Phys 141_HW Sol_Ch9_W08

Phys 141_HW Sol_Ch9_W08 - Cal Poly State University SLO...

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Cal Poly State University, SLO D. Niebuhr Physics Department Winter 2008 Phys 141 Homework Solutions Chapter 9 Assigned Problems: Ch. 9: 2, 9, 10, 13, 19, 21, 24, 26, 28, 38, 42, 49, 57 9.2. Model: Model the bicycle and its rider as a particle. Also model the car as a particle. Solve: From the definition of momentum, car bicycle p p = car car car bicycle bicycle bicycle car bicycle m m v m v v v m = = ( 29 1500 kg 5.0 m/s 75.0 m/s 100 kg = = Assess: This is a very high speed ( 168 mph). This problem shows the importance of mass in comparing two momenta. 9.9. Model: Use the particle model for the falling object and the impulse-momentum theorem. Visualize: Note that the object is acted on by the gravitational force, whose magnitude is mg . Solve: Using the impulse-momentum theorem, f i f i ( ) t y y y y t p p J F t dt - = = f i y y mv mv mg t - = - i f y y v v t g - ⇒ ∆ = ( 29 2 5.5 m/s 10.4 m/s 0.50 s 9.8 m/s - - - = = Assess: Since y F mg = - is independent of time, we have taken it out of the impulse integral. 9.10. Model: Model the tennis ball as a particle, and its interaction with the wall as a collision.
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D. Niebuhr Physics Department Winter 2008 Phys 141 Visualize: The force increases to F max during the first two ms, stays at F max for two ms, and then decreases to zero during the last two ms. The graph shows that F x is positive, so the force acts to the right. Solve: Using the impulse-momentum theorem f i x x x p p J = + , ( 29 ( 29 ( 29 ( 29 ( 29 6 ms 0 0.06 kg 32 m/s 0.06 kg 32 m/s x F t dt = - + The impulse is ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 6 ms max max max max 0 max 1 1 area under force curve 0.002 s 0.002 s 0.002 s 0.004 s 2 2 0.06 kg 32 m/s 0.06 kg 32 m/s 960 N 0.004 s x F t dx F F F F F = = + + = + = = 9.13. Model: Choose car + gravel to be the system. Visualize: Solve: There are no external forces on the car + gravel system, so the horizontal momentum is conserved. This means p f x = p i x . Hence, ( 29 ( 29 ( 29 ( 29 ( 29 f 10,000 kg 4,000 kg 10,000 kg 2.0 m/s 4,000 kg 0.0 m/s x v + = + f 1.43 m/s x v = 9.19. Model: The two cars are not an isolated system because of external frictional forces. But during the collision friction is not going to be significant. Within the impulse approximation, the momentum of the
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Phys 141_HW Sol_Ch9_W08 - Cal Poly State University SLO...

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