HW_09_Solutions_F05 - Physics 112 HW#9 Solutions 8-37[Collision on a Frozen Pond Conservation of momentum gives ms vsi = msvscos mavAcos mA vAi =

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Physics 112 – HW #9 Solutions Fall 2005 1 S A v si v Ai S A v A v s θ=37 φ=23 8-37 [Collision on a Frozen Pond] Conservation of momentum gives: m s v si = m s v s cos θ + m a v A cos φ m A v Ai = m s v s sin θ - m a v A sin φ which give: v si = (6.0 m/s) cos 37 ° + 50.0kg 80.0kg (9.0 m/s) cos 23 ° = 9.97m/s v Ai = 80.0kg 50.0kg (6.0 m/s) sin 37 ° + (9.0 m/s) sin 23 ° = 2.26m/s The energy loss is: E f - E i = ½ m s (v s 2 - v si 2 ) + ½ m A (v A 2 - v Ai 2 ) = ½ (80 kg) {(6.0 m/s) 2 - (9.97 m/s) 2 )} + ½ (50 kg) {(9.0 m/s) 2 -(2.26 m/s) 2 } = 640 J 8-38 [Blocks with Spring Bumpers] (a) At the instant when the spring is under maximum compression, the blocks are moving together with the same velocity, V. Use conservation of momentum to find the velocity of the blocks: m A v A,i = (m A + m B )V V = (m A v A,i ) m A + m B = (2.00 kg)(2.00 m/s) 2.00 kg + 10.0 kg = 0.333 m/s Now use conservation of energy to find how much energy U s is stored in the spring at that instant: K A,i = K A+B ½ m A v a,i 2 = U s + ½ (m A + m B )V 2 U s = ½ m A v a,i 2 - ½ (m A + m B )V 2 = ½ (2.00 kg)(2.00 m/s) 2 - ½ (12.00 kg)(3.33 m/s) 2 = 3.33 J (b) Since the spring bumpers are “ideal”, the collision is elastic . For an elastic collision, we can use Equs 8.24 and 8.25 in your text: v A,f = m A - m B m A + m B v A,i = 2.00 kg - 10.00 kg 2.00 kg + 10.00 kg (2.00 m/s) = -1.33 m/s (moving to left) v B,f = 2m A m A + m B v A,i = 2(2.00 kg) 2.00 kg + 10.00 kg (2.00 m/s) = 0.67 m/s (moving to right)
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Physics 112 – HW #9 Solutions Fall 2005 2 8.56 [Rocket in Space] (a) v(t) – v o = v ex ln m m 0 , so m 0 m = exp v - v 0 v ex = exp 8.00 x 10 3 m/s - 0 m/s 2100 m/s = 45.1 8-65 [Railroad Handcar] (a) Since the mass is thrown sideways out of the car in the car's reference frame, it's momentum component along the car's direction of motion doesn't change. So the car's momentum and speed don't change. v car f = 5.00 m/s East (b) The mass's final velocity with respect to the track is 0. Use Cons. of Momentum: Mv car o = m(0) + (M - m)v car f v car f = Mv car o M - m = (200 kg)(5.00 m/s) 200 kg - 25 kg = 5.71 m/s East (c) Cons. of Momentum again: Mv car o - mv throw = (M + m)v car f v car f = Mv car o - mv throw M + m = (200 kg)(5.00 m/s) - (25 kg)(6.00 m/s) 200 kg + 25 kg = 3.78 m/s East 8-94 [Walking in a Canoe] The keys here are: (1) the fact that the CM of the system doesn't move relative to the water; and (2) symmetry. Relative to the canoe, the woman's final position in the canoe's right side is symmetrically located about the canoe's center with respect to her starting position in the left side. So the final position of the canoe and woman relative to the water can be found simply by rotating the picture 180° about the system's CM.
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This homework help was uploaded on 04/21/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell University (Engineering School).

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HW_09_Solutions_F05 - Physics 112 HW#9 Solutions 8-37[Collision on a Frozen Pond Conservation of momentum gives ms vsi = msvscos mavAcos mA vAi =

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