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**Unformatted text preview: **Chapter G - Problems Blinn College - Physics 2425 - Terry Honan Problem G.1 A 0.12 kg ball dropped from a height of 2.0 m rebounds to 1.8 m. (a) What is the change in the ball's momentum during its collision with the floor? (b) Suppose the ball is in contact with the floor for 0.08 s. What is the average force of the floor on the ball? Solution to G.1 (a) We can relate the speed at the ground to the height of the ball by equating the energy at the top and the bottom. In both cs This gives the speeds before and after hitting the ground. v bottom 2 = v top 2 + 2 g h ï v = ± 2 g h Choose upward as the positive direction. v i = - 2 g h i = - 2 μ 9.8 μ 2 = - 6.2610 v f = + 2 g h f = 2 μ 9.8 μ 1.8 = 5.9397 D p = m I v f- v i M = 0.12 μ H 5.9397-- 6.2610 L = 1.464 kg m s (b)The impulse-momentum gives the average force F D t = D p ï F = 1.464 0.08 = 18.3 N Problem G.2 F H N L t H s L 0.05 0.03 80 The graph above is the force vs. time for some collision on a 0.25 kg ball that is initially at rest. (a) What is the impulse given to the ball? (b) What is the average force acting on the ball during the collision? (c) What is the velocity of the ball after the collision? Solution to G.2 (a) The impulse I = Ÿ t i t f F H t L „ t has the graphical interpretation of the area under the F vs. t graph. In this case I = 1 2 base μ height = 1 2 0.02 μ 80 = 0.8 N ÿ s . (b) I = F D t gives the average force. F = 0.8 0.02 = 40 N (c) The impulse momentum theorem gives the final velocity. I = D p = m I v f- v i M ï 0.8 = 0.25 I v f- M ï v f = 3.2 m s Problem G.3 A ball of mass m bounces elastically with a wall. The ball's speed is v before and after the collision and q , the angle the ball makes with the wall, is the same before and after the collision. It the ball is in contact with the wall for a time T , then what is the average fore of the wall on the ball? Solution to G.3 v sin q v sin q v ” i v ” f D v q q F D t = D p = m I v ” f- v ” i M Because the speeds and angles before and after the collision are the same the parallel components are unchanged. Consider the perpendicular components. 2 Chapter G - Problems F ¶ D t = D p ¶ = m I v f , ¶- v i , ¶ M ï F ¶ T = m H v sin q-- v sin q L ï F ¶ = 2 m v T sin q Problem G.4 A 2 kg mass has a velocity of X 4,- 3 \ m ê s and a 3 kg mass has a velocity of X- 1, 5 \ m ê s . What is the total momentum of the system....

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