Problem Sets: SolutionsJ.P. McCarthy February 4, 2010 1 Introduction to Applied Mathematics 1.1 Algebra 1.1.1 Problem Solve the simultaneous equations x − y = 0, (x + 2) + y 2 = 10. 2 Solution: Let x − y = 0 (A) (x + 2) + y 2 = 10 (B) 2 (A) ⇒ x = y. (B) : (x + 2)2 + x2 = 10 ⇒ x2 + 4x + 4 + x2 − 6 = 0 ⇒ 2x2 + 4x − 6 = 0 ⇒ x2 + 2x − 3 = 0 ⇒ (x + 3)(x − 1) = 0 ⇒ x = −3, or 1 ⇒ Sol. Set = {(1, 1), (−3, −3)}. 1 LC Applied Maths Problem Set Solutions 1.1.2 2 Problem Show that the following simplifies to a constant when x ̸= 2 3x − 5 1 + x−2 2−x Solution: 3x − 5 1 + , x−2 2−x 3x − 5 1 = − , x−2 x−2 3x − 5 − 1 3x − 6 (x − 2) = = =3 = 3. x−2 x−2 x−2 1.1.3 Problem √ √ −1 + 3 √ =2− 3 1+ 3 Show that Solution: √ √ −1 + 3 1 − 3 √ × √ 1+ 3 1− 3 √ √ −1 + 3 + 3 − 3 = 1−3 √ √ 2 3−4 = 2 − 3. = −2 1.1.4 Problem x2 − px + q is a factor of x3 + 3px2 + 3qx + r. (i) Show that q = −2p2 . (ii) Show that r = −8p3 . (iii) Find the three roots of x3 + 3px2 + 3qx + r = 0 in terms of p. Solution: A cubic function f (x) = x3 + bx2 + cx + d has factors (x − α)(x − β)(x − γ) where α, β and γ are the roots of f (x). Similarly a quadratic function g(x) = x2 − px + q has factors g(x) = (x − α1 )(x − α2 ) where α1 and α2 are the roots of g(x). If a quadratic g(x) is a factor of a cubic f (x) then the roots of g are roots of f and f (x) = g(x).(x − α3 ) where α3 is the third root of f (x). Hence let f (x) = x3 + 3px2 + 3qx + r and g(x) = x2 − px + q. f (x) = (x2 − px + q)(x − α3 ) ⇒ f (x) = x3 − px2 + qx − α3 x2 + α3 px − α3 q ⇒ f (x) = x3 + (−α3 − p)x2 + (q + α3 p) + (−α3 q) LC Applied Maths Problem Set Solutions 3 (i) Equating the x2 -coefficients: 3p = −α3 − p ⇒ α3 = −4p. Equating the x-coefficients: 3q = q + (−4p)p ⇒ 2q = −4p2 ⇒ q = −2p2 . (ii) Equating the constant coefficient: r = −α3 q ⇒ r = −(−4p)(−2p2 ) ⇒ r = −8p3 . (iii) α3 = −4p is one root. The other roots of f (x) = g(x)(x − α3 ) are the roots of g(x); ! g(x) = x2 − px − 2p2 = 0 ⇒ x2 − 2px + px − 2p2 = 0 ⇒ x(x − 2p) + p(x − 2p) = 0 ⇒ (x − 2p)(x + p) Hence the set of roots is {x : f (x) = 0} = {−4p, 2p, −p}. 1.2 Vectors 1.2.1 Problem s = 4i + 3j and t = 2i − 5j. Find |st| Solution: st = t − s ⇒ st = (2i − 5j) − (4i + 3j) = −2i − 8j. Where v = xi + yj; |v| = ⇒ |st| = √ √ x2 + y 2 . (−2)2 + (−8)2 = √ 68 = √ (1) √ 4(17) = 2 17. find the measure of the angle between a and a + b correct to the nearest degree. where k ∈ Z.2 4 Problem a = 2i + (2k + 3)j and b = k 2 i + 6j. ⇒ a·b=0 a⊥b ! ⇒ 2k 2 + 6(2k + 3) = 0 ⇒ 2k 2 + 12k + 18 = 0 ⇒ k 2 + 6k + 9 = 0 ⇒ (k + 3)2 = 0 ⇒ k = −3 (ii) ⇒ a = 2i − 3j k=−3 b = 9i + 6j ⇒ a + b = 11i + 3j (iii) By the properties of the Dot Product: a · (a + b) = a · a+a ·b |{z} |{z} =|a|2 =0 Also. a · (a + b) = |a||a + b| cos θ ⇒ |a|2 = |a||a + b| cos θ |a|2 |a| ⇒ cos θ = = |a||a + b| |a + b| √ √ 4+9 13 1 ⇒ cos θ = √ =√ =√ 121 + 9 130 10 √ −1 ◦ ⇒ θ = cos (1/ 10) = 71. where θ is the angle between a and b. write a + b in terms of i and j. (2) . a is perpendicular to b. (i) Find the value of k. (iii) Hence.5651 ≈ 72◦ . Solution: (i) a ⊥ b ⇔ a · b = 0.2.LC Applied Maths Problem Set Solutions 1. (ii) Using your value for k. sr · st = |sr||st| cos θ sr · st ⇒ cos θ = |sr||st| 21 + 4 ⇒ cos θ = √ sr=−rs 5 50 √ √ 25 1 5 25 1 ⇒ cos θ = √ = √ = √ = =√ 2 5 50 50 50 2 ◦ ⇒ θ = 45 . st and tr in terms of i and j.2. s = −4i − 2j and t = 3i − j. Solution: (i) rs = s − r = (−4i − 2j) + i − 2j −3i − 4j.3 Problem rst is a triangle where r = −i + 2j. ⇒ ∆rst right-angled at r (iii) Where θ := ∠rst. (ii) Show that the triangle rst is right-angled at r (iii) Find the measure of ∠rst. 5 .LC Applied Maths Problem Set Solutions 1. st = ts = 3i − j + 4i + 2j = 7i + j tr = r − t = −i + 2j − 3i + j = −4i + 3j (ii) For ∆rst to be right-angled at r: rs ⊥ tr: rs ⊥ tr ⇔ rs · tr = 0. (i) Express rs. rs · tr = (−3i − 4j) · (−4i + 3j) = 12 − 12 = 0. Hence B ≡ ax + 6y + 21 = 0 21 a ⇒B ≡y =− x− . (4) When a line is written in the form: L ≡ y = bx + c. x2 − x1 (3) 10 + 2 6 =− . −2). 1.1 6 Coordinate Geometry Problem The line B contains the points (6. Find the equation of the line perpendicular to L that contains the point (3. Find the value of the real number a. y1 ) and (x2 .3. The line A with equation ax + 6y + 21 = 0 is perpendicular to B. L ⊥ K ⇔ mL . y1 ) of slope m is given by: y − y1 = m(x − x1 ). Solution: L ≡ 14x + 6y + 1 = 0 14 1 ⇒L≡y =− x− 6 6 Hence mL = −7/3. y2 ) are two points on a line L. 6 6 A⊥B 6 ( a) ⇒− − = −1 5 6 ⇒ a = −5. (5) then b = mL .3. 7 7 (6) . −2) ∈ K. The equation of a line A containing a point (x1 .LC Applied Maths Problem Set Solutions 1.3 1. Solution: If (x1 .2 Problem The equation of the line L is 14x + 6y + 1 = 0. −4 − 6 5 For two lines L and K. 10). −2) and (−4. Let K ⊥ L and (3. 3 ⇒ K ≡ y + 2 = (x − 3) 7 3 9 3 9 14 ⇒K ≡y = x− −2= − − 7 7 7 7 7 3 23 ⇒K ≡y = x− . K ⊥ L ⇒ mK = 3/7.mK = −1. then the slope is given by: mL = ⇒ mB = y2 − y1 . 3. 2). y1 ). −6) and q has coordinates (−3. Determine the ratio in which the line 2x − 3y + 1 = 0 divides [pq].4 ( 13 2 ) ) 13 . 0 . Solution: The mid-point of [pq] where p = (x1 . q = (x2 . −4 . . 2 − 8(−4) − 71 = 39 + 32 − 71 = 0. Suppose a = (x1 . −3 − 7 5 4 ⇒ pq ≡ y − 2 = − (x + 3) 5 ⇒ pq ≡ 5y − 10 = −4x − 12 ⇒ pq ≡ 4x + 5y = −2. Find the point of intersection of pq and the line 2x − 3y + 1 = 0. −4 ∈ L ≡ 6x − 8y − 71 = 0. 0). ( ) 13 ⇒ . y2 ) is given by: ( ) x1 + x2 y 1 + y 2 . (7) 2 2 Hence the mid-point of [ab]: ( ⇒6 1. .LC Applied Maths Problem Set Solutions 1. y1 ). y2 ). Then the point that divides [a. Solution: 2+6 4 mpq = =− . b = (x2 . (8) m+n m+n .3 7 Problem Show that the line 6x − 8y − 71 = 0 contains the midpoint of [ab] where a has coordinates (8.3. To find the intersection between this line and the line 2x − 3y + 1 = 0 is the solution of the simultaneous equations: 4x + 5y − 2 (A) 2x − 3y = −1 (B) ⇒ 2x = 3y − 1 (B) ⇒ 6y − 2 + 5y = −2 (A) ⇒ 11y = 0 ⇒ y = 0 1 ⇒ 2x = −1 ⇒ x = − )2 ( 1 point of intersection = − . b] in the ratio s : t is given by: ( ) mx2 + nx1 my2 + ny1 . −6) and b has coordinates (5. −2). 2 Problem Find the equation of the line pq where p has coordinates (7. 2 Suppose 2x−3y +1 = 0 divides [pq] in the ratio m : n at (−1/2. 1 Problem Find the value of θ for which √ cos θ = − Solution: In the first instance cos 30◦ = 3 .LC Applied Maths Problem Set Solutions Thence ( 8 ) ( ) 1 m(−3) + n(y) m(2) − 6n ! − . 1. 2 m+n m+n ! ⇒ 2m − 6n = 0 ⇒ m = 3n 3n m ⇒ =m:n= = 3 : 1.4. Express tan B in the form a/b. find tan 2A without evaluating A.4. where A is an acute angle. 3 The addition formula for tan: tan(A + B) = tan A + tan B 1 − tan A tan B (10) . b ∈ N. cos is negative in the second quadrant: cos(180◦ − θ) = cos(180◦ ) cos(θ) + sin(180◦ ) cos θ | {z } | {z } =−1 =0 ◦ ⇒ cos(180 − θ) = − cos θ ⇒ θ = 180◦ − 30◦ = 150◦ . 2 √ 3/2.2 Problem If tan A = 1/2. 0◦ ≤ θ ≤ 180◦ . where a. given that tan(2A + B) = 63 . 16 Solution: The double-angle formula for tan: tan 2A = tan 2A = 2 1− 2 tan A .0 = .4 Trigonometry 1. n n 1. 1 − tan2 A 1 4 = 1 3 4 (9) 4 = . 300 12 ⇒ tan(2A + B) = 1. p ∈ R. where k.LC Applied Maths Problem Set Solutions 9 tan 2A + tan B ! 63 = 1 − tan 2A tan B 16 4 + tan B 63 ⇒ = 3 4 16 1 − 3 tan B ( ) ( ) 4 4 ⇒ 16 + tan B = 63 1 − tan B 3 3 ( ) ( ) 4 4 ⇒ 3(16) + tan B = 3(63) 1 − tan B 3 3 ⇒ 64 + 48 tan B = 189 − 252 tan B ⇒ 300 tan B = 125 125 5 ⇒ tan B = = .4. Find the values of A for which sin(135◦ − A) cos(135◦ + A) = − 3 4 where 0◦ ≤ A ≤ 180◦ . Solution: The subtraction formula for sin: sin(A − B) = sin A cos B − cos A sin B (11) sin(135◦ − A) = sin 135◦ cos A − cos 135◦ sin A 1 1 1 ⇒ sin(135◦ − A) = √ cos A + √ sin A = √ (cos A + sin A) 2 2 2 The addition formula for cos: cos(A + B) = cos A cos B − sin A sin B (12) cos(135◦ + A) = cos 135◦ cos A − sin 135◦ sin A 1 1 1 ⇒ cos(135◦ + A) = − √ cos A − √ sin A = − √ (sin A + cos A) 2 2 2 1 ⇒ sin(135◦ − A) cos(135◦ + A) = − (cos A + sin A)2 2 1 ◦ ◦ 2 2 ⇒ sin(135 − A) cos(135 + A) = − (cos A {z + sin A} + 2| sin A cos A})2 {z | 2 =1 =sin 2A 1 ⇒ sin(135◦ − A) cos(135◦ + A) = − (1 + sin 2A) 2 .3 Problem Express sin(135◦ − A) in terms of sin A and cos A. Express sin(135◦ − A) cos(135◦ + A) in the form k(1 + sin pA). 2.1 Accelerated Linear Motion Problem A lift decelerates from 3 m s−1 to rest during the last 6 m of its motion.1 Solution For this motion s=6 t =? u=3 v=0 a =? Using v 2 = u2 + 2as. v 2 − u2 2s 0 − 32 ⇒a= 12 3 ⇒ a = − m/s2 4 ⇒a= Using t= ⇒t= v−u a 0−3 4 3 = × =4s −3/4 4 3/4 Ans: Deceleration = 3/4 m/s2 and time taken = 4 s. .1.LC Applied Maths Problem Set Solutions 10 3 1 sin(135◦ − A) cos(135◦ + A) = − = − (1 + sin 2A) 4 2 3 1 ⇒ (1 + sin 2A) = 2 4 3 ⇒ 1 + sin 2A = 2 1 ⇒ sin 2A = ⇒ 2A = 30◦ 2 ⇒ A = 15◦ 2 2. Find the deceleration and the time taken. If the train continues to decelerate at the same uniform rate.LC Applied Maths Problem Set Solutions 2.2 11 Problem A train slows down from 70 m s−1 to 50 m s−1 over an eight-second time interval.2. how much further will it travel before it comes to rest? 2. Find the deceleration and the distance covered.1 Solution In the eight-second interval: s =? t=8 u = 70 v = 50 a =? Using a= ⇒a= v−u t 50 − 70 −20 5 = = − m/s2 8 8 2 Using ( ) u+v s= t 2 ( ) 70 + 50 ⇒s= 8 2 ⇒ s = 60 × 8 = 480 m Examining now the motion as the train decelerates from 70 m/s to rest: s =? t =? u = 70 v=0 5 a=− 2 Hence using: v 2 = u2 + 2as v 2 − u2 ⇒s= 2a 4900 0 − 4900 = = 980 m ⇒s= 2(−5/2) 5 . Find in metres/second2 the deceleration.1 Solution (a) km hr km 1000 m 1000 m ⇒ 72 = 72 = 72 hr 60 mins 60(60 s) km 72(1000) ⇒ 72 = = 20 m/s hr 3600 72 (b) In the first instance units must be converted to SI units: 72 km/hr = 20 m/s 2 40 48 km/hr = 72 km/hr = m/s 3 3 1 km = 500 m 2 Hence over the 500 m: s = 500 t =? u = 20 40 v= 3 a =? .LC Applied Maths Problem Set Solutions 12 However ∴ (distance travelled from 50 m/s to rest) = (distance travelled from 70 m/s to rest) − (distance travelled from 70 m/s to 50 m/s) ⇒ (distance travelled from 50 m/s to rest) = 980 − 480 = 500 m. and the time taken. 2. 2.3. If the train continues to decelerate at this rate find out how much further it will travel before it comes to rest.3 Problem (a) Convert 72 km/hour to metres per second (b) A train decelerates from 72 km/hour to 48 km/hour over a distance of 1/2 km. Ans: 500 m. s= ⇒s= v 2 − u2 2a 0 − 400 9 400 3600 = × = = 900 m −(4/9) 9 (4/9) 4 However ∴ (distance travelled from 48 km/hr to rest) = (distance travelled from 72 km/hr to rest) − (distance travelled from 72 km/hr to 48 km/hr) ⇒ (distance travelled from 48 km/hr to rest) = 900 − 500 = 400 m. v 2 − u2 2s 2 (40 /9) − 400 ⇒a= 1000 9 (402 /9) − 400 402 − 3600 −2000 ⇒a= × = = 9 1000 9000 9000 2 ⇒ a = − m/s2 9 a= Using t= ⇒t= v−u a 9 (40/3) − 20 120 − 180 (40/3) − 20 = × = = 30 s −(2/9) 9 −(2/9) −2 Examining now the motion as the train decelerates from 72 km/hr to rest: s =? t =? u = 20 v=0 2 a=− 9 Using v 2 = u2 + 2as.LC Applied Maths Problem Set Solutions 13 Hence using v 2 = u2 + 2as. . Ans: 500 m. 2s 70 70 2 Examining the motion from 15 m/s to rest: s =? t =? u = 15 v=0 a = −5/2 Using v 2 = u2 + 2as. s= ⇒s= v 2 − u2 2a 0 − 225 225 = = 45 m. The speed of a car is reduced from 72 km/hour to 54 km/hour over a distance of 35 m. assuming it is uniform throughout.LC Applied Maths Problem Set Solutions 2. ( ) 5 72 km/hr = 72 m/s = 20 m/s 18 3 ⇒ 54 km/hr = (72 km/hr) = 15 m/s 4 Examining the motion over the 35 m: s = 35 t =? u = 20 v = 15 a =? Using v 2 = u2 + 2as. 3600 18 1 km/hr = 1 First convert to SI units.4. Find the retardation. a= v 2 − u2 225 − 400 −175 5 = = = − m/s2 . how much farther will the car travel before coming to rest? 2.1 Solution Velocity is speed in a given direction Speed is the rate of change of distance with respect to time 1000 m 1000 m =1 60 mins 60(60) s 1000 5 ⇒ 1 km/hr = m/s = m/s. Show that a speed of 1 km/hour is equivalent to 5/18 m/s. If the retardation continues.4 14 Problem: LC OL 1984 Define velocity and speed. −5 5 . 1 Solution (i) Using a= ⇒a= v−u t 27 − 0 = 3 m/s2 9 (ii) In the decelerating part of the motion. Calculate: (i) the uniform acceleration (ii) the uniform deceleration (iii) the average speed of the car for the journey (iv) the two times that the velocity of the car will be 15 m/s 2.5 m/s 13 . ⇒ ⇒ t2 = t1 d t1 = a t2 a 3 =9 =4s d 6. v = 0. The graph is a triangle of height 27 and base 13: 1 total distance = (13)(27) = 175. The distance travelled is the area under the graph.5 ⇒ v¯ = = 13.5. Using v 2 = u2 + 2as v 2 − u2 ⇒a= 2s 0 − 729 ⇒ a= = −6.75 m/s2 2 27 =729 108 (iii) The average speed is given by: v¯ = distance travelled time taken (13) The motion is acceleration followed immediately by deceleration hence d : a = t1 : t2 (14) where t1 is the time accelerating.LC Applied Maths Problem Set Solutions 2. t2 the time decelerating.75 Therefore the time taken is 9+4=13 s. u = 27. The brakes are then applied and it comes to rest with uniform deceleration after travelling a further 54 m.5 15 Problem: LC OL 1983 A car starts from rest with a uniform acceleration and reaches a velocity of 27 m/s in 9 s.5 m 2 175. A car travelling towards p at a steady speed of 15 m/s. q and r. To find the first time. If |pr| = 980 m and the time from p to q was 40 seconds. By the theorem. p. or otherwise.LC Applied Maths Problem Set Solutions 16 (iv) Clearly the velocity will be 15 m/s once when t < 9 and once when 9 < t < 13. 4 2 . accelerated at a constant rate between p and q. Let acceleration from 15 m/s to 25 m/s take T s. This speed was maintained as far as r. calculate the acceleration. the area under the graph must equal to the distance travelled. Using. using t= v−u 15 = =5s a 3 To find the second time.1 Solution To draw the time-velocity graph note it has the rough shape of Fig 1. v = 15. draw a time-velocity graph of the motion and hence. Hence: Ans: 5 s and 10 97 s. along the line in that order.6 Problem: LC OL 1982 Consider three points on a line. 2. u = 27.75: t= 15 − 27 7 v−u = =1 s a 6. 2.75 9 But this represents the time after t = 9 s. for the time t = 0 to t = T ( ) ( ) u+v 15 + 25 s= t= T = 20T 2 2 ∴ 980 = 20T + 25(40 − T ) ⇒ 980 = 20T + 1000 − 25T ⇒ 5T = 20 ⇒T =4s Therefore the time-velocity graph is Fig 2. Using a= ⇒a= v−u t 25 − 15 5 = m/s2 . At q its speed was 25 m/s. consider the decelerating part of the motion.6. a = −6. A particle starts from rest with uniform acceleration 2 m/s2 . Calculate the deceleration.7 17 Problem: LC OL 1981 Define uniform acceleration in a straight line. acceleration is the rate of change of acceleration with respect to time. After how many seconds will its speed be 30 km/hr? How far from its starting point will the particle be when its speed is 60 km/hr? The particle is then brought to rest in 2 m.v=0 4 9 625 m/s2 ⇒d= 9 .1 Solution Uniform acceleration in a straight line is motion in a single direction with constant acceleration.LC Applied Maths Problem Set Solutions 2. 30 km 1000 25 = 30 m/s = m/s hr 3600 3 Using t= v−u 25/3 25 = = s a 2 6 60 km/hr = 50 m/s 3 Using v 2 = u2 + 2as v 2 − u2 ⇒s= 2a (50/3)2 2500/9 625 ⇒s= = = 4 4 9 625 ⇒s= m 9 Again using v 2 = u2 + 2as v 2 − u2 ⇒a= 2s −2500/9 625 ⇒ a= =− m/s2 s=2. 2.7. 8 18 Problem: LC OL 1980 p and q are points 162 m apart. At the same instant another body leaves q and travels towards p with initial speed 7 m/s and uniform acceleration 2 m/s2 . After how many seconds do they meet and what.1 Solution The particles meet after T s when the sum of the distance travelled by the particle at p-particle and the distance travelled by the q-particle is 162 m: particles meet ⇔ sp + sq = 162 Using 1 s = ut + at2 2 3 ⇒ sp = 5T + T 2 2 ⇒ sq = 7T + T 2 1 2 2=1 Hence solve for1 T sp + sq = 162 5 ⇒ 12T + T 2 = 162 2 ⇒ 24T + 5T 2 = 324 ⇒ 5T 2 + 24T − 324 = 0 ⇒ 5T 2 + 54T − 30T − 324 = 0 ⇒ T (5T + 54) − 6(5T + 54) = 0 ⇒ (T − 6) (5T + 54) = 0 | {z } see footnote ⇒T =6s The velocities of the p and q-particles after t s.LC Applied Maths Problem Set Solutions 2.8. then. refers to time when distance between them was 324 m (15) . using: v = u + at ⇒ vp = 5 + 3t = 23 m/s at t = 6 ⇒ vq = 7 + 2t = 19 m/s at t = 6 1 ignore t < 0. A body leaves p with initial speed 5 m/s and travels toward q with uniform acceleration 3 m/s2 . is the speed of each body? 2. and 1 s2 = (0)(2) + g(4) = 2g 2 But after 2 s. find the value of f . After 2 s. – The average speed is given by: d total distance = total time T where T = t1 + t2 is the total time. using 1 s = ut + at2 2 1 s1 = 4g(2) + (−g)(4) 2 ⇒ s1 = 6g . • – See Fig 1.LC Applied Maths Problem Set Solutions 2. Hence as s1 = 3s2 . At the same instant a second particle is let fall vertically from a point q directly above point p. 2. using v = u + at v1 = u − 2g v2 = 2g ⇒ u − 2g = 2g ⇒ u = 4g Let s1 and s2 be the distance travelled by the particles. Hence √ d d = 3 √T √ 3 = 3d ⇒T =d d v¯ = (16) (17) . – Draw a speed-time graph for the motion of the train √ – If the average speed for the whole journey is d/3.9 19 Problem: LC HL 2009 • A particle is projected vertically upwards from a point p. After 2s. |pr| = 3|rq|. The particles meet at a point r between them after 2s The particles have equal speeds when they meet at r Prove that |pr| = 3|rq| • A train accelerates uniformly from rest to a speed v m/s with uniform acceleration f m/s2 . s1 = |pr| and s2 = |rq|. It then declerates uniformly to rest with uniform retardation 2f m/s2 .9. The total distance travelled is d metres.1 Solution • Let v1 be the speed of the particle projected from p and v2 the speed of the particle dropped from q. 2 m/s. 3d 23 1 ⇒ d = f 3d = f d 3 ⇒ f = 1 m/s2 2. Find – the time taken to reach the maximum height – the distance travelled in 5 s .LC Applied Maths Problem Set Solutions 20 Also the total distance is equal to the area under the time-velocity curve.10 Problem: LC HL 2008 • A ball is thrown vertically upwards with an initial velocity of 39. in this case the triangle of width T and height v: 1 d = vT 2 (18) Since the motion is acceleration from rest immediately followed by declaration to rest: d : a = t1 : t2 (19) Hence 2f : f = t1 : t2 ⇒ t1 : t2 = 2 : 1 2 1 ⇒ t1 : t2 = : 3 3 2 ⇒ t1 = T 3 Using v = u + at v = 0 + f t1 2 ⇒ v = fT 3 Hence using this and (17) in (18): 1 d = vT 2 12 √ √ ⇒d= f 3d. Two minutes later Q passes P . and Q is then moving at 65.LC Applied Maths Problem Set Solutions 21 • Two particles P and Q. using 1 s = ut + at2 2 1 ⇒ s = 39.5 m/s.10. are moving in the same direction along parallel lines. Using v = u + at v = 39. Find – the acceleration of P and the acceleration of Q – the speed of P when Q overtakes it – the distance P is ahead of Q when they are moving with equal speeds 2.2 − gt 39.1 • Solution – The ball reaches the maximum when v = 0.5 uQ = 5.5 − 5.5g = g 2 2 • – Consider the motion of Q.2(5) − g(25) 2 15 25 ⇒ s = 4g(5) − g = 20g − 12.5 1 = m/s2 120 2 . each having constant acceleration.5 aQ =? sQ =? Using a= aQ = v−u t 65. respectively. t = 120 (120 s = 2 min) vQ = 65.2 ⇒ ts=max = =4s g – After 5 s.5 m/s. When P passes Q the speeds are 23 m/s and 5. 5) ⇒t= = 60 s 7 23 + Hence look at sP and sQ after 60 s. using v = u + at 5 vP = 23 + t 24 1 vQ = 5.5(120) − 23(120) + (120)2 2 4 ⇒ 240aP = 22 − 92 + 120 = 50 5 ⇒ aP = m/s2 25 – Using v = ua + at ⇒ vP = 23 + 5 (120) = 48 m/s 24 – After t s. That is P is 525 m ahead of Q when they are moving with equal speed. .5(60) + (60)2 = 1230 4 Hence after 60 s.LC Applied Maths Problem Set Solutions 22 ! After 120 s. using 1 s = ut + at2 2 1 5 sP = 23(60) + (602 ) = 1755 2 24 1 sQ = 5.5(120) + (120)2 4 ⇒ sP = sQ 1 1 ⇒ aP (120)2 = 5. Using 1 s = ut + at2 2 1 sP = 23(120) + aP (120)2 2 1 sQ = 5.5 24 (24)(17. sP > sQ by 525 m.5 + t 24 2 7 ⇒ t = 17.5 + t 2 For what t is vP = vQ ? 5 1 t = 5. sP = sQ . 9 m. using ( ) u+v s= t 2 ( ) 2u + 5g ⇒ 29. After 2 s and 3 s. In travelling a total distance d metres the train accelerates through a distance pd metres and decelerates through a distance qd metres.11.9 = 2 ⇒ 59. 2.4)4 + g(16) 2 ⇒ s = 21. Find – the value of u – the height of the tower • A train accelerates uniformly from rest with a speed v m/s.6 + 8g = 100 m . It takes the particle 4 s to reach the bottom of the tower.1 • Solution – The particle moves under acceleration a = g.4 m/s ⇒u= 2 – The height of the tower is s after 4 s.8 = 2u + 5g 59. p+q+b find the value of b. During the third second of its motion the particle travels 29.9 m where u = v(2) and v = v(3). Using 1 s = ut + at2 2 1 ⇒ s = (5.8 − 5g = 5.11 23 Problem: LC HL 2007 • A particle is projected vertically downwards from the top of a tower with speed u m/s. using v = u + at v(2) = u + 2g v(3) = u + 3g From t = 2 to t = 3. where p < 1 and q < 1. the particle travels 29.LC Applied Maths Problem Set Solutions 2. – Draw a speed-time graph for the motion of the train – If the average speed of the train for the whole journey is v . It continues at this speed for a period of time and then decelerates uniformly to rest. the area of the triangle with perpendicular height v and base t1 is: 1 t1 v = pd 2 2pd ⇒ t1 = v Similarly t2 = 2qd v Let tc be the time spent at constant speed. d .LC Applied Maths Problem Set Solutions • 24 – See figure 2 – Let T be the total time taken for the journey. Where d is the total distance travelled. From the time-velocity graph. Examining the time-velocity graph. the average speed v¯ is given by: v¯ = d T v d = p+q+b T ) ( v ⇒d=T p+q+b ⇒ pd + qd + bd = T v ⇒ bd = T v − pd − qd T v − pd − qd ⇒b= d (20) (21) Let t1 be the time spent accelerating and t2 the time spent decelerating. the distance travelled in t1 . 1 T = (2pd + 2qd + d − pd − qd) v 1 d ⇒ T = (pd + qd + d) = (p + q + 1) v v Substituting into (21): b= d(p + q + 1) − pd − qd = p + q + 1 − p − q = 1. d − pd − qd = d(1 − p − q) Hence the distance travelled in tc is the area under the curve: vtc = d(1 − p − q) d(1 − p − q) ⇒ tc = v Now T = t1 + t2 + tc . the distance travelled at constant speed is given by. each of length 79.3 m/s2 and the acceleration of Q is 0. meet at o. The total distance travelled is d and the total time taken is t. The acceleration of P is 0. – Draw a speed-time graph for the motion – Find d in terms of f and t • Two trains P and Q. when their speeds are 15 m/s and 10 m/s respectively.1 • Solution – See Figure 3 – Since the motion is uniform acceleration from rest followed immediately by uniform deceleration from rest: t1 : t2 = 3f : f 3 1 ⇒ t1 : t2 = : 4 4 3 ⇒ t1 = t 4 Using v = u + at ⇒ v = f t1 ( ) 3 3 t = ft ⇒v=f 4 4 The total distance d is the area under the graph: 1 d = vt ( 2) 1 3 ⇒d= ft t 2 4 3 ⇒ d = f t2 8 . – Hence.5 m.12 25 Problem: LC HL 2006 • A lift starts from rest. or otherwise.2 m/s2 .LC Applied Maths Problem Set Solutions 2. – Find the distance travelled by each train in t seconds.12. It then travels with uniform retardation 3f and comes to rest. calculate the value of t – How long does it take for 2/5 of the length of train Q to pass the point o? 2. moving in opposite directions along parallel lines. For the first part of its descent is travels with uniform acceleration f . It takes the trains t seconds to pass each other. √ −b ± √ b2 − 4ac 2a 1002 − 4(1)(−318) 2 √ −100 ± 11272 ⇒t= 2 ⇒ t = 3.LC Applied Maths Problem Set Solutions • 26 – Using 1 s = ut + at2 2 1 3 2 3 ⇒ sP = 15t + t = 15t + t2 2 10 20 1 11 2 ⇒ sQ = 10t + t = 10t + t2 25 10 – The trains pass each other when the distance they travel adds up to twice their length: 159 m.5 = 31. Ans: t = 3. 3 2 1 t + 10t + t2 20 10 2 2 ⇒ 300t + 3t + 200t + 2t = 3180 ⇒ 5t2 + 500t − 3180 = 0 ⇒ t2 + 100t − 636 = 0 ⇒ t2 + 106t − 6t − 636 = 0 ⇒ t(t + 106) − 6(t + 106) = 0 ⇒ (t + 106)(t − 6) = 0 ⇒t=6s 159 = 15t + The case t = −106 does not concern us.085 s.085 or − 103.8 = 10t + t2 10 ⇒ 318 = 100t + t2 ⇒ t2 + 100t − 318 = 0 Using 2 roots of ax + bx + c are x = t= Ignore t < 0.085 −100 ± (22) .8 m 5 1 ⇒ 31. – In this case t needs to be found such that: 2 sQ = 79. 5 s later car B starts to brake with a constant retardation of 3 m/s2 . Now B travels at 20 m/s for half a second before decelerating. Find (i) the distance travelled by car A before it comes to rest (ii) the minimum value of d for car B not to collide with car A 2.13. 2 . Therefore when B stops it must have travelled d + 100/3 m to just avoid a collision.LC Applied Maths Problem Set Solutions 2. In this half second it travels: 1 s = (20) = 10 m. At a certain instant car A starts to brake with a constant retardation of 6 m/s2 . −3 3 t= Now before decelerating B travels at constant speed for a half second. when at rest: u = 20 v=0 a = −3 t =? Using v−u a 0 − 20 20 ⇒t= = s.1 Solution (i) With respect to car A. With respect to B decelerating. Car A is at a distance d metres in front of car B. 0.13 27 Problem: LC HL 2005 [Part (a)] Car A and car B travel in the same direction along a horizontal straight road. Each car is travelling at a uniform speed of 20 m/s. when at rest: u = 20 v=0 a = −6 s =? Using v 2 = u2 + 2as v 2 − u2 ⇒s= 2a −400 100 ⇒s= = m −12 3 (ii) Car A has a greater deceleration than car B and also begins it deceleration before car B. therefore car A comes to rest before car B does. giving your answers correct to the nearest metre.14 Problem: LC HL 2004 [Part (a)] A ball is thrown vertically upwards with an initial velocity of 20 m/s.75 m/s. The balls collide after a further 2 seconds.1 Solution (i) Let s1 (t) be the height of the first particle and s2 (t) be the height of the second particle. One second later.75. 2. (i) Show that u = 17. For the particles to collide after 3 s: ! s1 (3) = s2 (3).14.LC Applied Maths Problem Set Solutions 28 Hence when stopped car B has travelled: ( ( ) ( )2 ) 20 3 20 s = 10 + 20 − 3 2 3 ⇒s= 230 3 and this must equal d + 100/3: 230 100 =d+ 3 3 130 ⇒d= m 3 2. (ii) Find the distance travelled by each ball before the collision. ⇒u= 2 (23) . another ball is thrown vertically upwards from the same point with an initial velocity of u m/s. 1 s1 (3) = 3(20) − g(32 ) 2 9 ⇒ s1 (3) = 60 − g 2 Particle 2 is motionless for one of these seconds and thus 1 s2 (3) = 2u − g(4) = 2u − 2g 2 9 ! ⇒ 2u − 2g = 60 − g 2 5 ⇒ 2u = 60 − g 2 60 − 5g/2 = 17. 508 m. using v = u + at ⇒ 0 = 20 − gt 20 ⇒t= ≃ 2.041 s g Hence the distance travelled up is given by. Hence to the nearest metre the first particle travels 25 m. Hence to the nearest metre the first particle travels 16 m.175 m. then falls down the total distance is the distance travelled going up plus the distance travelled going down. That is until.0356) = 0. using v = u + at 71 − gt ⇒ 0= 17.LC Applied Maths Problem Set Solutions 29 (ii) Take the distance to mean total distance in the sense that if a particle travels up. For the second particle the motion is up until v = 0.75=71/4 4 71 ⇒t= ≃ 1. using: 1 s = ut + at2 2 ( )2 1 71 ⇒s= g 2− 2 4g ⇒ s ≃ 4.9(0. stops. using: ) ( u+v s= t 2 ( ) 71 71 ⇒s= =≃ 16.811 s 4g Hence the distance travelled up is given by. using: ( s= ( ⇒s= 20 + 0 2 ) u+v 2 ) t 20 200 = ≃ 20.408 m g g The distance travelled on the way down is. using: 1 s = ut + at2 2 ( )2 1 20 ⇒s= g 3− 2 g ⇒ s ≃ 4.9(0.075 m 2(4) 4g The distance travelled on the way down is.920) = 4. For the first particle the motion is upwards until v = 0. . show that the closest he gets to the bus is 17.LC Applied Maths Problem Set Solutions 2.15. (b) A man runs at constant speed to catch a bus.5 m 2. where |pq| = |qr| = 125 m. A train passes point p with speed u m/s.1 (a) Solution (i) Considering the motion from p to q. q and r all lie on a straight line. The train is travelling with uniform retardation f m/s2 . The man just catches the bus 20 s later. using: 1 s = ut + at2 2 1 ⇒ 125 = 10u − f (100) 2 ⇒ 125 = 10u − 50f ⇒ 25 = 2u − 10f (24) Now considering the motion from p to r: 1 250 = 25u − f (625) 2 25 ⇒ 10 = u − f 2 25 ⇒ u = 10 + f 2 (25) Plugging into (24): 25 = 20 + 25f − 10f ⇒ 15 = 5f 1 ⇒f = 3 . At the instant the man is 40 m away from the bus. it begins to accelerate uniformly from rest away from him. Find s.15 30 Problem: LC HL 2003 (a) The points p. (i) Show that f = 1/3 (ii) The train comes to rest s metres after passing r. giving your answer correct to the nearest metre. The train takes 10 s to travel from p to q and 15 s to travel from q to r. (i) Find the constant speed of the man (ii) If the constant speed of the man had instead been 3 m/s. after 20 s. That is if sm is the distance travelled by the man and sb the distance travelled by the bus the condition to just catch the bus is u = vb when sm = sb + 40. a = 1/5 m/s2 .LC Applied Maths Problem Set Solutions 31 (ii) From (25). the particle comes to rest when v = 0. (b) (i) If the man just catches the bus then when his constant speed u is equal to that of the bus vb and he has travelled as far as the bus has plus the 40 m between them. sb + 40 = 25 10 . u = 10 + 25 85 = . the distance travelled by the man after t s is given by: sm = 3t As u = 4 above. using: v 2 = u2 + 2as v 2 − u2 ⇒s= 2a ⇒s= 02 − (85/6)2 ≃ 301. 1 s = ut + at2 2 ⇒ sm = 20u 1 ⇒ sb = a(400) = 200a = 10u 2 ⇒ 20u = 10u + 40 ! sm =sb +40 ⇒ u = 4 m/s (ii) If u = 3 m/s. With respect to the man. using: v = u + at ⇒ vb = 20a u ⇒ a= ! 20 vb =u Using. (26) Let a be the acceleration of the bus. (−2/3) The particle travels 250 m from p to r hence travels s = 51 m after passing r before coming to rest. after t s. 6 6 From p. the bus has travelled a distance sb + 40 away. t2 11 2 t + 40 = + 40. After 20 s.042 m. 2.5 m/s 2 . (i) Find the value of u (ii) Find the speed with which the stone hits the ground. the particle travels a distance p. the particle travels a distance r. The stone is under acceleration −g.16.LC Applied Maths Problem Set Solutions 32 Hence in terms of t. the distance between the man and bus is given by the distance travelled by the bus less the distance travelled by the man: t2 + 40 − 3t. the particle travels a distance q. During the time interval from t to 2t. (b) A particle. Using 1 s = ut + at2 2 1 ⇒ −30 = 5u − g(25) 2 25 ⇒ 5u = g − 30 2 5 ⇒ u = g − 6 = 18. (27) 10 To minimise this function differentiate with respect to t and solve equal to 0: t −3=0 5 ⇒ tmin = 15 s To show this is a min note the second derivative is 1 > 0 ⇒ t = 15 a local minimum.5 m. The stone hits the ground 5 s later. During the time interval from 2t to 3t. 5 Hence the minimum separation is: ⇒ t=15 225 + 40 − 45 = 17.1 (a) Solution (i) With respect to the point the stone was thrown with s = −30 after t = 5 s. with initial speed u. During the time interval from 0 to t.16 Problem: LC HL 2002 (a) A stone is thrown vertically upwards under gravity with a speed of u m/s from a point 30 m above the horizontal ground. (i) Show that 2q = p + r (ii) Show that the particle travels a further distance 2r − q in the time interval from 3t to 4t. moves in a straight line with constant acceleration. 10 2. (ii) Suppose the particle travels a distance s further in the next second. using 1 s = ut + at2 2 1 ⇒ p = ut + at2 2 In the first 2t seconds: In the first 3t seconds: (28) 4 p + q = 2ut + at2 2 (29) 9 p + q + r = 3ut + at2 2 (30) Now r = (30) − (29): 4 9 r = 3ut + at2 − 2ut − at2 2 2 5 ⇒ r = ut + at2 2 ⇒ p + r = 2ut + 3at2 p=(28) Now q = (29) − (28): 4 1 q = 2ut + at2 − ut − at2 2 2 3 2 ⇒ q = ut + at 2 2 ⇒ 2q = 2ut + 3at = p + r.LC Applied Maths Problem Set Solutions 33 (ii) Using v = u + at ⇒ v = 18. (b) Considering the motion in the first t seconds. Hence in the first 4t seconds: 16 p + q + r + s = 4ut + at2 2 7 ⇒ s = ut + at2 p+q+r=(30) 2 Now 3 10 2 at − ut − at2 2 2 7 2 ⇒ 2r − q = ut + at = s 2 2r − q = 2ut + .5 − g(5) ⇒ v = −30.5 m/s Hence the stone hits the ground with speed |v| = 30.5 m/s. (b) A particle is projected vertically upwards with an initial velocity of u m/s and another particle is projected vertically upwards from the same point and with the same initial velocity T seconds later. Using v 2 = u2 + 2as v 2 − u2 s= 2a 400 sa = = 200 m 2 −400 = 100 m sb = −4 Let sc be the distance travelled at constant speed v = 20 and tc be the time spent travelling at constant speed. Show that the particles (i) will meet ( T u + 2 g ) seconds from the instant of projection of the first particle (ii) will meet at a height of 4u2 − g 2 T 2 metres.17 34 Problem: LC HL 2001 (a) Points p and q lie in a straight line. 8g 2. respectively. Find the time it takes the train to go from p to q.1 Solution (a) Let sa be the distance travelled whilst accelerating and sd be the distance travelled while decelerating.LC Applied Maths Problem Set Solutions 2. coming to rest at q. It continues at this speed of 20 m/s and then decelerates at 2 m/s2 . assuming that the acceleration and deceleration remain unchanged at 1 m/s2 and 2 m/s2 .17. where |pq| = 1200 m. Starting from rest at p. a train accelerates at 1 m/s2 until it reaches the speed limit of 20 m/s. The total distance travelled is 1200 m: 1200 = 200 + 100 + 20tc 900 ⇒ tc = = 45 s 20 . Find the shortest time it takes the train to from rest at p to rest at q if there is no speed limit. LC Applied Maths Problem Set Solutions 35 To travel a distance from rest to rest in the shortest possible times implies acceleration followed by immediate deceleration such that if t1 is the time spent accelerating and t2 the time spent decelerating: t1 : t2 = d : a (31) ⇒ t1 : t2 = 2 : 1 2 1 ⇒ t1 : t2 = : 3 3 ⇒ t1 = 2T /3. and t2 = T /3 Also the maximum speed reached is given by. using v = u + at 2 ⇒v= T 3 Now distance travelled is the area under the graph: ! 1 1200 = vT ( 2) 1 2 T T ⇒ 1200 = 2 3 3(2)(1200) ⇒ T2 = = 3600 2 ⇒ T = 60 s (b) (i) If the particles meet at a time t after the first particle is emitted. then they will have equal heights at that time: ! s1 (t) = s2 (t) (32) Using 1 s = ut + at2 2 g ⇒ s1 (t) = ut − t2 2 The second particle is only in motion after a time T so in terms of t it is in motion for a time t − T : s2 (t) ≡ s2 (t − T ) g ⇒ s2 (t) = u(t − T ) − (t − T )2 2 g 2 g 2 ⇒ ut − t = ut − uT − (t − 2tT + T 2 ) ! 2 2 s1 =s2 T 2g 2 uT T 2g ⇒t= + gT 2gT ( ) T u ⇒t= + 2 g ⇒ uT = gtT − . where |pq| = 10 000 m. reaches its maximum speed 25 m/s by constant acceleration in the first 500 m and continues at this maximum speed for the rest of the journey. starting from rest and travelling from p to q on a straight level road.LC Applied Maths Problem Set Solutions 36 (ii) To find the height they meet at is to find s1 or s2 at the time t (s1 (t) = s2 (t)): ) ( ) ( )2 u T u 1 T u T + =u + − g + s1 2 g 2 g 2 2 g ( ) ( 2 ) 2 T u Tu u 1 T T u u2 ⇒ s1 + = + − g + + 2 2 g 2 g 2 4 g g ( ) 2 2 u Tu u gT T u u2 T + = + − − − ⇒ s1 2 g 2 g 8 2 2g ( ) 2 2 2 T 4T ug + 8u − g T − 4uT g − 4u2 u ⇒ s1 = + 2 g 8g ( ) T u 4u2 − g 2 T 2 ⇒ s1 + = 2 g 8g ( 2. reaches the same maximum speed by constant acceleration in the first 250 m and continues at this maximum speed for the rest of the journey. find which car is delayed and by how many seconds.18 Problem: LC HL 2000 (a) A stone projected vertically upwards with an initial speed of u m/s rises 70 m in the first t seconds and another 50 m in the next t seconds. also. A second car. (ii) If the start of one car is delayed so that they meet each other at exactly halfway between p and q. . Find the value of u. the distance travelled by each car in that time. (i) If the two cars start at the same time. starting from rest and travelling from q to p. after how many seconds do the two cars meet? Find. (b) A car. 1 = 40 s Similarly.1 37 Solution (a) Examining separately the motion in the first t seconds and in the first 2t seconds.2 = 20 s Therefore the motion of the cars in terms of t after they take off (because the cars certainly don’t meet in less than 40 s .2 2 ta. using: 1 s = ut + at2 2 1 ⇒ 70 = ut − gt2 2 ⇒ 120 = 2ut − 2gt2 ⇒ 2ut = 140 + gt2 (33) (34) (35) (33) ⇒ 120 = 140 + gt2 − 2gt2 (34) ⇒ gt2 = 20 √ 20 ⇒t= g √ 20 20 ⇒ 2u = 140 + g (35) g g √ √ √ g 100g ⇒u= (80) = 8 = 8 5g 20 20 (b) (36) (i) The cars meet when s1 (t) + s2 (t) = 10000 m (37) How long does it take car 1 to accelerate to 25 m/s? Using ( ) u+v s= t 2 ( ) 25 ⇒ 500 = ta.LC Applied Maths Problem Set Solutions 2.1 2 ⇒ ta.18. ( ) 25 250 = ta.s1 (40)+s2 (40) = 500+250+20(25) = 1250 m) is given by the distance whilst accelerating plus the distance travelled at constant speed 25 m/s for a time (t−the time spent accelerating): s1 (t) = 500 + 25(t − 40) s2 (t) = 250 + 25(t − 20) (38) (39) . Hence if car 2 is delayed by 10 s. correct to two decimal places. The particle then travels a further 55 metres at constant speed in 5t seconds.19 Problem: LC HL 1999 [Part (b)] A particle travels in a straight line with constant acceleration f for 2t seconds and covers 15 metres. (i) Draw a speed-time graph for the motion of the particle.LC Applied Maths Problem Set Solutions 38 ! Now (38) + (39) = 10000: ⇒ 750 + 25(2t − 60) = 10000 ⇒ 25(2t − 60) = 9250 ⇒ 2t − 60 = 370 ⇒ t = 215 s Also s1 (215) = 500 + 25(215 − 40) = 4875 m s2 (215) = 250 + 25(215 − 20) = 5125 m (ii) Car 1 travels 500 m in 40 s. At constant speed. using: s = vt s ⇒t= v 4750 ⇒t= = 190 s 25 So it takes car 2 20 + 190 = 210 s to travel 5000 m. a = 0. (ii) First examining the constant speed motion. (ii) Find the initial velocity of the particle in terms of t. At constant speed. 2. using: s = vt s 55 11 ⇒v= = = t 5t t . Similarly car 2 travels 250 m in 20 s. a = 0. Finally the particle is brought to rest by a constant retardation 3f . they will meet at half way. (iii) Find the total distance travelled in metres. using: s = vt s ⇒t= v 4500 ⇒t= = 180 s 25 So it takes car 1 40 + 180 = 220 s to travel 5000 m. is: d = 15 + 55 + 5. using: v 2 = u2 + 2as v 2 − u2 ⇒s= 2a 121 − 2 121 ⇒s= t = 2 ( 7 ) −6f t 6 2t2 121 ⇒s= ≃ 5.762 m 21 Hence the total distance travelled.LC Applied Maths Problem Set Solutions 39 (i) Now using: ( ) u+v s= t 2 2s ⇒u= −v t 15 11 ⇒ u= − t=2t t t 4 ⇒u= t (iii) Using v−u t 11 4 − t ⇒f = t 2t 7 7 ⇒f = t = 2 2t 2t a= Now for the decelerating motion.76 m .76 = 75. d. Similarly let t1 . moving with uniform acceleration 2b/9 m/s2 passes the same point with speed 5u m/s.4 m/s respectively.LC Applied Maths Problem Set Solutions 2.1 Remark I think Q. 2. constant speed and deceleration respectively. constant speed and deceleration respectively. s2 and s3 be the distances travelled at acceleration.2 Solution (a) Let s1 . B overtakes A when their speeds are 6. moving with uniform acceleration 3b/20 m/s2 passes a point p with speed 9u m/s.5 m/s and 5.20. Find (i) the value u and the value b (ii) the distance travelled from p until overtaking occurs. 2. 1 I’ve ever seen.1 in 1998 was particularly difficult. t2 and t3 be the time spent at acceleration.20 40 Problem: LC HL 1998 (a) A train accelerates uniformly from rest to a speed v m/s. If the average speed for the whole journey is 5v/6.20. It continues at this constant speed for a period of time and then decelerates uniformly to rest. Hence as the area under the time-velocity graph is the distance travelled: 1 s1 = vt1 2 s2 = vt2 1 s3 = vt3 2 Using: total distance total time 1 vt + vt2 + 12 vt3 5v 2 1 ⇒ = 6 t + t2 + t3 ( 1 ) 1 1 ⇒ 5(t1 + t2 + t3 ) = 6 t1 + t2 + t3 2 2 ⇒ 5t1 + 5t2 + 5t3 = 3t1 + 6t2 + 3t3 ⇒ t2 = 2t1 + 2t3 average speed = . Three seconds later car B. This is certainly the most difficult AM Q. find what fraction of the whole distance is described at constant speed. (b) Car A. vB (T ) = 6.4 m/s. where ( ) 2b vB (t − 3) = 5u + (t − 3) 9 For overtaking to occur after T seconds.16 = 81u2 + 10 ( ) 2b ! ⇒ vB2 (T ) = 6.65 = 2565u2 40×(41)+−27×(42) √ ⇒u= ⇒ u2 = 0.5 m/s and sA (T ) =: s := sB (T ) (40) Now using.01 = 1 100 1 1 1 =√ m/s = 100 10 100 (41) (42) .75 = −675u2 − 12bs ⇒ 25.4 = 81u + 2 s 20 3bs ⇒ 29. using.4 = 3240u2 + 12bs −27 × (42) = −1140. v =(u +)at 3b ⇒ vA (t) = 9u + t 20 Car B is stationary for three of these seconds: vB (t) ≡ vB (t − 3) .52 = 25u2 + 2 s 9 4bs ⇒ 42. vA (T ) = 5.LC Applied Maths Problem Set Solutions 41 Now the fraction travelled at constant speed: s2 s1 + s2 + s3 vt2 = 1 vt + vt2 + 12 vt3 2 1 2t2 = 2t2 + (t1 + t3 ) 2t2 = 1 (t2 =2t1 +2t3 ) 2t2 + t2 2 4t2 4t2 4 = = = 4t2 + t2 5t2 5 (b) (i) In terms of a t after car A starts moving. 2 v 2 = u( + 2as ) 3b ! 2 2 2 ⇒ vA (T ) = 5.25 = 25u2 + 9 40 × (41) = 1166. 16 = 81u2 + (43) .5 = 5u + (43) ⇒b=1 (ii) Note the distance from overtaking is s1 (T ) =: s := s2 (T ) Taking (41): 3bs 10 2 10 ⇒ s = (29.4.5 m 29.16 − 81(0.5 = 0.01)) 3 ⇒ s = 94.4 = 9u + Similarly 2b (T − 3) 9 2bT 6b ⇒ 6.LC Applied Maths Problem Set Solutions 42 Now note vA (T ) = 5.5 + + 9 9 2bT 6b − ⇒6= 9 9 ⇒ 54 = 2bT − 6b ⇒ 6b = 2(bT ) − 54 ⇒ 6b = 60 − 54 = 6 vB (T ) = 6. 3bT 20 3bT ⇒ 5.4 = 0.5 = 20 ⇒ bT = 30 vA (T ) = 5.16 − 81u ) 3b 10 ⇒ s = (29.9 + 20 3bT ⇒ 4. in terms of α.LC Applied Maths Problem Set Solutions 3 43 Projectiles 3. (ii) Show that the two possible directions of projection are at right angles to each other. The speed of projection is 14 10 m/s at an angle α to the horizontal.2= 4 2 (44) .1 Problem: LC HL 2009 [Part (a)] A straight vertical cliff is 200 m high. 3. (i) Find.1.1 Solution (i) Let T be the time when sx = 200 and sy = −200. The particle strikes the level ground at a distance 200 m from the foot of the cliff. using: 1 sy (t) = uy t − gt2 2 ) ( ) ( √ 10 √ 1000 1 ⇒ −200 = 14 10 sin α 10 sec α − g sec2 α 7 2 49 Now sin α sec α = tan α and sec2 α ≡ 1 + tan2 α g ⇒ −200 = 200 tan α − (1000(1 + tan2 α)) 98 1 ⇒ −200 = 200 tan α − (1000(1 + tan2 α)) 10 ⇒ −200 = 200 tan α − 100(1 + tan2 α) ⇒ −2 = 2 tan α − (1 + tan2 α) Let u := tan α: −2 = 2u − 1 − u2 ⇒ u2 − 2u − 1 = 0 Using the formula for the roots of a quadratic: √ √ 2± 4+4 2± 8 u= = 2 √ 2 √ 2±2 2 =1± 2 √ √ ⇒ √ √ tan α = 2 8= 4. Using s (t) = ux t √x ⇒ 200 = 14 (10 cos αT √ ) 10 200 ⇒ T = √ sec α × √ 14 10 10 10 √ ⇒T = 10 sec α 7 (ii) Now sy (T ) = −200. A particle is projected from the √ top of the cliff. the time taken for the particle to hit the ground. LC Applied Maths Problem Set Solutions 44 Now if mL and mK are the slopes of lines L and K then mL × mL = −1 ⇒ L ⊥ K (45) Equivalently. the velocity of projection being u at angle 45◦ to the horizontal. 3.2 (46) Problem: LC HL 2008 [Part (b)] A ball is projected from a point on the ground at a distance of a from the foot of a vertical wall of height b. two directions are at right angles if: tan α1 × tan α2 = −1 √ √ ⇒ tan α1 × tan α2 = (1 + 2)(1 − 2) ⇒ tan α1 × tan α2 = 1 − 2 = −1 ∴ the two directions are at right angles. Using vy = uy − gt ⇒ 0 = uy − gtmax uy ⇒ tmax = g ( 2) ( ) uy 1 uy ⇒ sy max = uy − g g 2 g2 u2y 1 u2y u2y ⇒ sy max = − = g 2 g 2g (47) √ Now uy = u sin 45◦ = u/ 2: sy max u2 = 4g (48) .2.1 Solution In the first instance maximum height is sy when vy = 0. If the ball just clears the wall prove that the greatest height reached is a2 . 4(a − b) 3. 1 Solution The range is sx when sy = 0.3 Problem: LC HL 2007 [Part (a)] √ A particle is projected with a speed of 7 5 m/s at an angle α to the horizontal. Find the two values of α that will give a range of 12. Using 1 sy = uy t − gt2 2 ( g ) ⇒ t uy − t = 0 2 2uy ⇒t= g ( ) 2uy ⇒ R = sx (t) = ux g 2 u ⇒ R = 2 sin α cos α g u2 ! 25 ⇒ R= sin 2α = 2 sin x cos x=sin 2x g 2 25g 1 25g = ⇒ sin 2α = 2 = 2u 2(49)(5) 2 ◦ ⇒ 2α = 30 or 150◦ ⇒ α = 15◦ or 75◦ .LC Applied Maths Problem Set Solutions 45 Now at a time. say T .5 m. sx (T ) = a and sy (T ) = b. Using: sx = ux t ◦ ⇒ a = u cos 45 √T 2a ⇒T = u (√ ) ( 2) u 2a 1 2a ⇒ b= √ − g sy (T )=b u 2 u2 2 ga2 u2 ga2 ⇒a−b= 2 u 2 ga ⇒ u2 = a−b 2 u ga2 = ⇒ sy max = 4g 4g(a − b) a2 ⇒ sy max = 4(a − b) ⇒b=a− 3. 3.3. LC Applied Maths Problem Set Solutions 3.4 46 Problem: LC HL 2006 [Part (a)] A particle is projected from a point o with velocity 9.8 i + 29.4 j m/s where i and j are unit perpendicular vectors in the horizontal and vertical directions, respectively. (i) Express the velocity and displacement of the particle after t seconds in terms of i and j. (ii) Find, in terms of t, the direction in which the particle is moving after t seconds. (iii) Find the two times when the direction of the particle is at right angles to the line joining the particle to o. 3.4.1 Solution (i) Noting first that 9.8 = g and 29.4 = 3g. The velocity vector is given by: v(t) = vx (t)i + vy (t)j (49) and using vx (t) = ux , and vy = uy − gt ⇒ v(t) = gi + (3g − gt)j The displacement vector is given by: r(t) = sx (t)i + sy (t)j (50) Using: sx (t) = ux t , and 1 sy (t) = uy t − gt2 2) ( 1 2 ⇒ r(t) = gti + 3gt − gt j 2 (ii) The direction the particle is travelling in is the slope of the tangent to the displacement at t; the tangent being given by the derivative: r′ (t) = vx i + vy j vy 3g − gt j-component r′ (t) = = =3−t ⇒ direction after t = ′ i-component r (t) vx g (iii) The line joining the particle to o has slope: sy sx For two lines L and K to be perpendicular the slopes mL and mK must satisfy mL × mL = −1. (51) LC Applied Maths Problem Set Solutions 47 Hence solve: sy ! × (3 − t) = −1 sx 3gt − 12 gt2 ⇒ (3 − t) = −1 gt ( ) 1 ⇒ 3 − t (3 − t) = −1 ÷gt 2 3 1 ⇒ 9 − t + t2 − 3t = −1 2 2 ⇒ 18 − 3t + t2 − 6t = −2 ⇒ t2 − 9t + 20 = 0 ⇒ t2 − 4t − 5t + 20 = 0 ⇒ t(t − 4) − 5(t − 4) = 0 ⇒ (t − 4)(t − 5) = 0 Ans: At t = 4s, and 5 s. 3.5 Problem: LC HL 2004: [Part (a)] A particle is projected from a point on the horizontal floor of a tunnel with maximum height of 8 m. The particle is projected with an initial speed of 20 m/s inclined at an angle α to the horizontal floor. Find, to the nearest metre, the greatest range which can be attained in the tunnel. 3.5.1 Solution The range R is sx when sy = 0. Solving this gives: u2 R= sin 2α g Next the angle of projection which gives the max height as the height of the tunnel, 8 m, is found. Max height is sy when vy = 0. Solving this gives: u2y u2 sy max = = sin2 α 2g 2g u2 sin2 α = 8 ⇒ 2g 16g ⇒ sin2 α = 400 √ √ 4 g g ⇒ sin α = = 20 5 √ ⇒ α = arcsin( g/5) ≈ 38.763◦ √ Now if α > arcsin( g/5) then max height is bigger than 8 m, so this motion is not in the √ tunnel as required. Hence α < arcsin( g/5). Now looking at the range: R= u2 sin 2α g LC Applied Maths Problem Set Solutions 48 For θ ∈ [0, 45◦ ], sin 2θ is increasing, as d sin 2θ = 2 cos 2θ > 0 , θ ∈ [0, 45◦ ] dθ √ √ Hence as 0 ≤ α ≤ arcsin( g/5), sin 2α attains it maximum at α = arcsin( g/5): R= 202 2 sin α cos α g Now the model triangle gives: √ √ √ g 25 − g 800 800 g √ R= 2 = 15.2 g 5 5 25 g 32 ⇒ R = √ (3.8987) = 39.85 ≃ 40 m g 3.6 Problem: LC HL 2003: [Part (a)] A particle is projected from a point on level horizontal ground at an angle θ to the horizontal ground. Find θ, if the horizontal range of the particle is five times the maximum height reached by the particle. 3.6.1 Solution The range, R, is sx when sy = 0: R= u2 sin 2θ g (52) The maximum height, sy max , of the particle is sy when vy = 0: sy max = u2 sin2 θ 2g (53) 66◦ 5 3. ÷g 2 ⇒ t − 8t sin α − 3 = 0 Using the formula for the roots of a quadratic.7.7 = 3g/2 and 39. t2 − t1 = 3. respectively. (i) Show that √ 64 sin2 α − 12 √ (ii) Find the value of α for which t2 − t1 = 20.7 Problem: LC HL 2002: [Part (a)] A particle is projected from a point on the horizontal ground with a speed of 39. t1 and t2 are times when sy = 14.7. letting t1 be the ‘+’ solution and t2 be the ‘-’ solution: √ 8 ± 64 sin2 α − 12 t= 2 ( ) ( ) √ √ 2 8 + 64 sin α − 12 8 − 64 sin2 α − 12 t1 − t2 = − 2 2 √ √ 64 sin2 α − 12 64 sin2 α − 12 −4+ ⇒ t1 − t2 = 4 + 2 2 √ ⇒ t1 − t2 = 64 sin2 α − 12 .2 m/s inclined at an angle α to the horizontal ground.7 m above the horizontal ground at times t1 and t2 seconds. hence are the solutions of the quadratic equation: 3 1 g = 4gt sin α − gt2 2 2 ⇒ 3 = 8t sin α − t2 ×2.2 = 4g.LC Applied Maths Problem Set Solutions 49 Hence θ is the angle such that R = sy max : 2 2 u2 ! 5u sin θ sin 2θ = g 2g 5 ⇒ 2 sin θ cos θ = sin2 θ u̸=0 2 5 ⇒ 2 cos θ = sin θ θ̸=0 2 sin θ 4 ⇒ = cos θ 5 4 ⇒ tan θ = 5 ( ) 4 ⇒ θ = arctan ≃ 38. The particle is at a height of 14.1 Solution (i) First note 14. Find the value of u. noting sin 45◦ = 1/ 2 = cos 45◦ : u sx = √ T 2 √ 21 2 ⇒T = T u Now sy (T ) = 1. cancelling ⇒ 20u2 = 212 g 212 g ⇒ u2 = 20 √ g ⇒ u = 21 = 14.7 m/s 20 .8 Problem: LC HL 2001: [Part (a)] A player hits a ball with an initial speed of u m/s from a height of 1 m at an angle of 45◦ to the horizontal ground.8. Suppose the catcher catches the ball at a time T .LC Applied Maths Problem Set Solutions 50 (ii) Hence α is such that: 64 sin2 α − 12 = 20 ⇒ 64 sin2 α = 32 1 ⇒ sin2 α = 2 1 ⇒ sin α = √ 2 ⇒ θ = 45◦ 3.1 Solution sy √ = 1 when sx = 21. 21 m away. 3. A member of the opposing team. catches the ball at a height of 2 m above the ground. using: 1 sy = uy t − gt2 2 √ u 21 2 1 (21)2 (2) ⇒1= √ − g 2 u2 2 u ⇒ u2 = 21u2 − 212 g 2 ×u . 1 ! sy = uy t − gt2 = 0 2 2uy ⇒t= g Now s x = ux t ( ) 2uy ⇒ R = ux g 2 2u sin β cos β ⇒R= g 2 sin β cos β ⇒ R = u2 g 2 u sin 2β ⇒ R= 2 sin x cos x=sin 2x g Taking u to be fixed.LC Applied Maths Problem Set Solutions 3. to maximise R vary β. Find the angle of projection which gives maximum range.9. This occurs when 2β = 90◦ (sin 90◦ = 1) ⇒ β = 45◦ for maximum range . Show that the particle hits the ground at a distance u2 sin 2β g from the point of projection.1 Solution Range.9 51 Problem: LC HL 2000: [Part (b)] A particle is projected with a velocity u m/s at an angle β to the horizontal ground. 3. The maximum value of sine is 1. R. is sx when sy = 0. = . The plane of projection is vertical and contains the line of greatest slope.10.LC Applied Maths Problem Set Solutions 3. Using vx = u cos θ − g sin 60◦ t u cos θ ⇒T = g sin 60◦ Figure 1: If the particle lands at right angles.10 52 Problem: LC HL 2009: [Part(b)] A plane is inclined at an angle 60◦ to the horizontal. the final velocity is entirely in the y-direction Using 1 sy = u sin θ t − g cos 60◦ t2 2 2u sin θ ⇒T = cos 60◦ u cos θ 2u sin θ ⇒ = ◦ g sin 60 cos 60◦ 1 cos 60◦ 1 ⇒ tan θ = . 13g 3. If the particle lands at right angles. A particle is projected up the plane with initial speed u at an angle θ to the inclined plane. Show that the range on the inclined plane is √ 4 3u2 .1 Solution Let T be the time of flight. sy (T ) = 0 and vx (T ) = 0. cot 60◦ ◦ 2 sin 60 2 1 ⇒ tan θ = √ 2 3 . The particle strikes the plane at right angles. . Using 1 sx = ux t − gx t2 2 ) ( ) ( 2 2 u cos θ 1 ◦ u cos θ sin 60 − g ⇒ R = u cos θ ◦ g sin 60 2 g2 sin2 60◦ u2 cos2 θ ⇒R= (2 − 1) 2g sin 60◦ u2 4(3) 2 ⇒R= .√ 2 g 13 3 √ 2 4 3u ⇒R= 13g .LC Applied Maths Problem Set Solutions 53 Figure 2: The model triangles for 60◦ and θ Now range means sx (T ). Using 1 sy = u sin 2θ t − g cos θ t2 2 2u sin 2θ ⇒T = g cos θ 4u sin θ ⇒ sin 2x=2 sin x cos x g Using 1 sx = ux t + gx t2 2 8 ) ( 2 sin2 θ 1 4u sin θ 16u + . R. The plane of projection is vertical and contains the line of greatest slope. means sx when sy = 0.1 Solution Range. Find the value of k.LC Applied Maths Problem Set Solutions 3. ⇒ R = u cos 2θ g sin θ g 2 g2 ⇒R= ⇒ cos 2x=cos2 x−sin2 x R= 4u2 sin θ (cos 2θ + 2 sin2 θ) g 4u2 sin θ (cos2 θ − sin2 θ + 2 sin2 θ) | {z } g =cos2 θ+sin2 θ ⇒2 cos2 x+sin x=1 R= 4u2 sin θ g ⇒k=4 . The range of the particle on the inclined plane is ku2 sin θ/g. 3.11.11 54 Problem: LC HL 2008: [Part(b)] A particle is projected down an inclined plane with initial velocity u m/s. The line of projection makes an angle 2θ◦ with the inclined plane and the plane is inclined at θ◦ to the horizontal. 12 55 Problem: LC HL 2006: [Part (b)] A particle is projected up an inclined plane with initial speed u m/s.1 Solution Range. g 2 g2 2 ◦ 2u tan 30 ⇒R= (cos 30◦ − sin 30◦ tan 30◦ ) g (√ ) 2 u2 1 3 1 1 ⇒ R= . Using 1 sy = u sin 30◦ t − g cos 30◦ t2 2 2u tan 30◦ ⇒T = g 2 ( ) 2u 1 4 u2 tan2 30◦ ⇒ R = u cos 30◦ tan 30◦ − g sin 30◦ . 3.LC Applied Maths Problem Set Solutions 3.12. the range of the particle on the inclined plane. The plane of projection is vertical and contains the line of greatest slope. Find in terms of u. means sx when sy = 0.√ − . R.√ using Figure 2 g 2 2 3 3 ( ) u2 3−1 √ ⇒R= √ 3g 3 2u2 ⇒R= 3g . The line of projection makes an angle 30◦ with the plane and the plane is inclined at 30◦ to the horizontal. Using 1 sy = u sin α t − g cos β t2 2 2u sin α ⇒T = g cos β 2 ( ) 2 2 2u sin α 1 4 u sin α ⇒ R = u cos α − g sin β g cos β 2 g2 cos2 β 2u2 sin α (cos α cos β − sin α sin β) g cos2 β u2 ⇒ R= 2 sin α cos(α + β) cos(x+y)=cos x cos y−sin x sin y g cos2 β u2 ⇒ R= (sin(2α + β) + sin(−β)) 2 sin x cos y=sin(x+y)+sin(x−y) g cos2 β u2 (sin(2α + β) − sin β) ⇒ R= sin(−x)=− sin x g cos2 β ⇒R= . α and β. A particle is projected up the plane with initial speed u at an angle α to the inclined plane. The plane of projection is vertical and contains the line of greatest slope. (i) Find the range of the particle on the inclined plane in terms of u. (ii) Show that for a constant value of u the range is a maximum when α = 45◦ − 3.1 β 2 Solution Range. R.13.LC Applied Maths Problem Set Solutions 3. means sx when sy = 0.13 56 Problem: LC HL 2005: [Part (b)] A plane is inclined at an angle β to the horizontal. R. means sx when sy = 0. The line of projection makes an angle α with the horizontal and the inclined plane makes an angle β with the horizontal. . 3. α and β. g. the range of the particle up the inclined plane. 2α + β = 90◦ ⇒ 2α = 90◦ − β β ⇒ α = 45◦ − 2 3. Sine has a maximum value of 1. namely sin 90◦ = 1.LC Applied Maths Problem Set Solutions 57 To maximise R the only term which may be varied is sin(2α + β). ( The plane of projection is vertical and contains the line of greatest slope). Hence for Rmax .14 Problem: LC HL 2003: [Part (b)] A particle is projected up the inclined plane with initial speed u m/s.1 Solution Range.14. Find in terms of u. show that the vertical height h of the particle above the plane after t seconds is √ 10 3 t − 4. 3. down a plane inclined at an angle of 30◦ to the horizontal. (The plane of projection is vertical and contains the line of greatest slope).1 Solution Using 1 sy = uy t − gy t2 2 ⇒ sy (t) = 15t − 4.15. or otherwise.LC Applied Maths Problem Set Solutions 58 Using 1 sy = u sin(α − β) t − g cos β t2 2 2u sin(α − β) ⇒T = g cos β 2 ( ) 2 2 2u sin(α − β) 1 4 u sin (α − β) ⇒ R = u cos(α − β) − g sin β g cos β 2 g2 cos2 β 2u2 sin(α − β) (cos(α − β) cos β − sin(α − β) sin β) g cos2 β u2 2 sin(α − β) cos (α − β + β) ⇒ R= | {z } cos(x+y)=cos x cos y−sin x sin y g cos2 β ⇒R= =α u2 ⇒ R= (sin(α − β + α) + sin(α − β − α)) 2 sin x cos y=sin(x+y)+sin(x−y) g cos2 β u2 ⇒ R= (sin(2α − β) − sin β) sin(−x)=− sin x g cos2 β 3.15 Problem: LC HL 1997: [Part (b)] A particle is projected from a point p with initial speed 15 m/s.9 cos 30◦ t2 √ 4. Find (i) the perpendicular height of the particle above the plane after t seconds and hence.9t2 (ii) the greatest vertical height it attains above the plane (i. the maximum value of h) correct to two places of decimals.9 3 2 t ⇒ sy (t) = 15t − cos 30◦ from Fig 2 2 . The direction of projection is at right angles to the inclined plane.e. = g 2 g g ⇒ hmax ≃ 15.31 m .9t2 To maximise h(t) note that it is a concave down2 quadratic and hence has a single local maximum when ⇒ hmax dh =0 dt √ ⇒ 10 3 − gtmax = 0 √ 10 3 ⇒ tmax = g ( √ ) ( ) √ 10 3 g 300 = h(tmax ) = 10 3 − g 2 g2 ⇒ hmax = 2 or sad 150 300 1 300 − .LC Applied Maths Problem Set Solutions 59 Due to the geometry. sy sin 60◦ = h sy ⇒h= sin 60◦ ( √ ) 4.9 3 2 2 √ h = 15t − t ⇒ sin 60◦ from Fig 2 2 3 30 ⇒ h = √ t − 4.9t2 3 √ ⇒ h = 10 3t − 4. The plane of projection is vertical and contains the line of greatest slope. Figure 3: If the particle lands horizontally then as alternate angles the landing angle is equal to the angle the plane makes with the horizontal.LC Applied Maths Problem Set Solutions 3.16. Show that tan θ = 2. The particle is moving horizontally when it strikes the inclined plane.16 60 Problem: LC HL 2007: [Part(b)] A plane is inclined at an angle 45◦ to the horizontal. The landing angle is given by: tan l = ⇒ −vy vx −vy = tan 45◦ = 1 vx ⇒ −vy = vx (54) (55) (56) . A particle is projected up the plane with initial speed u at an angle θ to the horizontal. 3.1 Solution If the particle lands horizontally then the landing angle is 45◦ . T 2 2 2u sin(θ − 45◦ ) ⇒T = g cos 45◦ ◦ ◦ 2u sin(θ − 45 ) 45 ⇒ vy = u sin(θ − 45◦ ) − g cos .and x-directions. vx are the final speeds in the y. ◦ g cos 45 ⇒ vy = −u sin(θ − 45◦ ) ( ) ◦ 2u sin(θ − 45 ) ◦ ◦ 45 g sin vx = u cos(θ − 45 ) − ◦ g cos 45 ⇒ vx = u(cos(θ − 45◦ ) − 2 sin(θ)) ⇒ sin(θ − 45◦ ) = cos(θ − 45◦ ) − 2 sin(θ − 45◦ ) (56) ⇒ 3 sin(θ − 45◦ ) = cos(θ − 45◦ ) 1 ⇒ tan(θ − 45◦ ) = 3 tan θ − 1 1 = ⇒ tan A−tan B 1 + tan θ 3 tan(A−B)= 1+tan A tan B ⇒ 1 + tan θ = 3 tan θ − 3 ⇒ 2 tan θ = 4 ⇒ tan θ = 2 . Using ⇒ √ sin 45◦ =1/ 2=cos 45◦ 1 sy (T ) = 0 = u sin(θ − 45◦ )T − g cos 45◦ .LC Applied Maths Problem Set Solutions 61 where vy . 2 tan β tan(2β − β) = 1 ⇒ 2 tan2 β = 1 1 ⇒ tan β = √ 2 . (i) Show that 2 tan β tan(θ − β) = 1 (ii) Hence. show that if θ = 2β. The particle strikes the plane at right angles. (The plane of projection is vertical and contains the line of greatest slope). (The plane of projection is vertical and contains the line of greatest slope.1 1 + 2 tan2 θ tan θ Solution If the particle strikes the plane at right angles vx (T ) = 0 where T is the time of flight.18.1 Solution (i) This is shown in four lines in the above solution.17 62 Problem: LC HL 2004: [Part(b)] A particle is projected up an inclined plane with initial speed u m/s. show that tan α = 3. or √ otherwise. The line of projection makes an angle α with the horizontal and the inclined plane makes an angle θ with the horizontal.18 Problem: LC HL 2002: [Part(b)] A particle is projected with speed u m/s at an angle θ to the horizontal. the range of the particle up the inclined plane 2 is u /(g 3) 3. up a plane inclined at an angle β to the horizontal. 2u sin(α − θ) g cos θ ( ) 2u sin(α − θ) ⇒ vx (T ) = u cos(α − θ) − g sin θ =0 g cos θ ⇒ cos(α − θ) = 2 tan θ sin(α − θ) T = ×1/u ⇒ 1 = 2 tan θ tan(α − θ) ) ( tan α − tan θ ⇒ 1 = 2 tan θ 1 + tan α tan θ ⇒ 1 + tan α tan θ = 2 tan θ tan α − 2 tan2 θ ⇒ tan α tan θ = 1 + 2 tan2 θ 1 + 2 tan2 θ ⇒ tan α = tan θ 3.LC Applied Maths Problem Set Solutions 3.17. (ii) If θ = 2β.) If the particle strikes the inclined plane at right angles. g 3 2 3 u2 ⇒R= √ g 3 . Range is sx (T ): 2 2u sin β 1 4 u2 sin2 β R = u cos β. g cos β 2 g2 cos2 β 2u2 sin β ⇒R= (cos2 β − sin2 β) g cos2 β ( ) 2u2 1 3 2 1 ⇒R= . − g sin β. − g 3 2 3 3 ( ) 2 u2 3 1 ⇒R= √ .√ .LC Applied Maths Problem Set Solutions 63 Figure 4: The Model Triangle for β. or 45◦ . (The plane of projection is vertical and contains the line of greatest slope).19.LC Applied Maths Problem Set Solutions 3. tan α = 3.20. 3. The particle strikes the plane at right angles. the landing angle is θ as the particle lands horizontally. As in Figure 18.1 Solution (a) Let α := tan−1 (1/3). Hence tan θ = −vy (T ) vx (T ) . (The plane of projection is vertical and contains the line of greatest slope).565◦ .19 64 Problem: LC HL 2000: [Part(b)] A particle is projected at an angle α = tan−1 3 to the horizontal up a plane inclined at an angle θ to the horizontal. (b)(i) If tan θ = 0. 3.17: 1 + 2 tan2 θ tan θ 1 + 2 tan2 θ ⇒3= tan θ ⇒ 3 tan θ = 1 + 2 tan2 θ ⇒ 2 tan2 θ − 3 tan θ + 1 = 0 ⇒ 2 tan2 θ − 2 tan θ − tan θ + 1 = 0 ⇒ 2 tan θ(tan θ − 1) − 1(tan θ − 1) = 0 ⇒ (2 tan θ − 1)(tan θ − 1) = 0 ⇒ θ = arctan(1/2) ≃ 26.20 Problem: LC HL 1999 [Parts (a) & (b)(i)] √ A particle is projected from a point p up an inclined plane with a speed of 4g 2 m/s at an angle tan−1 (1/3) to the inclined plane. The particle is moving horizontally when it strikes the plane at the point q. Find the two possible values for θ.1 Solution From Problem 3. The plane is inclined at an angle θ to the horizontal. (a) Find the two possible values for θ.5 then find the magnitude of the velocity with which the particle strikes the inclined plane at q. g cos θ ⇒ sx (T ) = u cos α − 2u tan θ sin α u sin α ⇒ tan θ = u cos α − 2 u tan θ sin α 2 ⇒ tan θ cos α − 2 tan θ sin α = sin α tan θ − 2 tan2 θ = 1 ⇒ ×1/ sin α tan α ⇒ 2 tan2 θ − 3 tan θ + 1 = 0 T = tan α=1/3 ⇒ 2 tan2 θ − 2 tan θ − tan θ + 1 = 0 ⇒ 2 tan θ(tan θ − 1) − 1(tan θ − 1) = 0 ⇒ (2 tan θ − 1)(tan θ − 1) = 0 ⇒ θ = arctan(1/2) ≃ 26. or 45◦ .565◦ . 2u sin α g cos θ ⇒ sy (T ) = −u sin α .LC Applied Maths Problem Set Solutions 65 where T is the time of flight. and 2u sin α sx (T ) = u cos α − g sin θ. (b)(i) The final speed of the particle is given by: √ |v(T )| = vx2 (T ) + vy2 (T ) Figure 5: The model triangles for θ & α (57) . 6 2g = ⇒ sin α = 28 28 ( ) 19.4◦ 28 (ii) The landing angle l is given by: tan l = −vy (T ) vx (T ) (58) . The particle √ is projected from a point on the slope and has an initial velocity of 7 2 m/s at an angle α to the inclined plane. vy (T ) = −u sin α √ 2 ⇒ vy (T ) = −4g √ 10 2 16g .6 ⇒ α = sin−1 ≃ 44. and ⇒ vy2 (T ) = 5 vx (T ) = u cos α − 2u tan θ sin α √ 3 √ 1 ⇒ vx (T ) = 4g 2 √ − 4g 2 √ 10 10 √ 4g 2 8g ⇒ vx (T ) = √ (3 − 1) = √ 10 5 64g 2 ⇒ vx2 (T ) = 5 √ 2 √ 80g ⇒ |v(T )| = = 16g 2 5 ⇒ |v(T )| = 4g 3.1 Solution (i) The particle hits the slope when sy = 0: 1 sy = u sin αt − g sin 45◦ t2 2 √ 1 ! ⇒ 0 = 7 2 sin α.21 Problem: LC HL 1999: [Part (b)] A particle is projected down a slope which is inclined at 45◦ to the horizontal. Find the value of α if (i) the particle first hits the slope after 2 seconds (ii) the landing angle with the slope is tan−1 (1/3) 3.21.(2) − g sin 45◦ (2)2 2 √ 2 ⇒ 0 = 14 2 sin α − √ g 2 19.LC Applied Maths Problem Set Solutions 66 Now from (a). Exercises 3.22.D 3(iii) [LC 1979] A plane is inclined at an angle α to the horizontal. Prove that the particle will strike the plane horizontally if tan θ = 3. The plane of projection is vertical and contains the line of greatest slope.and x-directions respectively. and tan−1 is one-to-one on [0◦ . tan−1 (tan l) = tan−1 (tan α) ⇒l=α . 90◦ ].1 sin α cos α (2 − cos2 α) Solution Let l be the landing angle.22 Problem: F. The time of flight is when sy = 0: 1 ! sy = u sin αt − g cos 45◦ t2 == 2 2u sin α ⇒T = g cos 45◦ ( ) 2u sin α ◦ 45 ⇒ vy (T ) = u sin α − g cos ◦ 45 g cos ⇒ vy (T ) = u sin α − 2u sin α = −u sin α and ) 2u sin α ◦ g cos 45 ⇒ vx (T ) = u cos α + 2u sin α ◦ 45 vx (T ) = u cos α + g sin ( Now if l = tan−1 (1/3): 1 −vy (T ) = 3 vx (T ) ⇒ vx (T ) = −3vy (T ) ⇒ u cos α + 2 u sin α = 3 u sin α ⇒ cos α = sin α ⇒ tan α = 1 ⇒ α = 45◦ ⇒ 3.LC Applied Maths Problem Set Solutions 67 where T is the time of flight and vy (T ) and vx (T ) are the final velocities in the y.M.A. A particle is projected up the plane with a speed u at an angle θ to the plane. tan l = −vy (T ) vx (T ) (59) Suppose tan l = tan α. As both l and α are greater than 0◦ and less than 90◦ . Now from before vy (T ) = −u sin θ vx (T ) = u cos θ − 2u tan α sin θ sin θ ⇒ tan l = cos θ − 2 tan α sin θ Now if tan θ = sin α cos α (2 − cos2 α) Figure 6: The model triangle for θ Hence.LC Applied Maths Problem Set Solutions 68 which implies that the particle strikes horizontally. 3. λ2 = sin α cos2 α + (2 − cos2 α)2 √ ⇒ λ = sin α cos2 α + (2 − cos2 α)2 ⇒ tan l = sin α cos α λ (2−cos2 α) 2 tan α sin α cos α − λ λ sin α cos α 2 − cos2 −2 sin2 α tan sin cos=sin2 sin α cos α ⇒ 2 tan l = 2 2 2 2 2 cos α + 2 sin α − cos2 α − 2 sin α cos + sin =1 sin α cos α sin α ⇒ tan l = = 2 cos α cos α ⇒ tan l = tan α ⇒ tan l = Hence the particle strikes the plane horizontally. . by Pythagoras Theorem.22. where λ is the hypotenuse of the model triangle. I wouldn’t expect anything of this difficulty in LC 2010 Q.3(b). This question is taken from LC 1979.2 Remark Questions of this subtlety rarely come up. Consider these statements P & Q. and shown to be equal to tan α. It is shown for l. Note the difference. S are rP = i − 2j. If one proved P ⇒ Q one would not receive full marks as the question has been fundamentally misunderstood.A. Thence l = α and implies that the particle lands horizontally.1 Problem: F. and this is straightforward.1 Solution Using rAB = rA − rB ⇒ rQP = rQ − rP ⇒ rQP = −5i + 3j Let ⇒ rT S rT = xi + yj ⇒ rT S = rT − rS = (x + 3)i + (y − 5)j Now for rT S = rQP . 4 Relative Velocity 4. tan θ = sin α cos α (2 − cos2 α) Ordinarily one is asked to show that P ⇒ Q (read P implies Q). Note our approach here. comparing components: ! −5i + 3j = (x + 3)i + (y − 5)j ⇒ rT = −8i + 8j . rS = −3i + 5j. Find in terms of i and j the position vector of T if rT S = rQP . rQ = −4i + j. 90◦ ]: If tan l = tan α. the displacement of Q relative to P .LC Applied Maths Problem Set Solutions 69 P. That is assuming P . α ∈ [0◦ .4.M. 4.1. the particle strikes the plane horizontally Q. then l = α and tan l is computed. Q. Find rQP . prove Q. However in this example one is asked to show Q ⇒ P .A. Q.6 With O as the origin the position vectors of P . 7 A train is travelling on a straight track with velocity 30j and a car. The velocity of P is 6i + 2j m/s. and the velocity of Q is −4i + 2j m/s.3 Problem: F.A.38◦ tan θ = Hence the direction of VCT is E 67. How much time will pass before the collision occurs? 4.4.4.LC Applied Maths Problem Set Solutions 4.1 Solution Clearly RP Q = −100i.2.A. 4.M.8 A particle P is 100 m due West of another particle Q. visible from the train.A. The direction of VCT is E θ S.M.38◦ S. Q.A. Using VP Q = VP − VQ ⇒ VP Q = 10i ∴ the particles are on collision course.3.1 Solution Using VCT = VC − VT ⇒ VCT = 10i − 24j √ √ 2 ⇒ |VCT | = 10 + 242 = 100 + 576 ⇒ |VCT | = 26 m/s Figure 7: VCT in the plane.2 70 Problem: F. Calculate the magnitude and direction of the car’s velocity as it appears to a person sitting on the train. Q. Show that P and Q are on collision course. 4. Now 24 12 = 10 5 −1 ⇒ θ = tan (12/5) ≈ 67. is travelling on a straight road with velocity 10i + 6j where speeds are measured in m/s. . Note θ. Using VKT = VK − VT ⇒ VKT = 3i + 4j Now consider the position of K relative to T .4.1 Solution Now RKT = −20i. Find the shortest distance between them in subsequent motion. Using relative distance relative speed 100 ⇒t= = 10 s 10 time = 4. Their velocities are i + 2j m/s and −2i − 2j m/s respectively. back along the i axis towards particle Q.4. in fact its position relative to particle Q is along the negative i-axis at −100i m.4 Problem: F.A. Therefore.M. The velocity of particle P relative to Q is +10i. RKT : Now tan θ = 4/3: Now from Figure 9: sin θ = ⇒d= d ! 4 = 20 5 80 = 16 m 5 . Q.9 K is a particle which is 20 m due West of another particle T .LC Applied Maths Problem Set Solutions 71 Figure 8: Particle P is due West of particle Q.A. they must be on collision course. 4. Find the velocity of K relative to T . between T and the path of K relative to T . 4.LC Applied Maths Problem Set Solutions 72 Figure 9: The shortest distance between K and T in the subsequent motion is the perpendicular distance. Find the velocity of P relative to Q.A.A. Show the positions of P and Q on a diagram and show the path of P relative to Q.M. Calculate this distance of O from this path. What does this distance represent? 4. P is at the point 119i when Q is at the origin O. d.10 A particle P is moving with velocity −8i+12j while another particle Q is moving with velocity 7i + 4j.5 Problem: F. Figure 10: The model triangle for θ. .1 Solution Using VP Q = VP − VQ ⇒ VP Q = −15i + 8j Now from Figure 11: d ! 8 = 119 17 8 ⇒ d = 119 = 56 17 sin θ = This distance represents the shortest distance between P and Q is subsequent motion.5.4. Q. d is the distance of O from the path. α = tan−1 (8/15). Sides 8 & 15 are given by α = tan (8/15). The √ model triangle hypotenuse is 82 + 152 = 289 = 17 .LC Applied Maths Problem Set Solutions 73 Figure 11: The position RP Q and path VP Q of P relative to Q. −1 Figure 12: The √ for α. find in terms of i and j its velocity if it is to intercept (collide with) M .6 74 Problem: F.11 (a) If |ti + 3j| = 5. and x + 2 > 0.6. Using t= relative distance relative speed 60 = 10 h ⇒t= 6 . Q.A. x > −2: √ x2 + 9 = 5 ⇒ x = ±4 ⇒ x=4 x>−2 ⇒ VK = 4i + 3j (ii) Now VKM = 6i.LC Applied Maths Problem Set Solutions 4. (Hint: K will have to travel with the same j-speed as ship M in order to keep on collision course.A.4.M. find the value of t > 0. y − 3 = 0.) (ii) When will collision occur? 4. t>0 (b) (i) RKM = −60i hence for collision VKM = ki with k > 0. (b) (i) A ship K is 60 km due West of another ship M which is travelling with velocity −2i + 3j km/hr. Let VK = xi + yj . with √ x2 + y 2 = 5 Using ⇒ VKM VKM = VK − VM = (x + 2)i + (y − 3)j For collision course. y = 3. √ |a| = x2 + y 2 √ ! ⇒ |ti + 3j| = t2 + 9 = 5 ⇒ t2 + 9 = 25 ⇒ t2 = 16 ⇒ t = ±4 ⇒ t = 4.1 Solution (a) For any vector a := xi + yj. t ∈ R. P is moving with speed 5 2 m/s in a NE direction. Q can travel at 13 m/s. As sin 45◦ = 1/ 2 = cos 45◦ .4 km due West of another ship Q.1 Solution Let VQ = xi + yj with |VQ | = 13 √ ! x2 + y 2 = 13 RP Q = −3400 i.M.LC Applied Maths Problem Set Solutions 4. If they are on collision course find the velocity of Q in terms of i and j. Now |VP | = 13: √ x2 + 52 = 13 ⇒ x2 = 132 − 52 = 144 ⇒ x = ±12 ⇒ x = −12 ! (5−x)>0 ⇒ VQ = −12 i + 5 j Now VP Q = 17 i. When will collision occur? 4. it is required that: VP Q = k i . hence for P to collide with Q. Q.7 75 Problem: F.4. VP = 5 i+5 j. Using t= relative distance relative speed ⇒t= 200 3400 17 = 200 s . k > 0 (60) Now VP : √ Figure 13: NE is the direction E 45◦ N as shown. Using ⇒ VP Q VP Q = VP − VQ = (5 − x)i + (5 − y)j Hence for collision.A.A.12 √ A ship P is 3.7. y = 5. also. T is travelling at 10 km/hr in a direction E 30◦ S.8 76 Problem: F. Find. Q.4. the velocity of Q. √ VT = 5 3 i − 5 j Consider now VQ : Figure 15: VQ = −20 cos 45◦ i + 20 sin 45◦ j.A. 4.A.8. the magnitude and direction of the velocity of T relative to Q and hence find the shortest distance between them in subsequent motion correct to one decimal place.LC Applied Maths Problem Set Solutions 4.13 Ship T is 100 km due West of ship Q. VT = 10 cos 30◦ i − 10 sin 30◦ j.1 Solution Consider VT : Figure 14: VT and the model triangle for 30◦ . Q is travelling at 20 km/hr in direction W 45◦ N.M. 1 1 VQ = −20 √ i + 20 √ j 2 2 √ √ 1 1 2 2 ⇒ VQ = −20 √ · √ i + 20 √ · √ j 2 2 2 2 20 √ 20 √ ⇒ VQ = − 2i + 2j 2 √ 2√ ⇒ VQ = −10 2 i + 10 2 j (61) . Find in terms of i and j the velocity of T . and the velocity of T relative to Q. 01 ⇒ d = 100(0.01◦ S. d 100 ⇒ d = 100 sin 40. Now RT Q = −100 i.772 km/hr √ a b2 x=ab x √ √ √ Figure 16: The direction of VT Q is given by θ = tan−1 ((5 (1 + 2 2))/(5 ( 3 + 2 2))) √ 1+2 2 √ ≈ 40.01◦ tan θ = √ 3+2 2 Hence the direction of VT Q is given by E 40.3 m sin 40. Now from Figure 17.01 = .LC Applied Maths Problem Set Solutions 77 Using VT Q = VT − VQ √ √ √ ⇒ VT Q = (5 3 + 10 2)i + (−5 − 10 2)j √ √ √ ⇒ VT Q = 5( 3 + 2 2)i − 5(1 + 2 2)j For a general vector a = x i + y j: |a| = √ x2 + y 2 √ √ √ √ 2 ( 3 + 2 2)2 + 52 (1 + 2 2)2 ⇒ 5 |V T Q| = (−x)2 =x2 √ √ √ ⇒ |VT Q | = 5 3 + 4 6 + 8 + 1 + 4 2 + 8 √ √ √ ⇒ |VT Q | = 5 20 + 4 2 + 4 6 √ √ √ ⇒ √ |VT Q | = 10 5 + 2 + 6 ≈ 29.64291) ≈ 64. Using VAB = VA − VB ⇒ VAB = 15 i − 20 j When both cars are 800 m from the intersection RA = −800 i and RB = −800 j. As a line.9 Problem: LC HL 2009: [Part(a)] (i) VA = 15 i m/s and VB = 20 j m/s. Relative to Q. 800) ∈ L := VAB . hence the shortest distance between T and Q in the subsequent motion is given by d. 4. Also (−800.LC Applied Maths Problem Set Solutions 78 Figure 17: Ship T is initially 100 km West of ship Q. . Using RAB = RA − RB ⇒ RAB = −800 i + 800 j Now Figure 18: The position of A relative to B and the subsequent motion of A relative to B. the slope of VAB is m = − tan θ = −4/3. T travels along VT Q . m1 − m2 1 + m1 m2 −1 + 4/3 −1/3 1 ⇒ tan θ = ± =± = ◦ 1 + 4/3 7/3 α∈[0.LC Applied Maths Problem Set Solutions 79 Using L ≡ y − y1 = m(x − x1 ) 4 ⇒ L ≡ y − 800 = − (x + 800) 3 ⇒ L ≡ 3y − 2400 = −4x − 3200 ⇒ L ≡ 4x + 3y + 800 = 0 Using. where d is the perpendicular distance between a point p(x1 . where θ is the angle between two lines of slope m1 and m2 .90 ] 7 tan θ = ± Hence Figure 19: tan α = d/D d 1 = 7 D ⇒ D = 7d = 1120 m tan α = Using relative distance relative speed 1120 224 = s ⇒ time = |VAB | |VAB |=25 5 time = . y1 ) and the line ax + by + c = 0. with respect to the angle α between the lines RAB (m1 = −1) and VAB (m2 = −4/3). d= |ax1 + by1 + c| √ a2 + b2 = x1 =0=y1 ⇒d= √ |c| a2 + b2 800 = 160 m 5 (ii) Using. Consider Figure 20: cos θ = L/100. L 3 = 5 100 300 ⇒L= = 60 m 5 cos θ = 80 .10 Problem: LC HL 2008: [Part(a)] (i) VC = 1. Now tan θ = −4/3 Figure 21: The model triangle for θ. using s = ut: 224 = 672 m .LC Applied Maths Problem Set Solutions Hence in this time. Using VCD = VC − VD ⇒ VCD = 1. sA = 15. = 896 m 5 Hence A is 128 m from the intersection and B is 96 m from the intersection. and 5 224 sB = 20.5 i m/s and VD = 2 j m/s.5 i − 2 j (ii) When D passes the intersection RCD = −100 i m. 4. and noting (−4 2.11 Problem: LC HL 2007: [Part(a)] (i) VB = −24 i km/hr and VA = 32 j km/hr. The distance between ship A and ship B is 8 km when the . : x ≥ −4 2 =: {px : x ≥ −4 2} 3 Now consider the following: V is the set of points {px }. −4 2) ∈ L. Using VAB = VA − VB ⇒ VAB = 24 i + 32 j (ii) If ship B is 8 km NE of ship A. ship A is 8 km SW of ship B: RAB = −8 cos 45◦ i − 8 sin 45◦ j 8 8 ⇒ RAB = − √ i − √ j 2 2 √ √ √⇒√ RAB = −4 2 i − 4 2 j × 2/ 2 √ √ Considered as a line.5(24) = 36 m ⇒ 64 m from the intersection 4. V := VAB has slope 32/24 = 4/3. using: L ≡ y − y1 = m(x − x1 ) √ √ 4 ⇒ V ≡ y + 4 2 = (x + 4 2) √ 3 √ ⇒ V ≡ 3y + 12 2 = 4x+!6 2 √ 4x + 4 2 ⇒V ≡y= 3 {( } √ ) √ √ 4x + 4 2 ⇒V ≡ x.LC Applied Maths Problem Set Solutions 81 Using relative distance relative speed 60 60 60 ⇒ time = =√ =√ = 24 s |VCD | 9/4 + 4 25/4 time = Using s = ut ⇒ sC = 1. The circle represents a circle of radius 8 km about B.396(60 min) = 23. 0).847 or − 5. The speed of A relative to B in this direction is 24 km/hr. initially and again at a later time. Using −b ± time = ⇒ time = relative distance relative speed 9.76 min ≈ 24 min 24 . Hence ship A travels 4 2+3.504 km in the i-direction. Twice ship A will be exactly 8 km from ship B.504 = 0.847 ≃ 9.657 km is the initial (−5.657 ≃ −4 2). is 8: v ( )2 u √ u 4x + 4 2 ! d = t(x − 0)2 + − 0 = |px | = 8 3 √ √ 2 + 32 2 x + 32 16x ! =8 ⇒ x2 + 9 √ 2 16x + 32 2 x + 32 ⇒ x2 + = 64 9 √ ⇒ 9x2 + 16x2 + 32 2 x + 32 = 576 √ ⇒ 25x2 + 32 2 x − 544 = 0 Using √ b2 − 4ac x± = 2a √ √ −32 2 ± 2048 + 54400 ⇒ x± = 50 √ −32 2 ± 237.LC Applied Maths Problem Set Solutions 82 Figure 22: R denotes RAB .588 ⇒ x± = 50 ⇒ x = 3.396 h = 0. distance from px to (0. d.657 km √ √ The x = −5. D is the distance A has to travel relative to B up to the instant the cars are d-close together. and RB = 136 i Using RAB = RA − RB ⇒ RAB = 84 i + 220 j Figure 23: R denotes RAB .LC Applied Maths Problem Set Solutions 4.12 83 Problem: LC HL 2005: [Part(b)] (i) VA = −p cos 45◦ i − p sin 45◦ j √ √ p 2 p 2 i− j √⇒√ VA = − 2 2 × 2/ 2 VB = −8 i Using ( ⇒ VAB = (ii) Initially VAB = VA − VB √ ) √ p 2 p 2 ! 8− i− j = −2 i − 10 j 2 2 √ p 2 ⇒ = 10 √ 2 √ 20 20 2 ⇒ p = √ √= √ = 10 2 2 × 2/ 2 2 √ RA = 220 2 cos 45◦ i + 220 sin 45◦ j = 220 i + 220 j . As a line VAB =: V has slope −10/ − 2 = 5. θ is the angle between V and R. (84. 220) ∈ V and using L ≡ y − y1 = m(x − x1 ) ⇒ V ≡ y − 220 = 5(x − 84) ⇒ V ≡ 5x − y − 200 = 0 . 90 ] 148 Now tan θ = d D 8 148 d 200 ⇒D= =√ . Using m1 − m1 tan θ = ± 1 + m1 m2 5 − 55/21 50/21 25 ⇒ tan θ = ± =± = ◦ 1 + 275/21 296/21 θ∈[0.LC Applied Maths Problem Set Solutions 84 Using d= |ax + by + c| |c| √ =√ 2 2 2 a +b a + b2 200 ⇒d= √ m 26 The slope of RAB =: R as a line is of slope 220/84 = 55/21. tan θ 25 26 √ 2 ⇒ D = 592 13 Using relative distance time = relative speed √ 2 1 296 ⇒ time = 592 .1 = 10.√ s = 13 4 + 100 13 Using √ ⇒ sA = 10 2 ( 296 13 s = ut ) ≃ 322. hence A is now 322 − 311.127 m from the intersection. hence (20. 40) ∈ V .9 m≈ 11 m from the intersection.006 m √ But A is initially 220 2 ≃ 311. Initially RQP = 20 i + 40 j.13 Problem: LC HL 2004: [Part(b)] (i) Using VAB = VA − VB ⇒ VQP = i − 8 j (ii) As a line VQP =: V has slope −8. Using L ≡ y − y1 = m(x − x1 ) ⇒ V ≡ y − 40 = −8(x − 20) ⇒ V ≡ y − 40 = −8x + 160 ⇒ V ≡ 8x + y − 200 = 0 . 4. 6 s ⇒ time = 2 65 time = 4.0) √ |c| + b2 a2 200 ⇒ d = √ ≈ 25 m 65 (iii) Figure 24: R denotes RQP . θ is the angle between V and R.LC Applied Maths Problem Set Solutions 85 Using d= |ax + by + c| √ a2 + b2 = (x. D is the distance Q has to travel relative to P up to the instant the particles are d-close together. d D d 3 ⇒D= = 25 tan θ 2 tan θ = Using relative distance relative speed 75 1 √ ≈ 4.14 Problem: LC HL 2003: [Part(b)] • VA = 7.5 i VB = 10 cos 60◦ i + 10 sin 60◦ j √ ⇒ VB = 5 i + 5 3 j ⇒ VAB = VA − VB √ ⇒ VAB = 2.90 ] 3 From Figure 24. m1 − m2 tan θ = ± 1 + m1 m2 −8 − 2 2 ⇒ tan θ = ± = ◦ 1 − 16 θ∈[0.y)=(0.5 i − 5 3 j . Using. noting as a line RAB =: R has slope 40/20 = 2. 52 + (5√3)2 √ 375/ 13 ⇒ time = √ 25 + 75 4 time = . √ √ √ From VAB = 2. √ 13 2. 1 D cos θ = √ = 375 13 375 ⇒D= √ m 13 Using relative distance relative speed 375 1 ⇒ time = √ .LC Applied Maths Problem Set Solutions 86 Figure 25: The model triangle for 60◦ • Initially RAB = −375 i Figure 26: D is the distance A has to travel relative to B up to the instant the particles are d-close together. From Figures 27 and 26.5 i − 5 3 j. tan θ = (5 3)/(5 /2) = 2 3. LC Applied Maths Problem Set Solutions 87 Figure 27: The model triangle for θ.4) = 4 m (ii) Consider the following diagram: Figure 28: Straight away it should be clear A = 30◦ .8) = 8 m sA = 10(0. √ √ √ (iii) From Figure 28. 4. h = 10 sin 60◦ = 10 3/2 = 5 3 and √ √ r =h−c=5 3− 7 . √ Now a = 10 sin 60◦ = 5 m (see Figure 25).15 Problem: LC HL 2006: [Part(b)] (i) Using s = ut sB = 10(0. c = 7. Hence b = 3 m and thus α = arctan( 7/3). h = 10 sin 60◦ and a = 10 cos 30◦ . 2 = i component of VGB 10 5 .LC Applied Maths Problem Set Solutions 88 Figure 29: The model triangle for α. 2 j 5 5 2 √4 7 3 ⇒ VG = − i − j 10 10 4 VB = − i 5 Using ( √ ) 7 j 10 √ 7 1 = VG − VB = i − j 2 10 VGB = VG − VB = ⇒ VGB 4 3 − 5 10 i− Now √ √ Figure 30: Clearly B = θ and so d = (5 3 − 7) cos θ. 2 i − . Now 2 2 VG = − cos α i − sin α j 5 5 √ 2 3 2 7 ⇒ VG = − . √ √ −j component of VGB 7 7 tan θ = = . 316 m 4 2 4.16 Problem: LC HL 2005: [Part (a)] To cross the river in the shortest time she must head straight across: VW C = u j VC = v i ⇒ VW = VW C + VC ⇒ VW = v i + u j The time taken to cross is: distance in the j direction speed in the j direction d ⇒ 10 = u ⇒ d = 10u In order to cross by the shortest path she must head upstream at an angle θ as shown to counteract the current: VW C = −u sin θ i + u cos θ j To counteract the current: u sin θ = v v ⇒ sin θ = u Now using distance in the j direction speed in the j direction d ⇒t= u cos θ 10u u ⇒ t = (10u) √ =√ u u2 − v 2 u2 − v 2 t= . √ ≈ 5. √ √ 5 d = (5 3 − 7).LC Applied Maths Problem Set Solutions 89 Figure 31: Model Triangle for θ. LC Applied Maths Problem Set Solutions 90 Figure 32: The woman must head upstream at a speed u at an angle θ to counteract the current. it must head at an angle θ against the wind as shown: VBW = 22 sin θ i + 22 cos θ j √ √ ⇒ VB = (22 sin θ − 9 2) i + (22 cos θ + 9 2) To counteract the current: √ 22 sin θ = 9√ 2 9 2 ⇒ sin θ = 22 √ ⇒ θ = arcsin(9 2/22) ≈ 35◦ .17 Problem: LC HL 2004: [Part (a)] (i) If the wind blows from SE then it blows in a NW direction: VW = −18 cos 45◦ i + 18 sin 45◦ j √ √ ⇒ VW = −9 2 i + 9 2 j In order for the bird to fly due North. 4. Figure 33: The model triangle for θ. LC Applied Maths Problem Set Solutions 91 Figure 34: The bird must head into the wind at a speed u at an angle θ to counteract the wind.15 s ⇒t= √ 322 + 9 2 t= .1507 ≈ 8. the adjacent side is √ 222 − 81(2) = 322 Now using distance in the j direction speed in the j direction 250 √ ⇒t= 22 cos θ + 9 2 250 √ ≃ 8. (ii) Figure 35: The √ model triangle for θ. Using Pythagoras Theorem. particle B is 1 m above ground and travelling at the same speed as particle A.LC Applied Maths Problem Set Solutions 5 92 Newton’s Laws and Connected Particles 5. . When particle A hits the ground the string is no longer taut and particle B travels freely under gravity.1 LC HL 2009: [Part (a)] (i) The force diagrams are: Figure 36: The force diagrams for particles A & B Hence 10g − T = 10a T − 5g = 5a ⇒ 5g = 15a g ⇒ a = m/s2 3 (62) (63) v 2 = u2 + 2as √ ⇒ v = u2 + √2as √ 2g ⇒ v = 2 (g/3) = 3 (65) (64) Using (ii) When particle A hits the ground. Hence the speed of B is a maximum when v = 0: v 2 − u2 s= 2a −2g/3 1 ⇒s= = m −2g 3 Hence particle B reaches a height 4/3 m above the ground. LC Applied Maths Problem Set Solutions 5. Hence 2T − mg = ma m1 g − T = 2m1 a ⇒ 2m1 g − 2T = 4m1 a ⇒ 2m1 g − mg = 4m1 a + ma ⇒ (2m1 − m)g = a(m + 4m1 ) (2m1 − m)g ⇒a= 4m1 + m (66) (67) (68) .2 93 LC HL 2008: [Part (a)] The force diagrams are as follows: Figure 37: Note that there are two tensions acting on pulley A. Also the acceleration of the particle is twice that of pulley A because if pulley A is raised a distance x. the particle will be lowered a distance 2x. LC Applied Maths Problem Set Solutions 5.3 94 LC HL 2006: [Part (a)] (i) The force diagrams: Hence 2 2 T− g= a 5 5 g a −T = 2 2 9 g ⇒ g= 10 10 g ⇒ a = m/s2 9 (69) (70) (ii) First find the speed when the 0.5 kg mass strikes the horizontal surface. Using v 2 = u2 + 2as (71) √ √ 2g ⇒ v = u2 + 2as = m/s 9 Once the 0.5 kg particle strikes the horizontal surface, the 0.4 kg particle acts as a projectile from this point and the string returns to taut when the 0.4 kg particle returns back √ down to this height. Now the constant√acceleration is −g, the initial speed is 2g/9 and by symmetry the final speed is − 2g/9. Now using v = u + at v−u ⇒t= √ √a − 2g 9 − 2g 9 ⇒t= −g √ √ 2 2g 8 9 ⇒t= √ = ≈ 0.30 s 9g g2 LC Applied Maths Problem Set Solutions 5.4 95 LC HL 2005: [Part (a)] (i) The force diagrams are: Figure 38: Note R = 4g and hence µR = g. Hence T − g = 4a 8g − T = 8a ⇒ 7g = 12a 7 ⇒ a = g m/s2 12 10 7 ⇒ T = g + 4a = g + g = g N 3 3 (ii) The force diagram is: Figure 39: Note R = 4g and hence µR = g. Now to add these put them in an i-j basis: F = −T i − T j 10 10 ⇒ F = − gi− gj 3 3 (72) (73) LC Applied Maths Problem Set Solutions 5.5 96 LC HL 2004: [Part (a)] (i) The force diagrams are: Hence 2mg − T = 2ma T − mg = ma ⇒ mg = 3ma g ⇒ a = m/s2 3 (74) (75) (ii) When the speed of the first particle is v, using v 2 = u2 + 2as v 2 − u2 ⇒s= 2a 3v 2 ⇒s= 2g However the second particle will travel the same distance up as the first particle travels this distance down: 3v 2 (76) ∴ vertical separation = g LC Applied Maths Problem Set Solutions 5.6 97 LC HL 2003: [Part (a)] The force diagrams are: Hence T −g =a S − T = 3a 6g − S = 6a ⇒ 5g = 10a g ⇒ a = m/s2 2 (77) (78) (79) (80) . Hence T − g = 5a 3g − T = 3a ⇒ 2g = 8a g ⇒a= 4 Using 1 s = ut + at2 2 g 1 (g ) 4= m ⇒s= 2 4 2 (81) (82) .7 98 LC HL 2000: [Part (a)] The force diagrams are: Figure 40: Note R = 5g and hence µR = g.LC Applied Maths Problem Set Solutions 5.