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Solutions to Chapter 6 Review Problems

# The floor also needs to support the weight of the

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The floor also needs to support the weight of the ball during impact. w = mg = (0.200 kg)(9.80 m/s 2 ) = 1.96 N. So the total force is 22.0 N + 1.96 N = 24.0 N upward . 26. (a) Hitting it back requires a greater force. Force is proportional to the change in momentum. When a ball changes its direction, the change in momentum is greater. If p = p 1 p 2 and p 1 and p 2 are opposite (hitting it back), then p = p 1 – ( p 2 ) = p 1 + p 2 . Had you caught the ball, then p 2 = 0 so p = p 1 . (b) The final velocity is opposite to the initial velocity (“hitting it back”). F avg = p t = mv mv o t = (0.45 kg)( 7.0 m/s 4.0 m/s) 0.040 s = 1.2 × 10 2 N = 1.2 × 10 2 N in direction opposite v o .

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27. F avg t = mv mv o = mv o , F avg = mv o t . So the magnitude is mv o t . F avg1 = (0.16 kg)(25 m/s) 3.5 × 10 3 s = 1.1 × 10 3 N ; F avg2 = (0.16 kg)(25 m/s) 8.5 × 10 3 s = 4.7 × 10 2 N . 28. (a) Impulse = area of the trapezoid = 1 2 (0.30 s + 0.14 s)(900 N) = 76.5 N = 77 N·s . (b) F avg = impulse t = 76.5 N·s 0.14 s = 5.5 × 10 2 N . (c) F avg t = mv mv o , so v = v o + F avg t m = 6.0 m/s + 76.5 N·s 3.0 kg = 32 m/s . 35. 40 km/h = 11.1 m/s, 2400 lb = 10 680 N. The force on the infant is opposite to the velocity. F avg t = mv mv o , t = mv mv o F avg = (55 kg)(0 11.1 m/s) 10 680 N = 0.057 s . 40.
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The floor also needs to support the weight of the ball...

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