction be east and the +y direction be south.y = (285 km/h)sin 45° − (75 km/h)cos 22° vPG.y = 132 km/h  230 km/h  θ = tan −1    132 km/h  θ = 60° The velocity of the plane relative to the ground is 2. W for the water. A for the wind (air). (b) vSG = ? vSG = vSW + vWG vSG.80 m/s [across the river] ∆t = 108s ∆x vWG = ∆t 54 m = 108s vWG = 0. and G for the ground.6 ×10 2 km/h  vPG.50 m/s vSG. Use the subscripts S for the swimmer.x = vWG = 0.y      = (230 km/h)2 + (132 km/h)2 vPG. Use the subscripts P for the plane.50 m/s The speed of the river current is 0.80 m/s [across the river] ∆y = 86 m ∆x = 54 m =? (a) v WG ∆y ∆t = vSW = 86 m 0. and G for the ground.x tan θ =   vPG.6 × 102 km/h [60° E of S].50 m/s.x = 230 km/h 2 2 vPG = vPG.80 m/s Copyright © 2003 Nelson Chapter 1 Kinematics 83 . 94 m/s  vSG.3h 2 vPG = 4.94 m/s [downstream.80 m/s)2 vSG = 0. Let the +x direction be south and the +y direction be east.50 m/s  θ = 58° The swimmer’s velocity relative to the shore is 0. (c) θ = ? vSG = vSW + vWG 2 2 2 vSW = vSG + vWG 2 2 2 vSG = vSW − vWG vSG = (0.80 m/s)2 − (0. vSG = vSG. 51° from the near shore.7 × 10 km/h [18° S of W] vPG = vPA + vAG vPA = vPG − vAG Using the relative velocity equation: vPG = vPA + vAG vPA = vPG − vAG 84 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .50 m/s)2 + (0.62 m/s  vSG tan θ =  v  WG      0. 35.62 m/s  θ = tan −1    0.5 × 103 km [18° S of W] = 5. and G for the ground. ∆ d = 2. Use the subscripts P for the plane.y 2 = (0.y tan θ =  v  SG.5 × 103 km [18° S of W] ∆t = 5.3 h vAG = 85 km/h [E] vPA = ? First we find the velocity of the plane relative to the ground that allows it to reach its destination on schedule. A for the wind (air).x 2 + vSG.50 m/s  θ = 51° The direction in which the swimmer should have aimed to land directly across from the departure position is upstream. ∆d PG vPG = ∆t 2. 58° from the initial shore].50 m/s) 2 vSG = 0.80 m/s  θ = tan −1    0.x      0. 7 s) + (9. Let +y be up. = vix ) = (21.y = (4.0 m/s [47° above the horizontal] ay = 9.0 m = 8.5 × 102 km/h [15° S of W].x 2 + vPA. the football will pass above the bar at a height of 12 m – 3.Using components with +x west and +y south: vPA.9 m.x = (4.7 s)2 2 ∆y = 12 m Since the horizontal bar of the goal post is 3.34 × 10 km/h  θ = 15° The velocity of the plane relative to the air is 5.45 × 102 km/h)2 vPA.8 m/s 2 )(1.4 m/s)(1.3 m/s ∆t = 1.y = 1.3 m/s Vertically (constant ay) viy = vi sin θ = (21.7 × 102 km/h) cos18° − ( −85 km/h) vPA.34 × 102 km/h)2 + (1.0 m/s)(cos 47°) Copyright © 2003 Nelson Chapter 1 Kinematics 85 .34 × 102 km/h vPA = vPA.x = 5.45 × 102 km/h  1.0 m/s)(sin 47°) viy = 15.7 s To calculate the vertical height of the football over the bar: 1 ∆y = viy ∆t + a y (∆t ) 2 2 1 = (15.4 m/s First we must determine the change in time: ∆x ∆t = vix 25 m 14.5 × 102 km/h  vPA. 36. vi = 21.x     = (5.8 m/s2 ∆x = 25 m ∆y = ? Find the horizontal and vertical components of the initial velocity: Horizontally (constant vix = vi cos θ vix = 14.0 above the field.7 × 102 km/h) sin18° − 0 vPA.y 2 vPA = 5.45 × 10 2 km/h  θ = tan −1   2  5.y tan θ =  v  PA. 3 cm  θ = sin −1   62 cm   θ = 5. In a complete experiment.50 m vi = 38. their calculations will show that the distance the target falls equals the difference between its height above the launch level and ∆y of the projectile. for background information. 38. (a) Since the only instrument allowed is a metre stick or a measuring tape. L = 62. is the effect of the wind and/or air resistance. A systematic source of error may occur if the ruler or metre stick has a worn end or if the calibration does not begin at 0.) 39. ∆x = 8. The angle (θ) of the ball’s initial velocity must be estimated. A source of error that could be either random or systematic.0° ) h = 5. The equation derived for the maximum horizontal range for a projectile that lands at the same level as its starting position can be applied. (b) a = ? Refer to Investigation 1.) ∆x = vi 2 = vi = vi 2 g sin 2θ g ∆x sin 2θ g ∆x sin 2θ (b) The biggest source of random error is in estimating the launch angle of the ball.Applying Inquiry Skills 37. and initial velocity students choose for their example. tan θ = target above the launch level: h = ∆x tan θ = (8.0 cm dU = 9. the student could measure the horizontal range of the ball that is observed to have the maximum range. page 49.0 cm. A systematic error may occur if the horizontal range is measured to a position below the level at launch. (See the text.2° above the horizontal. (c) The main random sources of error occur in using a ruler to measure the required distances. A specific example follows. Another source of random error is in measuring the horizontal range.0° above the horizontal] To satisfy the condition that the dart launcher is aimed at the stationary target.50 m )( tan 33. a = g sin θ = (9.9 cm dL = 4.0 m/s [33.89 m/s2.2°) a = 0. questions (c) and (j). No matter what values of the launch angle. (Notice that the sources of error mentioned here relate to the way in which the data were found and applied in this question.2° The angle of incline of the air table is 5. depending on the situation. horizontal range. where h is the level of the 86 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .9 cm − 4. the motion of the puck would be analyzed and there would be many more sources of error.89 m/s 2 The magnitude of the acceleration of the puck parallel to the inclined plane is 0.4.8 m/s 2 )(sin 5.1.3 cm (a) θ = ?  d − dL  sin θ =  U  L    9.52 m h ∆x . Copyright © 2003 Nelson Chapter 1 Kinematics 87 .17 m above the launch level.267 s ) − 4.9 m/s 2 ( )(0.50 m (38.0 m )(cos 33. the final result will be the same.) If different students choose different initial conditions. (b) At a higher speed there is more air resistance and more heat produced by the higher amount of friction in the moving parts of the car and its engines.349 m The target is thus 5. consider the vertical component of the dart’s displacement.0 m/s )(sin 33.52 m − 0. Finally. in the time interval that the target falls to a specific height above the launch level. If you have your students try this. More fuel per unit distance moved is required to compensate for these effects.17 h − 0.0 min The driver saves 2. using +y up: 1 ∆yD = viy ∆t + a y ∆t 2 2 1 = (vi sin θ ) ∆t + a y ∆t 2 2 = (38.0° )(0. (Rounding error resulted in a slight discrepancy in this example.0 minutes by breaking the speed limit.14 h ∆t = 0.18 m Thus.The time of flight can be found by using the horizontal component of the dart’s motion: ∆x vix = ∆t ∆x ∆t = vix = = ∆x vi cos θ 8.0° ) ∆t = 0. the dart’s motion results in the dart striking the target. the target falls by an amount ∆yT: 1 ∆yT = viy ∆t + a y ∆t 2 2 = 0 + 4.267 s ) 2 ( )(0.267 s ) 2 ∆yD = 5.03 h or 2. v1 = 125 km/h v2 = 100 km/h d = 17 km ∆t = ? (a) ∆t = t2 − t1 = = d d − v2 v1 17 km 17 km − 100 km/h 125 km/h = 0.9 m/s 2 ∆yT = 0.349 m = 5.267 s In that time interval. be sure they use a consistent +y direction throughout the entire solution. This problem can also be solved for general cases. Making Connections 40. ∆yT. This applies to both Earth and the asteroid.0 × 108 m/s tD = 0. and the subscript H represent the helicopter.50 m. (c) The explosive was beginning to fall back downward and was moving slightly to the right when detonation occurred.e. Three-dimensional positions. students are expected to confine their answers to concepts related to Chapter 1.4 s vS = 3. the subscript S represent the sonar signal. (Likely the true acceleration would be even greater than indicated here because most sports enthusiasts can run faster than 5.5 × 102 m/s vH = ? 88 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .0 × 108 m/s ∆t = 0.41. (b) The time delays mentioned are obvious in live television broadcasts involving large distances between interviewers. blood). (b) The explosive was fired vertically upward and detonated while still rising fairly quickly. Let d1 represent the distance the helicopter is from the cliff. ∆d1 = 7.0 m/s. At this stage. d = 4.5 g .55 s ttotal = 0.0 m/s) 2 2(0. we calculate the acceleration using magnitudes: vf 2 = vi 2 + 2a∆d 0 = vi 2 + 2 a∆d a= = −vi 2 2 ∆d −(5. 43.50 m) a = −25 m/s 2 Since this acceleration is more than 2. (a) The main principles are velocity and relative velocity of a flowing liquid (i. Typical estimated speed and stopping distance are 5. velocities. and accelerations of the past and present could be analyzed and used to predict the motion of an asteroid or other body in the future.) 44. Both sets of data could be stored in a computer for later analysis.) 42. Using these values.. (d) The explosive was moving upward and fairly quickly to the left when detonation occurred. Extension 46. and often the stopping distance is less than 50 cm. One design could be a tiny device that sends signals outside the body so its motion ( vsensor/artery ) could be monitored by a receiver at the same time that a sensor within the device measures the speed of the blood ( vblood/sensor ). respectively. it is recommended that the patient avoid a rigorous racket sport. Let ∆t represent the time it takes for the signal to travel from Earth to the satellite and tD represent the delay time.16 s) + 0.87 s The total time interval between sending the signal and receiving the return signal on Earth is 0.0 × 102 m ∆tS = 3. (Time delays are less when transmission is via cable rather than via satellite. (a) The explosive was fired vertically upward and detonated right at its maximum height. both of which are conservative values.16 s Now we can determine the total time: ttotal =∆t + tD + ∆t = 2(0.8 × 107 m c = 3.8 ×10 7 m = 3.55 s ttotal = ? (a) First we must calculate the change in time: d ∆t = c 4. 45.0 m/s and 0. The equations are: d vblood/artery = vblood/sensor + vsensor/artery v = blood and blood ∆t (b) Answers will vary.87 s. ∆dP = ∆dT. Let the subscript T represent the truck travelling at constant speed. the subscript P represent the police cruiser. (b) ∆t = ? ∆d T ∆t = vT = 2. ∆d T = vT ∆t ∆t = ∆d T vT 1 a P ( ∆t ) 2 2 ∆d P = viP ∆t + ∆d P = Substitute for ∆t: 1 a P ( ∆t ) 2 2 2 1  ∆d  ∆d P = aP  T  which is equal to 2  vT  ∆d P = = = 1  ∆d P  aP   2  vT  2 vT aP 2 2 2(18 m/s) 2 2.1 × 10 2 m 3. vT = 18 m/s aP = 2. we can determine the speed of the helicopter: ∆d H vH = ∆tS = 2. Copyright © 2003 Nelson Chapter 1 Kinematics 89 .2 m/s2 ∆d P = 2.9 ×10 2 m 18 m/s ∆t = 16 s The pursuit lasts 16 s.0 ×10 2 m) − 1.1 × 10 2 m Finally.First we must calculate the distance the sonar signal travels: ∆dS = vS ∆tS = (3.9 × 102 m before catching the truck.4 s vH = 62 m/s The speed of the helicopter is 62 m/s 47.2 × 103 m Next we calculate the distance the helicopter travels: ∆d H = 2∆d1 − ∆dS = 2(7.5 ×10 2 m/s)(3.2 × 103 m ∆d H = 2.4s) ∆dS = 1.2 m/s2 viP = 0 m/s (a) To calculate how far the cruiser travels before catching the truck.9 × 102 m The cruiser travels 2. 49.5° W of S]. then ay = −g = −9. Begin by determining an expression for ∆t using the vertical component of the motion: 1 ∆y = viy ∆t + a y ∆t 2 2 ∆y = vi sin θ∆t − 4. Let the +x direction be east and the +y direction be south.7 km/h ) θ = 7.0 (km/h)/s [7.5° The direction of the acceleration is [7.9 m/s ) ∆t 2 ( ) 2 − vi sin θ∆t + ∆y = 0 90 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . we can calculate the change in velocity: ∆v = = ( ∆vx )2 + ( ∆v y ) 2 ( −9.5° W of S] ∆t = 14 s The time interval of the acceleration is 14 s.48. (b) ∆t = ? From (a) above.3 km/h ) (70. ∆v = vf − vi ∆vx = vfx − vix 1 = 70.3 km/h tan θ = ∆v x ∆v y θ = tan −1 (9.5° W of S] = 5. Let +y be up.9 m/s 2 ∆t 2 (4.3 km/h )2 + (70.0 (km/h)/s (a) vfx =? v fy = ? vfx = vf cos θ = (100 km/h)(cos 45°) vfx = 71 km/h vfy = vf sin θ = (100 km/h)(sin 45°) vfy = 71 km/h The direction of the acceleration is the same as the direction of the change of velocity.7 km/h − 8.7 km/h )2 ∆v = 71 km/h We can now calculate the time interval: ∆v a= ∆t ∆v ∆t = a 71 km/h [7.8 m/s2. vi = 80 km/h [E] vf = 100 km/h [45° S of E] a = 5.0 × 10 km/h ∆vx = −9. 0 km downstream.0 km [downstream] = 0. Consider the motion in the frame of reference of the flowing river.0 km/h. the raft drifts for 30 min or 0. Let R represent the river and S represent the shore.9 m/s 2 ) ∆y 9. so it will take her exactly the same length of time to swim directly back to the raft. Refer to the SIN solutions book.8 m/s 2    Copyright © 2003 Nelson Chapter 1 Kinematics 91 .50 h vRS = 2. Thus.)  v sin θ ± i ∆x = vi cosθ     ( −vi sin θ )2 − (19.Using the quadratic formula to solve for ∆t: ∆t = −b ± b 2 − 4 ac 2a vi sin θ ± ∆t = ( −vi sin θ )2 − (19. ∆d vRS = ∆t 1. which moves at a constant velocity downstream relative to the shore. The sunbather swims directly away from the raft for 15 min at a constant speed.0 km/h [downstream] The speed of the current in the river is 2. question 75-10.9 m/s 2 ) ∆y   9.50 h while moving (relative to the shore) 1. (A more complex solution from the frame of reference of Earth or the shore provides the same answer.8 m/s 2 Substituting this equation for ∆t into the equation involving the horizontal component of the motion: ∆x = vix ∆t ∆x = vi cosθ∆t 50. (a) A is an applied force or tension force. Copyright © 2003 Nelson Chapter 2 Dynamics 93 . 4. If the mass is two times as great. 3. Thus. (a) (b) The vector sum of all the forces must be zero because the box is initially at rest and remains at rest.CHAPTER 2 DYNAMICS Reflect on Your Learning (Page 68) 1. C is the force of friction. (b) 2. The applied force on the two books would have to be two times as great as that on one book in order to achieve a constant velocity. B is a normal force. and D is the force of gravity on the loaded toboggan. then the normal force pushing upward on the books is two times as great. so the kinetic friction acting against the motion of the two books is two times as great. the force needed to overcome friction and maintain constant velocity must be two times as great. 1 FORCES AND FREE-BODY DIAGRAMS PRACTICE (Page 71) Understanding Concepts 1. (a) The orientations are (i) J. The readings on all 5 springs are equal. Table 1 Common Forces Name of Force gravity normal tension friction kinetic friction static friction air resistance applied force 2. In (b) and (c). 2. (ii) I. they should be checked to be sure they are properly zeroed. (a) Predictions may vary. Type of Force action-at-a-distance contact force contact force contact force contact force contact force contact force contact force Example in Daily Life force exerted by Earth on a ball dropped from your hand force exerted by the ground on your feet force exerted by a dog on a leash force acting between your shoes and the ground that allows you to walk without slipping force of the ice on a moving puck horizontal force of the floor against a stationary desk when a horizontal force is applied to the desk but the desk remains at rest friction of air molecules against a moving airplane force applied to an object.” 94 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . 3.5. Springs M and N must be zeroed while held horizontally. 9. such as a push to a swing The tension is still 18 N because the tension in a single cord is the same everywhere along the length of the cord. (b) Try This Activity: Predicting Forces (Page 69) If spring scales are used for this activity. One way of rephrasing the statement is: “A rope is useful to apply a force on an object only when it is under tension.8 N. the pulleys and strings should be as near to “frictionless” as possible. Answers may vary. (b) If any students had differences. and (iii) K. urge them to explain why. such as a push to a swing The tension is still 18 N because the tension in a single cord is the same everywhere along the length of the cord. Table 1 Common Forces Name of Force gravity normal tension friction kinetic friction static friction air resistance applied force 2. (a) The orientations are (i) J. (b) If any students had differences. (b) Try This Activity: Predicting Forces (Page 69) If spring scales are used for this activity. 9. One way of rephrasing the statement is: “A rope is useful to apply a force on an object only when it is under tension.1 FORCES AND FREE-BODY DIAGRAMS PRACTICE (Page 71) Understanding Concepts 1. Type of Force action-at-a-distance contact force contact force contact force contact force contact force contact force contact force Example in Daily Life force exerted by Earth on a ball dropped from your hand force exerted by the ground on your feet force exerted by a dog on a leash force acting between your shoes and the ground that allows you to walk without slipping force of the ice on a moving puck horizontal force of the floor against a stationary desk when a horizontal force is applied to the desk but the desk remains at rest friction of air molecules against a moving airplane force applied to an object. 3. Springs M and N must be zeroed while held horizontally. (a) Predictions may vary. Answers may vary. 2. they should be checked to be sure they are properly zeroed.5.” 94 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . and (iii) K. In (b) and (c). the pulleys and strings should be as near to “frictionless” as possible. urge them to explain why. (ii) I. The readings on all 5 springs are equal.8 N. 7. 5. 6. The FBD of the ball is the same in all three cases.PRACTICE (Page 73) Understanding Concepts 4. (a) Copyright © 2003 Nelson Chapter 2 Dynamics 95 . (b) The force of air resistance is upward and equal in magnitude to the downward force of gravity. Applying Inquiry Skills 8. If we estimate the mass of the skydiver and parachute apparatus to be about 102 kg.(b) (c) The first diagram is more convenient because only one force needs to have its components shown. we can apply the equation involving the gravitational field strength (which students learned about in Grade 11 physics): Fair = mg = (10 2 kg)(9. (a) Fapp = 3.8 × 10 2 N The air resistance is about 103 N [up]. PRACTICE (Page 75) Understanding Concepts 9.74 N Copyright © 2003 Nelson Unit 1 Forces and Motion: Dynamics .8 N/kg) Fair = 9.27 N [down] Fair = 0.74 N [up] Fg =3.0 N 96 Fapp.354 N [horizontally] ΣF = Fg + Fapp Let +x be the horizontal direction of the motion and +y be up. The components of Fapp are: Fapp.x = 0.y = 3. y = 0.15 N ΣFy = Fapp.354 N Thus.15 × 103 N [23.0° W of N] Fapp.0 N ΣFy = 0.27 N The components of Fair are: Fair.27 N ) + 0. (b) Fapp = 6382 N [28.x + Fgx + Fair.x = −0. The components of Fg are: Fgy = −3.59 N [53° above the horizontal].3°) Fapp.x ΣFx = −0. ΣF = ( −0.y + Fgy + Fair.2 = 478 N [36.3°) The components of Fg are: Fgx = 0.y = (6382 N)(sin 28.59 N ΣFy = Fapp.9° above the horizontal]. Copyright © 2003 Nelson Chapter 2 Dynamics 97 .1 = 412 N [27.3° above the horizontal] Fg = 538 N [down] ΣF = Fg + Fapp Let +x be opposite to the horizontal direction of the jumper’s motion and +y be up.47 0. ΣF = (5619 N) 2 + (2488 N) 2 ΣF = 6.y + Fgy + Fair.354 N Fair.47 N θ = tan −1 0.47 N)2 ΣF = 0.x = 5619 N Fapp.9° The net force on the jumper is 6.x = (6382 N)(cos 28. ΣF = Fg + Fapp Let +x be east and +y be north.y = 3.0 N The components of the net force are: ΣFx = Fapp. (c) Fapp.0 N Fgy = −538 N Fapp.74 N + ( −3.0 N Fgx = 0. The components of Fapp are: Fapp.y = 3026 N The components of the net force are: ΣFx = Fapp.0 N ΣFx = 5619 N Thus.x + Fgx = 5619 N + 0.y = 3026 N + ( −538 N ) ΣFy = 2488 N θ = tan −1 2488 N 5619 N θ = 23.354 N) 2 + (0.0° N of E].354 θ = 53° The net force on the bird is 0. ΣF = F1 + F2 The components of F1 are: F1.y = (−48 N)(sin 16°) F2. the sum of the tension forces in the two ropes is 92 N [E]. C = 180° − 16° − 16° C = 148° c 2 = a 2 + b2 − 2ab cos C c = a 2 + b 2 − 2ab cos C = (48 N) 2 + (48 N)2 − 2(48 N)(48 N)(cos148°) c = 92 N 98 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . and the corresponding opposite angles be A.x = (48 N)(cos 16°) The components of F2 are: F2. ΣF = (200 N) 2 + (648 N) 2 ΣF = 678 N ΣFy = Fapp1. In the triangle in that diagram.x = (−412 N)(sin 27.0°) The components of Fapp2 are: Fapp2. 10.y = (412 N)(cos 27.0°) Fapp2.0°) Fapp1.y = 281 N The components of the net force are: ΣFx = Fapp1.0°) Fapp2.x = −187 N Fapp1. B. and C.x = 46 N + 46 N ΣFx = 92 N ΣFy = F1.1° E of N].x = (478 N)(cos 36.y = 367 N + 281 N ΣFy = 648 N θ = tan −1 200 N 648 N θ = 17.y = (478 N)(sin 36.x = −187 N + 387 N ΣFx = 200 N Thus.x = 46 N F1.y + Fapp2.x = (48 N)(cos 16°) F2. side ΣF be c.x + F2.y = 367 N Fapp2.y = (48 N)(sin 16°) F1. Let the +x direction be to the east and the +y direction be to the north. let side F1 be a.1° The net force on the quarterback is 678 N [17.y = 13 N F2.y = 13 N + (−13 N) ΣFy = 0 N Thus.y + F2. The components of Fapp1 are: Fapp1.x + Fapp2. (a) F1 = 48 N [16° N of E] F2 = 48 N [16° S of E]. (b) To solve Sample Problem 5(a) using the cosine law. we need to know the angle opposite the resultant force in Figure 11 on page 75 in the text.x = 46 N F1.x = 387 N Fapp1.y = −13 N The components of the net force are: ΣFx = F1. side F2 be b. y FT2.x + FT2. ΣFapp = 56 N [16° S of E] FT1 = 27 N [E] ΣFapp = FT1 + FT2 Let +x be east and +y be south.0 N FT2. and the force of kinetic friction are the contact forces acting on the ruler.x − FT1.y = (56 N)(sin 16°) Fapp.x = 27 N ΣFT2 = (27 N) 2 + (15 N)2 ΣFT2 = 31 N ΣFapp.x = 27 N Thus.) Copyright © 2003 Nelson Chapter 2 Dynamics 99 .x = ΣFapp. (b) The electromagnetic force is responsible for the contact forces. The components of Fapp are: Fapp.y = 15 N θ = tan −1 15 N 27 N θ = 30° The tension in cord 2 is 31 N [30° S of E]. 11.x = 54 N − 27 N FT2. (The electric force is also an acceptable answer.y = 15 N FT1. the sum of the tension forces in the two ropes is 92 N [E].1 Questions (Page 76) Understanding Concepts 1. Section 2.Applying the sine law to find angle B: sin B sin C = b c b sin C sin B = c b sin C B = sin −1 c (48 N)(sin 148°) = sin −1 92 N B = 16° Thus.y + FT2.y = 0.y = 15 N − 0.x = FT1. ΣFapp. the applied force of the hand. x = (56 N)(cos 16°) Fapp.y = FT1.y − FT1.x FT2.0 N Fapp.y = ΣFapp.x = 54 N The components of FT1 are: FT1. (a) The normal force of the desk on the ruler. Gravity is the noncontact force. neglecting air resistance). the net force is 18 N [down] (the force of gravity.42 N Fgx = 0. 4.50 N)(sin 32°) Fair.50 N)(cos 32°) Fair.x = 0.50 N [32° above the horizontal] ΣF = Fg + Fair Let +x be in the horizontal direction of the air resistance and +y be up.y = 0.y = (0.5 N [down] Fair = 0. The force of gravity is balanced by the normal force of your hand on the book. The components of Fg are: The components of Fair are: Fair.x = (0. 3. (b) If you suddenly remove your hand.(c) 2.0 N Fgy = −1.27 N 100 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . Fg = 1.5 N Fair. (a) When the book is held stationary in your hand. the net force on the book is zero. 9 N ΣFx = −5.42N) 2 + ( − 1.6 N)(sin 28°) FAy = (−3. = −1. FA = 3.42 N 1.6 N [28° W of S] FB = 4.1 N)(sin 24°) FCx = 0.2 N The components of FC are: FCx = (2.1 N − 1.3 N)(sin 15°) FBy = 1. 5.2 N + 1.42 N Thus. using components Let +x be east and +y be north. The components of FA are: FAx = (−3.y = 0.7 N − 4.42 N ΣFx = 0.0N ΣFy = FAy + FBy + FCy = −3.9 N ΣFy = −4.6 N)(cos 28°) FAy = −3.9 N FAx = −1.7 N The components of FB are: FBx = (−4.0 N Copyright © 2003 Nelson Chapter 2 Dynamics 101 .22 N θ = 19° The net force on the ball is 1.5 N + 0.3 N θ = tan −1 0.24 N) 2 ΣF = 1.24 N ΣF = (0.The components of the net force are: ΣFx = Fgx + Fair.0 N + 0.1 N)(cos 24°) FCx = −1.9 N The components of the net force are: ΣFx = FAx + FBx + FCx = −1.2 N + 0.3 N [71° below the horizontal].2 N FBy = (4.1 N [24° E of S] (a) FA + FB + FC .3 N [15° N of W] FC = 2. using a vector scale diagram (b) FA + FB + FC .27 N ΣFy = −1.1N FCx = (−2.x ΣFy = Fgy + Fair.3 N)(cos 15°) FBx = −4. using a vector scale diagram ( ) (d) FA − FB .0N) 2 + ( − 4.9 N ΣFy ΣFx 4.5N Thus.2 N) ΣFx = 2.7 N − ( −4. (c) FA − FB = FA + − FB .3 N) 2 ΣF = 4. F1 = 36 N [25° N of E] F2 = 42 N [15° E of S] ΣF = 0 N 102 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .9 N [60° S of E]. let +x be east and +y be north.5N)2 + ( − 4.2 N − 1.4 N ΣFy ΣFx 4.Thus. ΣF = (−5.3 N 2.3 N θ = tan −1 = tan −1 θ = 60° (to two significant digits) 6.5 N ΣFy = FAy − FBy = −3.0 N θ = tan −1 = tan −1 θ = 39° The net force is 6.4 N [39° S of W]. The net force is 4.0 N 5.1 N ΣFy = −4. using trigonometry As in (b). ΣF = (2.0 N)2 ΣF = 6. The components of the net force are: ΣFx = FAx − FBx = −1. Thus. Understanding Concepts ( ) Copyright © 2003 Nelson Chapter 2 Dynamics 103 . The components of F1 are: F1x = (36 N)(cos 25°) F1y = (36 N)(sin 25°) F1y = 15 N F2 y = −(4. ΣFx = F1x + F2 x + F3 x = 0 F3 x = 0 N − F1x − F2 x = 0 N − 33 N − 11N F3 x = −44 N ΣF3 = (−44 N)2 + (26 N)2 ΣF3 = 50 N ΣFy = F1 y + F2 y + F3 y = 0 F3 y = 0 N − F1 y − F2 y = 0 N − 15 N + 41N F3 y = 26 N θ = tan −1 = tan −1 ΣFy ΣFx 26 N 44 N θ = 30° (to two significant digits) The force F3 that must be added is 5. The snowboarder will tend to maintain a constant velocity (according to the first law of motion) while the snowboard will slow down rapidly. the result would be much different because of air resistance. (a) The upward lift force on the plane: Flift = 6. In this case. the magnitude and direction of the object’s speed are continually changing. 5. we assume that air resistance on the ball is negligible.3 × 104 N [E] 2. the net force must be zero. The objects in (c) and (e) are not examples on Newton’s first law of motion. FR + F1 + F2 = 0 FR = − F1 + F2 4. 2.3 N)(cos 15°) F2 y = −41N F1x = 33 N The components of F2 are: F2x = (42 N)(sin 15°) F2x = 11N Thus.6 × 104 N [up] (b) The force due to air resistance on the plane: Fair = 1. these two objects are undergoing acceleration. In other words.Let +x be east and +y be north. Since Fapp = 38 N. Let the required force be FR . Thus.0 × 101 N [30° N of W]. (a) Let +x be east.2 NEWTON’S LAWS OF MOTION PRACTICE (Page 80) 1. the ball will not collide with the thrower. 3. the snowboarder will likely fall to the snow ahead of the board.) 6. the magnitude of the force of friction must also be 38 N since the desk is not moving. The ball is in motion horizontally and that horizontal component will continue (as stated in the first law of motion). In each case. (If the thrower tried the same experiment while on a motorcycle. In both cases. In other words. the magnitude and direction of the object’s speed are continually changing.0 × 101 N [30° N of W].3 × 104 N [E] 2. 2. FR + F1 + F2 = 0 FR = − F1 + F2 4. we assume that air resistance on the ball is negligible. the net force must be zero. Let the required force be FR . In this case. Thus. 3. The components of F1 are: F1x = (36 N)(cos 25°) F1y = (36 N)(sin 25°) F1y = 15 N F2 y = −(4. the magnitude of the force of friction must also be 38 N since the desk is not moving. The objects in (c) and (e) are not examples on Newton’s first law of motion. The ball is in motion horizontally and that horizontal component will continue (as stated in the first law of motion).Let +x be east and +y be north. (a) Let +x be east. The snowboarder will tend to maintain a constant velocity (according to the first law of motion) while the snowboard will slow down rapidly. In each case. the ball will not collide with the thrower. the snowboarder will likely fall to the snow ahead of the board. 5. In both cases. Understanding Concepts ( ) Copyright © 2003 Nelson Chapter 2 Dynamics 103 .6 × 104 N [up] (b) The force due to air resistance on the plane: Fair = 1.2 NEWTON’S LAWS OF MOTION PRACTICE (Page 80) 1. (a) The upward lift force on the plane: Flift = 6. (If the thrower tried the same experiment while on a motorcycle. Since Fapp = 38 N. these two objects are undergoing acceleration.3 N)(cos 15°) F2 y = −41N F1x = 33 N The components of F2 are: F2x = (42 N)(sin 15°) F2x = 11N Thus. Thus. ΣFx = F1x + F2 x + F3 x = 0 F3 x = 0 N − F1x − F2 x = 0 N − 33 N − 11N F3 x = −44 N ΣF3 = (−44 N)2 + (26 N)2 ΣF3 = 50 N ΣFy = F1 y + F2 y + F3 y = 0 F3 y = 0 N − F1 y − F2 y = 0 N − 15 N + 41N F3 y = 26 N θ = tan −1 = tan −1 ΣFy ΣFx 26 N 44 N θ = 30° (to two significant digits) The force F3 that must be added is 5.) 6. the result would be much different because of air resistance. 104 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .2 N = = − (6.9 N (sin15°) ) FRx = − ( F1x + F2x + F3x ) FRx = −2.0 N FR = FRx 2 + FRy 2 FR = 7.5 N (sin25°) − 4. (b) Let +x be east and +y be north. FR = 7.0 N )2 + ( −6.5 N + 3. FR = 54 N [36° S of W] .2 N [16° W of S] .9 N = − (6. (c) Let +x be east and +y be north. FR + F1 + F2 + F3 = 0 FR = − F1 + F2 + F3 ( ) FRy = − F1y + F2y + F3y FRy = −6.9 N )2 θ = tan −1 = tan −1 FRx FRy 2.5 N (cos25°) + 0 N + 3.9 N θ = 16° Therefore. FR = 143 N [W] . FR + F1 + F2 = 0 FR = − F1 + F2 ( ) FRy = − F1y + F2y FRy = −32 N = − (32 N + 0 N ) FRx = − ( F1x + F2x ) FRx = −44 N = − (0 N + 44 N ) ( ) FR = FRx 2 + FRy 2 FR = 54 N = ( −44 N )2 + ( −32 N )2 θ = tan −1 FRy FRx 32 N = tan −1 44 N θ = 36° Therefore.FRx = − ( F1x + F2x ) FRx = −143 N = − ( 265 N − 122 N ) Therefore.9 N (cos15°) ) ( ) ( −2.0 N 6. ) (b) Answers will vary.42 × 104 N [fwd] = 2. but it is negligible if the paper is pulled quickly. such as a toy Teddy bear.820 m/s2 [fwd]. The coin. crashing into the barrier. (a) Place the coin on the paper on the desk and quickly jerk the paper horizontally. One example is to place an object. (The force of static friction would be noticed if the paper is pulled slowly. m = 2. such as a collision of fast braking.0 m Vertically: θ = 4.42 × 104 N [fwd] ΣF a= m 2. will continue to move forward.16 kg ax = 32 m/s2 Fx = ma x = (0.6°) FT = 1. m = 30. PRACTICE (Page 83) Understanding Concepts 10. speed crashing into a brick barrier. The angle of the line with respect to the horizontal is: ∆y θ = tan −1 ∆x 0. Applying Inquiry Skills 8.820 m/s 2 [fwd] The truck’s acceleration while the force is applied is 0.8 × 103 N.16 kg)(32 m/s 2 ) Fx = 5. on a dynamics cart.6° ΣFy = 2 FT.95 × 104 kg ΣF = 2. The cart will stop but the Teddy bear. 2 The vertical distance is ∆y = 0. The horizontal distance from the pole to the middle of the line is ∆x = 10 m = 5.40 m. initially in motion. Making Connections 9.1N The magnitude of the force is 5. and could easily injure any person in its path.95 × 104 kg a = 0.0 m. any object in the rear window will continue moving forward.7. yet safe. will remain at rest relative to the desk because the net force acting on it is essentially zero.y − Fg = 0 2 FT sin θ − Fg = 0 FT = Fg 2(sin θ ) 294 N = 2(sin 4. 11.1 N. m = 0. and send the cart at a fairly high. initially at rest. Copyright © 2003 Nelson Chapter 2 Dynamics 105 .0 kg Let +y be up.8 ×103 N The magnitude of the tension in the clothesline is 1.40 m = tan −1 5. In any emergency. 1× 103 N [W] ΣF = 7. Repeat the experiment using the same stretch of the elastic band but adding a second cart on top of the one being pulled. objects of approximately equal size can be separated according to their densities by accelerating them on a specially-designed device. The method mentioned in the question is proposed in an abstract found at http://www.earthlink.html.27 kg) 1. are controlled to determine their effect on the acceleration.2 × 10–3 s  v − vi  ΣF = m  f   ∆t  (4. Care must be taken to ensure that the carts do not fall to the floor or crash into other objects. Then repeat with the third cart atop the others. m = 7.8 N/kg) Fg = 1.3 × 106 kg)(9.61 m/s [W] − 5.3 × 106 kg Fg = mg = (1. and then triple it while pulling on a cart of constant mass.1 × 103 N [E].8 N/kg) Fg = 24 N The magnitude of the weight of a horseshoe is 24 N.utexas. use one elastic band stretched a small amount. General discussion about mining on asteroids can be found at http://home. To help distinguish useful from non-useful minerals.2 × 10−3 s = −7.4 kg 16. the magnitude of the weight is Fg = Fg = mg .27 kg vi = 5.1× 103 N [E] The net force on the ball during the collision is 7. Making Connections 15. observe the cart’s acceleration.8 N/kg m = 1.78 m/s [W]) = (7.tsgc.htm. attach the elastic band to a single cart and. Fg = mg = (2. ΣF = ma v − vi a= f ∆t  vf − vi  ΣF = m    ∆t  13. To determine the effect of changing the force.4 kg)(9.8 N/kg m = 2.net/∼nrunner/trav/tirem/belter. then double that amount.edu/floatn/1999/99_fall/ teams/charleston. In each case. mass and net force. 106 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . (b) g = 9. 12. To determine the effect of changing the mass. with the elastic band stretched a convenient amount. PRACTICE (Page 84) Understanding Concepts (a) g = 9.3 × 107 N.78 m/s [W] vf = 4. The experiment involves measuring (or at least observing) the acceleration of an object as two variables.3 ×107 N The magnitude of the weight of an open-pit coal-mining machine is 1.61 m/s [W] ∆t = 1. Applying Inquiry Skills 14. (e) g = 9.77 × 10−8 N.50 g Fg = mg = (2. m = 76 kg g = 3.80 N/kg) Fg = 1.8 ×102 N [down] The weight of the astronaut is 2.77 × 10−8 N The magnitude of the weight of a speck of dust is 1.8 N/kg) Fg = 5.53 N [down] g = 9. 18.80 N/kg [down] m = 1.18 × 105 kg The mass of the payload capacity of a C-5 Galaxy cargo plane is 1.80 N/kg [down] Fg = mg Fg m= g 1.53 N [down] = 9. Copyright © 2003 Nelson Chapter 2 Dynamics 107 . (d) g = 9.45 × 10−2 N.(c) g = 9. 17.4 × 102 N The magnitude of the weight of a 55-kg student would be 5. (a) Fg = 1.50 × 10−3 kg)(9.80 N/kg) Fg = 2.16 × 106 N [down] 9.81 ×10−9 kg)(9. (b) Fg = 1.8 N/kg m = 55 kg Fg = mg = (55 kg)(9.4 × 102 N.80 N/kg m = 1.7 N/kg[down]) Fg = 2.56 ×10 −1 kg The mass of a field hockey ball is 1.8 × 102 N [down].80 N/kg m = 2.45 × 10 −2 N The magnitude of the weight of a table tennis ball is 2.18 × 105 kg.81 × 10−9 kg Fg = mg = (1.56 × 10−1 kg.80 N/kg [down] m = 1.16 × 106 N [down] g = 9.80 N/kg [down] Fg m= g = 1.7 N/kg [down] Fg = mg = (76 kg)(3. Answers will vary. (b) The action force is the downward force of the rotating propeller blades on the air. The reaction force is the upward force of the expanding gases on the engine (and thus the rocket). The reaction force is the force of the air eastward on the wall of the balloon. Since the net force is zero. a water rocket (for outdoor use only). Any action-reaction toys are acceptable. kg s  2 m kg  2  s  = m . (a) (b) There are two forces acting on the pencil. and those forces are equal in magnitude but opposite in direction.) 108 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . The vehicle moves forward and the cardboard moves backward. 21.Applying Inquiry Skills N m = 19. a toy projectile launcher that jerks backward when the projectile is launched forward. kg s2 PRACTICE (Page 86) Understanding Concepts 20. (In the last example. the forces can be shown more readily if the wheels are “wound up” before placing the vehicle gently down on a piece of cardboard that is resting on several straws or round pencils. pushing the balloon eastward. The reaction force is the upward force of the air on the rotating propeller blades. Examples include a balloon. (a) The action force is the downward force of the rocket engine on the expanding gases as they exit the bottom of the rocket. as long as they are demonstrated safely. and a toy car or truck with wind-up or battery-powered propulsion to show action-reaction forces between the wheels and the floor. no acceleration occurs. Applying Inquiry Skills 22. Thus. causing the rocket to accelerate forward. (c) The action force is the force of the interior walls of the east end of the balloon pushing westward on the air as it exits the west end of the balloon. The units N/kg and m/s are equivalent since 1 N = 1 kg  2  . 20 × 10 −15 N = 9. Copyright © 2003 Nelson Chapter 2 Dynamics 109 . Let +y be up.8 m/s 2 ) FN = 19 N FN = 19 N [up] The normal force acting on a stationary 1.20 × 10−15 N ax = ΣFx m 3.11 × 10−31 kg ΣFx = 3.0 × 10–3 s v − vi (a) a = f ∆t 0 m/s − 13 m/s[down] = 3.0 ms = 3.6 × 102 N [up]. me = 9. ΣFy = FN − Fg = 0 FN = Fg = mg = (1.8 m/s 2 ) FN = 6. m = 67 kg v = 85 cm/s [down] Let +y be up. a x = 3. Thus.Section 2.6 ×102 N FN = 6.6 × 102 N [up] The normal force exerted by the cherry picker on the worker is 6. Since the velocity is constant. 4. 2.51× 1015 m/s 2 The magnitude of the resulting acceleration of the electron is 3.3 ×103 m/s 2 [up] The acceleration of the hand is 4. Since the mallard duck is flying at a constant velocity.9 kg)(9. vf =0 m/s vi = 13 m/s [down] ∆t = 3. the net force acting on the duck must be 0 N. ΣFy = FN − Fg = 0 FN = Fg = mg = (67 kg)(9.11 ×10 −31 kg 3.0 × 10−3 s a = 4.9-kg carton of juice is 19 N [up].51 × 1015 m/s2.2 Questions (Page 87) Understanding Concepts 1. Let +x be the direction of the net force and the acceleration. ΣFy = 0 .3 × 103 m/s2 [up]. 5. a 55-kg student’s weight on Venus would decrease by 9% compared to its magnitude on Earth.65 kg ΣF = ma = (0.8 N/kg[down]) ΣF = 6.4 × 10 N Fg. (b) On Earth: Fg = m g = (55 kg)(9.9 ×102 N = 2 Fg. Thus.24 × 102 N 4. ΣF = mg = (65 kg)(9.0 g.00 – 0. which is exerted by the brick.91) × 100% = 9%. 110 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson 7.8 × 103 N [up] The net force acting on the hand is 2. the ratio of the magnitude of the force of the brick to the magnitude of the expert’s weight is 4. Since the weight on Venus is less than the weight on Earth.9 × 102 N.3 × 103 m/s 2 [up]) ΣF = 2.Venus 4. (a) The baking pan exerts a pulling force on the chef toward the oven.(b) m = 0.) (c) m = 65 kg. (Notice that the force of gravity on the hand has been ignored because it is negligible compared to the force of the brick on the hand.9 N/kg) Fg = 4.0 g The mass of the arrow is 28.43 × 103 m/s2 ΣF m= x ax = 1.9 N/kg m = 55 kg (assumed mass of student) (a) Fg = m g = (55 kg)(8.24 ×102 N ax = 4.8 × 103 N [up] = ΣFexpert 6.43 ×103 m/s 2 = 0.Venus The percentage change in weight is (1.4:1. (c) The water exerts a forward normal force on the hands.4 × 102 N [down] ΣFbrick 2. ΣFx = 1.0280 kg m = 28. .8 N/kg) Fg = 5. (b) Saturn exerts a gravitational force on the Sun toward Saturn. g = 8.9 ×10 2 N The magnitude of a 55-kg student’s weight on Venus is 4. 8.Venus = 0.4 ×102 N Fg.4 × 102 N [down] ΣFbrick = 4.Venus 5.65 kg)(4.8 × 103 N [up]. Let +x be the direction of the net force and the acceleration.4 ΣFexpert 6.91 Fg. 5 × 10−2 kg. (a) Answers will vary.200 kg)(9.gsfc.5 × 10−2 kg g = 9. m = 0. is undergoing constant free fall.80 N/kg) Fg = 1. two of which are listed here: http:/nasaexplores. (a) The astronaut. (e) The hailstone exerts a downward applied or normal force on the air. (d) To find out more about the inertial device. assume the value is 45 N [down].2 s ) 2 ∆d = 1. (b) m = 3.(d) The watermelon exerts an upward gravitational force on Earth.8 N/kg Fg = m g = (0. so there can be no normal force pushing up from a bathroom scale.8 N/kg ) ( ) (c) Answers will vary.96 N.nasa.5 m/s 2 [fwd] (1. or a wall.8 N/kg [down] Fg = mg Fg = 34 N [down] = 3.5 × 10−2 kg (9. a floor. Applying Inquiry Skills 10.5 m/s 2 [fwd] m = 58 kg The mass of the astronaut is 58 kg.2 s 1 2 ∆d = vi ∆t + a ( ∆t ) 2 1 2 = 0 + 1. like all objects aboard an orbiting vehicle. a chair.com/lessons/o1-044/9-12_1. (d) accepted value = 45 N [down] estimated value = 34 N [down] estimated value − accepted value % error = × 100% accepted value 45 N − 34 N = 45 N % error = 24% Making Connections 11.96 N This downward force of gravity on each bag is balanced by the tension in the string.200 kg g = 9. 9.1 m [fwd] The astronaut moved 1. For this example. 1. (b) ΣF = 87 N [fwd] a = 1.1 m. The example used here is 35 g or 3.gov/stargaze/Smass.html http://www-istp.5 m/s2 [fwd] ΣF = ma ΣF m= a 87 N [fwd] = 1. (c) ∆t = 1. This tension in turn equals the reading on the scale.htm ( ) Copyright © 2003 Nelson Chapter 2 Dynamics 111 . students can refer to a variety of NASA sites. 8 m/s 2 m = 3.74 × 103 N The upward (buoyant) force on the balloon. with each member discovering something different to share with the other members. 112 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .15 m/s 2 + 9. basket. ΣF = ma ΣFy = ma y Fapp + (− Fg ) = ma y Fapp − mg = ma y m= = Fapp ay + g 0. 2. ΣFy = ma y (b) FB + (− Fg ) = 0 FB = mg m= = FB g 2.5 × 10−2 kg 2.3 APPLYING NEWTON’S LAWS OF MOTION PRACTICE (Page 92) Understanding Concepts 1. (a) ΣF = ma ΣFy = ma y FB + (− Fg ) = −ma y FB − mg = −ma y FB = m( g − a y ) = (315 kg)(9.10 m/s2 [down] m = 315 kg Let +y be up. both in resource centres and on the Internet. The mass of the fork is 35 g.15 m/s2 [up] Let +y be up.12.10 m/s 2 ) FB = 2.80 m/s 2 − 1.35 N [up] a = 0. There are numerous resources on the life and times of Isaac Newton. Fapp = 0. a = 1.35 N 0. Students can work in groups.80 m/s 2 m = 280 kg (3 significant digits) The mass of the ballast that must be discarded overboard is 315 kg – 280 kg = 35 kg.74 × 103 N 9. and its contents is 2.74 × 103 N [up]. 3 APPLYING NEWTON’S LAWS OF MOTION PRACTICE (Page 92) Understanding Concepts 1. Fapp = 0.12.35 N 0.74 × 103 N 9. with each member discovering something different to share with the other members.74 × 103 N [up].5 × 10−2 kg 2.10 m/s2 [down] m = 315 kg Let +y be up. a = 1.10 m/s 2 ) FB = 2. ΣFy = ma y (b) FB + (− Fg ) = 0 FB = mg m= = FB g 2.80 m/s 2 − 1. (a) ΣF = ma ΣFy = ma y FB + (− Fg ) = −ma y FB − mg = −ma y FB = m( g − a y ) = (315 kg)(9.80 m/s 2 m = 280 kg (3 significant digits) The mass of the ballast that must be discarded overboard is 315 kg – 280 kg = 35 kg. 112 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . The mass of the fork is 35 g. and its contents is 2.15 m/s2 [up] Let +y be up.8 m/s 2 m = 3. basket. ΣF = ma ΣFy = ma y Fapp + (− Fg ) = ma y Fapp − mg = ma y m= = Fapp ay + g 0. There are numerous resources on the life and times of Isaac Newton.74 × 103 N The upward (buoyant) force on the balloon. 2. Students can work in groups. both in resource centres and on the Internet.15 m/s 2 + 9.35 N [up] a = 0. Let m be the unknown mass and m2 be the second mass. ΣFy = may Fg − Ff = ma y Ff = Fg − ma y = mg − ma y = m( g − a y ) = (35.37 m2 m = 0.7 N mA = 2.00 s ∆d = 3.80 m/s 2 − 1.7 kg)(9. For block B let +y be down and for block A let +y be to the right.2 kg.0 kg a2 = 0. ΣFx = m2 (0. (a) vi = 0 m/s ∆t = 2. (b) Let +y be down.2 kg 5.74 kg 0. The mass m is 1. ∆y = vi ∆t + ay = 1 a y ( ∆t ) 2 2 2 ∆y ( ∆t ) 2 2(3. m2 = m + 2.10 m [down] m = 35. Copyright © 2003 Nelson Chapter 2 Dynamics 113 .55 m/s 2 ) Ff = 295 N The upward force of friction exerted by the pole on the girl is 295 N.55 m/s2.37 a x ) 4.7 kg Let +y be down.3.0 kg) m − 0.37 m = 0. FK = 5.0 kg m = 1.00 s)2 a y = 1.55 m/s 2 The girl’s (constant) downward acceleration is 1.7 kg mB = 3.63m = 2.37a ΣFx = ma x ΣFx = m2 a2 x ma x = m2 (0.37(m + 2.7 kg Let the magnitude of the acceleration of the blocks be ay.37a x ) m = 0.10 m) = (2. 7 kg + 3.58s) vf = 0. (b) To calculate the tension in the string.8 m/s 2 The magnitude of the acceleration of the blocks is 4.1° + (17.37 m/s 2 ) Ff = 2. (a) ΣFy = 0 Fapp.37 m/s 2 )(0.y + Fg − FN = 0 FN = Fapp.1° + (17.9 N) sin 35.79 m/s.x − Ff = ma x Ff = Fapp. m = 17.37 m/s2 ∆t = 0.7 kg) a y = 4.80 m/s 2 ) FN = 194 N The magnitude of the normal force on the mower is 194 N.79 m/s The magnitude of the maximum velocity of the mower is 0. (b) ΣFx = ma x Fapp. 6. v − vi (c) a x = f ∆t vf = vi + ax ∆t = 0 + (1.9 kg)(9.1° above the horizontal] a = 1.8 m/s2.9 N [35.7 kg)(9.4 N. 114 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .9 N) cos 35.58 s vi = 0 m/s Let +x be in the horizontal direction of the applied force and +y be down.y + Fg = Fapp sin θ + mg = (32.4 N The magnitude of the frictional force on the mower is 2.(a) ΣFy = ma y FgB − FT = mB a y FT − Ff = mA a y FgB − Ff = (mA + mB )a y ay = = mB g − Ff mA + mB (3. use either equation in part (a): FT − Ff = mA a y FT = mA a y + Ff = (2.7 N (2.8 m/s 2 ) + 5.7 kg)(4.9 kg Fapp = 32.x − ma x = Fapp cos θ + ma x = (32.8 m/s 2 ) − 5.9 kg)(1.7 N FT = 19 N The magnitude of the tension is 19 N. m = 65 kg θ = 12° (a) Let the +x direction be downward parallel to the hillside and the +y direction be downward. (a) (b) With +y up.(d) ΣFx = 0 Fapp. ΣFx = 0 FA cos θ − Ff = 0 Ff = FA cos θ Copyright © 2003 Nelson Chapter 2 Dynamics 115 .8 m/s 2 ) sin12° FN = 6.x − Ff = 0 Fapp.0 m/s 2 The magnitude of the skier’s acceleration is 2. g.9 N. The magnitude of the force applied by the boy to maintain the constant velocity is 2. Applying Inquiry Skills 8. m. ΣFy = 0 mg cos θ − FN = 0 FN = mg cos θ = (65 kg)(9.0 m/s2.2 × 102 N. and θ is as follows: ΣFy = 0 FA sin θ + FN − Fg = 0 FN = mg − FA sin θ (c) With +x in the horizontal direction of the applied force.4 N = cos 35. the equation for the magnitude of the normal force on the block in terms of the given parameters FA.9 N Fapp 7.8 m/s 2 ) sin12° a x = 2. (b) ΣFx = ma x mg sin θ = ma x a x = g sin θ = (9.1° = 2.x = Ff Fapp cos θ = Ff Fapp = Ff cos θ 2. parallel to the hillside.2 × 102 N The magnitude of the normal force on the skier is 6. the equation for the magnitude of friction on the block in terms of FA and θ is as follows. the student’s apparent weight is 482 N and at 9. at 1. 116 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . (d) The term “weightless” is used because that is what a person feels during free-fall acceleration. The apparent weight is less than the true weight when the acceleration is downward. the force of gravity) acting on the body in free fall. PRACTICE (Page 94) Understanding Concepts 10.80 m/s 2 − 1.e. When the elevator is undergoing free fall.08 m/s 2 ) FN = 482 N 2 If a2 = 9. (e) Let +y be up.80 m/s [down]. a = 0.33 m/s2 [fwd] m1 = m2 = 3.80 m/s 2 − 9.80 m/s 2 ) FN = 542 N The reading on the scale is 542 N. m = 55. (c) Let +y be down.Making Connections 9. (b) The student’s true weight (Fg = mg = 542 N) is less than the apparent weight when the acceleration is upward.3 kg (a) Let +y be up.08 m/s2 [up] ΣFy = ma y FN − Fg = ma y FN = m( g + a y ) = (55. then FN = (55.1 × 104 kg Let the +x direction be the direction of the acceleration. the student’s apparent weight is 0 N.08 m/s [up] ΣFy = 0 FN − Fg = 0 FN = mg = (55.80 m/s 2 + 1. a1 = 1.08 m/s 2 ) FN = 602 N The reading on the scale is 602 N..3kg)(9.80 m/s2 [down].3kg)(9. v = 1.08 m/s2 [down].08 m/s2 [down] ΣFy = ma y Fg − FN = ma y FN = m( g − a y ) FN = (55. the apparent weight is zero since there is no normal force.3kg)(9. The term is not valid from the physics point of view because there is still weight (i. a = 1.3kg)(9.80 m/s 2 ) FN = 0 N Thus. 0 ×10 4 N The force exerted by the locomotive on the first car is 2.8 kg.21m/s 2 ) Fapp = 0. (b) ΣFx = m2 a x Fapp = (1.58 N a = 0.80 N The magnitude of tension in the lowest thread is 9.21m/s 2 ) 0.21m/s 2 m2 = 1.3 Questions (Pages 95–96) Understanding Concepts 1.0 × 104 N [fwd].(a) ΣFx = m2 a x Fapp = m2 a x = (3.00 kg)(9. Copyright © 2003 Nelson Chapter 2 Dynamics 117 .58 N − (1. (a) The bottom mass ΣFy = 0 FT − m3 g = 0 FT = m3 g = (1. (a) ΣFx = ( m1 + m2 ) ax Fapp = m1a x + m2 a x m2 = = Fapp − m1a x ax 0. 3.0 kg)(0.0 × 104 N The force exerted by the first car on the second car is 1. 11.1× 104 kg)(0.8 kg)(0.33m/s 2 ) Fapp = 1.8 kg The mass of the second book is 1. 2.1× 104 kg)(0.37 N. The net force acting on the shark is zero since the acceleration is zero (constant velocity). (b) ΣFx = ( m1 + m2 ) ax Fapp = 2m2 a x = 2(3. Fapp = 0.0 kg Let the +x direction be the direction of the acceleration.80 N.33 m/s 2 ) Fapp = 2. All the FBDs are identical since there is only one force acting on the basketball.0 × 104 N [fwd]. Section 2.21 m/s2 [horizontally] m1 = 1.37 N The magnitude of the force exerted by one book on the other is 0.80 m/s 2 ) FT = 9. 8 m/s 2 ) Fapp = 2.00 kg)(9.5 g ) = 1. (c) The top mass ΣFy = 0 FT − (m1 + m2 + m3 ) g = 0 FT = (m1 + m2 + m3 ) g = (5.80 m/s 2 ) FT = 29. The magnitude of tension in the highest thread is 78.4 N.80 m/s 2 ) FT = 78. This force is the reaction force to the action force of the engine pushing downward on the expanding gases.50 g [up] m = 2.9 × 107 N.4 N.4 N 4.(b) The middle mass ΣFy = 0 FT − (m2 + m3 ) g = 0 FT = ( m2 + m3 ) g = (2.0 × 106 kg. (b) The upward force is caused by the gases as they are expelled from the base of the rocket engine.9 × 107 N The approximate magnitude of the upward force on the shuttle is 2.4 N The magnitude of tension in the middle thread is 29. Let +y be up. ΣFy = ma y (a) Fapp − mg = ma y Fapp = m( g + a y ) = m( g + 0.00 kg +1. a = 0.00 kg +1.00 kg)(9.5(2. 118 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .0 × 106 kg)(9.00 kg +2. 8 m/s 2 ) (45 kg + 35 kg) a y = 1.9 × 102 N The magnitude of the tension in the rope is 3.7 N FgB = 2. Copyright © 2003 Nelson Chapter 2 Dynamics 119 .15 m.2 m/s 2 ) FT = 3. ΣFy = ma y (a) FT − m1 g = m1a y m2 g − FT = m2 a y Adding these two equations gives: m2 g − m1 g = m1a y + m2 a y ay = = (m2 − m1 ) g (m1 + m2 ) (45 kg − 35 kg)(9.8 N The magnitude of the tension in the horizontal rope is 1.15 m The magnitude of each box’s displacement is 0.5 N The magnitude of the tension in the vertical rope is 2.2 m/s 2 The magnitude of the acceleration of the boxes is 1.5.8 N FgA = 6. (b) Block A ΣFx = 0 FT2 − FfA = 0 FT2 = FfA FT2 = 1. (b) Substituting ay into either of the equations in part (a) gives: FT − m1 g = m1a y FT = m1 ( g + a y ) = (35 kg)(9.9 × 102 N.8 m/s 2 + 1. m1 = 35 kg m2 = 45 kg Let +y be up for m1 and down for m2 (as the rope slides over the metal rod).5s)2 2 ∆d = 0. FfA = 1. (c) ∆t = 0.5 N.5 N (a) Block B ΣFy = 0 FT1 − FgB = 0 FT1 = FgB FT1 = 2.50 s 1 ∆d = vi ∆t + a y (∆t ) 2 2 1 = (1.2 m/s 2 )(0.2 m/s2.8 N. 6. 1 kg Let +x be to the right.1N 7.744 m/s 2 The magnitude of the acceleration of the carts is 0.3 N) cos 21. The tension in the third rope is 3.2 kg m3 = 16.7 N. 120 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . (c) Point P ΣFy = 0 FT3 sin θ − FT1 = 0 FT3 = FT1 F = T2 sin θ cos θ F sin θ = T1 cos θ FT2 tan θ = 2.0 kg m2 = 13.5 N = sin 54° FT3 = 3.7 N The magnitude of the normal force acting on block A is 6. (a) ΣFx = ( m1 + m2 + m3 ) ax FT cos θ = ( m1 + m2 + m3 ) a x ax = = FT cos θ (m1 + m2 + m3 ) (35. m1 = 15.8 N FT1 sin θ ΣFx = 0 FT3 cos θ − FT2 = 0 FT3 = FT2 cos θ θ = 54° Substitute angle into either equation for FT3: F FT3 = T1 sin θ 2.ΣFy = 0 FNA − FgA = 0 FNA = FgA FNA = 6.5 N 1.1 N [54° above the horizontal].744 m/s2.0 kg + 13.0° (15.2 kg + 16.1 kg) a x = 0. 8°.2 kg + 16. ΣFx = ma x mg sin θ = ma x sin θ = ax g 9.5 m/s 2  θ = sin −1  2   9.076 m/s2 [fwd] (a) Let +y be up.(b) ΣFx = m3 a x FT3 = m3 ax = (16. ΣFx = ma FT − mX g sin θ = mX a FT = mX ( g sin θ + a ) Copyright © 2003 Nelson Chapter 2 Dynamics 121 . mX = 5.22 kg (a) For block X: Let the +x direction be defined as up the slope parallel to the slope. ax = 1.8 N. Fapp = 91 N [15° above the horizontal] FN = 221 N [up] a = 0.0 N The magnitude of the tension in the last cord is 12.8° The angle between the hill and the horizontal is 8.1kg)(0.8 m/s 2 m = 25 kg The mass of the chair is 25 kg. ΣFy = 0 FN + Fapp sin θ − mg = 0 m= FN + Fapp sin θ g 221 N + (91 N) sin15° = 9.8 m/s  θ = 8. ΣFx = max Fapp cos θ − Ff = max Ff = Fapp cos θ − max = (91N) cos15° − (25 kg)(0.1 kg)(0. 8. 10.0 N.5 m/s2 Let +x be downward. (c) ΣFx = ( m2 + m3 ) ax FT2 = (m2 + m3 )a x = (13. parallel to the hillside.744 m/s 2 ) FT3 = 12.  1.12 kg mY = 3.8 N The magnitude of the tension in the middle cord is 21.076 m/s 2 ) Ff = 86 N The magnitude of the friction force on the chair is 86 N. (b) Let +x be the direction of the acceleration.744 m/s 2 ) FT2 = 21. v −v a x = fx ix ∆t 0.12 kg)sin35. ΣFy = ma mY g − FT = mY a FT = mY ( g − a ) Since the two equations for FT are equal: mX ( g sin θ + a ) = mY ( g − a ) mX a + mY a = mY g − mX g sin θ a= = (mY − mX sin θ ) g mX + mY (3. m = 56 kg ∆t = 0.80 m/s 2 ) 5.22 kg a = 0.22 kg)(9. (b) Substitute a into either equation for tension: FT = mY ( g − a) = (3.7°)(9.0 m/s )(0.12 kg + 3.0 m/s 2 The magnitude of the skater’s acceleration is 1.0 m/s 2 ) Fx = 56 N The magnitude of the force the skater exerted on the boards is 56 N. 1 ∆d1 = vix ∆t1 + a(∆t1 )2 2 ∆d 2 = vfx ∆t2 1 2 2 = (1.84 m The magnitude of the skater’s displacement from the boards after 1.75s) 2 ∆d 2 = 0.80 m/s 2 − 0.273m/s 2 The magnitude of the acceleration is 0.75 m/s)(0. (d) The figure skater accelerates for ∆t1 = 0.75 s vfx = 75 cm/s vix = 0 cm/s (a) Let +x be the direction of the acceleration away from the boards.75 s.273 m/s 2 ) FT = 30. 11. (b) Let +x be opposite to the direction of the acceleration. 122 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .22 kg − (5.0 m/s2.273 m/s2.28 m + 0. (c) The magnitude of the force the boards exerted on the skater is 56 N.75s) = (0.56 m ∆d = 0.7 N.84 m.7 N The magnitude of the tension in the fishing line is 30. Fx = ma x = (56 kg)(1. toward the boards.50 s is 0.75 m/s − 0 m/s = 0.56 m ∆d1 = 0.28 m ∆d = ∆d1 + ∆d 2 = 0. that is.75s a x = 1.For block Y: Let the +y direction be defined as vertically down.75 s then travels at a constant velocity (vfx) for ∆t2 = 0. and they are capable of exerting a strong force on the driver or passenger in the front seat. (a) Acceleration versus mass: a ∝ m (b) Acceleration versus net force: a ∝ ΣF Making Connections 13. safety boots for roofing installers. resulting in a greater impact with the deployed airbag. cookie sheets).Applying Inquiry Skills 1 12. Table 1 lists examples of situations when it would be advantageous to have increased friction and decreased friction.geocities.htm Copyright © 2003 Nelson Chapter 2 Dynamics 123 . mountain climbing equipment. students can refer to various Web sites. the airbags deploy very quickly. For more information. Some students may want to try to crush the WintOGreen Lifesaver in their partially open mouth. racing bicycle wheels. airplanes Try This Activity: Observing Triboluminescence (Page 101) Viewing the flashes of light produced when the crystals are crushed is much better in a very dark room. 2. that is not recommended. mountain bike wheels.com/RainForest/9911/tribo. truck engine. In the case of a child. but for safety reasons. The light is caused by the excitation of molecules due to electric charge differences on the planes of the crystals as the crystals are crushed against each other. rollerblade wheels. Any passenger not wearing a seatbelt will be moving forward and will continue to do so after the collision.4 EXPLORING FRICTIONAL FORCES PRACTICE (Page 97) Understanding Concepts 1. it is the child’s head that may impact with the airbag. Table 1 Increased Friction basketball shoes. even at a relatively low speed. This could result in serious injury or even death. vehicle braking systems Decreased Friction cookware (cake pans. which would be even more dangerous than a chest impact. It is important that the viewer’s eyes be dark-adapted since the light emitted is very dim. In a front-end collision. one of which is given here: http://www. cookie sheets). rollerblade wheels. truck engine.geocities. vehicle braking systems Decreased Friction cookware (cake pans. even at a relatively low speed. which would be even more dangerous than a chest impact. It is important that the viewer’s eyes be dark-adapted since the light emitted is very dim. Table 1 Increased Friction basketball shoes. For more information. The light is caused by the excitation of molecules due to electric charge differences on the planes of the crystals as the crystals are crushed against each other. it is the child’s head that may impact with the airbag. resulting in a greater impact with the deployed airbag. Some students may want to try to crush the WintOGreen Lifesaver in their partially open mouth. that is not recommended. Any passenger not wearing a seatbelt will be moving forward and will continue to do so after the collision. and they are capable of exerting a strong force on the driver or passenger in the front seat. but for safety reasons. (a) Acceleration versus mass: a ∝ m (b) Acceleration versus net force: a ∝ ΣF Making Connections 13. safety boots for roofing installers.com/RainForest/9911/tribo.Applying Inquiry Skills 1 12. Table 1 lists examples of situations when it would be advantageous to have increased friction and decreased friction. In a front-end collision. mountain bike wheels.4 EXPLORING FRICTIONAL FORCES PRACTICE (Page 97) Understanding Concepts 1. students can refer to various Web sites. one of which is given here: http://www.htm Copyright © 2003 Nelson Chapter 2 Dynamics 123 . airplanes Try This Activity: Observing Triboluminescence (Page 101) Viewing the flashes of light produced when the crystals are crushed is much better in a very dark room. mountain climbing equipment. In the case of a child. This could result in serious injury or even death. 2. racing bicycle wheels. the airbags deploy very quickly. 124 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . ΣFx = max = 0 Fapp. x = FK Fapp.8 N/kg) FN = 225 N Now determine the magnitude of the maximum static friction: FS. ΣFx = ma x Fapp + (− FK ) = ma x FK = Fapp − ma x = 17 N − (5. the magnitude of the minimum horizontal force needed to set the mat in motion is 97 N.1 kg a = 0. Fapp = 17 N [W] m = 5. x − FK = 0 Fapp.x. This force is a reaction force caused as the tires apply a force of friction (the action force) northward against the road.8 m/s 2 ) µK = 0.43)(225 N) FS. max = µS FN = (0.1kg)(9.43 µK = 0. (b) Let +x be the direction of the applied force. max = 97 N Thus. Fapp. A horizontal force of magnitude 81 N will keep the mat moving at a constant velocity. 3.1 kg)(0.39 m/s2 [W] Let +x be the direction of the acceleration. x = 81 N 4.39 m/s 2 ) FK = 15 N µK = = FK FN 15 N (5.36)(225 N) Fapp. assuming the tires do not spin on the road surface.PRACTICE (Page 101) Understanding Concepts 2.36 (a) Let +y be up.30 The coefficient of kinetic friction between the case and the table is 0. ΣFy = ma y = 0 FN + (−mg ) = 0 FN = mg = (23 kg)(9. (b) The force is static friction. (a) The direction of the frictional force is southward on the tires. x = µ K FN = (0. m = 23 kg µS = 0.30. Copyright © 2003 Nelson Chapter 2 Dynamics 125 .5 m/s 2 9.8 m/s 2 µS = 0.5. = 6.max = ma x µS FN = max µS mg = max ma x µS = mg a = x g 2. (a) (b) Static friction between the two boxes causes the small box to accelerate horizontally. (c) Let +x be the direction of the acceleration of the boxes and +y be up.26. In the vertical direction: ΣFy = ma y = 0 FN − Fg = 0 FN = Fg FN = mg In the horizontal direction: ΣFx = ma x FS. and let m be the mass of the smaller box.26 The smallest coefficient of static friction is 0. m = 47 kg θ = 23° µK = 0. Another major source of random error is that it is very difficult to judge when an object moving down the board experiences a constant speed. The observed results depend partly on the temperatures of the water baths and the viscosities of the oils chosen. each group of 3 or 4 students needs a stopwatch. = Applying Inquiry Skills 8. the bubbles in lower viscosity oils (e. In general.11  FT = 53 N Thus. (a) Using a straight board that can be raised at one end. Two sets of 3 test tubes containing the 3 different samples of motor oil can be set up. SAE20) travel faster than the bubbles in higher viscosity oils (e.) (b) A major source of random error is that the runner tends to get “stuck” and “unstuck” in an irregular fashion as it moves down the inclined board. (a) One way to increase friction on the lid is to place a rubber pad over the lid or use a high-friction gloves or mitts. cooking oils. grease. Butter. 2 beakers (one with cold water and the other with water from the hot water tap). 6 test tubes with solid rubber stoppers.. SAE40). 126 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . Another source of random error is parallax error in measuring the rise and run of the board when it is at an angle to the desk or floor. (b) Nonstick surfaces of frying pans and other cooking utensils decrease friction. a test tube rack. ΣFx = 0 ΣFy = 0 FN + FT sin θ + (−mg ) = 0 FN = mg − FT sin θ Equate two equations for the normal force: mg − FT sin θ = FT cos θ µK FT cos θ + (− Ff ) = 0 FT cos θ = µK FN FN = FT cos θ µK  cos θ  + sin θ  = mg FT   µK  FT = mg  cos θ  + sin θ   µ  K  (47 kg)(9. Try This Activity: Oil Viscosity (Page 102) To perform this activity.g.7. labelled appropriately.8 m/s 2 )  cos 23°  + sin 23°    0.11 Let +x be the direction of the velocity and +y be up. A minor systematic error occurs if the metre stick has a worn end. (Refer to Sample Problem 2 on page 100 of the textbook for the derivation of the run corresponding equation for the coefficient of static friction. They can then use the metre stick to determine the rise and run of the board and calculate the coefficient of kinetic friction using rise the relation µK = tan θ = . and cooking sprays also decrease friction. and 3 grades of motor oil. the magnitude of the tension in the rope is 53 N. students can place the runner at one end of the board and raise that end gradually until they discover an angle at which the runner moves at a constant speed down the board. and stored for future use..g. Making Connections 9. Another way is to use two hands to apply forces to the lid while someone else holds the jar steady. Some liquids. is reduced. including the windscreen that prevents the air from striking the rider directly. listed in order of lowest to highest viscosity. the cans move toward each other. (a) Most of the components are smooth and curved. (c) All components are smooth and curved. where the speed of the air is low or nil. PRACTICE (Page 105) Understanding Concepts 10. 12. (b) The implication is that blood has a higher viscosity than water. the procedure begins by determining the time it takes to fill a container of measurable volume with water flowing from a nozzle of measurable diameter.0 × 103 cm3 time interval to collect water = ∆t = 85 s Copyright © 2003 Nelson Chapter 2 Dynamics 127 . The speed of the syrup at the top of the bulge is higher than the speed where the syrup leaves the jar. As soon as the air moves quickly through the gap between the cans. are vinegar.48 cm)2 = 0. diameter of nozzle = 0. while the pressure on the outside of the cans. with blood relatives more closely knit than non-relatives. The phrase refers to the tendency for family members to defend one another at all costs.Try This Activity: How Will the Cans Move? (Page 105) The air blown between the empty cans should travel horizontally and fairly close to the straws. so radius = 0. (d) The front of locomotives is curved and smooth. where the speed of the air is high. 14. the air pressure above the top is less than the air pressure in the car. (b) Moving air across the top of the chimney reduces the pressure there while the pressure in the room at the fireplace remains higher. so the top bulges upward. The viscosity of molasses increases as the temperature drops. 15. (a) The speed of the air above the convertible top is higher than the speed of the air below (relative to the top). This occurs because the pressure between the cans. 13. Thus. just as molasses flows when it is at a low temperature. The pressure difference causes the forces on the cans. Applying Inquiry Skills 16.72 cm2 volume of water collected = 3800 mL = 3. resulting in a better draft. in this case downward. This provides an example of laminar flow as the liquid flows smoothly with the layers nearest the jar experiencing more friction than layers farther from the jar.48 cm area of nozzle = πr2 = π(0. A specific example will show how to use the data to calculate the speed of the water. 11. According to the hint. hair shampoo.96 cm. and hair cream rinse. (a) This phrase describes something moving very slowly. The diagram below shows that the ball will curve away from the high pressure toward the low pressure. and above the cab the air deflector reduces the impact of the air resistance on the trailer behind the cab. is greater. Answers will vary. The pressure difference causes the air to move more readily up the chimney. (b) The rockets are pointed and smooth. tomato juice. 3 × 103 N [horizontally] Let +x be the direction of the applied force and +y be up.62 m/s.8 m/s 2 ) Ff = 3. ΣFy = 0 FN + (−mg ) = 0 FN = mg Ff = µK FN = µ K mg = (0. which in turn causes an increase in thermal energy.4 Questions (Pages 106–107) Understanding Concepts 1. The rubbing involves kinetic friction between the hands.) m = 2.1× 103 kg)(9. The burrowing animal is applying Bernoulli’s principle. air will rise there.To find the speed: v= d ∆t (volume ÷ area) = ∆t 3 (3.7 ×103 N The magnitude of the frictional force is 3.72 cm2 ) = 85 s v = 62 cm/s Making Connections 17. The speed of the water from the nozzle in this example is 62 cm/s or 0. With a lower pressure above the front entrance. 128 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . the speed of the air moving across the top of the mound is likely higher than the speed along the ground surrounding the lower entrance.8 ×10 cm ) ÷ (0.1 × 10 3 kg µK = 0. 2. causing a circulation from the rear entrance to the front entrance. causing agitation of the surface molecules. (Students have studied the kinetic molecular theory at previous grade levels.18)(2.18 Fapp = 5.7 × 103 N. As shown in the diagram below. Section 2. φ.76 m/s 2 The magnitude of the acceleration of the sled is 0.3 × 103 N − 3. µS = 0.76 m/s2.ΣFx = ma x Fapp + (− Ff ) = max ax = Fapp − Ff m 5.1× 103 kg a x = 0. and µK can be derived as follows: ΣFy = 0 FN + (−mg cos φ ) = 0 FN = mg cos φ ΣFx = ma x mg sin φ + (− Ff ) = ma x ax = Substitute for FN: mg sin φ − µ K FN m mg sin φ = µK FN FN = mg sin φ µK mg sin φ − µ K FN m mg sin φ − µ K mg cos φ = m a x = g (sin φ − µ K cos φ ) (c) The skier’s mass has no affect on the magnitude of the acceleration. m cancels out. 4. As shown in (a) and (b) above.7 × 103 N = 2. 3. Let +x be parallel to and down the hillside and +y be perpendicular to and up from the hillside. (a) The equation for the coefficient of kinetic friction in terms of the angle φ can be derived as follows: ΣFx = 0 ΣFy = 0 mg sin φ + (− Ff ) = 0 FN + (−mg cos φ ) = 0 FN = mg cos φ Equate the two equations for the normal force: mg sin φ mg cos φ = µK sin φ = µK cos φ µ K = tan φ (b) The equation for the magnitude of the acceleration in terms of g.88 (a) ax = Copyright © 2003 Nelson Chapter 2 Dynamics 129 . so the terminal speed of the ball in glycerine would be higher at 60°C than at 20°C. (c) The experimental values can be found by using Sample Problem 2 on page 100 of the textbook as reference. (Students may present a more complete explanation by considering the law of conservation of energy. and pumping stations are needed to get the speed back up to an acceptable level. For the fingers. A good estimate would be between 0. the greater must be the difference in the slow and fast speeds in the tube. Considering the vertical forces: ΣFy = 0 FN + (−mg ) = 0 FN = mg Considering the horizontal forces: ΣFx = max Ff = max ax = µS FN m µS mg = m = µS g = (0. Applying Inquiry Skills 8. it experiences both internal friction and friction with the interior walls of the pipe.88)(9.6 m/s 2 The magnitude of the maximum acceleration is 8. The friction causes the speed of the gas to slow down.8 m/s 2 ) 5. low speed along the bottom of the text’s cover. As the gas moves through the pipeline.58 130 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . 7. (a) Estimates will vary. 6. estimates will vary.(b) Let +x be the direction of the acceleration and +y be up.3 and 0. The difference in pressures causes the liquid mercury in the flowmeter tube to drop where the pressure above it is high and rise where the pressure above it is low. the speed of the liquid increases and the pressure decreases. Where the diameter of the tube is large. the speed of the liquid is low and the pressure is high. rise µK = run 14 cm = 24 cm µK = 0. The viscosity of a liquid decreases as the temperature increases.6. However.5 and 0. The difference is that to measure the coefficient of kinetic friction. Like a river than narrows. where the diameter of the tube becomes small. The greater the difference in heights.6 m/s2. The important thing here is to estimate a lower value than in part (a).4. which they have studied in previous classes. This is somewhat difficult because the mass of the book is fairly high. students must try to get their fingers and then their fingernails to move at a constant. rise = 14 cm and run = 24 cm. which coincides with Bernoulli’s principle. (b) Again. but a good estimate would be between 0. it is easier if the lower end of the book is held still or propped up against some barrier. The device can be calibrated to indicate actual speeds. a x = 8. rise µK = run 10 cm = 26 cm µK = 0. and provides high efficiency and quiet performance in airline and other transportation applications. The end surfaces of the blocks are manufactured with extreme precision. assuming that the experimental values are the “accepted” values. With air moving at a high speed across the top of the straw.htm 11.001 or even less in dry inert gases.com/forefl/9710/wiren9710. It is caused by a club swing and impact that is at an angle to the target line forcing the ball to spin clockwise as it travels away from the tee.com/sds/Pages/partout.html Copyright © 2003 Nelson Chapter 2 Dynamics 131 . Near-frictionless carbon (NFC) is an extremely hard carbon film that resembles smooth diamond films but has a much lower coefficient of friction.com/ixaccuracy http://www. The following Web sites provide details of the importance of considering friction in the design of running shoes: http://scire.ucsc.edu/~josh/5A/book/forces/node21. to gain proficiency in estimating coefficients of friction.edu/home/baez/physics/General/golf. as low as 0. The air pressure on the surface of the water in the beaker remains higher.html http://stravinsky.floridagolfing.html 12. In simple terms. Gauge blocks are end gauges that can be wrung together to provide a variety of specified lengths. it sprays onto the paper.anl. The ones listed here provide more detail about different types and causes of slices and hooks. One example. simulating the action of a paint sprayer. whereas today’s outsoles for track shoes have moulded plastic plates with interchangeable sets of spikes chosen for different events. like the one described in this question.html http://www.mrgolf. running shoes are much more advanced than in the past.ucr. taken from the first Web site listed below.html 13. More information can be found at the following Web site: http://www.se/trading_eng/broschrer/701/passbitar. rise = 10 cm and run = 26 cm. The hard film is durable and relatively easy to deposit on surfaces. The following Web site provides more details about NFC as well as links to other related sites: http://www. There are many Web sites devoted to this topic. This is a simple (and somewhat wet!) demonstration of Bernoulli’s principle.For the fingernails. The last site listed describes details of why golf balls are dimpled. Technically. pushing down on the water in the beaker and thus upward on the water in the straw.38 Students can compare their estimated and experimental values by finding the percent error. as guidelines for measuring tools. and improvements continue to be made. is that early track event running shoes were simple leather shoes with nails driven through them. as well as ways of preventing them. These blocks are used to check the dimensions of items in a workshop or manufacturing process. 9.izgolf. estimated value − experimental value × 100% % error = experimental value (d) It takes just a few activities. the air pressure there is low.com/slice. a slice is a golf shot that carries the ball to the right of the target line.html http://math.gov/techtour/nfc-faqs.cej. and to calibrate other gauge blocks. A hook carries the ball to the left of the target line. As the water exits the top of the straw. allowing the blocks to stick together as they are built up to obtain the desired length.techtransfer. http://www. Making Connections 10. and is caused by a swing and impact that sends the ball spinning counterclockwise after it leaves the tee. Ffict . ∑ Fy = ma y = 0 FT cos θ − Fg = 0 FT cos θ = Fg where Fg = mg FT = mg cos θ 132 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . (a) The ball remains stationary relative to the bus because the net force acting on it is zero. and once the puck is in motion it remains in motion because the net force acting on it is zero. with +y up. Once your hand no longer touches the puck. the ball moves at a constant velocity of 12 m/s [E] relative to the road. (b) The FBD is the same in each frame of reference. The bus is accelerating but it is the noninertial frame of reference. 3. (c) As the bus accelerates forward relative to the road. The truck is an inertial frame of reference. m = 25 g θ = 13° (a) Consider the vertical components of the forces. 2. the ball begins to roll backward relative to the bus. you will observe the puck moving at a constant velocity relative to the truck.2. experiencing acceleration.5 INERTIAL AND NONINERTIAL FRAMES OF REFERENCE PRACTICE (Page 110) Understanding Concepts 1. the same as the velocity of the bus relative to the road. However. (d) The FBD in the noninertial frame of reference has a fictitious force. 25 N The magnitude of the tension in the string is 0. (b) (i) The beads remain at rest in their lowest position. The mass of the rubber stopper is needed for this calculation. away from the 0° mark to some angle that depends on the magnitude of the acceleration. The mass of the rubber stopper is not needed since the force of tension depends on the mass and. 2. and can be described as a frame of reference in which Newton’s law of inertia does not hold. (a) The plane of the accelerometer must align in the north-south plane. cancels during the calculation.Substitute into the equation for the horizontal components. (iii) The beads drop down to the 0° mark and stay there as long as the velocity is constant.8 m/s 2 ( )(tan 13 ) a x = 2.5 Questions (Page 111) Understanding Concepts 1.025 kg)(9. thus. away from the 0° mark to some angle that depends on the magnitude of the acceleration.3 m/s 2 Thus. (c) In the frame of reference of the road: Copyright © 2003 Nelson Chapter 2 Dynamics 133 . with +y up. (b) Consider the vertical components of the forces.8 m/s 2 ) = cos13° FT = 0.25 N. ΣFy = 0 FT cos θ + (−mg ) = 0 mg cos θ (0. FT = Section 2. (ii) The beads move backward. A noninertial frame of reference is also called an accelerating frame of reference. (iv) The move forward.3 m/s2 [E]. with +x east: ∑ Fx = ma x FT sin θ = ma x  sin θ  a x = ( FT )    m   mg   sin θ  =    cos θ   m   sin θ  = g   cos θ  = g tan θ = 9. the acceleration of the train is 2. Consider the vertical components of the forces: ΣFy = ma y = 0 FN cos θ − Fg = 0 FN cos θ = Fg FN = mg cos θ Next consider the horizontal components of the forces: ∑ Fx = ma x FN sin θ = ma x  sin θ  a x = ( FN )    m   mg   sin θ  =    cos θ   m   sin θ  = g   cos θ  = g tan θ = 9.8 m/s 2 ( )(tan 11 ) a x =1.9 m/s2.2 × 10−3 kg)(9.2 g = 2.2 × 10−2 N.8 N/kg) = cos11° −2 FN = 2.(d) In the frame of reference of the vehicle: (e) Let +y be up and +x be the direction of the acceleration.2 × 10−3 kg From the vertical components considered in (e): mg FN = cos θ (2. 134 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .2 × 10 N The magnitude of the normal force acting on the middle bead is 2.9 m/s 2 The magnitude of the acceleration is 1. (f) m = 2. a ruler can be used to perform the measurement. but if you shift your frame of reference to the road (i. the forces acting on the object are gravity and tension. you are moving backward.e. (a) (b) The FBD of the ornament in the frame of reference of the road can be used to determine the acceleration. but it is not necessary. Relative to that car’s frame of reference.. Copyright © 2003 Nelson Chapter 2 Dynamics 135 . Since the only measurement required is tan of the angle to the vertical. and θ is the angle from the vertical that the pendulum ornament hangs from the vertical during acceleration. (A protractor could be used to measure that angle. pages 109 and 110. (a) The observation that your car is moving backward results from the adjacent car move forward slightly. Earth’s frame of reference). you soon realize that you are stationary and the adjacent car is moving.Applying Inquiry Skills 3. (b) If you are seated in a theatre seat with very dim lighting surrounded by images of moving objects. you will get the sensation that you are moving. we have a x = g tan θ . where ax is the magnitude of the acceleration. You place yourself in the frame of reference of the moving images rather than in the frame of reference of the theatre and the seat. Making Connections 4.) Using the equation derived in Sample Problem 2. g is the magnitude of the acceleration due to gravity. As shown in the second diagram above. are the horizontal components of FT1 and FT2 equal? Within experimental error. with +x to the right: ΣFx = max = 0 − FT1 cos a + FT2 cos d + 0 = 0 ? ? ? FT1 sin a + FT2 sin d = F3 ? 2. an object in static equilibrium is at rest and has a net force of zero acting on it. does the sum of the upward components of FT1 and FT2 equal the downward component of FT3 ? Within experimental error.2. Following is a start to the calculations. the answer is yes. d. students use vertical and horizontal components of vectors to test the condition of static equilibrium of a common point. Students will likely recall from Grade 11 physics that they can calculate the force of gravity on a mass by using the gravitational field strength. Procedure 1. respectively. they can perform the calculations as shown below. be FT1 . The data collected are analyzed as described in #1 above. and m3. Let the tensions in the strings attached to masses m1.98 N Materials See page 112. In the horizontal plane. referred to as the origin. with +y up: ΣFy = ma y = 0 FT1 sin a + FT2 sin d − F3 cos f = 0 ? ? FT2 cos d = FT1 cos a 136 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . the answer is yes. FT2 . and FT3 . After setting up the apparatus according to the instructions and measuring angles a and d. In the vertical plane with +y up: ΣFy = ma y = 0 FT1 sin a + FT2 sin d − F3 = 0 ? ? In other words.CHAPTER 2 LAB ACTIVITIES Activity 2. Thus. page 79. 3. Considering the vertical components of the forces. for example. In this activity. the students can perform the calculations and compare the components. to two significant digits. After the students follow the instructions and measure angles a. the magnitude of the force of gravity on a 100-g mass.1: Static Equilibrium of Forces (Page 112) As stated in the text. In other words. m2. is: Fg = mg = (0.10 kg)(9.8 N/kg) Fg = 0. and f. In that question only kinetic friction was investigated. a summary table of data and calculations (a sample of which is shown below). and question 8 on page 107. Evaluation. with +y to the right: ΣFx = max = 0 FT2 cos d + FT3 sin f − FT1 cos a = 0 Students should find that in each case the components add to zero. (d) The student report should include the Investigation title. Question (a) A typical question is: How do the coefficients of static and kinetic friction for a running shoe on wood compare to each other and to other objects on the same wood? Hypothesis/Prediction (b) For each set of surfaces in contact.4. Students should be sure that any breakable object tested does not fall off the raised board or the lab bench.1: Measuring Coefficients of Friction (Page 113) The intent of this investigation is to have students design and carry out their own controlled experiment to determine the coefficients of static and kinetic friction. if there is friction involving the strings holding m1 and m2. Thus. and Synthesis. Evaluation (c) The following ways would help to improve accuracy: • use very smooth strings and pulleys to reduce friction to a minimum • check carefully to be sure the circular protractor is properly aligned horizontally • eliminate parallax error as much as possible when checking to be sure the origin of the protractor coincides with the common point of the three strings Investigation 2. Experimental Design (approved by the teacher). Analysis. because rubbery substances tend to stick to other surfaces more.) Other parts of the text that relate directly to this investigation are Sample Problem 2 on page 100. calculations (samples of which are shown below). question 8 on page 101. and thus affect the calculations. especially wooden or plastic blocks or books. and the other to lower the end of the board slightly until the object appears to move down the slope at a constant velocity (to determine the coefficient of kinetic friction).0.Considering the horizontal components of the forces. Most running shoes on wood likely have higher coefficients of friction than other objects. Analysis (a) Static equilibrium occurs for a stationary object when the net force acting on that object is zero. Question. ? ? 4. Also. students should ensure that the raised board does not drop onto any other object or somebody’s fingers. The data collected are analyzed as described in question 3 above. The safety concerns for this investigation are rather simple. Hypothesis/Prediction. Copyright © 2003 Nelson Chapter 2 Dynamics 137 . students must work together carefully to obtain two sets of data for each pair of surfaces tested. Experimental Design (c) Refer to the answer to Practice question 8 earlier in this manual (and text page 101) with sight changes. whereas here both static and kinetic friction are investigated. then m3 would have to be larger than is predicted by theory to achieve static equilibrium. the coefficient of static friction should be greater than the coefficient of kinetic friction because objects tend to move more easily once their motion has begun. In step 1. For example the coefficient of static friction of rubber on wood may approach 1. one in which one end of the wooden board is raised until the object in it begins to slide down the board (to determine the coefficient of static friction). for example. (Notice that all of the Inquiry Skills are highlighted. (b) Friction between the strings and the pulleys can affect the tension in the strings. Evaluation. In that question only kinetic friction was investigated. the coefficient of static friction should be greater than the coefficient of kinetic friction because objects tend to move more easily once their motion has begun. especially wooden or plastic blocks or books. then m3 would have to be larger than is predicted by theory to achieve static equilibrium. Copyright © 2003 Nelson Chapter 2 Dynamics 137 . Experimental Design (approved by the teacher). if there is friction involving the strings holding m1 and m2. a summary table of data and calculations (a sample of which is shown below). (d) The student report should include the Investigation title. Evaluation (c) The following ways would help to improve accuracy: • use very smooth strings and pulleys to reduce friction to a minimum • check carefully to be sure the circular protractor is properly aligned horizontally • eliminate parallax error as much as possible when checking to be sure the origin of the protractor coincides with the common point of the three strings Investigation 2. In step 1. Question (a) A typical question is: How do the coefficients of static and kinetic friction for a running shoe on wood compare to each other and to other objects on the same wood? Hypothesis/Prediction (b) For each set of surfaces in contact. with +y to the right: ΣFx = max = 0 FT2 cos d + FT3 sin f − FT1 cos a = 0 Students should find that in each case the components add to zero. The data collected are analyzed as described in question 3 above. ? ? 4. and thus affect the calculations. students must work together carefully to obtain two sets of data for each pair of surfaces tested. Experimental Design (c) Refer to the answer to Practice question 8 earlier in this manual (and text page 101) with sight changes.0.Considering the horizontal components of the forces. Also. because rubbery substances tend to stick to other surfaces more. (b) Friction between the strings and the pulleys can affect the tension in the strings. question 8 on page 101. one in which one end of the wooden board is raised until the object in it begins to slide down the board (to determine the coefficient of static friction). and the other to lower the end of the board slightly until the object appears to move down the slope at a constant velocity (to determine the coefficient of kinetic friction). Students should be sure that any breakable object tested does not fall off the raised board or the lab bench. and Synthesis. students should ensure that the raised board does not drop onto any other object or somebody’s fingers. Analysis. (Notice that all of the Inquiry Skills are highlighted. Question.1: Measuring Coefficients of Friction (Page 113) The intent of this investigation is to have students design and carry out their own controlled experiment to determine the coefficients of static and kinetic friction. whereas here both static and kinetic friction are investigated. Analysis (a) Static equilibrium occurs for a stationary object when the net force acting on that object is zero. Hypothesis/Prediction. Most running shoes on wood likely have higher coefficients of friction than other objects. and question 8 on page 107. for example. The safety concerns for this investigation are rather simple.) Other parts of the text that relate directly to this investigation are Sample Problem 2 on page 100. Thus.4. calculations (samples of which are shown below). For example the coefficient of static friction of rubber on wood may approach 1. 47 m = 1.24 2.max and µK = K mg mg = 138 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .84 0.67 Run (m) 1.901 µK = 0.81 m. (ii) How do the coefficients of static and kinetic friction relate to the materials in contact? Answer: Rubber-soled shoes have much higher coefficients of static and kinetic friction than wood.754 µS = 0. and the force needed to keep the object moving at a constant velocity can be used to determine the coefficient of kinetic friction.93 m = 1.35 µK = 0.Sample calculations for a running shoe on a wooden board: For the coefficient of static friction. Analysis (e) Questions and answers may vary.42 µK = 0. the rise of the board = 1.754 × 100% 0.81 1. (i) For each object tested.30 2.34 Coefficient of Friction µS = 0.47 m and run of the board = 1.95 0. plastic.95 m. The percentage drop in the coefficient for the running shoe is: µ − µK × 100% % drop = S µS 0.63 1.901 For the coefficient of kinetic friction. (iii) What procedure could be used to check the validity of the coefficients of friction found in this investigation? Answer: The coefficients of static and kinetic friction for the same sets of surfaces can be found by pulling horizontally on the moveable object with a force sensor or a force meter attached and the board set up horizontally. rise µS = run 1. rise µS = run 1.47 0.95 m µS = 0. how does the coefficient of kinetic friction compare to the coefficient of static friction? Answer: In all cases.901 % drop = 16.901 − 0. Three examples are given here.29 2.37 µS = 0.754 The following is a sample data table for objects on a wooden board: Object Tested running shoe math text varnished wood Type of Friction static kinetic static kinetic static kinetic Rise (m) 1. the rise of the board = 1. The force required to just start the stationary object moving can be used to calculate the coefficient of static friction.3% For the other two examples given. or most other materials on the same surface (wood in this investigation).95 2.81 0.63 m and run of the board = 1.61 m µS = 0. the corresponding drops are 12% and 17%.29 Materials See page 113 of the text. The equations needed are: F F µS = S. the coefficient of kinetic friction was less than the coefficient of static friction. but the FBDs and basic facts should resemble what is depicted below. Students can verify the statement experimentally by carrying out the same steps they used in this investigation. page 100. they could secure two books together and repeat the measurements. the error can be added to the measured quantities. Copyright © 2003 Nelson Chapter 2 Dynamics 139 . Urge the students to add enough detail to help them understand and remember the contents of the chapter. but adding more mass to the object’s tested. Then they should analyze the forces by considering the components perpendicular and parallel to the inclined plane. As shown there. as shown in Sample Problem 2.Evaluation A major source of random error is that the running shoe tends to get stuck and unstuck in an irregular fashion as it moves down the inclined board. Synthesis (g) Students can use a mathematical derivation to verify that the statement is true. Figure 6 on page 100. They should begin with a set of diagrams like those in the text. but with the force of kinetic friction rather than static friction. mg cancels out. Another source of random error is parallax error in measuring the rise and run of the board when it is at an angle to the desk or floor. Another major source of random error is that it is very difficult to judge when an object moving down the board experiences a constant velocity. A minor systematic error occurs if the metre stick has a worn end. If this is the case. CHAPTER 2 SUMMARY Make a Summary (Page 114) Individual summaries may vary somewhat. The effects of these errors can be minimized by taking an average of several readings. This error can be minimized by looking straight at the metre stick and checking to be sure the metre stick is as vertical as possible when measuring the rise. (For example. Urge the students to add enough detail to help them understand and remember the contents of the chapter. The effects of these errors can be minimized by taking an average of several readings. This error can be minimized by looking straight at the metre stick and checking to be sure the metre stick is as vertical as possible when measuring the rise. A minor systematic error occurs if the metre stick has a worn end. Students can verify the statement experimentally by carrying out the same steps they used in this investigation. CHAPTER 2 SUMMARY Make a Summary (Page 114) Individual summaries may vary somewhat. page 100. mg cancels out. Another source of random error is parallax error in measuring the rise and run of the board when it is at an angle to the desk or floor. Another major source of random error is that it is very difficult to judge when an object moving down the board experiences a constant velocity. but the FBDs and basic facts should resemble what is depicted below. As shown there. (For example. If this is the case. they could secure two books together and repeat the measurements. They should begin with a set of diagrams like those in the text. Then they should analyze the forces by considering the components perpendicular and parallel to the inclined plane. the error can be added to the measured quantities. Copyright © 2003 Nelson Chapter 2 Dynamics 139 . but with the force of kinetic friction rather than static friction. Figure 6 on page 100. as shown in Sample Problem 2. Synthesis (g) Students can use a mathematical derivation to verify that the statement is true.Evaluation A major source of random error is that the running shoe tends to get stuck and unstuck in an irregular fashion as it moves down the inclined board. but adding more mass to the object’s tested. the net force on the person must be zero because the person is moving with a constant velocity.8 m/s 2 ) Fapp = 61N 17.18 because it is independent of the force of gravity acting against the skier’s motion. T 10. (b) A skier of mass m is sliding down a snowy slope that is inclined at an angle φ above the horizontal.0 N [W] on the single cart. T 11. (a) During the time interval that the spring acts on the double cart.5 kg)(9. 18. 140 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .CHAPTER 2 SELF QUIZ True/False 1. T F The magnitude of the tension in the string is 16 N. (b) The magnitude of the minimum horizontal force that must be applied is 61 N using the following equations. F Maximum static friction (i.. 9. 2. (d) The reaction force is the upward force gravity on Earth exerted by the can. F Fictitious forces must be invented to explain observations in an accelerating frame of reference. 14. 3. 8. T 7. 12. With +y defined as perpendicular to and upward from the hillside. the magnitude of the normal force on the skier is: ΣFy = 0 FN + (−mg cos φ ) = 0 FN = mg cos φ 15. With +y up: ΣFy = 0 FN + (− Fapp sin θ ) + (−mg ) = 0 FN = Fapp sin θ + mg F The coefficient of kinetic friction still equals 0. starting friction) tends to be greater than kinetic friction. (b) During the spring interaction. 5. F It is possible for an object to be travelling eastward while experiencing a net force that is westward.e. The object would be decelerating. 4. F One possible SI unit of weight is the newton. 6. 16. (d) A force exerted by the air and the downward force of gravity are acting on the football.0 N [W]. F The magnitude of the normal force is now greater than 155 N because the y-component of the applied force would add to the force due to gravity.65)(9. the double cart exerts an equal but opposite force of 2. the net force acting on the single cart is 2. 13. with +y up and +x the direction of the applied force: ΣFy = 0 FN − mg = 0 FN = mg ΣFx = 0 Fapp + (− Ff ) = 0 Fapp = µS FN = µS mg = (0. (a) According to Newton’s first law of motion. 2.0 kg a = 1. it moves in the direction of the force applied to it. Newton’s second law of motion can be applied to determine the mass of an object in interstellar space. A properly located headrest helps to limit how far back a person’s head goes after jerking backward. Then the mass is calculated: ΣF = ma ΣF m= a The physics is not good but the drama is greater than it would be with proper physics. Since the elevator and people in it undergo free fall at the same time. Simultaneously. it must have a net force of zero acting on it.0 N [E] = 2.8 m/s 2 The magnitude of the acceleration of the cars down the incline is 5.8 m/s2.20 s) vf = 0. 7. This can only happen if there are no forces on the object or if the sum of all the forces acting on it is zero. 4.25 cm/s [E]. and then jerks backward as the overstretched neck muscles pull backward (according to Newton’s second law of motion). Thus. your feet must exert a force on the inside base of the canoe that is downward and backward (the reaction force). (b) During the interaction.8 m/s 2 ) sin 36° a x = 5. but greater than zero. Copyright © 2003 Nelson Chapter 2 Dynamics 141 . In order for the people’s heads to be pressed against the ceiling. the acceleration of the double cart is 1. In order for an object to be at rest.38 × 10−2 N [horizontal] 3. the canoe exerts a force on you that is upward and forward (the reaction force).20 s is less than 0. the elevator would have to be accelerating downward faster than the people in it. 5. 6. θ = 36° Let +x be the direction of the acceleration down the hillside. Shoulder restraints and airbags help to limit the person’s forward motion after a collision. Since the canoe is relatively free to move in the water. a x = g sin θ = (9. (e) The velocity of the cart at 0. A known net force is applied to the object and the object’s acceleration is measured. Whiplash occurs when a vehicle stops suddenly and the head of a person in the vehicle continues to move forward (according to Newton’s first law of motion). the net force acting on the single cart is zero. vf = vi + a ∆t = 0 m/s + (1.0 m/s 2 [E] 20. there cannot be just one force since the sum would not be zero. (e) After the spring interaction is complete and the carts are separated.0 m/s2 [E]: ΣF = ma ΣF a= m 2.20 m/s [E] 21.0 m/s 2 [E])(0. they should be observed falling together. namely in the opposite direction to your motion. CHAPTER 2 REVIEW (Pages 117–118) Understanding Concepts 1. m1 = 113 g m2 = 139 g Fapp = 5.19. In order to move toward the front of the canoe. CHAPTER 2 REVIEW (Pages 117–118) Understanding Concepts 1. there cannot be just one force since the sum would not be zero. Newton’s second law of motion can be applied to determine the mass of an object in interstellar space.19.20 m/s [E] 21. your feet must exert a force on the inside base of the canoe that is downward and backward (the reaction force). m1 = 113 g m2 = 139 g Fapp = 5.20 s is less than 0. In order for an object to be at rest.20 s) vf = 0. and then jerks backward as the overstretched neck muscles pull backward (according to Newton’s second law of motion). the elevator would have to be accelerating downward faster than the people in it. 5. (b) During the interaction. vf = vi + a ∆t = 0 m/s + (1. A properly located headrest helps to limit how far back a person’s head goes after jerking backward.0 kg a = 1. Thus. A known net force is applied to the object and the object’s acceleration is measured. Then the mass is calculated: ΣF = ma ΣF m= a The physics is not good but the drama is greater than it would be with proper physics. Copyright © 2003 Nelson Chapter 2 Dynamics 141 . This can only happen if there are no forces on the object or if the sum of all the forces acting on it is zero. Since the elevator and people in it undergo free fall at the same time. Shoulder restraints and airbags help to limit the person’s forward motion after a collision. (e) After the spring interaction is complete and the carts are separated. the canoe exerts a force on you that is upward and forward (the reaction force). 7. it moves in the direction of the force applied to it. 6.8 m/s2. 4. it must have a net force of zero acting on it. a x = g sin θ = (9.25 cm/s [E]. Simultaneously.38 × 10−2 N [horizontal] 3.0 N [E] = 2.8 m/s 2 ) sin 36° a x = 5. but greater than zero. the acceleration of the double cart is 1.0 m/s 2 [E])(0.8 m/s 2 The magnitude of the acceleration of the cars down the incline is 5. Whiplash occurs when a vehicle stops suddenly and the head of a person in the vehicle continues to move forward (according to Newton’s first law of motion). θ = 36° Let +x be the direction of the acceleration down the hillside. 2. namely in the opposite direction to your motion. the net force acting on the single cart is zero. Since the canoe is relatively free to move in the water.0 m/s2 [E]: ΣF = ma ΣF a= m 2. (e) The velocity of the cart at 0. In order to move toward the front of the canoe.0 m/s 2 [E] 20. they should be observed falling together. In order for the people’s heads to be pressed against the ceiling. and for m3 y is down..213 m/s 2 ) Fapp = 2.4 m/s2. for m1. let the positive direction for the system be clockwise. Consider m1: ΣFy = m1a FTL − m1 g = m1a FTL = m1 ( g + a ) Consider m3: ΣFy = m3 a m3 g − FTR = m3 a FTR = m3 ( g − a) Consider m2: ΣFx = m2 a FTR − FTL = m2 a Substitute for the tensions in the two strings FTR and FTL: m3 ( g − a ) − m1 ( g + a ) = m2 a (m3 − m1 ) g = (m1 + m2 + m3 )a a= = (m3 − m1 ) g (m1 + m2 + m3 ) (41kg − 26 kg)(9. 142 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . we can determine the force applied by m2 on m1. Thus. ΣFx = (m1 + m2 )a x Fapp = (m1 + m2 )a x ax = = Fapp (m1 + m2 ) 5.41 × 10−2 N.8 m/s 2 ) (26 kg + 38 kg + 41kg) a = 1.41× 10−2 N The magnitude of the force exerted by each burger on the other burger is 2.213 m/s2.139 kg) a x = 0.(a) Let +x be the direction of the applied force and the acceleration. To find the magnitude of the forces.e.38 × 10 −2 N (0. ΣFx = m1a x Fapp = m1a x = (0. and let a be the magnitude of the acceleration of all three masses.213m/s 2 The magnitude of the acceleration of the two-burger system is 0. thus. for m1 y is up.113kg + 0.113 kg)(0. Also. 8.4 m/s 2 The magnitude of the acceleration of the blocks is 1. (b) The two forces named are equal in magnitude but opposite in direction (i. they are an action-reaction pair). for m2 y is to the right. m1 = 26 kg m2 = 38 kg m3 = 41 kg (a) Let the subscripts TL and TR represent the tension in the left and right strings. 0° above the horizontal]. +y be up.4 × 102 N.0°) = (cos 27.4 m/s 2 ) FTL = 2. and FA be the force applied by the cliff on the climber.9 × 102 N. The force of the cliff on the climber’s feet has both a vertical component and a horizontal component.8 m/s 2 − 1.0° Substituting this angle into either of the first equations: F cos θ FA = T cos φ (729 N)(cos 27. The magnitude of the tension in the right string is 3. Let +x be to the left. and the vertical component of the applied force is equal in magnitude to the vertical component of the tension.(b) Substitute the magnitude of the acceleration into the equations for the tensions developed in (a): FTL = m1 ( g + a ) = (26 kg)(9. FTR = m3 ( g − a) = (41kg)(9.0°) φ = tan −1 = tan −1 φ = 27.0°) (729 N)(cos 27.8 m/s 2 + 1.9 ×10 2 N The magnitude of the tension in the left string is 2.4 m/s 2 ) FTR = 3. Copyright © 2003 Nelson Chapter 2 Dynamics 143 . Considering the vertical components of the forces: ΣFy = ma y = 0 FT sin θ + FA sin φ − Fg = 0 FA sin φ = mg − FT sin θ FA = mg − FT sin θ sin φ Considering the horizontal components of the forces: ΣFx = ma x = 0 FA cos φ − FT cos θ = 0 FA cos φ = FT cos θ FA = FT cos θ cos φ Equating these equations for FA: FT cos θ mg − FT sin θ = cos φ sin φ sin φ mg − FT sin θ = FT cos θ cos φ tan φ = mg − FT sin θ FT cos θ mg − FT sin θ FT cos θ (67.4 ×102 N 9.80 N/kg) − (729 N)(sin 27.0°) FA = 729 N The force of the cliff on the climber’s feet is 729 N [27. Note: This question can also be solved by applying the fact that the horizontal component of the applied force is equal in magnitude to the horizontal component of the tension.5 kg)(9. 8 m/s 2 ) sin14.1N The magnitude of the normal force on the wagon is 70.3 N The magnitude of the force applied by the child is 18.3 N.38 kg(6.3° FN = 70.45 cm/s2 [up the hill] (a) Let +x be the direction of the acceleration.1 N. m = 7. ΣFx = max Fapp + (− mg sin θ ) = max Fapp = m(a x + g sin θ ) Fapp = 7.3°) = 18. Let +y be up. 11.10.3° a = 6.38 kg θ = 14.45 × 10−2 m/s 2 + (9. The diagrams below show the sagging and tightly stretched clotheslines.8 m/s 2 ) cos14. For the sagging clothesline: ΣFy = ma y = 0 2 FT1 sin θ − mg = 0 FT1 sin θ = mg 2 144 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . ΣFy = 0 FN + (−mg cos θ ) = 0 FN = mg cos θ = (7.38 kg)(9. (b) Let +y be perpendicular to and up from the hillside. 61 Since φ < θ. (b) Fapp = 109 N ∆d = 75 cm ΣFx = ma x Fapp + (− Ff ) = max ax = Fapp − µK mg m 109 N − 96 N = 16 kg a x = 0. 12. so small steps must be taken. especially if some of the terrain is rocky. Some sports would be difficult or perhaps unsafe if the athletes wore high-friction shies. spikes are a disadvantage. and the hammer throw. Examples are the trampoline. ice hockey. Let +y be up and +x be the direction of the applied force. spiked shoes provide an advantage. aikido. 13. Newton’s third law of motion is applied in order to walk. bandy. the backward component of the action force must be small to prevent slipping and falling. shoes must not have heels. and no shoes are worn in judo. in events in which the athlete rotates within the throwing circle. kendo. and luge tobogganing. FT2 sin θ Thus. curling. roller hockey. In most track and field events. the discuss throw. (a) Considering the vertical components of the forces: ΣFy = ma y = 0 FN − mg = 0 FN = mg Considering the horizontal components of the forces: ΣFx = 0 Fapp + (− Ff ) = 0 Fapp = µ K FN = µ K mg = (0. sports acrobatics (especially with group work).8 m/s 2 ) Fapp = 96 N The magnitude of the applied force is 96 N. Field events in this category are the shot put. 14. However. On a slippery surface. and jiu jitsu. so T1 < 1 and FT1 < FT2 . karate. buckles.For the tightly stretched clothesline: ΣFy = ma y = 0 2 FT2 sin φ − mg = 0 FT2 sin φ = mg 2 Equating these quantities: FT1 sin θ = FT2 sin φ FT1 sin φ = FT2 sin θ F sin φ < 1 .84 m/s 2 Copyright © 2003 Nelson Chapter 2 Dynamics 145 . or nailed soles. spikes are a disadvantage. the tension is greater in the clothesline that is tightly stretched. beach volleyball. The action force is the foot pushing downward and backward on the walking surface. In some cross-country running events. In wrestling. and it is more likely to break. m = 16 kg µK = 0. and the reaction force is the surface pushing upward and forward on the foot. speed skating.61)(16 kg)(9. 4 kg. the table would take 1.75 m) 0. Considering the vertical components of the forces: ΣFy = 0 FN + (−mg cos θ ) = 0 FN = mg cos θ 146 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .8 m/s m = 4.55 µK = 0.4 kg The smallest possible mass of the box if it remains at rest is 4. FT = 21 N [31° above the horizontal] µS = 0. θ = 4.55   = 2 9.3 s to travel 75 cm.11 vf = 0 m/s Let +x be down the hillside and +y be perpendicular to and up from the hillside. Considering the vertical components of the forces: ΣFy = 0 FT sin θ + FN + (−mg ) = 0 FN = mg − FT sin θ Considering the horizontal components of the forces: ΣFx = 0 FT cos θ + (− Ff ) = 0 FT cos θ = µS FN FT cos θ = µS (mg − FT sin θ )  cos θ  + sin θ  FT   µS  m= g  cos 31°  + sin 31°  (21N)  0. 16.84 m/s 2 ∆t = 1.∆d = vi ∆t + ∆d = 1 a x ( ∆t ) 2 2 1 a x ( ∆t ) 2 2 2 ∆d ∆t = ax = 2(0.7 m/s [down] µK = 0.50 Let +y be up and +x be in the direction of the horizontal component of the tension force.7° vi = 2.3s Thus. 15. 11cos 4. Considering the vertical components of the forces: ΣFy = ma y FN − mg = 0 FN = mg Copyright © 2003 Nelson Chapter 2 Dynamics 147 .27 m/s 2 vf 2 = vi 2 + 2ax ∆d vf 2 − vi 2 2a (0 m/s) 2 − (2.27m/s 2 ) ∆d = 13 m The skier will slide 13 m down the slope before coming to rest.7° − 0.Considering the horizontal components of the forces: ΣFx = ma x mg sin θ + (− Ff ) = ma x ma x = mg sin θ − µ K FN ma x = mg sin θ − µ K mg cos θ a x = g (sin θ − µ K cos θ ) = 9. 17.7 m/s)2 = 2(−0.7°) a x = −0.8 m/s 2 (sin 4. (a) In Earth’s frame of reference: ∆d = (b) In the train’s frame of reference: (c) Let +y be up and +x be the direction of the passenger’s acceleration. 00 cm Let R be the rate at which the water ejects from the hose: 2. the speed of the air near the ball’s surface is slower on the left side.7 cm3 /s = 3.8 m/s 2 ) a x .6 m/s 2 The magnitude of the maximum acceleration of the train is 4.00 × 103 mL = 2.14 cm 2 v= R A 66.max = 4. ∆t = 30.max a x .00 × 103 cm3 r = 1. Applying Inquiry Skills 20. making the pressure there greater. 19. Thus.2 cm/s Thus. The pressure difference forces the ball toward the right.6 m/s2.max µS mg = max .14 cm 2 v = 21.67 × 10−2 L / s R = 66. divide the rate by the cross-sectional area of the hose: A = π r2 = π (1 cm)2 A = 3. 18. the air moves westward relative to the ball.0 s V = 2. the speed of the water being ejected from the hose is 21. making the pressure there lower. Simultaneously. With the ball spinning clockwise when viewed from above.0 s = 6. the ball’s surface drags air near it eastward on the left and westward on the right. 148 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .max = µS g = (0.7 cm3 /s To determine the speed of the water. the speed of the air is greater on the right side. or southward.00 L R= 30.47)(9.Considering the horizontal components of the forces: ΣFx = ma x Ff = ma x .max µS FN = max .00 L = 2. As the ball travels eastward through the air.2 cm/s. Considering the horizontal components of the forces: ΣFx = ma x = 0 Fapp − FN = 0 Fapp = FN Considering the vertical components of the forces: ΣFy = ma y = 0 FS − Fg = 0 FS = Fg µS FN = mg µS Fapp = mg µS = mg Fapp mg .0370 kg Using unit analysis. mg µS = Fapp (ii) Determine the minimum horizontal force needed to start an object moving on a horizontal surface. 21.0 kg The mass of the loaded wagon is 27. Fapp The coefficient of static friction is µS = (b) At this stage.0370 kg −1 m = 27.0 kg m/s 2 −1 ( ) slope = 0. students should be able to determine the coefficient of static friction using these three methods: (i) Determine the minimum applied horizontal force needed to prevent an object from sliding down a vertical surface. µS = Fapp mg rise run Copyright © 2003 Nelson Chapter 2 Dynamics 149 .0 kg. it is evident that the mass must be the reciprocal of the slope.85 m/s 2 50. 1 m= slope 1 = 0. µS = (iii) Determine the angle of incline of a ramp such that an object just begins to slide down the ramp. (a) Let +x be the direction of the applied force and +y be up.The magnitude of the slope of the line is: ∆a slope = ∆ΣF = 1. 0 m/s ∆d = 92 cm = 0. (b) The force of gravity is so much smaller than the average force applied by the club.5 × 10 4 m/s 2 ) Fav = 2. The other two methods require measuring both force and mass. 22. When these movies are viewed at lower speeds. vi = 4.5 × 10−2 kg v − vi (a) aav = f ∆t 65 m/s − 0.0 m/s vf = 65 m/s m = 45 g = 4. The air moves across the top or curved part of the wing more quickly than beneath the wing because the air has a greater distance to cover to reach the far end of the wing. and the pressure difference between the bottom and top of the wing causes an upward force on the wing.8 m/s 2 )(0. (a) m = 72 kg vi = 0.2 m/s.5 × 10−2 kg)(6. The design in Figure 5(b) does not produce lift on the paper wing. Where the speed of the air is high.35 m 150 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .2 m/s The gymnast’s speed at impact is 4.9 × 103 N The magnitude of the average force exerted by the club on the ball is 2.0 ms and to determine a speed of 65 m/s.0 ms vi = 0.6 × 103 :1. Making Connections 23. a very short time interval between two frames would be needed to observe a contact time of about 1.8 N/kg ) ) = 6. ∆t = 1. However. the design in Figure 5(c) produces a noticeable lift when air blows across it. Extension 24. frame-by-frame analysis can be done with very short time intervals between frames.92 m vf 2 = vi 2 + 2a∆d vf = 2a∆d = 2(9.5 ×10 2.The advantage of using a ramp is that the only measurements needed are lengths. the air pressure is low. A disadvantage of using a vertical surface is that the object may twist around slightly rather than begin moving downward as the applied force is gradually reduced during experimentation. The demonstration described is a simple but very effective way of demonstrating Bernoulli’s principle.9 × 103 N.2 m/s vf = 0 m/s ∆d = 35 cm = 0.92 m) vf = 4. (b) Let +y be down. In the case of the golf ball. The ratio of the magnitudes of the forces is: Faverage Faverage = Fg mg = Faverage Fg (4.9 ×103 N −2 kg (9.0 (c) High-speed movies are taken with many frames of the film per second.5 × 10 4 m/s 2 Fav = maav = (4.0 m/s = 1 ×10 −3 s aav = 6. m = 22 kg θ = 45° µS = 0. 25.7 ×10 2 N The magnitude of the largest force that can be applied upward is 2. ΣFx = 0 Ff + (− mg sin θ ) = 0 µS FN = mg sin θ mg sin θ FN = µS Copyright © 2003 Nelson Chapter 2 Dynamics 151 .78 cos 45° + sin 45°) Fapp = 2.9 × 103 N.9 × 103 N The magnitude of the force exerted by the floor is 1.2 m/s) 2 = 2(0.7 × 102 N.35 m) a y = −26 m/s 2 ΣF = ma y = (72 kg)(−26 m/s 2 ) ΣF = −1.vf 2 = vi 2 + 2a y ∆d ay = −vi 2 2∆d −(4. (b) The FBD for this situation is shown below.8 m/s 2 )(0. ΣFy = 0 FN + (−mg cos θ ) = 0 FN = mg cos θ ΣFx = 0 Fapp + (− Ff ) + (− mg sin θ ) = 0 Fapp = µS FN + mg sin θ = µS mg cos θ + mg sin θ = mg ( µS cos θ + sin θ ) = (22 kg)(9.65 (a) The FBD for this situation is shown below.78 µK = 0. 0 m/s This problem must be separated into two parts: determine the projectile motion in the region ABCD.8 m/s 2 )  − cos 45°   0.0 × 10−2 m 2.78  = 43 N Fapp The magnitude of the smallest force that can be applied onto the top of the box is 43 N.0 cm to the screen.51× 1015 m/s 2 Determine the distance below the initial axis (∆y1) and the final velocity as the electron exits the region ABCD: Horizontally (constant velocity): ∆x ∆t1 = vx = 3. Fapp 3.3 × 10−9 s)2 2 ∆y1 = 3. 26.11 × 10−31 kg vix = 2.11× 10−31 kg a y = 3.1× 10 −3 m vfy = viy + a y ∆t1 = (3.51× 1015 m/s 2 )(1.51× 1015 m/s 2 )(1.3 × 10−9 s) vfy = 4.25 × 107 m/s ∆t1 = 1.ΣFy = 0 FN + (− Fapp ) + (−mg cos θ ) = 0 Fapp = FN − mg cos θ = mg sin θ − mg cos θ µS  sin θ  = mg  − cos θ   µS   sin 45°  = (22 kg)(9.67 × 106 m/s 152 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . with +y down and +x to the right: ΣFy ay = m Fapp = m 3.25 × 107 m/s viy = 0.20 × 10−15 N [down] m = 9.20 × 10−15 N = 9.3 × 10 −9 s Vertically (constant acceleration): ∆y1 = viy ∆t1 + = 1 a y (∆t1 ) 2 2 1 (3. For the projectile motion. and determine the constant velocity for the remaining 13. 0 kg a x = 1. Considering the horizontal components of the forces on Tarzan: ΣFx = m1a FT = m1a (Equation 1) Considering the vertical components of the forces on Jane: ΣFy = m2 a mg − FT = m2 a Adding Equations 1 and 2: m2 g = m1a + m2 a m2 g = a (m1 + m2 ) a= m2 g m1 + m2 (Equation 2)   50 kg = g  100 kg + 50 kg  1 a= g 3 1 Tarzan’s acceleration is a = g .0 m/s 2 Copyright © 2003 Nelson Chapter 2 Dynamics 153 .78 × 10−9 s) ∆y2 = 2.70 × 10−2 m ∆y = 3.When the electron exits the region ABCD. for Tarzan +x is to the right and for Jane +y is down). m1 = 2.0 kg m2 = 1.70 × 10 −2 m Determine the total vertical distance below the axis of entry: ∆y = ∆y1 + ∆y2 = 3.0 cm (1. the electron is 3.e. Determine the vertical distance (∆y2) as the electron travels the remaining horizontal distance of 13. m1 = 100 kg m2 = 50 kg Let the positive direction of the acceleration of the system of masses be clockwise (i.1× 10−3 m + 2.67 × 106 m/s)(5.30 × 10−3 m) to the screen: ∆x ∆t 2 = vx = 1.02 × 10−2 m Thus.0 N 3.78 ×10 −9 s ∆y2 = vfy ∆t2 = (4. 27.30 × 10 −3 m 2. 3 28.02 × 10−2 m below the axis of entry when it hits the screen. it travels with constant velocity..0 N ax = = ΣFx m1 + m2 3.0 kg F = 3.25 ×107 m/s ∆t2 = 5. the force of contact between the blocks is 1. Here. Although there is only one clothesline.For m2: ΣFx = m2 ax = (1.0 kg)(9. we can determine FT1 by substituting FT2 into Equation 1: FT1 sin θ + FT2 sin φ = mg mg − FT2 sin φ sin θ (2.8 N/kg) − (17.0 N.0 kg)(9.8 N/kg) cos 45° tan 30° + sin 45° = 17.0 N The force of m1 on m2 = the force of m2on m1 = 1. with magnitudes of FT1 and FT2. 29.57 N) sin 45° = sin 30° FT1 = 14 N Thus. This differs from the situation in which there is no friction between the object and the clothesline.57 N FT2 = 18 N To show that this tension is the limiting factor. FT1 = 154 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .0 m/s 2 ) ΣFx = 1. the chicken’s claws grab onto the line. Thus. the minimum breaking strength of the line is 18 N. there are actually two tensions in the line. The diagram below shows the FBD of the situation.0 N. Considering the vertical components of the forces: ΣFy = ma y = 0 FT1 sin θ + FT2 sin φ − mg = 0 Considering the horizontal components of the forces: ΣFx = ma x = 0 FT2 cos φ − FT1 cos θ = 0 FT1 = FT2 cos φ cosθ (Equation 2) (Equation 1) Substitute Equation 2 into Equation 1:  FT2 cos φ   cosθ  (sin θ ) + FT2 sin φ = mg    cos φ sin θ  + sin φ  = mg FT2   cos θ  FT2 (cos φ tan θ + sin φ ) = mg FT2 = = mg (cos φ tan θ + sin φ ) (2.0 kg)(1. ) (b) This is not an easy question to analyze the first time it is seen. rather it is moving in a horizontal arc. Thus.CHAPTER 3 CIRCULAR MOTION Reflect on Your Learning (Page 120) 1. The choice of positive directions is explained in (b) and (c) below. it is acceptable to draw the diagram with +x down the inclined plane and +y perpendicular to that axis. (a) The FBD is shown in below. The truck is not sliding down a hill. (a) The system diagram in the noninertial frame of reference The normal force acting on the truck is The FBD of the middle bead in the noninertial frame of reference Copyright © 2003 Nelson Chapter 3 Circular Motion 155 . cos φ (c) The net force acting on the truck is in the same direction as the acceleration. as described in (b) above. that choice would not be convenient for analysis of the truck’s acceleration. Now: ΣFy = ma y = 0 FN cos φ − mg = 0 FN = mg cos φ mg . 2. experiencing acceleration toward the centre of radius of that arc. which means that the +y direction is up. However. It is caused by the horizontal component of the normal. (At this stage. we will let the direction of the horizontal acceleration be called +x. referred to as the Stanford torus (or doughnut). The neat feature of this design is that in moving from one cylinder to another. a microwave turntable. (b) One example is a set of two huge cylinders tied together. the only way to get the marbles to stay in the upper two holes is to rotate the device in the horizontal plane. most students will realize the connection to other rotating devices.1 UNIFORM CIRCULAR MOTION PRACTICE (Page 122) Understanding Concepts 1. astronauts pass through a zone of zero gravitational effects. it rotates to create artificial gravity. a rotating lawn sprinkler. Otherwise they might lodge in the two upper holes for the “wrong reason. a coffee grinder. the word “uniform” means that the radius of the circle remains constant and the motion is at a constant speed. (a) Answers may vary. 3. (b) Answers will vary. a vinyl record on a record player. Try This Activity: A Challenge (Page 121) The device shown in Figure 4 of the text must be made so the marbles experience negligible friction. Examples are the wheel of an exercise bike. 156 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . Like other designs. Another design featured in older science fiction movies is the wheel-shaped design. The normal force of the wall on the occupants causes a sensation similar to standing on Earth. and a chicken on a rotating spit on a barbeque. A common way is to set the spacecraft rotating at an appropriate speed so that the components inside tend to move toward the inside of the outer wall. A car moving at a constant speed can be accelerating at the same time if the direction of the velocity is changing.” (a) Assuming the device is well made. The cylinders rotate around each other to create artificial gravity. named in honour of a Stanford University research group. such as a centrifuge and a rotating lettuce drier. (b) After realizing that the device is based on rotation.(b) The system diagram at a faster rotation The system diagram at a smaller distance 3. 2. (a) In uniform circular motion. a rotating spherical mirror on the ceiling of a ballroom. (b) Answers will vary. Examples are the wheel of an exercise bike. a microwave turntable. it rotates to create artificial gravity.1 UNIFORM CIRCULAR MOTION PRACTICE (Page 122) Understanding Concepts 1. named in honour of a Stanford University research group. (a) Answers may vary. 156 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . a coffee grinder. The neat feature of this design is that in moving from one cylinder to another. a vinyl record on a record player. The normal force of the wall on the occupants causes a sensation similar to standing on Earth. Otherwise they might lodge in the two upper holes for the “wrong reason. (a) In uniform circular motion. and a chicken on a rotating spit on a barbeque. 2.” (a) Assuming the device is well made. A common way is to set the spacecraft rotating at an appropriate speed so that the components inside tend to move toward the inside of the outer wall. a rotating spherical mirror on the ceiling of a ballroom. the word “uniform” means that the radius of the circle remains constant and the motion is at a constant speed. referred to as the Stanford torus (or doughnut). such as a centrifuge and a rotating lettuce drier. a rotating lawn sprinkler. the only way to get the marbles to stay in the upper two holes is to rotate the device in the horizontal plane. most students will realize the connection to other rotating devices. (b) After realizing that the device is based on rotation. Another design featured in older science fiction movies is the wheel-shaped design. The cylinders rotate around each other to create artificial gravity. 3. (b) One example is a set of two huge cylinders tied together.(b) The system diagram at a faster rotation The system diagram at a smaller distance 3. Like other designs. astronauts pass through a zone of zero gravitational effects. Try This Activity: A Challenge (Page 121) The device shown in Figure 4 of the text must be made so the marbles experience negligible friction. A car moving at a constant speed can be accelerating at the same time if the direction of the velocity is changing. (a) v = 2.18 × 106 m/s d = 1. v2 1 . (a) The direction of the instantaneous acceleration at every point in the arc is toward the centre of the circle.5 × 10 4 m ac = 0. (a) From ac = r r of the centripetal acceleration of the ball will be half its original value. 8.5 × 104 m v = 180 km/h = 50 m/s (to two significant digits) v2 ac = r (50 m/s) 2 = 2. The direction of centripetal acceleration is toward the centre of the circle. and the direction of the radius vector is south. if an object moving with uniform circular motion reverses direction. the acceleration vector is north.80 m ac = 2. (a) The direction of the velocity vector is east. 4.30 × 10−11 m r= 2 Copyright © 2003 Nelson Chapter 3 Circular Motion 157 .0 m/s v2 ac = r (4.0 m/s) 2 = 0.06 × 10−8 m = 5. the direction of the centripetal acceleration stays the same. Thus.10 m/s2.PRACTICE (Page 123) Understanding Concepts 3.10 m/s 2 The magnitude of the centripetal acceleration of the particles that make up the wind is 0.0 × 101 m/s2. (b) r = 0. the magnitude 6.06 × 10−10 m 1. the magnitude of the r centripetal acceleration of the ball will be four times its original value. so if the radius of the circle remains constant but the speed doubles. r = 25 km = 2. PRACTICE (Page 126) Understanding Concepts 5. ac ∝ . so if the speed of the ball remains constant but the radius of the circle doubles.0 × 101 m/s 2 The magnitude of the centripetal acceleration is 2.80 m v = 4. (b) The velocity and acceleration directions are shown on the sketch below. ac ∝ v 2 . 7. v2 (b) From ac = . v2 r (2.79 × 1010 m) 4.0 d  T = 7.0 h   1.13 m f = 33 1 rpm = 0.6 m/s2.97 × 1022 m/s2. (c) r = 13 cm = 0.56 s −1 ) 2 ac = 1.0 × 10−2 m/s2 4π 2 r ac = 2 T T= = 4π 2 r ac 4π 2 (5.0 m)(15 m/s 2 ) v = 5.3 m 4π 2 r ac = 2 T 4π 2 (4.18 × 106 m/s)2 = 5.2 s r = 4.6 m/s 2 The magnitude of the centripetal acceleration is 1. (b) T = 1.3 m) = (1.2 × 102 m/s2.5 m/s. r = 5.13 m)(0.2 ×10 2 m/s 2 The magnitude of the centripetal acceleration is 1. r = 2. T = 7.97 × 1022 m/s 2 The magnitude of the centripetal acceleration is 8.56 s–1 3 ac = 4π 2 rf 2 = 4π 2 (0.6 × 106 s or 88 d. 10.0 m ac = 15 m/s2 v2 ac = r v = rac = (2.6 ×106 s     3600 s   24 h  T = 88 d Mercury’s period of revolution around the Sun is 7.6 ×106 s Converting:  1.5 m/s The speed of the ball is 5.30 × 10−11 m ac = 8.0 × 10−2 m/s 2 9. ( ) 158 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .2 s)2 ac = ac = 1.79 × 1010 m ac = 4. 19 m/s 2 The magnitude of the centripetal acceleration is 9.6 × 104 s)2 ac = 3.2 m/s2. Since ac ∝ . (b) Some examples are a drill.1 Questions (Page 127) Understanding Concepts 1.38 × 106 m T = 24 h = 8. an electric screwdriver. (c) Some examples are a car travelling at a constant speed around a smooth curve in a highway. a microwave or oven turntable.6 × 104 s 4π 2 r ac = 2 T 4π 2 (6. a lettuce drier. (a) v = 7.2 m/s 2 The magnitude of the centripetal acceleration is 5.38 × 106 m) = (8.2 × 102 m v2 ac = r (25 m/s)2 = 1.4 × 10−2 m/s2. the magnitude of the centripetal acceleration of the shorter string is three times the magnitude of the r centripetal acceleration of the longer string. a coffee grinder. 1 2. (b) v = 25 m/s r = 1. and an aircraft in an air show at the bottom of an arc where the speed is constant. a food processor. a juicer. a racket or bat in a small part of its swing during which the speed is constant. a circular saw.77 × 103 m/s) 2 = 6. a meat grinder.19 m/s2. Copyright © 2003 Nelson Chapter 3 Circular Motion 159 .4 × 10 −2 m/s 2 The magnitude of centripetal acceleration due to the daily rotation of an object at the Earth’s equator is 3. a rotisserie. a looping roller coaster in a portion of the loop or going around a corner where the speed is constant for at least a short time interval. a rotating sander. (a) Some examples are an egg beater.57 × 106 m v2 ac = r (7. a router. 3. (a) r = 6.2 ×102 m ac = 5. and all electric motors. 4.57 × 106 m ac = 9.77 × 103 m/s r = 6.Applying Inquiry Skills 11. a paint stirrer. an electric meat slicer. a lathe. Section 3. and a dial clock. 1 in which students discover or derive the relationships between the variables. They are centripetal acceleration.7 × 10 −3 m/s 2 ) = 4π 2 8 r = 3. radius.3 d = 2.) 5.2 of the textbook.3 m/s2 v2 ac = r v2 r= ac = (25 m/s) 2 8. pages 135–136. and the period. ac = 25 m/s2 5. Applying Inquiry Skills 8.(b) The centripetal acceleration of magnitude 0.7 × 10−3 m/s2 4π 2 r ac = 2 T T 2 ac r= 4π 2 (2. and period. and verify that the following equations yield the same result: v2 4π 2 r ac = 4π 2 rf 2 ac = ac = 2 r T 160 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . the frequency.50 Hz The Rotor revolves with a minimum frequency of 0.5 m 2 ac = 4π 2 rf 2 f = = ac 4π 2 r 25 m/s 2 4π 2 (2. we begin by considering the variables involved.36 × 106 s ac = 2.034 m/s2 is much smaller than the magnitude of the acceleration due to gravity. r = 75 m The radius of the curve is 75 m. which is approximately 9.50 Hz. Students can twirl a rubber stopper in a horizontal circle of known radius and determine the time for 20 complete revolutions. They can calculate the speed. frequency. (The reason for the reduction is described in more detail in Section 3.5 m) 6.8 × 10 m The average distance from Earth to the Moon is 3.8 × 108 m. (a) Students are asked to describe an experiment to verify the relationships between the variables.) Thus. T = 27.1. f = 0. it does cause a very small weight reduction as the rotation causes a person to tend to move away from Earth’s centre. the speed of motion. v = 25 m/s ac = 8.0 m r= = 2.36 × 106 s)2 (2.8 m/s2. However.3m 7. (This differs from Investigation 3. count 3−2−1−0 and simultaneously start the stopwatch at 0).4 × 105 g.4 cm = 8.8 m/s 2 ac = 3. like those used in car cooling systems. vcircular = 1. Making Connections 9. affecting the direction of the tension in the string attached to the stopper.(b) The biggest source of random error occurs in trying to keep the stopper moving at a constant speed. (a) FN = 2mg r = 12 m ΣF = mac mv 2 r mv 2 2mg + mg = r v = 3 gr FN + mg = = 3(9. To keep the sources of error within reasonable bounds.0 × 104 rev/min = 1. and to determine molecular weights.) 3.4 × 10–2 m f = 6.0 × 103 s–1. (Centrifugal pumps.8 m/s 2 )(12 m) = 19 m/s v = 68 km/h The speed required by a coaster is 19 m/s or 68 km/h.3 × 106 m/s 2 ac 3. Copyright © 2003 Nelson Chapter 3 Circular Motion 161 . Systematic error occurs because it is almost impossible to twirl the stopper in the horizontal plane.e..4vclothoid = (1.3 × 106 m/s 2 = g 9.4 × 10−2 m)(1.4)(68 km/h) vcircular = 95 km/h Thus. (a) r = 8. (b) Some examples are to prepare serums and blood sample. to study organic molecules. This takes practice and coordination. as the stopper revolving in a circle moves past a specific location. are rotating pumps that operate in a way similar to a centrifuge. Other sources of random error include measuring the radius of the circle and the time for 20 revolutions of the stopper.4 × 105 g The magnitude of the centripetal acceleration is 3.2 ANALYZING FORCES IN CIRCULAR MOTION PRACTICE (Page 132) Understanding Concepts 1. to separate milk from cream. a coaster on a circular loop would have to travel at a speed of 95 km/h. to separate U-238 from U-235 for nuclear material enrichment. (b) Using the equation on page 132.0 × 103s −1 )2 ac = 3. students must perform several trials and then average the results. Friction may also affect the results in a systematic way. ac = 4π 2 rf 2 = 4π 2 (8. Gravity pulls the stopper downward slightly. Another idea is to use the “count-down” method to know when to start timing the motion (i. 4 × 10−2 m)(1.4vclothoid = (1. (b) Using the equation on page 132.0 × 103s −1 )2 ac = 3. students must perform several trials and then average the results.4 cm = 8. count 3−2−1−0 and simultaneously start the stopwatch at 0). ac = 4π 2 rf 2 = 4π 2 (8. Gravity pulls the stopper downward slightly. (a) FN = 2mg r = 12 m ΣF = mac mv 2 r mv 2 2mg + mg = r v = 3 gr FN + mg = = 3(9. are rotating pumps that operate in a way similar to a centrifuge. Systematic error occurs because it is almost impossible to twirl the stopper in the horizontal plane.. to separate U-238 from U-235 for nuclear material enrichment. This takes practice and coordination.4 × 10–2 m f = 6.e. to study organic molecules.(b) The biggest source of random error occurs in trying to keep the stopper moving at a constant speed. like those used in car cooling systems.8 m/s 2 ac = 3. Another idea is to use the “count-down” method to know when to start timing the motion (i. Other sources of random error include measuring the radius of the circle and the time for 20 revolutions of the stopper.0 × 103 s–1. Making Connections 9.8 m/s 2 )(12 m) = 19 m/s v = 68 km/h The speed required by a coaster is 19 m/s or 68 km/h.4)(68 km/h) vcircular = 95 km/h Thus. (a) r = 8.) 3.4 × 105 g The magnitude of the centripetal acceleration is 3.3 × 106 m/s 2 ac 3.3 × 106 m/s 2 = g 9. a coaster on a circular loop would have to travel at a speed of 95 km/h. affecting the direction of the tension in the string attached to the stopper. Copyright © 2003 Nelson Chapter 3 Circular Motion 161 .0 × 104 rev/min = 1. vcircular = 1. (b) Some examples are to prepare serums and blood sample. and to determine molecular weights. To keep the sources of error within reasonable bounds. Friction may also affect the results in a systematic way.4 × 105 g. to separate milk from cream. (Centrifugal pumps. as the stopper revolving in a circle moves past a specific location.2 ANALYZING FORCES IN CIRCULAR MOTION PRACTICE (Page 132) Understanding Concepts 1. 1 Earth years.8 m/s  25.42 × 1021 N = 2.211 kg r = 25. or 84.6 m  F = 5.80 × 103 m/s m = 8. 162 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .7 m/s ΣFy = mac (a) F + (− mg ) = mv 2 r  v2  F = m g +   r     (21. (a) The force that causes the centripetal acceleration is gravity.87 × 1012 m) 1.PRACTICE (Page 133) Understanding Concepts 2.42 × 1021 N. 4.65 × 109 s.87 × 1012 m v = 6.95 N     The magnitude of the upward lift on the bird’s wings at the bottom of the arc is 5.211kg)  + 9. 4π 2 mr (c) Fg = T2 T= = 4π 2 mr Fg 4π 2 (8. m = 0.95 N.80 × 1025 kg)(2. The cause of the centripetal acceleration in each case is: (a) the force of gravity of Earth on the Moon (b) the electric force (or electromagnetic force) of the proton in the nucleus on the electron (c) the difference between the normal force and Earth’s force of gravity on the snowboarder 3. (b) r = 2.65 × 109 s T = 84.80 ×103 m/s) 2 = 2.7 m/s) 2 2 = (0.1a The orbital period of Uranus is 2.87 × 1012 m Fg = 1.80 × 1025 kg)(6.42 ×10 21 N The magnitude of the force of gravity is 1.6 m v = 21.80 × 1025 kg mv 2 Fg = r (8. m = 2. r = 1. Copyright © 2003 Nelson Chapter 3 Circular Motion 163 .3°.00 m 2.00 kg r = 4.(b) The centripetal acceleration is caused by the difference between the upward normal force of the air on the bird and the downward for of gravity.00 kg)(4.0 g mv 2 mac = r v2 r= ac = = v2 7g (150 m/s) 2 7(9.400s)2 7.8 m/s 2 )(1. r = 450 m v = 97 km/h = 27 m/s Let +x be the direction of the acceleration.3 × 102 m.3° The proper banking angle for a car travelling at 97 km/h is 9.  v2  θ = tan −1   gr       (27 m/s) 2 = tan −1   (9.400s 5.00 m) = (0.8 m/s 2 )(450 m)     θ = 9.50 × 103 m ΣFy = mac mg = mv 2 r v = gr = (9. 5. page 130: ΣFx = mac mg tan θ = mv 2 r 6. v = 540 km/h = 150 m/s (a) a = 7.00 revolutions FT = 4π 2 mr T2 4π 2 (2.00 s T= = 0.50 km = 1.50 ×103 m) = 121 m/s v = 436 km/h The speed is the plane at the top of the loop is 121 m/s or 436 km/h. As shown in Sample Problem 2.97 × 103 N The magnitude of tension in the rope is 1.8 m/s 2 ) r = 3. 8.97 × 103 N. FT = 1.3 × 10 2 m The minimum radius of the plane’s circular path is 3. so wear and stress are much less on tracks with banked curves. the sideways force comes from the horizontal component of the normal force on the train. (a) The banking angles for on. (b) It is dangerous to travel around a curve with a relatively small radius at a high speed. the tracks are banked. The demonstrator can feel the difference in the tensions.7° r = 5. (This is verified in Sample Problem 3 for an object travelling at a constant speed in a vertical circle. If the tracks are not banked.(b) m = 82 kg Let +y be up. (If the pendulum were attached to the ceiling such that the observer would have to look straight upward to see it. When the speed is slow enough that the string becomes slack at the top of the circle. 11. gravity is still downward but the tension is upward.8 m/s 2 ) FN = 6. Making Connections 10. it is obvious in the demonstration that the tension is much greater at the bottom. If.) (b) At the equator.8 m/s 2 ) tan 5. the globe must be rotated counterclockwise when viewed from above the North Pole. acting against gravity. especially for trucks and other vehicles with a high centre of mass. the motion will eventually get to the stage where the tension in the string approaches zero at the top of the circle. gravity and tension are both downward. ΣFy = mac FN + (−mg ) = mac FN = m( g + ac ) = m(8 g ) = 8(82 kg)(9.4 ×103 N A force of 6. it must experience a sideways centripetal acceleration caused by a sideways force. In Earth’s frame of reference. the tension must be greater at the bottom of the circle. the Foucault pendulum’s motion would remain fixed relative to Earth’s frame of reference. the relationship shown is v = gr tan θ . Applying Inquiry Skills 9. that sideways force must come from the horizontal force of the tracks on the wheels of the train. Try This Activity: The Foucault Pendulum (Page 136) (a) As shown in the text. a person at the North Pole looking down at a Foucault pendulum would observe that its path of swing appears to move clockwise.) (b) If the stopper is twirled in a vertical circle starting with a fairly high (but safe) speed and gradually sowing down. (a) At the top of the vertical circle. Tracks are much more stable in the plane perpendicular to the surface than parallel to it. At latitudes between the equator and the North Pole.and off-ramps must be designed with a smaller radius of curvature than a gradual highway turn. the motion would appear to be counterclockwise. however.and off-ramps are larger than banking angles for gradual highway turns. 23 h and 56 min relative to the Sun’s frame of reference) to complete one rotation.4 × 103 N is applied at the lowest point of the pullout. the motion observed is between 23 h. Since the on. 56 min at the North Pole and infinity at 0° 164 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .5 × 102 m ΣFx = mac mg tan θ = mv 2 r v = rg tan θ = (5. This is even more crucial when the road conditions are slippery. In Sample Problem 2. taking close to 24 h (specifically.5 × 10 2 m)(9. (a) θ = 5.7° v = 23 m/s The ideal speed for a train rounding a curve is 23 m/s. Thus. it is evident from this relationship that θ must increase (tan θ increases as θ increases). At the bottom of the circle. especially at the part of the curve where vehicles may be travelling almost as fast as highway speeds. (b) In order for the train to travel in an arc. latitude. The force that causes this acceleration is static friction exerted by the floor of the ride on your feet. For example. This force is equal in magnitude but opposite in direction to the force of static friction. (b) In the accelerating frame of reference of the merry-go-round. so the net horizontal force is zero. (a) In Earth’s frame of reference. 56 min to complete one rotation. (The time interval 23. (a) The system diagram (as viewed from above) Copyright © 2003 Nelson Chapter 3 Circular Motion 165 . The tendency you feel to move away from the centre of the rotating ride is simply an example of Newton’s first law of motion: you are in motion and would continue in a straight line at a constant speed if the static friction did not force you toward the centre of the circle. at 45° North latitude. which can be found using the relationship ∆t = . you are experiencing centripetal acceleration toward the centre of the merry-go-round.93 h would be about 34 h. PRACTICE (Pages 136–137) Understanding Concepts 12. which is southern Ontario’s latitude. 13. you feel a force we can call the fictitious force or centrifugal force pushing you away from the centre of rotation. the pendulum will appear to rotate clockwise when viewed from above and would take more than 23 h.) sinθ (c) The easiest way to set up this demonstration is to place a globe with its axis oriented vertically beneath a pendulum that vibrates in the fixed frame of the classroom as the globe rotates. 9 m T= 4.1 s Considering the vertical components of the forces: ΣFy = ma y = 0 FT cos θ − mg = 0 FT cos θ = mg FT = mg cos θ (Equation 1) Considering the horizontal components of the forces: 4π 2 mr ΣFx = ma x = T2 2 4π mr FT sin θ = T2 mg 4π 2 mr sin θ = cos θ T2 4π 2 mr mg tan θ = T2 4π 2 r θ = tan −1 gT 2 = tan −1 4π 2 (2.9 m) (9. 166 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .8 m/s 2 )(4.1 s)2 θ = 35° The angle the string makes with the vertical is 35°.(b) The FBD in Earth’s frame of reference (c) The FBD in the frame of reference of the ride (d) r = 2. 14. Applying Inquiry Skills 15.34% less than g.37 × 10−2 m/s 2 ( ) weight difference = 2.5 m f = 0.78 m/s 2 % of g = 0. (b) r = 4. (a) The reduction is the acceleration due to gravity is caused by Earth’s rotation.) Relating this value to the acceleration due to gravity at the equator (as stated in Table 1.64 × 104 s 4π 2 r ac = 2 T = 4π 2 3.37 ×10−2 m/s 2 The magnitude of the centripetal acceleration is 3.(e) m = 45 g = 4. this is the magnitude of the centrifugal acceleration. depending on the mass of the student. text page 33): a % of g = c ×100% g 3.e. page 776: rE = 6.64 ×10 s ) 4 ( ) 2 ac = 3.4 × 10−2 N.38 × 106 m T = 8. Copyright © 2003 Nelson Chapter 3 Circular Motion 167 .8 N/kg) = cos 35° FT = 5.37 × 10−2 m/s2.5 × 10−3 kg)(9.. (In Earth’s accelerating frame of reference. If m = 65 kg: weight difference = mac = = (65 kg ) 3.45 Hz ac = 4π 2 rf 2 = 4π 2 ( 4. (a) You would hold the accelerometer just as it appears in that figure (i.38 × 106 m (8. (b) Answers will vary.45 Hz ) 2 ac = 36 m/s 2 The magnitude of the centripetal acceleration is 36 m/s2.5 × 10−3 kg From Equation 1 in part (d): mg FT = cos θ (4.34% The magnitude of the acceleration of the object is 0.2 N for a person of mass 65 kg.4 ×10−2 N The magnitude of the tension in the string is 5.2 N The difference in weight is only 2.37 ×10 −2 m/s 2 ×100% 9. (c) The FBD of the bead is shown below. From Appendix C. perpendicular to the direction of your instantaneous velocity).5 m )(0. From the beginning it was evident that the motion of the pendulum was changing relative to Earth’s frame of reference.5 kg. A key word search on the Internet (using “Foucault Pendulum Guelph”) will yield much more information about the University of Guelph pendulum and links to other related sites. Making Connections 16. the net force is created by static friction only. (d) m = 1.2 Questions (Page 138) Understanding Concepts 1. With the flat curve. With the banked curve. one in the bob and the other beneath it in line with the point of suspension.1× 10−3 kg)(9. help prevent unwanted deflections of the bob. suspended a 27-kg pendulum from the domed ceiling of the Pantheon in Paris.8 N/kg) = cos 75° FT = 4. As the swinging continued. the bob touches a ring that surrounds the setup. Section 3. In order for the car to experience acceleration toward the centre of curvature. especially in slippery conditions when static friction is greatly reduced. a record of the path was recorded on a ring of sand beneath the pendulum.1 g = 1. This is a great advantage. permanent magnets. When he burned the cord.1× 10 −2 N The magnitude of the normal force is 4. The banked curve is a better design. for example. Information about Foucault pendulums can be found in encyclopedias and astronomy reference books and on the Internet. The length of the pendulum was more than 60 m. 168 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .1 × 10−2 N.8 m/s 2 θ = 75° The angle from the vertical is 75°. a French physicist. Foucault pendulums on display at universities and science centres are not nearly as long as the one built by Foucault. the pendulum began swinging without any sideways influences. The one in the physics department at the University of Guelph.1 × 10−3 kg mg FN = cos θ (1. the net force is created by a component of the normal force of the road on the car. In 1851 Jean Foucault. is 83 cm long with a bob of mass 4.Considering the vertical components of the forces: ΣFy = ma y = 0 FN cos θ − mg = 0 FN cos θ = mg FN = mg cos θ Considering the horizontal components of the forces: ΣFx = max FN sin θ = mac mg sin θ = mac cos θ g tan θ = ac θ = tan −1 = tan −1 ac g 36 m/s 2 9. there must be a net force in that direction. He pulled the bob to one side and secured it with a cord. With each swing. v = 22.80 m/s 2 ) FT = 1.0 m) = (0.96 N The magnitude of the tension in the string when the pendulum is at rest is 1.00 kg r = 1.3 × 10 22 m/s 2 The magnitude of the acceleration of the electron is 9.4 × 10−8 N.4 ×10 −8 N The magnitude of the electric force acting on the electron is 8.3 × 10−11 m m = 9.5 × 10−16 s)2 ac = 9. Copyright © 2003 Nelson Chapter 3 Circular Motion 169 .00 ×10 2 N)(1.5 × 10−16 s 4π 2 r (a) ac = 2 T 4π 2 (5.2.20 kg)(10.4 m/s The maximum speed the stone can attain without breaking the string is 22.50 s 10 rotations ΣFx = max FT = mac 4π 2 mr T2 4π 2 (0.00 m) 1. m = 1. 4.12 m m = 0.2 ×102 N The magnitude of the tension in the string is 3. r = 5.1 × 10−31 kg T = 1. m = 0.3 ×10 −11 m) = (1.50s) 2 = FT = 3.4 m/s.0 s T= = 0. r = 1. 5.3 × 10 22 m/s 2 ) F = 8.200 kg (a) Let +y be up.2 × 102 N. ΣFy = 0 FT + (− mg ) = 0 FT = (0.00 m string FT = 5.1× 10−31 kg)(9.3 × 1022 m/s2. (b) F = mac = (9.20 kg r = 10.00 × 102 N ΣFx = max FT = mv 2 r FT r v= m = (5.00 kg 3.0 m 5.200 kg)(9.96 N. (b) v = 1.20 m/s ΣFy = mac FT + (− mg ) = mv 2 r  v2 FT = m  g +  r       (1.20 m/s)2  2 = (0.200 kg)   9.80 m/s +   1.12 m   FT = 2.22 N The magnitude of the tension at the bottom of the swing is 2.22 N. 6. (a) m = 15 g r = 1.5 m FT = 0 N Let +y be down. ΣFy = mac mg − FT = mv 2 r mv 2 mg = r v = rg = (1.5 m)(9.8 m/s 2 ) v = 3.8 m/s The speed of the stopper is 3.8 m/s. (b) If the mass of the stopper doubles, the speed will not change since speed is independent of the mass. 7. m = 0.030 kg r = 1.3 m v = 6.0 m/s ΣFy = mac FT + (± mg ) = mv 2 r  v2  FT = m   ± g   r  8.  (6.0 m/s)2 2 = (0.030 kg)   1.3m ± 9.80 m/s     FT = 1.1N or 0.54 N The maximum tension in the string is 1.1 N and the minimum tension in the string is 0.54 N. The statement is wrong. In Earth’s frame of reference, the child is moving in a circle and must be experiencing an acceleration toward the centre of that circle. This centripetal acceleration caused by the net, nonzero force toward the centre of the circle. Applying Inquiry Skills 9. (a) One force is the force of gravity and the other force is the normal force of the seat on the rider. (b) Let +y be down. 170 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson ΣFy = mac mv 2 r mv 2 2mg + mg = r v = 3 gr FN + mg = = 3(9.8 m/s 2 )(15 m) v = 21m/s The speed of the coaster at the top of the loop is 21 m/s. (c) One force is gravity and the other is the tension in the accelerometer’s spring. (d) At the top of the loop, the accelerometer is inverted (in Earth’s frame of reference), and the spring pulls downward on the bob. If students draw an FBD of the bob, both the tension and gravity are downward. If the coaster is moving such that the tension is zero, then the force on the bob would be 1mg, and the accelerometer would read 0g. However, the coaster is moving at a speed that causes the net force on the bob to be 3mg, as discussed in (b) above. Thus, the tension in the spring is 2mg and the reading on the accelerometer is 2g. (A shortcut answer can be stated by realizing that the accelerometer is calibrated to indicate the same value felt by the rider due to the normal force, which in this case is 2mg for force and 2g for acceleration.) (e) There are several random sources of error in using a vertical accelerometer on a roller coaster: • Coaster rides are fast and jerky, and the bob gets tossed around a lot. • It is very difficult to judge when you, as an excited and sometimes frightened rider, reach a particular location in the ride. • It is almost impossible to remember readings for more than a small number of locations on the ride. • It is difficult to hold the accelerometer so it is vertical relative to the ground. • Values between the calibration marks must be estimated. There are some sources of systematic error: • The calibration markings may be inaccurate. • The spring may be overstretched or tangled, causing the bob to be at the wrong position when at equilibrium. Making Connections 10. Answers will vary. Examples in the home are a clothes washer during the spin cycle, a food processor, a weed whacker, and an electric drill. In a car, the pulleys linked by fan belts rotate rapidly when the engine is running. Examples in the workplace are a centrifuge, a meat slicer, electric fans, and all electric motors. The precautions are specific to the examples. For instance, when operating a weed whacker, a person should wear goggles and protective shoes and gloves. 11. Numerous reference sites can be found on the Internet. Because there is so much information, students can share the research by choosing one of the categories of uses suggested. For example, each small group can begin their research by conducting a key word search using centrifuge and one of the following choices: blood analysis; DNA; proteins; dairy products; geology. Some examples of sites are: http://www.iptq.com/bloodanalysis.htm http://www.azduiatty.com/DUIBloodIssues.htm http://hdklab.wustl.edu/lab_manual/yeast/yeast3.html 3.3 UNIVERSAL GRAVITATION PRACTICE (Page 141) Understanding Concepts 1. 2. Both the third law of motion and the law of universal gravitation relate to two objects. If the action force is the force of gravity of object 1 on object 2, then the reaction force is the force of gravity of object 2 on object 1. The forces are equal in magnitude but opposite in direction, which is what the third law indicates. The direction of the force of A on B is toward A. Copyright © 2003 Nelson Chapter 3 Circular Motion 171 ΣFy = mac mv 2 r mv 2 2mg + mg = r v = 3 gr FN + mg = = 3(9.8 m/s 2 )(15 m) v = 21m/s The speed of the coaster at the top of the loop is 21 m/s. (c) One force is gravity and the other is the tension in the accelerometer’s spring. (d) At the top of the loop, the accelerometer is inverted (in Earth’s frame of reference), and the spring pulls downward on the bob. If students draw an FBD of the bob, both the tension and gravity are downward. If the coaster is moving such that the tension is zero, then the force on the bob would be 1mg, and the accelerometer would read 0g. However, the coaster is moving at a speed that causes the net force on the bob to be 3mg, as discussed in (b) above. Thus, the tension in the spring is 2mg and the reading on the accelerometer is 2g. (A shortcut answer can be stated by realizing that the accelerometer is calibrated to indicate the same value felt by the rider due to the normal force, which in this case is 2mg for force and 2g for acceleration.) (e) There are several random sources of error in using a vertical accelerometer on a roller coaster: • Coaster rides are fast and jerky, and the bob gets tossed around a lot. • It is very difficult to judge when you, as an excited and sometimes frightened rider, reach a particular location in the ride. • It is almost impossible to remember readings for more than a small number of locations on the ride. • It is difficult to hold the accelerometer so it is vertical relative to the ground. • Values between the calibration marks must be estimated. There are some sources of systematic error: • The calibration markings may be inaccurate. • The spring may be overstretched or tangled, causing the bob to be at the wrong position when at equilibrium. Making Connections 10. Answers will vary. Examples in the home are a clothes washer during the spin cycle, a food processor, a weed whacker, and an electric drill. In a car, the pulleys linked by fan belts rotate rapidly when the engine is running. Examples in the workplace are a centrifuge, a meat slicer, electric fans, and all electric motors. The precautions are specific to the examples. For instance, when operating a weed whacker, a person should wear goggles and protective shoes and gloves. 11. Numerous reference sites can be found on the Internet. Because there is so much information, students can share the research by choosing one of the categories of uses suggested. For example, each small group can begin their research by conducting a key word search using centrifuge and one of the following choices: blood analysis; DNA; proteins; dairy products; geology. Some examples of sites are: http://www.iptq.com/bloodanalysis.htm http://www.azduiatty.com/DUIBloodIssues.htm http://hdklab.wustl.edu/lab_manual/yeast/yeast3.html 3.3 UNIVERSAL GRAVITATION PRACTICE (Page 141) Understanding Concepts 1. 2. Both the third law of motion and the law of universal gravitation relate to two objects. If the action force is the force of gravity of object 1 on object 2, then the reaction force is the force of gravity of object 2 on object 1. The forces are equal in magnitude but opposite in direction, which is what the third law indicates. The direction of the force of A on B is toward A. Copyright © 2003 Nelson Chapter 3 Circular Motion 171 3. F1 = 36 N m1′ = 2m1 r′ = 3r Gm1′ m2 2 F2 = r′ Gm F1 1 m2 r2 F2  m1′   r 2    = m  F1  r ′2     1  2m   r 2  1  F2 = F1   m    (3r )2     1 2 = 36 N   9 F2 = 8.0 N The magnitude of the force would be 8.0 N. rM = 0.54rE mM = 0.11mE FE = 6.0 × 102 N GmmM FM r 2 = M GmmE FE rE 2 FM  mE = 2 FE   rM  rE 2   m  E      r 2  E  m  E         4.  0.11m E FM = FE   (0.54r )2 E   0.11 = 6.0 × 102 N   (0.54 )2  5. FM = 2.3 × 10 2 N Thus, the magnitude of the force of gravity on a body on Mars is 2.3 × 102 N. F1 = 14 N r1 = 8.5 m F2 = 58 N Gm1m2 Gm1m2 F1 = and F2 = 2 r1 r2 2 F1 r2 2 = F2 r12 r2 = = r12 F1 F2 (8.5 m)2 (14 N) 58 N r2 = 4.2 m The centres of the masses are 4.2 m apart. 172 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson Applying Inquiry Skills 6. Making Connections 7. Pluto is very small and extremely far away from Earth, making its discovery more than 70 years ago all that more amazing. At the beginning of the twentieth century, some astronomers who had observed very slight perturbations in the orbits of Neptune and Uranus, began searching for the cause of the perturbations. Percival Lowell, an American astronomer, searched for the unknown planet from 1906 until his death in 1916. By 1929, his brother donated newer instruments to continue the search. Each photograph of the search portion of the sky had about 300 000 stars, so a device called a blink microscope was used to compare photos taken several days or even weeks apart. Finally in February 1930, Clyde Tombaugh, comparing photos made on January 23 and 29 of that year, discovered the planet. It was named after Pluto, the god of the underworld. Toward, the end of the twentieth century, astronomers began discovering other bodies orbiting the Sun beyond Pluto, causing some astronomers to question whether these newly observed bodies should be called planets, or perhaps Pluto should be downgraded to something less than a planet. Some of the reasons for the controversy are: • Pluto is small than several solar system moons. • Pluto’s own moon is larger in proportion to the size of the planet than any other moon–planet example in the solar system. • Pluto’s orbit is unusual when compared to the orbits of the other planets. • All the other planets far from the Sun are gas giants. • There are more than 100 objects discovered beyond Pluto that have properties that resemble Pluto’s, yet they are not classified as planets. Despite the controversy, Pluto remains a sentimental favourite of many astronomers and others, so its classification is unlikely to change, at least for now. For more information, refer to the Internet. One suitable Web site is http://science.nasa.gov/newhome/headlines/ast17feb99_1.htm. PRACTICE (Page 143) Understanding Concepts 8. m = 1.8 × 108 kg r = 94 m Gm 2 r2 (6.67 × 10 −11 N ⋅ m 2 /kg 2 )(1.8 × 108 kg) 2 = (94 m)2 FG = 9. FG = 2.4 × 102 N The magnitude of the force of gravitational attraction is 2.4 × 102 N. r = 6.38 × 106 m m = 50.0 kg mE = 5.98 × 1024 kg Copyright © 2003 Nelson Chapter 3 Circular Motion 173 15 × 107 m GmmJ mg = r2 Gm g = 2J r (6.67 × 10−11 N ⋅ m 2 /kg 2 )(555 kg)(5.95 ×107 m − 6.0 kg)(5.98 × 1024 kg) = (6.98 × 1024 kg GmmE FG = r2 GmmE r= FG = (6.0 kg) ( 2 m)2 FG3 = 4.95 × 107 m The vehicle is 2.8 m/s 2 The magnitude of the acceleration due to gravity on Jupiter is 24.67 × 10−11 N ⋅ m 2 /kg 2 )(50.90 × 102 N. (b) h = r − rE = 2.0 kg)(1.0 kg m3 = 4.67 × 10−11 N ⋅ m 2 /kg 2 )(1.8 m/s2.0 kg Gm1m2 FG2 = r12 2 = (6.FG = GmmE r2 (6.0 m)2 FG2 = 2. 12.0 kg m2 = m4 = 1.0 × 10−10 N 174 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .95 × 107 m from the centre of Earth. (a) FG = 255 N m = 555 kg mE = 5. 11.31× 107 m The vehicle is 2. mJ = 1.90 × 1027 kg) = (7.90 × 102 N The magnitude of the force of gravity on the student is 4. 10. m1 = 3.15 ×107 m) 2 g = 24.0 ×10 −10 N FG3 = = Gm1m3 r132 (6.0 kg)(4.0 kg) (1.38 × 106 m)2 FG = 4.67 × 10−11 N ⋅ m 2 /kg 2 )(3.31 × 107 m above the surface of Earth.98 × 1024 kg) 255 N r = 2.90 × 1027 kg r = 7.67 × 10−11 N ⋅ m 2 /kg 2 )(3.38 × 106 m h = 2. 0 N + 4.0 × 10−10 N (sin 45° ) + 2.0 × 10 −10 N ΣFGy . it r2 is evident that although FG might approach zero as r approaches infinity.8 N The magnitude of the net gravitational force on m1 is 6.1 = ΣFGx. it never reaches zero.Considering the horizontal components of the force on m1.3 Questions (Page 144) Understanding Concepts 1.8 N. the statement is true.4 = 0.0 N ΣFGx .80 m/s 2 )(6.1 ) + ( FGy.1 ) 2 2 ( 4. (c) Knowing Earth’s mass accurately helps scientists determine the density and thus the composition of Earth’s interior.1 = ΣFGy .2 + ΣFGx. It helps in determining factors that influence the launching and orbits of space vehicles. with +x to the right: ΣFGx .83 N ΣFGy . F1 = 26 N 1 r′ = r 2 m2′ = 3m 2 Copyright © 2003 Nelson Chapter 3 Circular Motion 175 .2 + ΣFGy . It also helps scientists understand more about our place in the solar system.0 × 10−10 N (cos 45° ) + 0. Thus. (a) r = 6.83 N FG1 = = ( ) ( ) ( FGx. 2. Considering the fact that FG ∝ 1 where r is the distance between the centres of any two distant objects in the universe.4 = 2.3 + ΣFGy . Section 3.38 × 106 m)2 = (6.83 N ) 2 FG1 = 6.83 N ) 2 + ( 4. Making Connections 13.67 × 10 −11 N ⋅ m 2 /kg 2 ) mE = 5.0 ×10−10 N + 4.3 + ΣFGx .38 × 106 m mg = GmmE r2 gr 2 mE = G (9.1 = 4.1 = 4.98 × 10 24 kg (b) The first relatively accurate calculation of Earth’s mass could not have been made until after 1798 when Henry Cavendish determined a fairly accurate value of the universal gravitation constant. 11 × 10−31 kg r = 5.e.06 × 10 −47 N The magnitude of the force of gravitational attraction is 4. 176 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .67 × 10−27 kg me = 9.11× 10−31 kg)(1. at rE) and F2 represent your weight at an altitude r where  GmmE    (r + rE )2  F2  =1 = F1 2  GmmE    2    rE  rE 2 (r + rE ) 2     F2 1 = . ( 2 − 1 rE ) At an altitude of ( 2 − 1) rE . Let F1 represent your weight on Earth’s surface (i.0 × 10 −11 m)2 FG = 4. F1 2 = 1 2 (r + rE )2 = 2rE 2 r 2 + 2rE r − rE 2 = 0 This is a quadratic equation with solution r = r= −2rE ± 4rE 2 + 4rE 2 −b ± b 2 − 4ac where only the positive root applies. your weight is half of your weight on the surface.67 × 10−27 kg) = (5.67 × 10−11 N ⋅ m 2 /kg 2 )(9.Let F2 represent the force of attraction when m2 is tripled and the distance between m2 and m1 is halved.0 × 10−11 m Gme mp FG = r2 (6.1 × 10−47 N.. mp = 1. Gm1m2′ 2 F2 = r′ Gm1m2 F1 r2 F2  m2′   r 2   =  F1  r ′2   m    2      3m2   r 2  F2 = F1   1 2     m2  r    2     = (26 N)(12) F2 = 3.1 × 102 N. 3. 2a 2 8 rE = −rE ± 2 2 2  = − 1  rE  2    r= 4.1 × 10 2 N The magnitude of the force of attraction is 3. 84 × 105 km mM = 0.27 × 10−7 N  θ = tan −1   −7    5. (b) Let +x be west and +y be south.68 m)2 FABx = 5. 6.95 × 10 −7 N − 5. mA = 55 kg mB = 75 kg mC = 95 kg rAB = 0.8 × 10−8 N [W]. the net force acting on B is 7. FAB = 5.27 × 10−7 N  5. (a) rEM = 3. Then the distance of the object from the Moon’s centre is ( rEM − r ) .27 × 10−7 N [W] ΣFx = FABx + FCBx = 5. (a) Let +x be west.27 × 10−7 N ΣFy = 5.95 × 10 −7 N − 0.27 × 10−7 N ΣFx = 6.68 m rBC = 0. the net force acting on B is 6.95 × 10−7 N FCBx = = −GmC mB rCB 2 −(6.95 × 10 −7 N ΣF = ΣFx 2 + ΣFy 2 = (5. FABx = = GmA mB rAB2 (6. Copyright © 2003 Nelson Chapter 3 Circular Motion 177 .95 × 10 N  θ = 42° Thus.9 × 10−7 N tanθ = ΣFy ΣFx ΣFy = FABy + FCBy = 0.95 m.9 × 10−7 N [42° S of W].0 N + 5.27 ×10−7 N ΣFx = FABx + FCBx = 5.67 × 10−11 N ⋅ m 2 /kg 2 )(95 kg)(75 kg) (0.8 × 10−8 N Thus.95 × 10−7 N [W] FCB = 5.0 N ΣFx = 5.5.95 × 10−7 N) 2 + (5.67 × 10−11 N ⋅ m 2 /kg 2 )(55 kg)(75 kg) (0.95 m)2 FCBx = −5.27 × 10−7 N) 2 ΣF = 7.012mE Let r be the distance between the object and Earth’s centre. FE = FM = GmmE r2 GmmM (rEM − r ) 2 GmmM ( rEM − r )2 0.012mE (rEM − r ) 2 FE = FM GmmE r2 mE r 2 = = 0.012r 2 = (rEM − r )2 0.012r 2 = rEM 2 − 2rEM r + r 2 0.988r 2 − 2rEM r + rEM 2 = 0 This is a quadratic equation with solution r = r= −b ± b 2 − 4ac . 2a 2rEM ± 4rEM 2 − 4(0.988)rEM 2 2(0.988) 2r ± 0.219rEM = EM 1.976 r = 1.12rEM or 0.90rEM Since the distance must be less than the Earth–Moon distance (rEM = 3.84 × 105 km): r = 0.90rEM = 0.90(3.84 × 105 m) r = 3.5 × 105 m At a distance of 3.5 × 105 m from Earth’s centre the net gravitational force exerted on an object by Earth and the Moon is zero. There is no other such point located between Earth and the Moon. (The location represented by the other root of the equation lies beyond the Moon.) (b) There is a location beyond the Moon, at r = 1.12rEM , where the magnitudes of the two forces are equal. However, these forces are in the same direction, which can be shown on an FBD of the object. Applying Inquiry Skills 7. The values on the graph depend on the student’s mass. 178 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson For the sample graph, the mass used is 82 kg. At rE, the force is: GmS mE FG = rE 2 (6.67 ×10 = −11 N ⋅ m 2 /(kg) 2 (82 kg ) 5.98 × 1024 kg (6.38 × 10 m ) 6 ) ( ) 2 FG = 8.0 × 102 N The data for the remaining points on the graph are: r FG (N) rE 8.0 × 10 2 3rE 89 5rE 32 7rE 16 Making Connections 8. (a) A geosynchronous satellite must have the same period as Earth to remain at the same location above Earth’s equator: T = 1 day = 8.64 × 104 s. (b) FG = Fc GmmE r 2 = r3 = r= (c) mE = 5.98 × 10 kg T = 8.64 × 104 s r= = 3 24 4π 2 mr T2 GmET 2 4π 2 3 GmET 2 4π 2 GmET 2 4π 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(8.64 × 104 s) 2 4π 2 3 r = 4.23 × 107 m The satellite is 4.23 × 107 m above Earth’s centre. (d) The satellite dishes on Earth’s surface that receive the electromagnetic signals from a satellite must remain aimed at that satellite. If the satellite moved across the sky (relative to the dishes), then the dishes would have to continually track it. That would be impractical, and would be impossible once the satellite “set” below the horizon. (e) The spacing of satellites in a geosynchronous orbit directly above the equator is limited to 0.5° per satellite. This means that in a full circle of 360°, the maximum number of such satellites is 720. If the spacing decreases too much, the signals from the adjacent satellites can interfere with each other. If students research these types of satellites on the Internet, inform them that some sites distinguish geostationary and geosynchronous satellites. Both types take 24 h to travel once around Earth, but a geostationary satellite is in an orbit directly above the equator. Some of the sites are: http://www.msu.edu/course/tc/850/scripts/script8.htm http://www.spaceconnection.org/satellites http://www.geo-orbit.org/sizepgs/geodef.htm http://www2.crl.go.jp/ka/control/Kansyo01/konzatu-e.html http://www.cs.wpi.edu/∼cs4514/b98/week2-physical.html Copyright © 2003 Nelson Chapter 3 Circular Motion 179 3.4 SATELLITES AND SPACE STATIONS PRACTICE (Pages 147–148) Understanding Concepts 1. (a) From the equation v = GmE 1 , so as the radius increases the speed , the speed of the satellite is proportional to r r decreases in a nonlinear fashion. (b) rM = 3.84 × 105 km = 3.84 × 108 m mE = 5.98 × 1024 kg GmE vM = rM = (6.67 × 10 −11 N ⋅ m 2 /kg 2 )(5.98 ×1024 kg) 3.84 × 108 m vM = 1.02 ×103 m/s The speed of the Moon is 1.02 × 103 m/s. The speed of the HST is 7.46 × 103 m/s (from Sample Problem 1, page 146). Thus, as the distance of a satellite from Earth’s centre increases, the speed of the satellite decreases. 2. r = 4.50 × 105 m + 6.38 × 106 m = 6.83 × 106 m (a) mE = 5.98 × 1024 kg v= = GmE r (6.67 × 10 −11 N ⋅ m 2 /kg 2 )(5.98 ×10 24 kg) 6.83 ×106 m v = 7.64 × 103 m/s The speed of the ISS is 7.64 × 103 m/s. (b) Using the data from (a) above: 2π r T= v 2π (6.83 ×106 m) = 7.64 × 103 m/s = 5.62 × 103 s T = 1.56 h It takes the ISS 1.56 h to make one revolution around Earth. The force of gravity causes the centripetal acceleration. GmS mE mSv 2 = r r2 GmE r= 2 v 3. 180 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson Since v = 2π r , we can substitute for v: T GmE r= 2  2π r     T  r= r3 = r= GmE T 2 4π r T GmE 2 2 2 ( ) 4π 2 3 T 2GmE 4π 2 4. (a) Since a digital TV satellite follows a geosynchronous orbit, it has a period of 24 hours or 8.64 × 104 s. (b) mE = 5.98 × 1024 kg rE = 6.38 × 106 m Using the equation derived in question 3(b): r= = 3 T 2 GmE 4π 2 (8.64 × 104 s) 2 (6.67 ×10 −11 N ⋅ m 2 /kg 2 )(5.98 × 10 24 kg) 4π 2 3 r = 4.22 × 107 m h = r − rE = 4.22 × 107 m − 6.38 × 106 m h = 3.59 × 10 4 m The altitude of the orbit above the surface of Earth is 3.59 × 104 km. Applying Inquiry Skills 5. (a) Speed versus mass: v ∝ m (b) Speed versus radius: v ∝ 1 r Copyright © 2003 Nelson Chapter 3 Circular Motion 181 Making Connections 6. (a) r = 5.7 × 1017 m v = 7.5 × 105 m/s v= m= Gm r v2r G (7.5 ×105 m/s) 2 (5.7 ×1017 m) = 6.67 ×10 −11 N ⋅ m 2 /kg 2 m = 4.8 × 1039 kg The mass of the black hole is 4.8 × 1039 kg. (b) mS = 1.99 × 1030 kg m 4.8 ×1039 kg = mS 1.99 × 1030 kg m = 2.4 × 109 mS Thus, the ratio of the black hole’s mass to the Sun’s mass is 2.4 × 109:1. According to the estimate calculated here, this black hole has a mass equivalent to 2.4 billions Suns. This incredibly huge mass implies than the black hole grew more and more massive by sucking in the materials from numerous surrounding stars, so its makeup is likely stellar and interstellar material. (c) Any body that is relatively small yet is so massive is also extremely dense. Probably the word “hole” does not describe an object with such high density. PRACTICE (Pages 150–151) Understanding Concepts 7. m = 56 kg Let the +y direction be upward. ΣFy = ma y FN − mg = ma y FN = m( g + a y ) (a) ay = –3.2 m/s 2 FN = 56 kg(9.8m/s 2 − 3.2m/s 2 ) FN = 3.7 × 10 2 N The magnitude of the apparent weight of the student when the acceleration is downward is 3.7 × 102 N. (b) ay = 3.2 m/s2 FN = 56 kg(9.8m/s 2 + 3.2m/s 2 ) FN = 7.3 × 10 2 N 8. The magnitude of the apparent weight of the student when the acceleration is downward is 7.3 × 102 N. The astronauts are in free fall toward Earth’s centre, just as the ISS and all other objects in it, so they appear to float in space. 182 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson 38 × 106 m (a) m = 64 kg FG = = GmmE (r + rE )2 (6. Copyright © 2003 Nelson Chapter 3 Circular Motion 183 . Since the ISS is so small and has no artificial gravity. for example. health-conscious diet. plenty of stretching and strengthening exercises. If the frequency of rotation is too high. the exercises can be done along the interior surface of the craft where the artificial gravity provides the normal force for many of the exercises.5 × 102 N. (b) Using the data from (a): 2π r v= T 2π r T= v 2π (1. an exercise program there must involve machines that provide resistance to forces exerted by the astronaut’s arms.9. etc. legs. Alternatively.38 × 10 6 m) 2 FG = 5. you could measure your own weight.24 × 103 m = 1.5 × 102 N The magnitude of the gravitational force is 5. On a large.62 × 103 m)(9. Of course. Applying Inquiry Skills 11. if you had a spring scale with the same calibration. however. (a) r = 2 g = 9.e. and adjustments to the frequency would be required. Making Connections 12. as it would be on Earth (i.8 N/kg).8 s. 3. Any experiment that would be done on Earth’s surface to determine your weight could be done on the interior surface of the rotating spacecraft. In each case. exercise machines could still be used.62 × 103 m) = 126 m/s T = 80.87 mg Therefore.8 N/kg) = 5. r = 450 km = 4.80 m/s 2 ) v = 126 m/s The speed of an astronaut would be 126 m/s. respiratory system. Thus.. F 5.5 × 10 2 N 6. the force calculated in (a) is 87% of the astronaut’s Earth-bound weight.5 × 10 2 N (b) G = mg (64 kg)(9. the program would involve a balanced. the artificial gravitational field would be greater than 9. For example.3 × 10 2 N FG = 0. the exercise program must consider the person’s bones.98 × 1024 kg) (4. muscles.5 × 105 m rE = 6.0-kg mass. if you had a bathroom scale calibrated in newtons.5 × 105 m + 6. 9.62 × 103 m 10. rotating spacecraft to Mars.8s The period of rotation of the spacecraft is 80. and circulatory system.8 N/kg. and plenty of cardiovascular exercise. and some of them could resemble ones used here on Earth’s surface. you could find the weight of a standard 1.80 m/s2 mv 2 = mg r v = rg = (1.67 × 10−11 N ⋅ m 2 /kg 2 )(64 kg)(5. 00 The ratio of the speed of an artificial satellite in orbit around Earth to the speed of a similar satellite in orbit around the Moon is 9. 184 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .4 Questions (Page 151) Understanding Concepts 1. 3. Thus.93 ×107 s v = 2. the ISS.23 v1 100 = v2 1.) GmE r GmM r mE mM 2. distant body (e. A space station is a satellite when it revolves around a central body (e.28 × 1011 m 2π r (a) v = T 2π (2.93 × 107 s r = 2.28 × 108 km = 2.02:1. when it is on a mission to Mars). a geosynchronous satellite.23 v1 9.g.42 × 104 m/s The orbital speed of Mars (relative to the Sun) is 2. (The speed to several significant digits is left in the calculator for the next part of the question. 1. the ISS travelling in orbit around Earth)..02 = v2 1. where v2 is the speed of r r the satellite in orbit around the Moon. A space station is not a satellite when it is travelling from one body toward a second. mM 1. v1 = The ratio of the speeds is: v1 = v2 v1 = v2 Since mE 100 = . 1 The order of decreasing speed corresponds to increasing distance from Earth’s centre (since v ∝ ).23 mM = mE 100 Since gravity causes the centripetal acceleration: GmS mE mSv 2 = r r2 GmE GmM . T = 1.88 Earth years = 5. the required r order is: a weather-watch satellite. Thus.Section 3. the Moon.28 ×1011 m) = 5.. 4.00.g. where v1 is the speed of the satellite in orbit around Earth and v2 = .42 × 104 m/s. 3 × 10−3 Hz The frequency of rotation should be 8.2 × 102 s f = 8.05 × 104 km Each satellite’s distance from the surface of Earth is 4.05 × 104 km.98 × 1024 kg GmE m mv 2 = r r2 GmE r= 2 v (6.05 × 104 km/h = 2. 5.38 × 103 km d = 4.8 m/s 2 T = 1. Copyright © 2003 Nelson Chapter 3 Circular Motion 185 .38 × 103 km Let d represent the distance from the surface of Earth of each satellite.69 × 104 km. or 4.99 × 1030 kg.2 × 102 s. r = = 1. Thus. (b) Using the period from (a) above: 1 f = T 1 = 1.68 × 104 km The orbital radius of each satellite is 4.92 × 103 m/s mE = 5. (b) rE = 6.38 × 106 m = 6.2 ×10 2 s The period of rotation should be 1.67 ×10 −11 N ⋅ m 2 /kg 2 )(5.4 ×103 m 2 mv 2 = mg (a) r 2π r v = rg = T 2 2 4π r rg = T2 T= = 4π 2 r g 4π 2 (1. (a) v = 1.42 × 104 m/s)2 (2.69 × 104 km – 6.4 × 103 m) 3.68 × 107 m.92 ×103 m/s) 2 r = 4.98 × 1024 kg) = (2.8 km 6. d = orbital radius of satellite – rE = 4.67 × 10−11 N ⋅ m 2 /kg 2 = mS = 1.3 × 10−3 Hz.28 × 1011 m) = 6. 2.(b) GmS m mv 2 r r2 2 v r mS = G (2.99 × 1030 kg The mass of the Sun is 1. r (b) The answer in (a) does not explain the temperature differences between summer and winter. (b) There are many Web sites suitable for this research.nasa.2. 17294-0850 (1994). The reasoning is that students will understand centripetal acceleration much better if they experience it in a hands-on activity. the investigation prepares the students for Section 3. http://spacelink. Earth must be closer to the Sun in our winter than in our summer 1 because v ∝ . ISBN # 0-8306-4533-0 (hardcover) and ISBN # 0-8306-4534-9 (paperback).Materials/Curriculum. for example) • jumping or flipping toys • physics demonstration toys A highly-recommended resource for this topic is Toys in Space: Exploring Science with the Astronauts. The second site listed shows wonderful photos of many of the toys.asp Making Connections 8.gov/Instructional. one at a time. (a) Since the speed is slightly greater during our winter. resulting in less direct sunlight and lower temperatures. It is highly recommended that the instructions be changed from one semester to the next so that the period of revolution replaces the frequency. The temperature differences between winter and summer depend mainly on the tilt of Earth’s axis of rotation to its orbital plane of revolution around the Sun.Support/. The first example listed below is also a source of videos showing many of the experiments with toys conducted by astronauts in space. published by TAB Books.index. CHAPTER 3 LAB ACTIVITIES Investigation 3. But the best way to analyze the variables in a controlled way is to include force. the radius of the circle.spintastics. (a) Answers will vary. Blue Ridge Summit. our winters would be slightly warmer than our summers.1. a division of McGraw-Hill.com/HistoryofYoYo.) Furthermore. it is a good idea to demonstrate the apparatus and show how the motion of the rubber stopper can be controlled and measured. If students have a difficult time getting started. One neat example of the experiments conducted in space is demonstrating longitudinal pulses on a coiled spring (such as a Slinky toy). the pulse travels easily along the axis. but its value is easily determined by calculating the force of gravity on the mass attached to the string. and the mass of the object) are varied. if Earth’s surface temperature depended solely on the distance to the Sun.htm http://www. When introducing the investigation to the students.2. Students will discover soon enough that the procedure requires a lot of cooperation within the group.Applying Inquiry Skills 7. The coil just lies still in free fall. which is not discussed until Section 3. The Northern Hemisphere faces away from the Sun in winter. Notice that this investigation involves force. by Dr. and when a longitudinal pulse is generated. PA. ask them to brainstorm the following categories of toys: • games • colliding objects • spinning objects • vibrating objects • magnetic toys • toys with wheels • flying toys • toys that move through a fluid (air or water. 186 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . Carolyn Sumners.1: Analyzing Uniform Circular Motion (Page 152) This important investigation is a controlled experiment in which the frequency of an object in circular motion is measured as the independent variables (the net for on the object.html http://aerospacescholars.org/Cirr/SS/L6/toys. (The force on the stopper in this investigation is tension in the string. In fact. The second site listed shows wonderful photos of many of the toys. When introducing the investigation to the students.nasa. (a) Since the speed is slightly greater during our winter. Earth must be closer to the Sun in our winter than in our summer 1 because v ∝ .Applying Inquiry Skills 7.2. (The force on the stopper in this investigation is tension in the string. The Northern Hemisphere faces away from the Sun in winter.1.html http://aerospacescholars. CHAPTER 3 LAB ACTIVITIES Investigation 3.com/HistoryofYoYo. The reasoning is that students will understand centripetal acceleration much better if they experience it in a hands-on activity. it is a good idea to demonstrate the apparatus and show how the motion of the rubber stopper can be controlled and measured. But the best way to analyze the variables in a controlled way is to include force. the investigation prepares the students for Section 3. (a) Answers will vary. if Earth’s surface temperature depended solely on the distance to the Sun.htm http://www.1: Analyzing Uniform Circular Motion (Page 152) This important investigation is a controlled experiment in which the frequency of an object in circular motion is measured as the independent variables (the net for on the object.Materials/Curriculum. our winters would be slightly warmer than our summers.asp Making Connections 8. but its value is easily determined by calculating the force of gravity on the mass attached to the string. The coil just lies still in free fall.Support/.gov/Instructional. and the mass of the object) are varied. for example) • jumping or flipping toys • physics demonstration toys A highly-recommended resource for this topic is Toys in Space: Exploring Science with the Astronauts. and when a longitudinal pulse is generated. ISBN # 0-8306-4533-0 (hardcover) and ISBN # 0-8306-4534-9 (paperback). One neat example of the experiments conducted in space is demonstrating longitudinal pulses on a coiled spring (such as a Slinky toy). the radius of the circle. which is not discussed until Section 3. the pulse travels easily along the axis. by Dr. PA.org/Cirr/SS/L6/toys. 17294-0850 (1994). The temperature differences between winter and summer depend mainly on the tilt of Earth’s axis of rotation to its orbital plane of revolution around the Sun. The first example listed below is also a source of videos showing many of the experiments with toys conducted by astronauts in space. resulting in less direct sunlight and lower temperatures. It is highly recommended that the instructions be changed from one semester to the next so that the period of revolution replaces the frequency. http://spacelink. a division of McGraw-Hill.2. If students have a difficult time getting started. one at a time. 186 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . In fact. Notice that this investigation involves force. Blue Ridge Summit. Students will discover soon enough that the procedure requires a lot of cooperation within the group. (b) There are many Web sites suitable for this research. r (b) The answer in (a) does not explain the temperature differences between summer and winter. published by TAB Books.spintastics. ask them to brainstorm the following categories of toys: • games • colliding objects • spinning objects • vibrating objects • magnetic toys • toys with wheels • flying toys • toys that move through a fluid (air or water.) Furthermore.index. Carolyn Sumners. pages 751 to 755.47 5 0. The materials needed for the analysis depend on which method of analysis you intend the students to apply. Materials The materials listed for each group are straightforward.Before performing this investigation. which would require more revolutions per second. for whatever reason.96 Time for 20 Cycles (s) Frequency of Revolution (Hz) 5 0. based on the instructions in steps 4 to 7. is shown below. space is left for three trials for each controlled test. A typical data table. so the frequency should decrease. • The frequency must decrease as the radius of the circle increases because the object will have a greater distance to cover with each revolution and will need more time to cover it. In the table. The curved lines represent the (correct) relationships indicated by the variation statements in (a). A typical response is • The frequency must increase as the force increases because a greater force means a greater acceleration. but the force is controlled to be constant. For example. as long as they attempt an explanation. Question Refer to the text. The straight lines represent a simple translation of the relationships described in words in (a).75 0. Step 4 Trial 1 2 3 average 1 2 3 average 1 2 3 average Mass of Stopper (kg) Radius of Circle (m) 0. For details on analyzing data.75 Weight of Hanging Mass (N) 1. • Students who.75 1. Procedure 1.98 Copyright © 2003 Nelson Chapter 3 Circular Motion 187 . students should be shown how to plot lines on the graph and just how convenient it is to determine the relationships between the variables plotted. refer to Appendix A in the text. if log-log graphing is used. Hypothesis/Prediction (a) Accept whatever the students predict. it is important that the students understand thoroughly at least one method of analyzing data in a controlled experiment involving four variables. • The frequency will decrease with an increase in mass because pulling on a greater mass would require a greater force to maintain the same frequency. apply mathematical relationships may predict the following relationships: 1 1 f ∝ ΣF f ∝ f ∝ r m (b) The two lines required are shown on each of the graphs below. 75 0.98 0. r is the radius of the circle.60 0.0 1. the relationships they discover should be as follows: 1 1 f ∝ FT f ∝ f ∝ r m where f is the frequency of revolution. students would plot and replot the data until they obtain a straight line.60 0. A good way to ensure a rewarding analysis for the students is to have them check their data with you or with a computer program you can write to check their data.3 4 5 5 6 6 7 7 Analysis (c) Whichever method the students use to analyze the data. 0.014 0.9 1. 20. If their value is not within 10% of the calculated value. .96 2.75 0.75 1. and m is the mass of the stopper.014 kg) for each stopper. 2. The “count-down” method is recommended for counting a predetermined number of revolutions. FT is the magnitude of the tension in the string.028 0. It is unrealistic to expect good results with every trial. have the students repeat the trial. If log-log graphing techniques are used. 3. As the stopper is twirled at its (attempted) constant speed. such as 20.6 13.0 11. m is the mass of the stopper. Step Mass of Stopper (kg) 0.75 1. r is the radius of the circle.014 0.1 13. counts 3. which equals the force of gravity on the suspended mass.5 1. the person counting.6 6 7 7 1 2 3 average 1 2 3 average 1 2 3 average 1 2 3 average 0.7 2.75 Weight of Hanging Mass (kg) 1.98 0. and then determine the proportionality statements.014 0. the slopes of the lines on the graph 188 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . Using this equation. For example.0 15.014 0. Students should not assume that the total mass of two rubber stoppers is twice that of one rubber stopper. using a mass of 14 g (0. The time for 20 complete revolutions is:  mr  t = 20T = 20   2π F   T   where T is the period of revolution. 1.96 Time for 20 Cycles (s) 9. starting the stopwatch at 0 and stopping it at 20.96 1.75 0. .96 1.2 1.75 0. .47 0. 2. 4–7. the average results of a typical investigation are given below.5 1. and FT is the tension in the string. The masses of rubber stoppers of the same size tend to differ by unpredictable amounts.45 0.042 Radius of Circle (m) 0. 3. If regular graphing techniques are used.98 1.45 0. 1. who is likely the person with the stopwatch. you can use the equation stated below to check the time interval for 20 revolutions of each trial without telling then students how you are evaluating their data.6 10.9 Frequency of Revolution (Hz) 2.98 0.98 0.96 0.2 10.014 0. the equations are the same. so it is not truly horizontal.75 m ) 1. experimental results may differ somewhat.2 s −1 (0.0256 FT = 39mrf 2 The equation given in the text is: ΣF = 4π 2 mrf 2 ΣF = 39mrf 2 Assuming the tension in the string is the net force causing the stopper to undergo its centripetal acceleration. Copyright © 2003 Nelson Chapter 3 Circular Motion 189 .014 kg )(0.96 N ( ) (0.96 kg ⋅ m/s 2 k = 0. Of course.0256 T mr mrf 2 FT = 0. but it has a small downward component due to gravity. Then they can write the equation relating the variables: F f = 0. The tension force is close to being horizontal.2 Hz ) = 2.014 kg )(0. (The slope of the graph of f versus FT is + 1 other two relationships is − .16 Notice that the units cancel out.can be used to write the proportionality statements.) 2 (d) Considering the proportionality statements in (c): FT f ∝ m r f ∝ f =k FT mr FT mr 1 and the slope for the 2 Evaluating the constant k using data from the first trial: mr k= f FT = ( 2.75 m ) 1. (e) Starting with the equation derived in (d): F f = 0. Students should calculate an average of at least three different trials to verify the constant. so k is dimensionless. The following example shows that the force of gravity on the stopper is smaller than the tension in the string.16 T mr Students can check this equation using the data for the remaining trials. (f) The FBD is shown below.16 T mr F f 2 = 0. The distance measured gives an approximate value of the radius of the circle.8 N/kg ) Fg. if the action force is the string pulling on the stopper toward the centre of the circle. the tension is closer to being horizontal at higher frequencies.8 N/kg ) Fg. Activity 3. the object accelerates in the direction of the net force.014 kg )(9. fingers wrapped gently around the string help Sources of systematic error are: • keeping the plane of motion as horizontal as possible. 0. Values will vary.10 kg )(9. Times will vary. Procedure 1.stopper = mstopper g = (0.98 N = 7. then the reaction force is the stopper pulling on the string away from the centre of the circle.0 :1.98 N The ratio of these forces is 0.4. 2. the motion becomes more horizontal. Thus. there is a simultaneous reaction force equal in magnitude but opposite in direction.14 N The tension is caused by the force of gravity on the hanging mass: Fg.weight = 0. this takes practice • friction between the string and glass tube. At higher frequencies. the effect of gravity pulling down on the stopper is quite noticeable.Fg. it helps to have the same student counting and using the stopwatch Synthesis (i) The three laws of motion are illustrated as follows: • An object maintains its state of rest or constant velocity only if the net force is zero.0 s.stopper = 0. • For every action force. A typical time interval for five revolutions is 10.0 s. • If the net force on an object is not zero. the net force on the hanging mass is zero. (h) There are several sources of random error in this investigation: • trying to keep the stopper moving at a constant speed. Copyright © 2003 Nelson 190 Unit 1 Forces and Motion: Dynamics .1: Simulating Artificial Gravity (Page 154) Materials See page 154. When this investigation is conducted properly.0 . a very smooth string helps Sources of human error are: • counting the number of revolutions. 3. so it maintains its state of rest. A typical time interval for five revolutions is 6. the net force on the stopper is toward the centre of the circle. an average of at least three trials should be taken • keeping the vertical part of the string holding the metal mass from swinging. which causes the centripetal acceleration of the stopper. the distance should be measured to the middle of the rubber stopper • measuring the time for 20 revolutions of the stopper.14 N Evaluation (g) With a low frequency (or a low speed of motion). For example. the setup should be checked constantly • measuring the radius of the circle. In this case. this takes practice and coordination • using the incorrect radius of the circle whenever the paperclip touches the bottom of the glass tube.weight = mweight g = (0. the count-down method described above helps • reaction time using the stopwatch. depending on the distance between the student’s shoulder and the bottom of the bucket. this takes practice and coordination • using the incorrect radius of the circle whenever the paperclip touches the bottom of the glass tube. the tension is closer to being horizontal at higher frequencies.1: Simulating Artificial Gravity (Page 154) Materials See page 154.10 kg )(9. the setup should be checked constantly • measuring the radius of the circle. At higher frequencies. A typical time interval for five revolutions is 10.98 N = 7. the effect of gravity pulling down on the stopper is quite noticeable. depending on the distance between the student’s shoulder and the bottom of the bucket. A typical time interval for five revolutions is 6. • If the net force on an object is not zero.stopper = 0.weight = 0. When this investigation is conducted properly.8 N/kg ) Fg. • For every action force. The distance measured gives an approximate value of the radius of the circle. fingers wrapped gently around the string help Sources of systematic error are: • keeping the plane of motion as horizontal as possible.014 kg )(9.4. Times will vary. there is a simultaneous reaction force equal in magnitude but opposite in direction. an average of at least three trials should be taken • keeping the vertical part of the string holding the metal mass from swinging. 3.14 N The tension is caused by the force of gravity on the hanging mass: Fg.Fg. the object accelerates in the direction of the net force.8 N/kg ) Fg. then the reaction force is the stopper pulling on the string away from the centre of the circle. 0. Activity 3. it helps to have the same student counting and using the stopwatch Synthesis (i) The three laws of motion are illustrated as follows: • An object maintains its state of rest or constant velocity only if the net force is zero. the motion becomes more horizontal.0 . (h) There are several sources of random error in this investigation: • trying to keep the stopper moving at a constant speed.weight = mweight g = (0. if the action force is the string pulling on the stopper toward the centre of the circle. which causes the centripetal acceleration of the stopper. Thus. a very smooth string helps Sources of human error are: • counting the number of revolutions. the count-down method described above helps • reaction time using the stopwatch. Copyright © 2003 Nelson 190 Unit 1 Forces and Motion: Dynamics . Procedure 1.98 N The ratio of these forces is 0.14 N Evaluation (g) With a low frequency (or a low speed of motion).stopper = mstopper g = (0.0 s. the net force on the hanging mass is zero.0 :1. 2.0 s. In this case. this takes practice • friction between the string and glass tube. the net force on the stopper is toward the centre of the circle. Values will vary. so it maintains its state of rest. the distance should be measured to the middle of the rubber stopper • measuring the time for 20 revolutions of the stopper. For example. Analysis (a) For Step 2 (b) Using typical values: r = 1.80 Fg Thus. the ratio of the apparent weight to the Earth-bound weight is 2.8:1. FA mac = Fg mg = = ac g 27 m/s 2 9. for the example shown.0.2 s v= = 2π r T 2π (1.2 m/s ac = = v2 r (5. depending on the values in (b). Let FA represent the apparent weight and Fg the Earth-bound weight.0 m T = 1.0 m ) 1. Copyright © 2003 Nelson Chapter 3 Circular Motion 191 .2 m/s )2 1. (c) Answers will vary.2 s v = 5.8 m/s 2 FA = 2.0 m ac = 27 m/s 2 The magnitude of the centripetal acceleration in this example is 27 m/s2. To see why the theoretical value is 9.1 m/s ac = = v2 r (3.9 m/s2. A typical value for five revolutions is 10.0 s. The main weakness of the model is that it is done in a gravitational field. 2π r v= T 2π (1.0:1. so the situation involving ac = 9.8 m/s2 occurs when the normal force on the object is zero rather than 9. so T = 2.9 m/s 2 The magnitude of the centripetal acceleration in this example turns out to be 9. consider that FN = 0. the ratio of the apparent weight to the Earth-bound weight is 1.8 m/s2.0 s. Using an estimated radius of 1. which is certainly within experimental error of the theoretical value of 9.1 m/s )2 1.0 m.8 N/kg as it would be in outer space where g = 0.0 m ) = 2.0 m ac = 9. the situation in step 3 involves a normal force of zero.0 s v = 3.8 m/s2. CHAPTER 3 SUMMARY Make a Summary (Page 156) Legend PIF: passenger on a merry-go-round (inertial frame) PNF: passenger on a merry-go-round (noninertial frame) B: bead in a horizontal accelerometer RT: rubber stopper at top of vertical circle RB: rubber stopper at bottom of vertical circle 192 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .(d) The FBD is shown below. Another strength is that it is a very simple model to demonstrate. Evaluation (e) One strength of this model is that it illustrates in a concrete way that artificial gravity created by a normal force on an object as it undergoes centripetal acceleration increases as the frequency of revolution of the motion increases.0. ΣFy = ma y FN + mg = mac mg = mac ac = g Thus. 0 m ac = 9. which is certainly within experimental error of the theoretical value of 9.8 m/s2. The main weakness of the model is that it is done in a gravitational field.8 N/kg as it would be in outer space where g = 0.1 m/s )2 1. Another strength is that it is a very simple model to demonstrate.0:1. To see why the theoretical value is 9.0 s v = 3. Using an estimated radius of 1.0 s. A typical value for five revolutions is 10.9 m/s2. the ratio of the apparent weight to the Earth-bound weight is 1. so the situation involving ac = 9.8 m/s2.0.1 m/s ac = = v2 r (3.0 m.9 m/s 2 The magnitude of the centripetal acceleration in this example turns out to be 9. so T = 2. 2π r v= T 2π (1. ΣFy = ma y FN + mg = mac mg = mac ac = g Thus.(d) The FBD is shown below. Evaluation (e) One strength of this model is that it illustrates in a concrete way that artificial gravity created by a normal force on an object as it undergoes centripetal acceleration increases as the frequency of revolution of the motion increases. consider that FN = 0.0 m ) = 2.8 m/s2 occurs when the normal force on the object is zero rather than 9.0 s. the situation in step 3 involves a normal force of zero. CHAPTER 3 SUMMARY Make a Summary (Page 156) Legend PIF: passenger on a merry-go-round (inertial frame) PNF: passenger on a merry-go-round (noninertial frame) B: bead in a horizontal accelerometer RT: rubber stopper at top of vertical circle RB: rubber stopper at bottom of vertical circle 192 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . CHAPTER 3 SELF QUIZ (Pages 157–158) True/False 1. F Centripetal force is a net force that applies to all objects. 3.RS: rubber stopper at side of vertical circle C: car on a banked curve G: geosynchronous satellite A: astronaut on the inside of a rotating space station VT: vertical accelerometer at inside top of a loop (Earth’s frame) VB: vertical accelerometer at bottom of a dip (Earth’s frame) Sample diagrams are shown below. 9. 6. T F Only the centripetal force is real. 7. T F The magnitude of your weight calculated using the equation F = mg equals the magnitude of the force of gravity GmmE between you and Earth calculated using the equation FG = . The centrifugal force is fictitious. both natural and human-made. 4. 5. and is made up in the accelerating object’s frame of reference. in circular motion. 2. 8. T F The centrifugal force on an object in uniform circular motion is a fictitious force in the accelerating frame of reference of the object and is directed away from the centre of the circle. T F Centripetal acceleration is in a direction that is toward the centre of the circle. r2 Chapter 3 Circular Motion 193 Copyright © 2003 Nelson . CHAPTER 3 SELF QUIZ (Pages 157–158) True/False 1.RS: rubber stopper at side of vertical circle C: car on a banked curve G: geosynchronous satellite A: astronaut on the inside of a rotating space station VT: vertical accelerometer at inside top of a loop (Earth’s frame) VB: vertical accelerometer at bottom of a dip (Earth’s frame) Sample diagrams are shown below. 2. 6. in circular motion. 7. 8. both natural and human-made. and is made up in the accelerating object’s frame of reference. 3. F Centripetal force is a net force that applies to all objects. T F Centripetal acceleration is in a direction that is toward the centre of the circle. T F The magnitude of your weight calculated using the equation F = mg equals the magnitude of the force of gravity GmmE between you and Earth calculated using the equation FG = . 9. 5. T F The centrifugal force on an object in uniform circular motion is a fictitious force in the accelerating frame of reference of the object and is directed away from the centre of the circle. r2 Chapter 3 Circular Motion 193 Copyright © 2003 Nelson . The centrifugal force is fictitious. 4. T F Only the centripetal force is real. (c) At the top of the circle. ac ∝ v 2 . 20. the net force that causes acceleration is vertically downward and greater in magnitude than mg. the tension in the string is zero. The direction of the centripetal acceleration of the car is vector 9. (b) From ΣFc = max . ∆t → 0 ∆t No. ΣFy = ma y FT − mg = ma y Since the acceleration is upward (i. r CHAPTER 3 REVIEW (Pages 159–160) Understanding Concepts 1. T2 2. 18.. (d) From v = . Centripetal acceleration is an instantaneous acceleration whose direction toward the centre of the circle is constantly changing as the object travels in a circle (or arc) of constant radius. the net force toward the centre of the circle is vertical and equal in magnitude to mg. T 11. 13. (c) The direction of the centripetal force is the same as vector 12. which is the direction of the instantaneous acceleration toward the centre of the circle. v ∝ mplanet . (d) The direction of the normal force acting on the car is vector 11. Since the acceleration is eastward. and the centripetal force must be opposite in direction. FT −mg must be upward. The equation defining the instantaneous acceleration ∆v is a = lim . 16. (c) From FG = 22. r2 v2 . which is vertically upward. T Multiple Choice 12. in other words downward. (b) The direction of the skier’s instantaneous velocity is to the left. FG ∝ 1 . (c) At the bottom of the circle. and the direction of the net force acting on the skier is the same as the direction of the centripetal acceleration. 17. Let +y be down. 14. 194 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . the centripetal force must be eastward. as seen in the equation ac = 4π 2 r .10. the parts farther from the centre of the circle experience a greater centripetal acceleration because ac ∝ r for a constant period of revolution of the circular motion. toward the centre of the circle). the net force that causes acceleration is vertically upward and greater in magnitude than mg. (e) At the top of the circle. At the instant describe. (a) The direction of the centrifugal force you feel is west. or westward. ΣFc ∝ m .e. At the lowest position of the swing. Let +y be up. the direction of the acceleration is toward the centre of the circle. 21. 19. (e) From ac = Gm1m2 r2 . which is perpendicular to the banked curve. (e) The direction of the acceleration of the tip is the same as vector 10. or vector 6. ΣFy = ma y FT + mg = ma y mg = ma y 15. r Gmplanet 23. If FT is the tension in the string and +y is down: ΣFy = ma y FT + mg = ma y The net force is FT + mg. which is horizontal and directed toward the centre of curvature of the curve. or vector 9. so FT must be greater than mg. . toward the centre of the circle). 16. r Gmplanet 23. Let +y be down. (c) The direction of the centripetal force is the same as vector 12. T2 2. and the direction of the net force acting on the skier is the same as the direction of the centripetal acceleration. (d) The direction of the normal force acting on the car is vector 11. the net force that causes acceleration is vertically downward and greater in magnitude than mg. 17. in other words downward. v ∝ mplanet . which is horizontal and directed toward the centre of curvature of the curve. the tension in the string is zero. so FT must be greater than mg. T 11. r2 v2 . which is perpendicular to the banked curve. ac ∝ v 2 . (b) From ΣFc = max . (c) At the bottom of the circle. the parts farther from the centre of the circle experience a greater centripetal acceleration because ac ∝ r for a constant period of revolution of the circular motion. (e) The direction of the acceleration of the tip is the same as vector 10. 14. and the centripetal force must be opposite in direction. (c) At the top of the circle. 19. At the instant describe. FT −mg must be upward. 18. the direction of the acceleration is toward the centre of the circle. Since the acceleration is eastward. (b) The direction of the skier’s instantaneous velocity is to the left. the centripetal force must be eastward. (e) At the top of the circle. 13. The equation defining the instantaneous acceleration ∆v is a = lim . which is vertically upward. ΣFy = ma y FT + mg = ma y mg = ma y 15. r CHAPTER 3 REVIEW (Pages 159–160) Understanding Concepts 1. (a) The direction of the centrifugal force you feel is west. At the lowest position of the swing. (e) From ac = Gm1m2 r2 . Centripetal acceleration is an instantaneous acceleration whose direction toward the centre of the circle is constantly changing as the object travels in a circle (or arc) of constant radius. (d) From v = . (c) From FG = 22.e. 21. or vector 6. FG ∝ 1 .10. the net force toward the centre of the circle is vertical and equal in magnitude to mg. or westward. Let +y be up. 20. the net force that causes acceleration is vertically upward and greater in magnitude than mg. or vector 9. 194 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . which is the direction of the instantaneous acceleration toward the centre of the circle. ΣFy = ma y FT − mg = ma y Since the acceleration is upward (i. ∆t → 0 ∆t No. T Multiple Choice 12. ΣFc ∝ m . If FT is the tension in the string and +y is down: ΣFy = ma y FT + mg = ma y The net force is FT + mg. as seen in the equation ac = 4π 2 r . The direction of the centripetal acceleration of the car is vector 9. 4.4 ×10 2 m The minimum radius of curvature of this curve is 1. Copyright © 2003 Nelson Chapter 3 Circular Motion 195 . 5.8 cm ) = (60 s)2 ac = 0.4 m/s 2 r = 1.4 × 10−5 cm/s2. (b) For the minute hand: r = 8.6 × 103 s)2 ac = 2. the particle’s acceleration will be at an angle away from the line to the centre of the circle. (a) For the second hand: r = 9.6 × 103 s 4π 2 r ac = 2 T 4π 2 (8. ac = 4. The net acceleration is the vector addition of the acceleration toward the centre of the circle and the tangential acceleration perpendicular to that acceleration.8 cm T = 60 s 4π 2 r ac = 2 T 4π 2 (9.3.11cm/s 2 The magnitude of the centripetal acceleration of the tip of the second hand is 0.4 × 102 m. If the speed of a particle in circular motion is increasing.4 × 10−5 cm/s 2 The magnitude of the centripetal acceleration of the tip of the minute hand is 2.11 cm/s2.0 cm T = 60 min = 3.0 cm ) = (3.4 m/s2 v = 25 m/s ac = v2 r v2 r= ac = (25 m/s)2 4. (c) For the hour hand: r = 6.0 cm T = 12 h = 4.3 × 104 s ac = 4π 2 r T2 4π 2 (6.0 cm ) = (4.3 × 10 4 s) 2 6. ac = 1.3 × 10 −7 cm/s 2 The magnitude of the centripetal acceleration of the tip of the hour hand is 1.3 × 10−7 cm/s2. r = 16 cm = 1.6 × 10–1 m ac = 0.22 m/s2 4π 2 r ac = 2 T T= = 4π 2 r ac 4π 2 (0.16 m ) 0.22 m/s 2 T = 5.4s The period of rotation of the plate is 5.4 s. 7. The force(s) causing the centripetal acceleration is (are): (a) static friction of the road on the tires The truck: +x is toward the centre of curvature of the curve. (b) horizontal component of the normal force of the road on the bus The bus: +x is toward the centre of curvature of the banked curve. (c) force of gravity of the Sun on the planet The planet: +x is toward the Sun. 196 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson (d) force of gravity of Earth on the satellite The satellite: +x is toward Earth 8. m = 0.65 kg r = 26 cm = 0.26 m f = 4.6 Hz (a) The force that causes the centripetal acceleration is the normal force exerted by the wall of the rotating tub on the wet towel. (b) v = 2π rf = 2π (0.26 m)(4.6 Hz) v = 7.5 m/s The speed of the towel is 7.5 m/s. mv 2 (c) ∑ F = r (0.65 kg)(7.5 m/s)2 = 0.26 m ∑ F = 1.4 × 102 N The magnitude of the centripetal force on the towel is 1.4 × 102 N. 9.0 ×1012 m = 4.5 × 1012 m 9. r = 2 mN = 1.0 × 1026 kg FG = 6.8 × 1020 N m v2 (a) FG = N r FG r v= mN = (6.8 ×10 20 N)(4.5 × 1012 m) 1.0 ×10 26 kg v = 5.5 × 103 m/s The speed of Neptune is 5.5 × 103 m/s. 2π r (b) v = T 2π r T= v 2π (4.5 × 1012 m) = 5.5 ×103 m/s = 5.1 × 109 s T = 1.6 × 102 a Thus, Neptune’s period of revolution around the Sun in Earth years is 1.6 × 102 a. Copyright © 2003 Nelson Chapter 3 Circular Motion 197 10. At the specific points, the force(s) causing the centripetal acceleration is (are): A: the difference between the normal force and the force of gravity B: the difference between the normal force and the component of the force of gravity parallel to the normal force C: the normal force D: the sum of the normal force and the component of the force of gravity parallel to the normal force E: the sum of the normal force and the component of the force of gravity parallel to the normal force 11. m = 45.7 kg r = 3.80 m v = 2.78 m/s Let the positive direction be upward. The tension in each of the two vertical support chains is equal. Since mv 2 , then: ΣF = 2 FT + (− Fg ) and ∑ F = r mv 2 2 FT = − (−mg ) r (45.7 kg)(2.78 m/s) 2 = + (45.7 kg)(9.80 m/s 2 ) 3.80 m 2 FT = 541 N FT = 2.70 ×102 N The magnitude of the tension in each of the two vertical support chains is 2.70 × 102 N. 12. m = 2.1 × 103 kg r = 275 m v = 26 m/s (a) Considering the vertical components of the forces, with +y up: ΣFy = ma y = 0 FN − mg = 0 FN = mg 198 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson Considering the horizontal components of the forces, with +x the direction of the acceleration: ΣFx = max FS,max = mv 2 r mv 2 µS FN = r mv 2 µS mg = r 2 v µS = gr = (9.8 m/s )(2.75 ×10 m ) 2 2 ( 26 m/s )2 µS = 0.25 The minimum coefficient of static friction between the tires and the road that will allow the SUV to go around the curve without sliding is 0.25. (b) As can be seen in (a) above, the mass m cancels out, so increasing the mass has no effect on the minimum coefficient of static friction. 1 (c) In (a) above, µS ∝ , so decreasing the radius would cause an increase in the minimum coefficient of static friction. r 13. m = 0.23 kg r = 75 cm = 0.75 m (a) (b) v = 3.6 m/s Let the +y direction be toward the centre of the circle. At the top of the circle: ΣFy = mac FT + mg = mv 2 r mv 2 FT = − mg r (0.23 kg)(3.6 m/s)2 = − (0.23 kg)(9.8 N/kg) 0.75 m FT = 1.7 N Copyright © 2003 Nelson Chapter 3 Circular Motion 199 At the bottom of the circle: ΣFy = mac mv 2 r mv 2 FT = + mg r (0.23 kg)(3.6 m/s) 2 = + (0.23 kg)(9.8 N/kg) 0.75 m FT = 6.2 N The magnitude of the tension in the string at the top of the circle is 1.7 N. The magnitude of the tension in the string at the bottom of the circle is 6.2 N. (c) The minimum speed of the ball at the top of the path would be the speed required for Fg to provide the force that causes the centripetal acceleration (tension would be zero). Thus, ΣFy = mac FT − mg = mg = mv 2 r v = gr = (0.75 m)(9.8 N/kg) v = 2.7 m/s The minimum speed of the ball at the top of the path is 2.7 m/s. GmA mB may be applied in situations (a) and (c), but not in situations (b) and (d). The equation 14. The equation FG = r2 applies only to two spherical objects and to situations in which at least one of object is of a smaller size than their separation distance. The objects in (b) and (d) are not spheres and are large compared to the separation distance. r 15. 1 = 3.9 r2  GmE m    2 F2  r2  = F 1  GmE m    2  r1  F2  r1  =  F 1  r2  2 = (3.9) 2 F2 = 15 F1 The force between Earth and the meteor increases by a factor of 15 times. 200 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson 19.84 × 108 m r2 = 1.67 ×10 −11 N ⋅ m 2 /kg 2 (5.08 × 108 km = 1. Thus.84 × 1011 m GmE mM FG1 = r2 = (6.67 ×10 = −11 N ⋅ m 2 / kg 2 (1. 18.84 ×108 m)2 ) FG1 = 1.08 × 1011 m Since gravity causes the centripetal acceleration: GmS mV mV ac = r2 Gm ac = 2 S r (6.35 ×1022 kg) (3. m1 = m2 = 1.35 × 1022 kg r1 = 3.21× 10 −10 N The magnitude of the gravitational force between two bowling balls is 4.98 ×10 24 kg)(7.62 kg) (0.49 × 108 km = 3.16.645 m) 2 ) FG = 4.83 × 1024 kg r = 1.  GmmE   2  F2  (r + rE )  = F1  GmmE    2  rE   r  0.17)rE r = 5.17  r + rE  0.17 r = (1 − 0. mS = 1.028 =  E   r + rE  2 F2 = 0.028.99 × 1030 kg mE = 5.13 ×10−2 m/s 2 The magnitude of the centripetal acceleration of Venus is 1.13 × 10−2 m/s2. mS = 1.645 m Gm1m2 FG = r2 (6.99 × 1030 kg) (1.84 × 105 km = 3.21 × 10−10 N. Let F1 be the gravitational force at Earth’s surface and F2 be the gravitational force at distance r.67 ×10 = −11 N ⋅ m 2 / kg 2 (1.0rE The required distance above Earth’s surface would be 5.62 kg)(1.62 kg r = 64.98 × 1024 kg mM = 7.99 × 1030 kg mV = 4. F1  rE    = 0.0 rE.5 cm = 0.08 ×10 m) 2 11 ) ac = 1. 17.99 × 1020 N Copyright © 2003 Nelson Chapter 3 Circular Motion 201 . 16 × 105 kg mE = 5.39 × 1020 N FG = FG12 + FG2 2 = (1. the force of gravity on the mass is not noticed.92 × 105 N The magnitude of the force of gravity acting on the maximum load that the arm can move is 9.67 ×10 −11 N ⋅ m 2 / kg 2 (5.5 Hz f2 = 3. the large mass is experiencing free fall. 20.50 m f1 = 1.98 ×1024 kg)(1.49 ×1011 m)2 ) FG2 = 4.99 × 1020 N) 2 + (4.0 units Fc1 = 4π 2 m1r1 f12 and Fc2 = 4π 2 m2 r2 f 2 2 202 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .92 × 105 N.35 ×1022 kg) (1.82 × 1020 N at an angle of 24.80 × 103 kg mmax load = 1.82 × 1020 N The direction of the net gravitational force is found by applying trigonometry: F θ = tan −1 G1 FG2 = tan −1 1.50 × 105 m + rE = 6.4° from the line to the Sun.99 × 10 20 N 4.83 × 106 m GmE mM (a) FG = r2 = (6.75 m r2 = 1. m1 = 1 ball m2 = 3 balls r1 = 0.67 ×10 r2 −11 N ⋅ m 2 /kg 2 (1.4° The net gravitational force on the Moon caused by the gravitational forces of Earth and the Sun is 4. (b) Like the robotic arm and everything moving along with the ISS.83 ×106 m)2 ) FG = 9.39 × 10 20 N) 2 FG = 4. The only force that affects the components of the arm is the force needed to cause the mass to accelerate slightly as it either starts to move or comes to a stop. m = 1.FG2 = = GmS mM (6. Applying Inquiry Skills 21.99 ×1030 kg)(7. Since the arm and the mass are falling toward Earth together.39 × 1020 N θ = 24.0 Hz Fc1 = 8.98 × 1024 kg rE = 6.38 × 106 m r = 4.16 ×105 kg) (6. 5 m)(1.) (b) The centripetal acceleration is caused by the horizontal component of the tension.0 Hz)2 (1 ball)(1. as shown in the equations in the remaining solutions.9 × 102 units The new value for the centripetal force is 1. the mass of the bob.5°)( tan 27. and radius are interrelated. with +x the direction of the centripetal acceleration: ΣFx = ma x FT = mv 2 r mv 2  mg    sin θ = r  cos θ  FT sin θ = g tan θ = v2 r 2 v = gr tan θ v = gr tan θ = g ( L sin θ ) tan θ = (9.15 m )(sin 27. FT sin θ. Copyright © 2003 Nelson Chapter 3 Circular Motion 203 .Fc2 4π 2 m2 r2 f 2 2 = Fc1 4π 2 m1r1 f12 = (3 balls)(3.5°) 2 v = 1. one at a time.5 Hz)2 Fc2 = 24 Fc1 Therefore. Fc2 = 24 Fc1 = 24(8. (a) Students should describe a controlled experiment in which the dependent variable (the frequency of revolution) is measured as the independent variables (the length of the pendulum. we begin by considering the vertical components of the forces. with +y up.0 units) Fc2 = 1.8 m/s ) (1.9 × 102 units.65 m/s. ΣFy = ma y = 0 FT cos θ − mg = 0 mg cos θ Considering the horizontal components of the forces.65 m/s The speed of the bob is 1. 22. angle. (c) To determine an equation for the speed of the conical pendulum. (Students would discover that the frequency is independent of the mass and that the variables length. and either the radius of the circle or the angle of the string) are varied in a controlled way.0 m)(3. Monday. 2001. In most races. The decision to cancel the race mentioned was made by CART’s officials about 75 min before the scheduled start.com/News/Article.forsthe-racing. 2001.com/News/Article. The force of gravity helps the marbles follow the grooves. in space where the marbles are in free fall with everything else in the ISS. Evidently the track had been designed for slower-speed NASCAR races. ΣFM = Fg = mac GmM mE rEM 2 = 4π 2 mM rEM TM 2 4π 2 rEM 3 (Equation 1) GTM 2 Next consider that the force of universal gravitation exerted by Earth on an object of mass m on Earth’s surface equals the object’s weight: Fg = mg GmmE rE 2 GmE rE 2 = mg =g grE 2 G (Equation 2) mE = mE = 204 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .asp?ID=587 http://www. page C12.) The race was called the Firestone Firehawk 600.(d) Using the data from (c): v = 2π rf f = = v 2π r 1. 23. TM = 27 d 8 h = 2. the device would work in any orientation.38 × 106 m rEM = 60.htm http://dallas. and there had been too little testing of the track prior to this particular CART race.3616 × 106 s rE = 6. explaining they felt ready to pass out. On Earth’s surface.bizjournals. May 1. At racing speeds of about 380 km/h. To operate in space. with clear plastic barriers) to prevent the marbles from floating away.5° ) f = 0. Making Connections 24.howstuffworks.forsthe-racing. Apr.15 m )(sin 27. At such a steep angle the drivers experienced accelerations up to 5.1 rE Consider first that the force that causes the centripetal acceleration of the Moon around Earth is the force of gravity.html Extension 25.5g for nearly 18 s in each lap. the device shown in the text must be rotated at a fairly high frequency in order for the marbles to move up the grooves and lodge in the holes.65 m/s 2π (1.html http://dallas. However. even if it were rotated slowly.567 Hz The frequency is 0. (CART stands for Championship Auto Racing Teams.4-km oval track with a banking angle of approximately 24°. The following references and Web sites provide more information: “CART cancels Texas race” The Toronto Star. “CART assailed by money woes” The Toronto Star. the cars took approximately 22 s to cover the 2.com/dallas/stories/2001/10/15daily18. Tuesday. 30. the accelerations would be around a much safer 3g.com/question633. page C8. Drivers felt the effects of such high accelerations after several practice laps.bizjournals. http://www.asp?ID=603 http://www. the sides would have to be covered (for example. located at the Texas Motor Speedway.567 Hz.com/dallas/stories/2001/05/07daily17. she gains an amount of kinetic energy equal to the amount of gravitational potential energy she loses. r = 24 m FT = 2mg vi = 0 r = 85 m Note: If you assign this question at this stage. give your students the hint that they must apply the law of conservation of energy. Copyright © 2003 Nelson Chapter 3 Circular Motion 205 . The second figure shows the corresponding FBD of the teacher at that instant. 27. The first figure shows the system diagram after the teacher has swung downward and the rope is at an angle θ to the horizontal.) Let +y be up.13 6.80 m/s2.38 × 106 m    ( )( ) g = 9.3616 × 106 s  ( ) 2   60.Equating Equations 1 and 2: grE 2 4π 2 rEM 3 = G GTM 2  4π 2 g = T 2  M  4π 2 =  2  TM   rEM 3     2     rE    (60. As the teacher in the question drops a vertical distance ∆y. which they studied in the previous grade. ΣFy = ma y FN − Fg = FN − mg = 4mg − mg = 4g − g = 3g = r= = mv 2 r mv 2 r mv 2 r v2 r v2 r v2 3g (50 m/s )2 3 9.8 m/s 2 ( ) The radius of the loop is 85 m.80 m/s 2 The value of g at Earth’s surface using the motion of the Moon is 9. v = 180 km/h = 50 m/s (Assume two significant digits. 26.1rE )  2    rE 3      4π 2  3 =  2  60.1 rE T  M  ( )  4π 2 =   2. This component of the acceleration is toward the right at the instant shown in the text. it has a centripetal acceleration toward the centre of the circle. ∆y = r sin θ 2 = (24 m)   3 ∆y = 16 m Thus. 2 3 206 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . 28. P. The net acceleration of these two components is in the direction of vector D. Figure 6(b) on page 161.Considering the y-components of the forces: ΣFy = ma y mv 2 r mv 2 2mg − mg sin θ = r 2 v 2 g − g sin θ = (Equation 1) r FT − mg sin θ = Applying the law of conservation of energy: EK = ∆EP mv 2 = mgr sin θ 2 v2 = 2 g sin θ (Equation 2) r Substituting Equation 2 into Equation 1: 2 g − g sin θ = 2 g sin θ 2 − sin θ = 2 sin θ 2 = 3sin θ sin θ = Finally. As the ball moves in its circular path.0 m above the ground. the ball is experiencing tangential acceleration downward due to gravity. the teacher has dropped by 16 m. and so is 8. At the same instant. With any of the three options. for a summary of technological problem solving. test. Option 1: A Model Roller Coaster This option involves numerous principles of forces and motion. Option 3: Centrifuges This option relates most directly to the principles presented in Chapter 3. especially friction and air resistance (discussed in Section 2. It is an open-ended task in which students exhibit creativity as they design. page 138) in the Solutions Manual. especially the choice of option. For a list of Internet sites.com. Options 1 and 2 involve building devices. Analysis The answers to the Analysis questions depend on various factors. More information about this competition can be found at www. but Option 2 is much more open-ended than Option 1. Each year of this competition the rules and the dimensions of the ride are altered slightly to promote student creativity and originality.g. Option 3 involves research and analysis of existing devices. Evaluation Here the students evaluate their own task. for a summary of technological problem solving. students are urged to create their own questions and answer them. this task is set up for ease of student assessment. It is recommended that you discuss the task. 2) • a maximum allowable mass Some of the qualitative criteria are: • originality/creativity • thrill level of the ride • aesthetic appeal • wise use of materials Students can refer to Appendix A4. Two models are required to show the effects of different amounts of friction or air resistance. Answers will depend on the option chosen. Copyright © 2003 Nelson Unit 1 Performance Task 207 . and revamp two different models of the same toy. refer to the answer to question 11 (text. Students can refer to Appendix A4. and the assessment with the students early in the unit. Internet sites.UNIT 1 PERFORMANCE TASK APPLYING PRINCIPLES OF MOTION AND FORCES (Pages 162–163) Like the other performance tasks in the textbook. and then have them work on the task as the unit progresses. and local experts who use centrifuges. a feature that helps to make the task more openended.4).canadaswonderland. It is important to decide on the criteria used to evaluate the success of the model before the students begin the task. Students can find numerous resources. The three options within the task are quite different in design. build. especially those involving circular motion (Chapter 3). Option 2: Toys That Apply Physics Principles This option applies the concepts of motion and forces. The task is designed so that it can stand on its own or be used in conjunction with the model roller coaster contest held at Paramount Canada’s Wonderland each May on Physics/Math Day. text page 767. Some of the quantitative criteria used to compare coasters of the same outside dimensions are: • total length of the track • number of vertical loops • a minimum number of horizontal loops (e. including the school’s resource centre.. text page 767. the options. 4. of magnitude mg. T T F The horizontal component of the velocity of the ball remains constant during the entire projectile motion. 12. To compare the tensions at positions 1 and 4. T T ∆v approaches the instantaneous acceleration as ∆t approaches zero. 7. the stopper’s instantaneous velocity will have an eastward component only. 2. T F The net force on the ball at the top of the flight is the force of gravity. pulling downward on the ball. 9. 5. The centripetal acceleration is independent of the mass of the object. At position 1: ΣFy = ma y FT1 − mg = mv 2 r mv 2 FT1 = + mg r At position 4: ΣFy = ma y FT4 + mg = FT4 Now:  mv 2   mv 2  + mg  −  − mg  FT1 − FT4 =   r   r  FT1 − FT4 = 2mg 10. 8. 14. If the ball is released at position 1. The equations for centripetal acceleration are: v 4π 2 r ac = 2 = 2 = 4π 2 rf 2 r T mv 2 r mv 2 = − mg r Multiple Choice 13. The magnitude of the tension at position 1 exceeds the magnitude of the tension at position 4 by 2mg. T 11.UNIT 1 SELF QUIZ (Pages 164–166) True/False 1. (c) The acceleration is down the hillside and its magnitude is g sin θ . 3. (e) With vi = 0: v 2 = vi 2 + 2ax ∆x = 0 + 2 g sin θ L v = 2 Lg sin θ v = 2 Lg sin θ 2 208 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . so the change of position when this F The quantity ∆t occurs is from position 4 to position 5. 6. let +y be toward the centre of the circle in each case. Notice that the speed v is constant. (c) The forces are equal in magnitude but opposite in direction. the acceleration is simply the acceleration due to gravity. (b) Using the +y and +x directions shown in the text: ΣFy = ma y = 0 vav = FN − mg cos θ = 0 FN = mg cos θ Thus. mg cos θ .1 (v + vi ) 2 1 = 2 Lg sin θ 2 Lg sin θ vav = 2 15. (b) Since there is nor air resistance. the magnitude of the average velocity is vav = At halfway: v 2 = vi 2 + 2ax ∆x v 2 = 0 + 2 g sin θ v 2 = Lg sin θ v = Lg sin θ Thus. so the magnitude of the force of the toboggan on the hill is the same. This is an action-reaction pair of forces. (a) Using the +y and +x directions shown in the text: ΣFy = ma y = 0 FN − mg cos θ = 0 FN = mg cos θ ΣFx = ma x Fg sin θ − Ff = ma x mg sin θ − µ K FN = ma x mg sin θ − µ K mg cos θ = ma x g sin θ − µ K g cos θ = a x L 2 a x = g (sin θ − µ K cos θ ) 19. (c) From question 14.8 m/s2 [down]. 9. the magnitude of the force of the hill on the toboggan is mg cos θ . vav < v . 20. Copyright © 2003 Nelson Unit 1 Self Quiz 209 . 16. (e) vi = 0 ax = g sin θ ∆x = L 1 ∆x = vi ∆t + ax ∆t 2 2 1 L = g sin θ∆t 2 2 2L ∆t 2 = g sin θ ∆t = 2L g sin θ Lg sin θ . 18. 2 17. 65 Let +x be the direction of the applied force. (e) The force of gravity toward Earth is the only force acting on the satellite.65 )(9. (a) m2 = m 2 r1 = 3rE r2 = 6rE  GmmE    2 F1  r1  = F2  Gm2 mE    2  r2   m   r2 2  =  2      r1   m2   m  r2  =    m2  r1  2 m  6rE  =    m   3rE    2 = ( 2 )( 2 ) F1 =8 F2 2 2 The ratio is 8:1. (a) 109 km/h = 30.19 × 106 m/s (c) 3. A windsock indicates direction and relative magnitude of the wind.7 cm/(ms)2 = 5. (a) 3 (b) 3 (c) 2 (d) 2 26.4 mm/s2 = 3. (d) m = 9. ΣFy = mac mg = mv 2 r v2 g= r v = gr 24. 23.49 µS = 0.21. To start the box moving.3 m/s (b) 7.5 kg )(9. FS. the minimum force needed is the force needed to overcome the force of static friction.5 kg µK = 0.max = 61 N Completion 25.8 N/kg ) FS.4 × 10–3 m/s2 (d) 5. (d) Let +y be down.16 × 104 km/min = 1. 22.max = µS FN = µS mg = (0.28 × 10–3 m/s2 27. 210 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .7 × 104 m/s2 (e) 4.62 × 10−3 (km/h)/s = 1. 30. A passenger of mass m is standing on an elevator that has an acceleration of magnitude a. The acceleration of an object falling vertically through the air at terminal speed is zero. we must invent fictitious forces to explain an observed acceleration. The law of inertia is also known as Newton’s first law of motion. vCE = vCD + vDE 33. 43. If the frame is rotating. 46. An accelerating frame of reference is also known as noninertial. 40. As the speed of a flowing river increases.28. (b) (g) (e) (a) (a) (c). your mass would be the same as on Earth’s surface. the invented force is called centrifugal force. (a) the slope of a line on a velocity-time graph is LT–2 (b) the area under the line on an acceleration-time graph is LT–1 (c) weight is MLT–2 (d) the universal gravitation constant is L3M–1T–2 (e) gravitational field strength is LT–2 (f) coefficient of static friction is dimensionless (g) frequency is T–1 (h) the slope of a line on an acceleration-force graph is M–1 36. 45. the direction of the centripetal acceleration is unchanged because it is still toward the centre of the circle.0. 35. On the Moon’s surface. and the direction of the instantaneous acceleration is down. 42. gas pedal. 39. 29. 37. 31. If vLM is 26 m/s [71° W of S]. You are facing southward when suddenly a snowball passes in front of your eyes from left to right. The three controls a car driver has for producing acceleration are brake pedal. The normal force acting on the passenger has a magnitude m(a + g) if the acceleration is upward and m(a − g) if the acceleration is downward. The direction of the instantaneous velocity is westward and down. Matching 41. If the direction of an object undergoing uniform circular motion is suddenly reversed. 44. but your weight would be reduced by a factor of 6. 32. and steering wheel. the pressure in flowing water decreases. In such a frame. The horizontal acceleration of a projectile is zero. 34. The snowball had been thrown earlier with an initial horizontal velocity. then vML is 26 m/s [71° E of N]. (d) Copyright © 2003 Nelson Unit 1 Self Quiz 211 . 38. 7 cm 2 To determine the length of each side of the square. striking the floor.71 m/s 212 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . (a) The object starts with a high velocity in the positive direction. but the Moon has no atmosphere and thus there is no air resistance. the positive direction would be down. The component of a vector is always less than or equal to the magnitude of the vector because the component is a projection of the vector along the axis of a rectangular coordinate system. The average speed exceeds instantaneous speed during the time when an object is slowing down and moving slower than its average speed. the object changes direction. consider the right-angled triangle. (c) The object is moving in one direction with a uniform positive acceleration until it reaches maximum velocity. starting with a high velocity and then moving with uniform acceleration in the same direction as the first acceleration. reaching zero velocity and then undergoing uniform acceleration in the opposite direction until reaching a velocity equal in magnitude to the initial velocity. (b) The object starts with a high velocity in the negative direction. In this case. (12 m ) 2 + (12 m ) 2 BY = 17 m vav = = d ∆t 2 (17 m ) 48 s vav = 0.) (d) The object starts at a high velocity in one direction and undergoes uniform negative acceleration. slows down uniformly until it comes to a brief stop. slows down uniformly. At that instant. and bouncing back upward with no loss in energy. (a) (b) (c) (d) 2. 6. the circumference of the circle is 2πr and half the circumference is: C =πr 2 = π (12 cm ) C = 37. (b) Average speed is less than instantaneous speed when an object is speeding up and moving faster than its average speed. coming to a brief stop before accelerating uniformly in the opposite direction until reaching the starting position. A parachute would not work on the Moon. and then accelerates uniformly in the positive direction until reaching the same position it started from. (An example of this motion would be a ball vertically falling from rest.UNIT 1 REVIEW (Pages 167–171) Understanding Concepts 1. BY 2 = BC 2 + CY 2 BY = BC 2 + CY 2 = (a) For the first person: 3. (a) The object is moving southward and is increasing its speed. The object is moving southward and is decreasing its speed. The object is moving northward and is decreasing its speed. 5. Since the radius of the circle is 12 cm. (c) Average speed equals instantaneous speed for any object moving at a constant velocity. BCY. A parachute relies on air resistance. 4. The object is moving northward and is increasing its speed. with downward defined as the positive direction.50 m/s [45° N of E] vav = 7.71 m/s for the person walking along the edge of the square. The area under each line on the graph represents the distance fallen.79 m/s for the person walking along the circumference. (b) Both average velocities are the same. so all three areas should be the same.For the second person: d ∆t 37. The graph is shown in below.8 × 10−2 h) 2 ∆d = 0. m = 1. Copyright © 2003 Nelson Unit 1 Review 213 .1 times the horizontal range on Earth.2 × 103 kg vi = 42 km/h vf = 105 km/h ∆t = 21 s = 5.7 m = 48 s vav = 0.50 m/s [45° N of E]. The average velocity of each person is 0. 9.1 ∆xE The horizontal range on the Moon is 6. ∆d vav = ∆t 24 m [45° N of E] = 48 s vav = 0.43 km The car travels 0.6 m/s 2 8. page 49:  vi 2 sin 2θ    ∆xM  g M  = ∆xE  vi 2 sin 2θ    gE   gE = gM = 9.8 × 10–2 h 1 (a) ∆d = (vi + vf ) ∆t 2 1 = ( 42 km/h + 105 km/h ) (5.43 km in this time interval.8 m/s 2 1. and 0. Using the equation derived for horizontal range derived in the text. ∆xM = 6.79 m/s The average speeds are 0. 2 × 103 kg) ΣF = 1.60 m/s.4s vav = 0.44 m d total = vav ∆t vav = d total ∆t 1.12 m [S] = 0.0 × 103 N The magnitude of the average force needed to cause this acceleration is 1.44 m[S] − 0. ∆ d1 = 0.88 m [N] + 0.32 m [N] The final position of the ball is 0.83 m/s2 ΣF = ma = (0.(b) a = vf − vi ∆t 105 km/h − 42 km/h = 21s a = 3.5 cm/s The average speed of the tip is 1.0 × 103 N. ∆d (c) vav = ∆t 0.0 (km/h)/s.44 m [S] ∆ d 2 = 0.88 m + 0.32 m [N] of its original position.5 cm/s. 10. 214 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .12 m [S] ∆t = 2. 11.12 m d total = 1. (a) r = 14 cm ∆t = 60.0 km/h/s = 0.44 m[S] + 0.32 m [N] = 2.88 m [N] ∆ d3 = 0.12 m [S] = −0.13 m/s [N].88 m [S] + 0.4s vav = 0.60 m/s The average speed of the ball is 0.83m/s 2 )(1.4 s (a) d total = d1 + d 2 + d3 = 0.0 s d vav = ∆t 2π r = ∆t 2π (14 cm ) = 60.0 s vav = 1.44 m + 0.13 m/s [N] The ball’s average velocity is 0. (c) a = 3.44 m = 2.32 m [S] ∆d = 0.0 (km/h)/s The magnitude of the car’s average acceleration is 3. (b) ∆d = ∆d1 + ∆d 2 + ∆d 3 = 0. The speed of the projectile is least at the top of the flight when the vertical component of the velocity is zero. 14.2 cm/s [down]. The motion of the balloon results from an action-reaction pair of forces.15 m/s2 ∆t = 95 s v −v a= f i ∆t vfx = vix + ax ∆t = 23m/s + (0 m/s 2 )(95 s) vfx = 23m/s vf = vfx 2 + vfy 2 vf = 27 m/s tan θ = vfy vfx = (23 m/s) 2 + (14. the second hand moves through 4 h × 30°/h = 120° in 20 s. the angle subtended is 12 At the 6:00 o’clock position: v = 1.2 cm/s [down] The average velocity of the tip is 1. Copyright © 2003 Nelson Unit 1 Review 215 . with a magnitude of c: c 2 = a 2 + b 2 − 2ab cos θ c = a 2 + b 2 − 2ab cos θ = (14 cm ) + (14 cm ) − 2 (14 cm )(14 cm )(cos120°) c = 24 cm ∆d vav = ∆t 24 cm [down] = 20 s vav = 1. and the reaction force is the expelled air pushing eastward on the balloon.25 m/s  θ = tan −1    23 m/s  θ = 32° The train’s velocity after this acceleration is 27 m/s [32° S of E]. As the tip sweeps past each hour on the clock. Let +x be east and +y be south. vix = 23 m/s viy = 0 ax = 0 ay = 0.5 cm/s and a direction tangent to the circle at that instant.25 m/s  14. 13.(b) The instantaneous velocity has a constant magnitude of 1. 360° = 30° . The action force is the force of the inner walls of the balloon pushing westward on the expelled air. The speed of the projectile is greatest just after launch and just before landing.25 m/s) 2 vfy = viy + a y ∆t = 0 m/s + (0.5 cm/s [30° right from up] (c) From the 1:00 o’clock position to the 5:00 o’clock position. 12.15 m/s 2 )(95 s) vfy = 14.5 cm/s [horizontally left] At the 10:00 o’clock position: v = 1. In these cases. the velocity has a (constant) horizontal component and a vertical component of the same magnitude. Its displacement during this time interval is straight downward. 9° ) T = 9. Let F1 be the gravitational force at Earth’s surface and F2 be the gravitational force at distance r above Earth’s surface.5 × 102 N The magnitude of the average horizontal force on the runner is 2.0 m/s ∆t = 2.424  r + rE  0.5 × 102 N.15.0 s a= vf − vi ∆t 8. which is a reaction to the action force of the runner’s feet on the ground. F Thus. 216 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .0 m/s − 0 m/s = 2. m = 63 kg vi = 0 m/s vf = 8.0 m/s 2 ΣF = ma = (63 kg)(4.18.4rE. This force is exerted by the ground on the runner’s feet.18 =  E   r + rE  2  rE    = 0.424)rE r = 1.7 × 10 2 N The magnitude of the tension in the rope is 9.0 m/s 2 ) ΣF = 2. 16. with +y up: ΣFy = ma y = 0 2T sin θ − mg = 0 T= mg 2sin θ 6.9° Considering the vertical components of the forces on the trapeze artist.4rE The required distance above Earth’s surface would be 1. 17.3 × 10 2 N = 2 (sin18. F1  GmmE   2  F2  (r + rE )  = F1  GmmE    2  rE   r  0. We begin by determining the angle of the rope from the horizontal: ∆y sin θ = L ∆y θ = sin −1 L 2.0s a = 4. 2 = 0.7 × 102 N.1 m θ = 18.424r = (1 − 0.3 m = sin −1 7. 1 × 102 m.92 × 107 s Copyright © 2003 Nelson Unit 1 Review 217 .49 m/s 2 r = 1. 19.27 ×1023 kg v = 2. the particle’s acceleration will be at an angle away from the line to the centre of the circle.49 m/s2 v = 22 m/s ac = v2 r v2 r= ac (22 m/s) 2 4.42 × 104 m/s. The net acceleration is the vector addition of the acceleration toward the centre of the circle and the tangential acceleration perpendicular to that acceleration.42 × 104 m/s) TM = 5.63 × 1021 N TE = 3.1×10 2 m = The minimum radius of curvature of the curve is 1. ac = 4.16× 107 s m v2 (a) FG = M r FG r v= mM = (1.28 ×1011m) 6. rM = 2.42 × 104 m/s The speed of Mars is 2.63 × 1021 N)( 2. If the speed of a particle in circular motion is decreasing.28 ×1011 m) (2.27 × 1023 kg FG = 1. 20.28 × 1011 m mM = 6. 2π r (b) vM = TM TM = = 2π r vM 2π (2.18. 45 × 106 m 2 4. the magnitude of the instantaneous velocity is the same and the direction is tangent to the circle at that instant. (a) T = 80.36 × 106 s ∆t = = 1.80 × 103 s v = 8.02 × 103 m/s 2. r = 6.36 × 106 s v = 1.80 × 103 s ∆t = = 2.16 × 107 s TM = 1.92 × 107 s = TE 3. (b) T = 2.0 min)(60 s/min) T= = 2.44 × 103 m/s) = 2.02 km/s = 1.44 × 103 m/s) − (−8.04 m/s2. 22.5s v = 16 m/s (a) The velocity of a passenger at the 3:00 o’clock position would be 16 m/s [down].04 m/s 2 The magnitude of the average acceleration is 7. (b) The velocity of a passenger at the 6:00 o’clock position would be 16 m/s [left].44 × 103 m/s ∆v aav = ∆t vf − vi aav = ∆t (8.88 TE The period of revolution of Mars around the Sun is 1.18 × 106 s 2 218 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .5 s 72 rev 2π r v= T 2π (6.40 × 103 s 2 2π r v= T 2π (6.88 Earth years.TM 5.45 × 106 m) = 4.40 × 103 s aav = 7.80 × 103 s 1.29 × 107 m r= = 6.3 m) = 2. 21.0 min = 4. In each case.3 m (3. (c) The velocity of a passenger at the 7:00 o’clock position would be 16 m/s [30° up from left]. 73 × 10–3 m/s2.2 × 103 N. the magnitude of the horizontal force on the male skater is the reaction force. m1 = 55 kg r = 1.2 × 103 N.2 × 103 N The magnitude of the force causing the female skater to maintain her circular motion is 3. which equals 3.88 Hz) 2 ∑ F = 3. ∆v aav = ∆t vf − vi aav = ∆t (1.18 × 106 s aav = 1.02 × 103 m/s) − (−1. 23.9 m m2 = 88 kg f = 0. Thus.2 × 103 N.02 × 103 m/s) = 1.9 m)(0. 24.73 × 10−3 m/s 2 The magnitude of the average acceleration is 1. (b) The force causing the female skater to maintain her circular motion is the action force of 3.88 Hz (a) ∑ F = 4π 2 mrf 2 = 4π 2 (55 kg)(1. (a) In Earth’s frame of reference (b) In the car’s frame of reference (c) Considering the vertical components of the forces: ΣFy = ma y = 0 FT cos θ − Fg = 0 FT = mg cos θ Copyright © 2003 Nelson Unit 1 Review 219 . m = 27 kg v = 1.12 × 102 N [27° above the horizontal] Let +y be up and +x be in the direction of the horizontal component of the applied force.47 The coefficient of kinetic friction between the garbage can and the sidewalk is 0.87 m/s2 [E]. Consider Box B and define the +y direction as down: ΣFy = mB a mB g − FT = mB a FT = mB ( g − a ) Consider Box A and define the +x direction as up the slope: ΣFy = 0 FN − mA g cos θ = 0 FN = mA g cos θ ΣFx = mA a FT − FK − mA g sin θ = mA a mA a = FT − µ K FN − mA g sin θ 220 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .8 m/s Fapp = 1.1°) a x = 0. 26.8 m/s 2 2 ( ) ( tan 5.12 × 102 N) (sin 27° ) (1.54 Let the general direction of the acceleration of the boxes be clockwise around the pulley.5 kg µK = 0.12 ×102 N) (cos 27° ) µK = 0. 25.8 m/s 2 ) − (1. ΣFy = 0 FN + Fapp sin θ − mg = 0 FN = mg − Fapp sin θ ΣFx = 0 Fapp cos θ − FK = 0 µ K FN = Fapp cos θ µK = = Fapp cos θ mg − Fapp sin θ (27 kg)(9.5 kg mB = 5.Considering the horizontal components of the forces: ΣFx = ma x FT sin θ = ma x ax = FT sin θ m mg sin θ = m cos θ = g tan θ = 9.87 m/s The acceleration of the car is 0.47. mA = 2. 4° ) (9.Substitute for FN and FT: mA a = FT − µK FN − mA g sin θ = mB ( g − a ) − µK mA g cos θ − mA g sin θ (mA + mB )a = (mB − µ K mA cos θ − mA sin θ ) g a= = (mB − µK mA cos θ − mA sin θ ) g (mA + mB ) (5.8 m/s 2 ) FN = 6. for a student living in Ottawa.1 × 102 N.8 m/s 2 + 2.5 × 102 N.54(2.5 m/s 2 ) FN = 7.5 ×102 N The magnitude of the apparent weight of the person is 4.5 kg − 0.5 m/s 2 ) FN = 4. (c) a = 0. Answers will vary. (a) Assume the following values: distance to the capital city: 500 km = 5 × 105 m diameter of Ferris wheel: 20 m Copyright © 2003 Nelson Unit 1 Review 221 . (b) a = 2.1 × 10 2 N The magnitude of the apparent weight of the person is 6.0 m/s2 v = 2.6 ×102 N The magnitude of the apparent weight of the person is 7. the closest capital city of an adjacent province is Quebec City. ΣFy = ma FN − mg = ma FN = m( g + a ) = 62 kg(9. ΣFy = 0 FN − mg = 0 FN = mg = 62 kg(9. m = 62 kg (a) a = 2.5 m/s2 [down] Let the +y direction be down. For example.6 × 102 N.5 kg) cos 25.8 m/s 2 − 2.9 m/s2.8 m/s2 ) (2. 27.5 m/s [up] Let the +y direction be up. depending on the student’s location. Applying Inquiry Skills 28.5 kg + 5.5 kg) 2 a = 3.4° − (2.9 m/s The magnitude of the acceleration of the boxes is 3.5 m/s2 [up] Let the +y direction be up. ΣFy = ma mg − FN = ma FN = m( g − a ) = 62 kg(9.5 kg) sin 25. (c) Safety considerations are fairly obvious. depending on the variables tested. we have aav = 2 .65 m/pace ) tan 30° y = 21 m The estimated diameter of the wheel is 21 m. d N= C / rev d = (π D ) / rev = 5 × 105 m π ( 20 m ) / rev N = 8 × 103 rev The estimated number of revolutions in this example is 8 × 103 revolutions. and the circumference of the wheel C. the distance travelled d. Observations: 1 pace = 0. For example. (a) There are many acceptable choices of objects.Let the number of rotations be N. The distance (d) from the base of the wheel can be estimated using paces or measured using a metric measuring tape. this equation reduces to ∆d = aav ∆t 2 . Isolating the average 2 2 2 ∆d acceleration.65 m d = 56 paces θ = 30° y = x tan θ = (56 paces )(0. Since ∆d and vi are known and ∆t can be measured. The following sample calculation is made with the assumption that the observer’s eyes are at the same height at the bottom of the wheel. depending on students’ interests. With an initial velocity of zero. 29. the mass of a simple parachute/load combination could be kept constant while altering the surface area of the parachute and determine the time for the combination to fall a known distance to the floor. The angle θ can be determined by using a horizontal accelerometer or a large protractor. ∆t 30. 222 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson . Another example is different surface areas of cloth with string and a small weight attached to form a parachute. (b) The following diagram shows the situation for measuring the diameter (y) of the Ferris wheel. The falling objects should not be breakable and they should not be allowed to fall onto anything breakable. (b) Answers will vary. One example could be different sizes or shapes of coffee filters. the equation used to find the average acceleration is 1 1 ∆d = vi ∆t + aav ∆t 2 . then the readings on the two sensors will be virtually the same. Copyright © 2003 Nelson Unit 1 Review 223 . The following diagram shows the sine curve. For the most obvious results. The setup shown in the text can be used to demonstrate Newton’s third law of motion in a concrete way assuming that the readouts from the force sensors can be displayed simultaneously. Consider the FBDs of the magnet and washers shown below. Sensor readout involving the magnetic forces only. so they are labelled simply x and y. According to the graph. The axes depend on the setup. refer to the article “An Extraordinary Demonstration of the Third Law” The Physics Teacher. so g ∆x ∝ sin 2θ . any change recorded by one sensor and balanced by a change in the other sensor is observed immediately. however the computer-generated graphs will show the force as a function of time.31. sin 60° and sin 120° are both 0. which is a single maximum value.866).g. The magnet and the washers attract one another. The dual graph nicely illustrates the action-reaction pair of magnetic forces. However. the exception is 45° because sin 2θ = sin 90° = 1 . so the shape will r be similar if the motion suggested is done smoothly. The graph illustrates that the force is 1 proportional to 2 . 32. for angles of 30° and 60°. it is best to ensure that the mass of the washers is negligible compared to the mass of the magnet. That way. with sensor B registering positive values and sensor B registering negative values. there are two values of sin 2θ that are equal on either side of 90° (e. However. The equation for the horizontal range of a projectile is ∆x = vo 2 sin 2θ . For more information on a similar demonstration. (b) The following graph illustrates the results if the sensor holding the magnet was zeroed with the magnet suspended from it and the mass of the washers is negligible compared to the mass of the magnet. pages 392–393. if the sensors are zeroed before the washers are brought close to the magnet. the action-reaction forces on the readouts are more obvious if sensor A is zeroed with the magnet suspended from it before the washers are brought near. October 2001.. Volume 39. (a) The answer depends on how the apparatus is set up. The force registered by sensor A in this case is much greater than the force registered by sensor B because the force of gravity on the magnet must be balanced. and the magnetic force of attraction of the magnet on the washers is equal in magnitude but opposite in direction to the magnetic force of attraction of the washers on the magnet. Furthermore. 34.95 m/s 2 7.28 kg −1 Using unit analysis. 1 m= slope 1 = 0.28 kg −1 m = 3. The acceleration-force graph is shown below. The magnitude of the slope of the line is ∆a slope = ∆ΣF = 1.6 kg.33.6 kg The mass of the cart is 3. The frequency graphs The period graphs The relationships are as follows: f ∝ FT f ∝ f ∝ 1 m 1 r T∝ 1 FT T∝ m T∝ r 224 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .0 kg m/s 2 ( ) slope = 0. it is evident that the mass must be the reciprocal of the slope. 45 m ay = 9.45 m ) 1 a y (∆t )2 − vfy ∆t + ∆y = 0 2 This is a quadratic equation with solution ∆t = ∆t = = vfy ± vfy 2 − 2a y ∆y ay 11m/s ± (11m/s) 2 − 2(9.8 m/s 2 ∆t = 0.Making Connections 35.8 m/s2.042 s. Let +y be down and let ∆t1 be the time interval for the stone to drop to the water level and ∆t2 be the time interval for the sound to travel back to the top of the well. the swimmer experiences an increasing downward velocity with a uniform acceleration of 9.8 m/s 2 )(0.30 s.0 m) vfy = 11m/s ∆y = vfy ∆t − 1 a y ( ∆t ) 2 2 = 2 ( 0.042s The time interval for the last 45 cm before reaching the mat is 0.45 m ay = 9. 36.0 m.8 m/s2 [down].40 × 102 m/s ∆ttotal = 4. The velocity then slows down rapidly with an acceleration that is upward and greater than 9. Air resistance is negligible in this short falling distance. (c) The time interval for a short distance at the top of the jump is approximately 7 times longer than the time interval for the same distance at the bottom of the jump. Let +y be down. which explains why the jumper appears to be in slow motion at the top of the jump. Starting from rest.30 s The time interval for the first 45 cm is 0. The swimmer reaches maximum velocity at the instant prior to touching the water. vfy 2 = viy 2 + 2a y ∆ytotal vfy = 2a y ∆ytotal = 2(9.8 m/s 2 )(6. 37.68 s ay = 9.8 m/s 2 −b ± b 2 − 4ac 2a ∆t = 0.8 m/s2 1 ∆y = viy ∆t + a y ∆t 2 2 1 2 ∆y = a y ∆t 2 2 ∆y ∆t = ay 9.80 m/s2 Copyright © 2003 Nelson Unit 1 Review 225 . (a) viy = 0 ∆y = 0. (b) viy = 0 ∆y = 0.45 m) 9.8 m/s2 Solve for the final velocity upon reaching the mat using ∆ytotal = 6. viy = 0 m/s vs = 3. 8 m/s2 1 2 ∆y = vfy ∆t1 + a y ( ∆t1 ) 2 2 ∆y ∆t1 = −a y = 2(0.90 m/s ) ∆t = (3. considering the sound’s motion: ∆y = vs ∆t2 ( = (3. (a) Let ∆t1 be the time to reach the maximum height.85 m ay = −9.Considering the stone’s motion: 1 ∆y = viy ∆t1 + ay ∆t12 2 1 2 ∆y = ay∆t1 2 Considering the sound’s motion: ∆y vs = ∆t 2 ∆y = vs ∆t2 Equating the expressions for ∆y: 1 a y ∆t12 = vs ∆t2 2 1 a y ∆t12 = vs ( ∆ttotal − ∆t1 ) 2 2 2 1 2 (4.40 × 103 m/s ( ∆ttotal − ∆t1 ) 3 ) m/s ) ( 4. The total time is ∆t = 2∆t1.40 s Finally. 38.68 s − 4.40 × 102 m/s ± = (3.0 m The well is 95.83s The player is in the air for 0.90 m/s ) ∆t + (3.40 × 10 = 3.59 × 103 m = 0 ) ( 4.40 ×10 2 2 1 2 m/s ∆t1 − 1. 226 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .42 s ∆t = 2∆t1 = 2(0.8 m/s 2 ∆t1 = 4.40 ×10 m/s ( 4.59 ×10 m ) 3 9.90 m/s 2 ( )(−1.68 s − ∆t1 ) ) Applying the quadratic formula: ∆t1 = −b ± b 2 − 4 ac 2a −3.85 m) − 9.40 ×10 2 m/s ) 2 − 4 4.8 m/s 2 ∆t1 = 0. vfy = 0 m/s ∆y = 0.40 s ) ∆y = 95.42 s) ∆t = 0.0 m deep.83 s. vfy 2 = viy 2 + 2a y ∆y ∆y = vfy 2 2a y vfy = 10.8 m/s2 Let +y be down.8 m/s 2 ) vfy = 5. viy = 0 m/s ay = 9.1m The practice platforms would need to have a range of heights from 1. 40.(b) Consider only the first half of the motion.78 s ) ∆x = 5. vfy 2 = viy 2 + 2a y ∆y viy = −2a y ∆y = −2(−9.1 m.1 m/s.8 m/s2 vix = 2.3 m to 5.8 m/s 2 ) ∆y = 1.0 m/s )( 2. The prevailing winds in much of Canada move from west to east.8 m/s 2 )(0. = 2 (38 m ) Copyright © 2003 Nelson Unit 1 Review 227 .0 m/s Let +y be down and +x be outward from the top of the falls. 41.6 m away from the falls.3m ∆y = 5. viy = 0 ∆y = 38 m ay = 9. so airplanes travelling westward have a headwind that slows them down relative to the ground.85 m) viy = 4.0 m/s) 2 2(9. Considering the vertical component of the motion: 1 ∆y = viy ∆t + a y ∆t 2 2 1 2 ∆y = a y ∆t 2 2 ∆y ∆t = ay 9. while airplanes travelling eastward have a tailwind that helps them increase their speed relative to the ground.0 m/s)2 2(9.1m/s The player’s vertical take-off speed is 4. 39.6 m The walkway should be built up to 5.0 m/s ∆y = (10.8 m/s 2 ∆t = 2.8 s Considering the horizontal component of the motion: ∆x = vix ∆t = ( 2.0 m/s ∆y = (5. ∆t. demonstrating a venturi flow meter (page 107.6 × 104 h The speed of the satellite is 2. 46. (c) Because of Earth’s curvature. (a) Converting the units: m   1 km   3600 s   v =  7. The pressure difference causes the observed force on the curtain.8 m/s [N] vf = 0 m/s a = 4. So the air pressure beyond the shower curtain is greater than the air pressure inside the shower. 45. (b) Earth takes 24 h to rotate once on its axis. to take a similar photograph people in the Southern Hemisphere would have to aim their cameras toward the southern sky aligned with Earth’s axis. and demonstrating the way in which lift can be created in an airplane wing (page 118.0 h  ∆t = (144° )    15°  ∆t = 9. rather it should be designed to offer maximum air resistance for its size and mass. distant objects in the sky appear to revolve around the North Star. Thus. the stars and planets are in motion relative to Earth. (a) The principle involved is Bernoulli’s principle. vi = 75. Examples from the text are throwing a curve ball (page 105).6 h. (a) Earth’s axis of rotation is aligned with the North Star.  1. 228 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .80 m/s 2 ) ∆d = 45. blowing air between two empty pop cans causing them to move toward each other (page 105).0 m – ∆d = 2.8 m/s ∆t = 0. 43.80 m/s2 [S] Solve for the distance ∆d needed to stop: vf 2 = vi 2 + 2a∆d ∆d = vi 2 2a (20. so the rate of rotation is 360°/24 h. question 7).13 s. the air pressure is reduced. so as Earth rotates. the kite should not be streamlined. If we choose an obvious arc and measure the angle it subtends in the photo.0 km/h [N] = 20. The reaction time is ∆d r ∆t = vi = 2.6 h The exposure time is approximately 9.2 ×103    s   1000 m   1 h   km v = 2. (b) Answers may vary. 44. question 22).8 m 20. 42. In the region where the air is moving quickly because it is being dragged downward by the water.e.8 m/s)2 = 2(4. However. Since Earth is in motion (i.. rotating) relative to the stars and planets. people in Australia and other parts of the Southern Hemisphere are unable to view the North Star.13s The reaction time is 0.8 m.2 m The reaction distance ∆dr is the distance travelled at constant speed vi before applying the brakes. ∆dr = 49. we can estimate the exposure time. or 15°/h. The kite must “catch” the wind in order for the force on the surfer to be sufficiently large.6 × 104 km/h. in the order of a year in each direction. with a simulation of what might happen if a tire blew out. 49.1 m ) 2 v = 4. terminal speeds. and other body components deteriorate if they do not have the resistance forces common in a gravitational field. and artificial satellites. The passengers would get knocked around more. banking angles. and environmental conditions at a specific track. even at the top of the loop. Copyright © 2003 Nelson Unit 1 Review 229 . (b) The most common way of creating artificial gravity on a mission to Mars will likely be in a rotating space station.7 × 106 m − 6. Answers will vary. Newton’s laws of motion. Students might think of applying the physics principles related to the acceleration due to gravity here and on other solar system bodies. and the likelihood of objects falling out of pockets would increase.3 × 103 m or 1. checking agriculture production. 48. These objects could become dangerous to other passengers. 47. Some practical examples are checking weather and climate conditions such as moisture content and water temperatures. One way to achieve this is to make the loop lower in maximum height so the coaster does not lose as much of its kinetic energy to gravitational potential energy. (c) Remote sensing is a means of monitoring certain variables from a distance. ∆Fy = mac mv 2 r mv 2 mg = r 2 v g= r v = gr Fg = = (9.38 × 106 m = 1.8 m/s2 Let +y be down. bones. Human muscles.67 ×10 = −11 N ⋅ m 2 / kg 2 5.2 ×10 3 )( m/s ) ) 2 r = 7.7 × 106 m The distance above Earth’s surface is 7. Artificial gravity helps to overcome the negative effects of space travel in a low gravitational field.5 m/s The speed of the model coaster at the top of the loop is 4.5 m/s. tracking pollution. frictional forces. (a) r = 2. Bernoulli’s principle. banked curves. roller coasters. and researching potential areas for mining.98 × 1024 kg (7.1 m g = 9.(b) mE = 5. observing forest fires.3 × 103 km.8 m/s ) (2. The coaster design should be changed so the passengers would feel a normal force.98 × 1024 kg GmE r GmE v2 = r Gm r = 2E v v= (6. Astronauts and objects along the inside of the outer circumference of the rotating station would experience a normal force that would depend on the radius of the station and the rate of rotation. projectile motion. dimensions. noninertial frames of reference. (b) It is dangerous to have a zero normal force on the passengers in this situation. One specific example could be a realistic speed car race created using images. (a) A trip to Mars would take a long time. relative velocity. 4 m below.1 m 4. with +y up: ay = −g = −9. Considering the horizontal component of the motion. = ( ) 230 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson .8 m/s Let +y be up and +x be toward the basket.5 m/s ∆t = 1.8 m/s2 ∆t = 1. which is constant velocity: vx = vo cos θ = 7.4 m Since the basket is 1.1 m vo = 7.36 s ) 2 ∆y = −0. ∆x = 6.36 s ) + −9.8 m/s 2 (1. the ball misses the basket by 1.2 m above the starting position and the final position is 0.Extension 50.6 m.5 m/s The time interval for the motion is: ∆x ∆t = vx 6.8 m/s (sin 55° )(1.36 s 1 ∆y = viy ∆t + a y ∆t 2 2 1 2 = 7.36 s Considering the vertical component of the motion.8 m/s ( cos 55° ) vx = 4.