Physics 121.6 2007/2008 Assignment 11- Solutions 1. Chapter 11, Problem 36. A puck of mass 80.0 g and radius 4.00 cm slides along an air table at a speed of 1.50 m/s as shown in Figure P11.36a. It makes a glancing collision with a second puck of radius 6.00 cm and mass 120 g (initially at rest) such that their rims just touch. Because their rims are coated with instant-acting glue, the pucks stick together and spin after the collision (Fig. P11.36b). (a) What is the angular momentum of the system relative to the center of mass? (b) What is the angular speed about the center of mass? Solution: (a) At the instant the rims the situation looks like… So m y + m2 y2 yCM = 1 1 m1 + m2 y1 = 4.00 cm 0 CM y2 = −6.00 cm 1.50 m/s (80 g)(4.00 cm) + (120 g)(−6.00 cm) (80 g + 120 g) = −2.00 cm Therefore, just before they stick, the angular momentum relative to the centre of mass is L = r1m1v1 = (6.00 × 10 −2 m)(80 × 10−3 kg)(1.50 m/s) = 7.20 × 10−3 kg ⋅ m 2 /s = (b) After they stick together the angular momentum about the CM will be the same. Since L = Iω , to find ω we must find I relative to the CM. For a disk through its centre the moment of inertia is 12 MR 2 so we can use the parallel axis theorem to find the moment of inertia relative to the CM. Therefore total moment of inertia is 2 2 2 2 I = 12 m1r1 + m1d1 + 12 m2 r2 + m2 d 2 = 12 (80 × 10 − 3 kg)(4.00 × 10− 2 m) 2 + (80 × 10 − 3 kg)(6.00 × 10− 2 m) 2 + 12 (120 × 10− 3 kg)(6.00 × 10− 2 m) 2 + (120 × 10− 3 kg)(4.00 × 10− 2 m) 2 = 7.60 × 10− 4 kg ⋅ m 2 Therefore ω = L 7.20 × 10 −3 kg ⋅ m 2 /s = = 9.47 rad/s 7.60 ×10 − 4 kg ⋅ m 2 I -1- P12.16). To change the direction of the axis of rotation of a spinning object.Solutions 2. Thus if the direction of L is to change. Thus the change in angular momentum is in the same direction as the torque. (D) Any of (A). Lost-aLot's mass combined with his armour and steed is 1 000 kg. (B) or (C).00 m long and has mass 2 000 kg.Assignment 11 . Sir Lost-a-Lot dons his armour and sets out from the castle on his trusty steed in his quest to improve communication between damsels and dragons (Fig. Since the torque is τ = 3. Changing the moment of inertia does not change L at all. The lift cable is attached to the bridge 5.0 m above the bridge. assume his weight acts on the bridge at 1. and to a point on the castle wall 12. -2- .00 m from the end of the bridge. Solution: dL . Lost-a-Lot and his horse stop when their combined center of mass is 1. (B) or (C) will work.00 m from the end. [i. (B) change the moment of inertia about the axis of rotation (C) apply a torque about an axis that is not the axis of rotation. Determine (a) the tension in the cable and the (b) horizontal and (c) vertical force components acting on the bridge at the hinge. Problem 16.e. one must (A) apply a torque about the axis of rotation. Unfortunately his squire lowered the drawbridge too far and finally stopped it 20.0° below the horizontal.00 m from the hinge at the castle end. the torque vector cannot be in the same direction as L. Thus the torque must be applied about an axis that does not coincide with the direction of L.] The uniform bridge is 8. (E) None of (A). in a time interval Δt the change in angular momentum is dt ΔL = τΔt . Chapter 12. Therefore the answer is C. 00 m)sin(51.00 m .1° 4. Mass of bridge = mB = 2000 kg. B y + 12.00 m .0 m x Ry Rx O 20° T lB/2 lT In order find the components of T. Mass of knight.0 m tan θ = = ⇒ θ = 71. we need to find the angle the tension force makes with the horizontal (θ) and with the bridge (ϕ).71 m b = lT cos 20° = (5.1°) = 3.70 m b ⇒ ϕ = θ − 20° = 51. lK = 7. lT = 5. O: ∑τ = Td − mB g (lB 2)cos 20° − mK gl K cos 20° = 0 …e -3- T ϕ θ .Assignment 11 . lK From the right angle triangles in the diagram at right: a = lT sin 20° = (5. horse and armour = mK = 1000 kg.71 m + 12.Solutions Solution: FBD of forces on bridge: Length of bridge lB = 8.00 m )sin (20°) = 1.1° The moment arm for T is d = lT sin ϕ = (5. and the torque due to T.89 m ∑F ∑F mBg mKg h d O lT a 20° b x = Rx − T cosθ = 0 …c y = Ry + T sin θ − mB g − mK g = 0 …d Take torques about hinge.70 m a + h 1.00 m )cos(20°) = 4.00 m . 80 m/s 2 )(4.0°) = 3.1°) = 1.19 × 103 N Thus this component points downward.80 m/s 2 )(7.15 × 104 N (c) From d Ry = mB g + mK g − T sin θ = (2000 kg)(9.1°) = −4.55 × 10 N (b) From c Rx = T cosθ = (3.Solutions (a) From e m g (l 2 )cos 20° + mK gl K cos 20° T= B B d (2000 kg)(9.55 × 104 N)cos(71.80 m/s2 ) − (3.0°) + (1000 kg)(9.89 m 4 = 3.80 m/s 2 ) + (1000 kg)(9.00 m) cos(20.55 × 104 N)sin(71.00 m) cos(20. -4- .Assignment 11 . 0 g)(5.00 cm) 2.0 g)(3. Problem 8.00 cm) = 9. light strings.00 cm) − m1 g (4. and (c) m3.0 g) g (3.00 cm) = 0 … d (12.0 g) g (5. and beach souvenirs as shown in Figure P12.5 g) g Consider forces on top bar: T3 ∑τ O = T2 (4.Assignment 11 .0 g) g − m1 g = 0 … c = (12.00 cm + O (12.00 cm) = m3 g (6.0 g (6.00 cm O 5.0 g) g Consider forces on second lowest bar: T From d m1 = y 2 ∑F ∑τ y O = T2 − m2 g − T1 = 0 … e = m2 g (2.00 cm 6.00 cm) ⇒ m3 = = 49. Chapter 12.00 cm) = 0 … f From f m2 g (2.00 cm) − T1 (5.00 cm) = 52.00 cm O From g (73. Determine the masses of the objects (a) m1.00 cm m2g (21.5 g) g + (21.Solutions 4.00 cm 4. Solution: (a) Consider forces on lowest bar: y T1 3.00 cm) = 0 … g 4.00 cm) T + x T1 ⇒ m2 = 2 (73.8.0 g) g m1g ∑F ∑τ y O x = T1 − (12.0 g) g = (73. A mobile is constructed of light rods.00 g) g = (21.00 cm) = (21.0 g + 9.00 cm) − m3 g (6.00 cm) (b) From c T1 = (12.5 g (2.5 g)(4.0 g) g + m1 g = (12.5 g) g (4.00 cm) -5- y + x m3g . (b) m2.00 cm) (c) From e T2 = m2 g + T1 = (52.00 g (4. Problem 39. and θ. Also the sign is supported by the two attachment points so half its weight is supported by each.39). From the triangle: l = (d + 2 L) sin θ (a) From e T (d + 2 L) sin θ = 12 Fg (2d + 2 L) = Fg (d + L) ⇒T = Fg (d + L) (d + 2 L) sin θ (b) From c Rx = T cosθ = Fg (d + L) cosθ (d + 2 L) sin θ and from d Ry = Fg − T sin θ = Fg − = Fg (d + L) (d + 2 L) tan θ Fg (d + L) (d + 2 L) sin θ sin θ d+L ⎞ ⎛ ⎛ d + 2 L − (d + L) ⎞ = Fg ⎜1 − ⎟ = Fg ⎜ ⎟ d + 2L ⎝ d + 2L ⎠ ⎝ ⎠ FL = g d + 2L -6- .Assignment 11 . (Alternatively we could consider its whole weight to be supported in its middle.) We do not know the direction of the reaction force of magnitude R. Note that it is a “light” beam so we ignore its weight. ∑ Fx = Rx − T cosθ = 0 … c ∑F ∑τ y O l R T + θ O ½ Fg d ½ Fg 2L = Ry + T sin θ − 12 Fg − 12 Fg = 0 … d = Tl − 12 Fg d − 12 Fg (d + 2 L) = 0 … e where l is the lever arm for the tension force. in terms of Fg. L. A uniform sign of weight Fg and width 2L hangs from a light horizontal beam hinged at the wall and supported by a cable (Fig. d. Chapter 12.Solutions 5. Determine (a) the tension in the cable and (b) the components of the reaction force exerted by the wall on the beam. Solution: FBD of forces on beam. P12. Both the Earth and the Moon orbit the Sun.67 ×10 −11 N ⋅ m 2 /kg 2 )(5. and Sun all lie on the same line.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)(7.Assignment 11 .36 × 1022 kg) = 4. MS = 1. ( ) -7- . rEM Moon Earth (a) Force exerted by Sun on Moon: GM S M M (6. Earth. the Moon.36 × 10 22 kg) FEM = = = 1. So rSM = rSE − rEM = 1.98 × 10 24 kg)(7.496 × 10 m) (d) The force of the Sun on the Moon is greater than the force of the Earth on the Moon.991 × 1030 kg.496 × 1011 m − 3.36 × 1022 kg. Chapter 13.492 × 10 m (b) Force exerted by Earth on Moon: GM E M M (6.39 × 1020 N = FSM = 2 2 11 rSM 1. with the Moon between the Earth and the Sun.98 × 1024 kg) = 3.98 × 1024 kg.84 × 108 m.99 × 1030 kg)(5. (a) What force is exerted by the Sun on the Moon? (b) What force is exerted by the Earth on the Moon? (c) What force is exerted by the Sun on the Earth? (d) Compare the answers from parts (a) and (b). Sun MM = 7. Problem 6. The Sun has captured the whole Earth-Moon system. rEM = 3.492 × 1011 m.55 × 1022 N = FSE = 2 2 11 rSE (1. ME = 5.67 × 10−11 N ⋅ m 2 /kg 2 )(1. In fact the orbit of the Moon is concave towards the Sun at all points with a little wobble due to the Earth presence (and the same could be said for the Earth).84 × 108 m = 1.84 ×10 m ) (c) Force exerted by Sun on Earth: GM S M E (6.496 × 1011 m.Solutions 6. Why doesn’t the Sun capture the Moon away from the Earth? Solution: rSM From textbook: rSE rSE = 1. During a solar eclipse.99 × 10 20 N 2 2 8 rEM (3. 5 m/s2. is greater than 9.8 m/s2. about 4.Assignment 11 . 2 RE And so at the position of the meteor: g M = 1 GM E g GM E GM E = = = 2 2 (2 RE ) 4 RE 2 4 RM 9. So g M ≈ -8- . ρ Moon ρ Earth .8 m/s 2 ≈ 2.667 ρ Earth (g E / RE ) g E RM g E ( RE / 4) 3 So ρ Earth = 8.Solutions 7. a value that depends on how the meteor is moving. Solution: For the Earth: ρ Earth = And g E = ME ME = 3 4 VE 3 πRE GM E g R ⇒ ME = E E 2 G RE 2 2 ME 3g R 3 gE = E 3E = 3 4 πRE 4πRE G 4πG RE 3 3 gM Similarly for the Moon: ρ Moon = 4πG RM We know: g M = g E / 6 and RM = 0. find the ratio of their average densities. The free-fall acceleration on the surface of the Moon is about one-sixth that on the surface of the Earth.250 RE.5 m/s 2 4 Answer A. If the radius of the Moon is about 0. 9. Solution: g= GM E at Earths surface. Problem 11. Chapter 13. The magnitude of the acceleration of a meteor at a height above the Earth’s surface equal to the radius of the Earth is (A) (B) (C) (D) (E) about 2.9 m/s2.250 RE = RE / 4 So: ρ Moon (g M / RM ) g M RE ( g E / 6) RE 2 = = = = = 0.8 m/s2. Solutions 9.99 × 1030 kg. These observations place limits on the size of these objects. Chapter 13. Chapter 13.Assignment 11 . Determine the greatest possible angular speed it can have so that the matter at the surface of the star on its equator is just held in orbit by the gravitational force. and the distance between the nose and the center of the black hole is 10. (b) What is the difference in the gravitational fields acting on the occupants in the nose of the ship and on those in the rear of the ship. P13. A spacecraft in the shape of a long cylinder has a length of 100 m and its mass with occupants is 1 000 kg. Many rotate very rapidly. The nose of the spacecraft points toward the black hole. MSun = 1.0 km.Thus: GMm = mac = mrω 2 2 r GM ⇒ω = r3 Now M = 2MSun.0 km ω= (6. 10.63 × 104 rad/s = 2.22). Suppose that the mass of a certain spherical neutron star is twice the mass of the Sun and its radius is 10. -9- . Problem 22.00 × 104 m)3 = 1. Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions.59 × 103 revolutions/s If the neutron star rotated faster than this. Solution: For the matter near the surface to be in orbit: Fg = mac .99 × 1030 kg) (1. Problem 18. the matter near its surface would be flung off! Many neutron stars (pulsars) have been found with very high revolution rates. It has strayed too close to a black hole having a mass 100 times that of the Sun (Fig.0 km.67 × 10−11 N ⋅ m 2 /kg 2 )2(1. r = 10. (a) Determine the total force on the spacecraft. farthest from the black hole? This difference in accelerations grows rapidly as the ship approaches the black hole. It puts the body of the ship under extreme tension and eventually tears it apart. Problem 26.31 × 1017 N r (1.0 km + 50 m = 1.000 × 104 m.16 × 107 m So height above Earth is h = rf − RE = 3.10 - −1 . To what height will it rise? Ignore air resistance and the rotation of the Earth.67 × 10 N ⋅ m /kg )(100)(1.37 × 106 m = 2. rfront = 1.010 × 104 m. At the Earth's surface a projectile is launched straight up at a speed of 10.0 × 103 m/s)2 ⎟⎟ − rf = ⎜⎜ 6 2 2 24 −11 ⎝ (6. and g back = g front = 2 2 (rfront ) (rback ) Δg = g front − g back ⎛ 1 ⎛ (rback )2 − (rfront )2 ⎞ 1 ⎞⎟ ⎜ ⎟ = GM − = GM ⎜ ⎜ (r )2 (r )2 ⎟ ⎜ (r r )2 ⎟ back front back ⎝ front ⎠ ⎝ ⎠ ( ) ( )( ) ⎛ 1. Total force on Spacecraft is: GMm (6.99 × 1030 kg)(1000 kg) Fg = 2 = = 1. 00 10 ⎝ ⎠ 12 12 2 ⇒ Δg = 2. .005 × 104 m. Solution: Conservation of Energy: Ki + U i = K f + U f h RE GM E m 1 GM E m 1 2 2 ⇒ mvi − = mv f − RE rf 2 2 rf 1 2 GM E GM E ⇒ vi − =− 2 RE rf 2 ⇒ vf = 0 vi 1 1 vi = − rf RE 2GM E ⎛ ⎞ 1 (10.Solutions Solution: (a) The distance of the centre of mass of the spacecraft from the black hole is r = 10.005 × 104 m)2 GM GM .67 × 10−11 N ⋅ m 2 /kg 2 )(100)(1.00 × 10 4 m 2 ⎞ ⎟ ⇒ Δg = (6.52 × 107 m. Chapter 13. −11 2 2 30 (( )) 11.62 × 10 m/s A very large difference in the acceleration of the front of the spacecraft relative to the back! Thus the spaceship will be torn apart.99 × 10 kg)⎜ 4 4 2 ⎜ ⎟ × × 1 . M = 100MSun.01 × 104 m 2 − 1.Assignment 11 .0 km/s. rback = 1.98 × 10 kg) ⎠ = 3.16 × 107 m − 6.67 × 10 N ⋅ m /kg )(5. 01 10 m 1 .37 × 10 m) 2(6.62 × 10 N/kg = 2. Find its distance away from the center of the Earth when it reaches the other end of the ellipse. (a) What is the speed v0 of the satellite? Suddenly. Solution: (a) Circular orbit: Mass of satellite is ms = 5m. Fg = ms ac GM E ms v ⇒ = ms 0 2 r r GM E ⇒ v0 = r v0 r 2 ms ME (b) Conservation of momentum: 5mv0 = m(0) + 4mvi Before After vi v0 ⇒ vi = 54 v0 4m 5m m (c) The initial position will be where the satellite piece is closest to Earth. (b) What is the speed vi of the larger piece immediately after the explosion? (c) Because of the increase in its speed.11 - rf vf r ME vi 4m . A satellite moves around the Earth in a circular orbit of radius r. Immediately after the explosion the smaller piece of mass m is stationary with respect to the Earth and falls directly toward the Earth. Chapter 13. Problem 38.Solutions 12. this larger piece now moves in a new elliptical orbit. with masses m and 4m. an explosion breaks the satellite into two pieces.Assignment 11 . Using conservation of angular momentum: r 4mvi = rf 4mv f ⇒ rf v f = rvi …c Conservation of Energy: K i + U i = K f + U f 1 2 (4m)vi − 2 ⇒ vi − 2 GM E 4m 1 GM E 4m 2 = 2 ( 4m)v f − r rf 2GM E 2GM E 2 …d = vf − r rf From c v f = ⇒ vi − 2 r vi rf 2GM E r 2 2 2GM E = 2 vi − r rf rf . so: v0 = 16 16 r 25 GM E 2GM E r 2 25 GM E 2GM E ⇒ − = 2 − r rf 16 r rf 16 r Also from (b) and (a): vi = 2 ⇒ r 25 25 r 2 −2= −2 2 rf 16 16 rf r 25 r 2 7 ⇒ −2 + =0 2 rf 16 16 rf ⇒ 7rf − 32rrf + 25r 2 = 0 2 − (−32r ) ± (−32r ) 2 − 4(7)(25r 2 ) 32r ± 324r 2 32 ± 18 = = r 2(7) 14 14 50 Thus: rf = r or 14 r . v2 r1 ⇒ v1 = v1 r2 = = 9r2 v2 Answer B. If Martian Orbiter I is sailing around the planet in a circle which has an orbital radius nine times that of Martian Orbiter II. what is the speed of Martian Orbiter I? (A) 19 v2 (B) 13 v2 (C) v2 (D) 3v2 (E) 81v2 Solution: For a circular orbit for Martian Orbiter I: GM M m1 v12 = m1ac = m1 where v1 is its speed. who’s speed is v2. v1 r = 2 and since r1 = 9r2. 13. r12 r1 GM M r1 Similarly for Martian Orbiter II: GM M ⇒ v2 = r2 Therefore.12 - . the first result is the initial value so the final value is ⇒ rf = rf = 25 7 r.Assignment 11 . m1 is its mass and r1 is its orbit radius.Solutions 25 2 25 GM E . 1 3 ⇒ v1 = 13 v2 . 98 × 1024 kg)(2.98 × 1024 kg La = L p va perihelion ⇒ ra mva = rp mv p ⇒ va = va = rp ra Sun rp ra aphelion vp vp (1.m 2 /kg 2 )(1. Determine (a) the Earth’s orbital speed at aphelion [do not use energy conservation to do this part].027 × 104 m/s.471 × 1011 m.521 × 1011 m) = 2. But.56 × 1033 J Ua = − (6. and the distance of closest approach (at perihelion) is 1.Solutions 14.Assignment 11 .98 × 1024 kg)(3.027 × 10 4 kg) 2 = 2.521 × 1011 m. since the mass of the Sun is so much larger than the mass of the Earth.m 2 /kg 2 )(1.13 - . . its speed is small and so its kinetic energy is negligible.) Solution: (a) Conservation of Angular Momentum Let m = mass of Earth = 5. The Sun does have some motion as the Sun-Earth system actually rotates about the centre of mass of the Sun-Earth system.027 × 104 m/s) (1.471 × 1011 m)(3.927 × 104 m/s (b) At perihelion: K p = 12 mv 2p = 12 (5.99 × 1030 kg)(5.67 × 10−11 N. The maximum distance from the Earth to the Sun (at aphelion) is 1.99 × 1030 kg)(5. so we see that the total mechanical energy is indeed conserved as we expect since there are no external forces doing work on the Sun-Earth system.98 × 1024 kg) GM S m Up = − =− = −5.22 × 1033 J (1.40 × 1033 J 11 (1.521 × 1011 m) ra Total energy at perihelion: E p = K p + U p = −2.66 × 1055 J They agree. Is the total energy of the system constant? (Ignore the effect of the Moon and other planets.67 × 10−11 N. The Earth’s orbital speed at perihelion is 3.98 × 1024 kg) GM S m =− = −5.927 × 104 kg) 2 = 2.66 × 1055 J Total energy at aphelion: Ea = K a + U a = −2. (b) the kinetic and potential energies of the EarthSun system at perihelion.471 × 10 m) rp (c) At aphelion: K a = 12 mva2 = 12 (5. Problem 52.74 × 1033 J (6. Chapter 13. Note: We have ignored the kinetic energy of the Sun in this calculation. and (c) the kinetic and potential energies at aphelion. This is a clear illustration that on the Earth it is there air pressure that pushes the water up the straw. Problem 14.80 m/s 2 ) P = PA (b) On the moon there is no atmosphere.] Solution: We assume that the pressure at the Earth’s surface is the total weight of the atmosphere spread over the surface area of the Earth. Chapter 14. In this case the pressure at the top is zero.3 m ρg (1000 kg/m 3 )(9. not the vacuum that ‘pulls’ it up. and atmospheric pressure at the Earth’s surface is 1.14 - . where m is to total mass of the atmosphere. Find the difference between the water levels inside and outside the straw. So mg P0 = 4πr 2 4πr 2 P0 4π (6. the Man of Steel repeats his attempt on the Moon.e. which has no atmosphere. throughout the height of the atmosphere.01×105 Pa PA h = = 10. Figure P14.14 shows Superman attempting to drink water through a very long straw.37 × 106 m)2 (1. With his great strength he achieves maximum possible suction. (b) What If? Still thirsty. (a) Find the maximum height through which he can lift the water. Problem 4. If the radius of the Earth is r. What is the total mass of the Earth’s atmosphere? (The radius of the Earth is 6.013 × 105 Pa. and assuming g does not change A much with height. ⇒h= . A = 4πr 2 .013 × 105 Pa) ⇒m= = = 5.3 × 1018 kg 2 (9. P = P0 + ρgh ⇒ PA = 0 + ρgh P0 = 0 1. Pressure at the bottom is atmospheric pressure PA. The atmosphere is not very thick compared to the radius of the Earth so you may assume that the acceleration due to gravity is approximately constant. mg P0 = .) [This is an estimate only. i. On the Moon there is no atmosphere to do the pushing. at 9.80 m/s ) g 16.Solutions 15.Assignment 11 . so both P and P0 are zero.8 m/s2. The walls of the tubular straw do not collapse. Solution: (a) With maximum possible suction there will be a vacuum at the top of the tube.37 × 106 m. Chapter 14. Thus h = 0. You can ignore the mountains and treat the Earth as a sphere. Assignment 11 . what distance h does the mercury rise in the left arm? Solution: (a) Mass of water column: m = ρV = ρA2 hw where hw = height of water column.18a. In the right hand column of water: P = P0 + ρghw …d In the left hand column of mercury: P = P0 + ρ Hg gh1 …e ⎛A ⎞ A1 h + h = ⎜⎜ 1 + 1⎟⎟h A1 ⎝ A2 ⎠ Equating d and e and inserting h1: ⎛ A + A2 ⎞ ⎛A ⎞ ⎟⎟h ρghw = ρ Hg gh1 = ρ Hg g ⎜⎜ 1 + 1⎟⎟h = ρ Hg g ⎜⎜ 1 ⎝ A2 ⎠ ⎝ A2 ⎠ Noting that from c h1 = h2 + h = ρhw (1. Chapter 14.0 cm2.00 m/cm3 )(5.6 g/cm )⎜⎜ 2 ⎝ 5. So A1h = A2 h2 …c The pressure at the top of the mercury in the right had column (P) must be equal to the pressure in the mercury at the same level in the left hand column.Solutions 17.00 cm 2 ) P0 A2 A (b) The right hand column of mercury has been P0 1 depressed by a height h2 and the left hand column h h1 hw original of mercury has risen by height h. Mercury is poured into a U-tube as in Figure P14. (100 g) m ⇒ hw = = = 20.00 cm2.18b.0 cm ρA2 (1.00 cm ⎠ ⎝ A2 ⎠ . This must be equal to the additional volume of mercury in the left hand column ( A1h ). (b) Given that the density of mercury is 13.490 cm 2 ⎛ A1 ⎞ ⎞ 3 ⎛ 10. and the right arm has a cross-sectional area A2 of 5. (a) Determine the length of the water column in the right arm of the U-tube.6 g/cm3.00 g/cm3 )(20. A volume h 2 height ( A2 h2 ) has been displaced by water in the right P P hand column.0 cm + 1⎟⎟ ρ Hg ⎜⎜ + 1⎟⎟ (13.0 cm) ⇒h= = = 0. The left arm of the tube has crosssectional area A1 of 10. Problem 16. One hundred grams of water are then poured into the right arm as in Figure P14. ρ = density of water.15 - . (B) Some liquid stays in the straw. but its level initially drops. Solution: Some liquid will begin to run out of the straw.Assignment 11 . Soon the pressure in this volume will be low enough so that the pressure difference between the bottom and the top of the liquid in the straw will be enough to support it. (D) The result depends on the type of liquid. As it does so volume between the top of the liquid and your finger increases. (A) The liquid will always run out of the straw. then raise the straw up and out of the liquid.Solutions 18. (C) The liquid will all stay in the straw. tightly cover the upper end with a finger. and so the pressure in that volume will decrease since no air can enter. Suppose you stick a drinking straw into a deep cup of liquid. Patm .16 - . Try it! P < Patm straw h Answer B.