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- University of Kansas
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- Physics 111
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- Physics- Ch 6 solutions.pdf

Ashley B.

PHSX 111 Chapter 6 Exercise and Problem Solutions S 09 Exercises 5, 15, 23, 28, 34, 48 5. Ropes that stretch easily are better for mountain climbing because they extend the time over which an impulse occurs, thereby reducing the force a climber feels in jumps, falls, etc. Compare the effect of jumping with a bungy cord versus with a steel cable. 15. A fire fighter is rocked onto his heels in reaction to the force exerted by the large volume of water from his hose. 23. The loose couplings allow the cars to start sequentially from the front. In this way, each car starts individually so that the impulse needed to start the train is divided into a series of smaller impulses and spread over far greater time. In this way the force needed to start the train is greatly reduced. 28. No. The momentum of the system comprised of you and the ball is 0 before you throw and must be 0 afterward. If you make as if to throw but do not, no momentum is imparted to the ball and therefore none can be imparted to you. 34. The component of gravity acting down the incline produces an impulse on the box that causes it to move down the inclined plane. 48. The force is equal and opposite and the impulses and changes in momentum are the same. The Ford experiences the greater acceleration because of its smaller mass. Problems 1, 4, 5, 7, 8 1. Given: mass (m) = 50 kg, speed (v) = 4 m/s and stopping time (∆t) = 3 s. We are asked to find the force of friction (F). Solution: we know F = ma and a = (∆v)/∆t. So F = m(∆v)/∆t = 50 kg 4 m/s/3 s = 66.7 kg m/s2 = 66.7 N. Units check, numbers check. 4. Given: mass (m) = 1000 kg, speed of hitting the ground (v) = 30 m/s. We are asked: a) What was the impulse? Answer: F∆t = ∆(mv) = 1000 kg 30 m/s = 30000 N s. b) What is the force of impact on the car? Answer: We do not know because we do not know ∆t. 5. Given: mass of baseball (m) = 0.15 kg, falls straight downward into the hands of a fan at (v) = 40 m/s. We are asked: a) What is the impulse required to stop the ball? Answer: As above F∆t = ∆(mv) = .15 kg 40 m/s = 6.0 N s. b) If the ball stops in .03 s, what is the average force on the fan’s hand? Answer: F = ∆(mv)/∆t = 6.0 kg m/s/.03s = 200 N Units and numbers check. 7. Given: a diesel engine weighs m and a freight car weighs m/4. The engine coasts into the freight car at v = 5 km/hr. We are asked to find (vf) the speed of the coupled pair after the collision. Answer: pi = mv + m/4vfc = pf = (m + m/4)vf. But the speed of the freight car (vfc) is 0. So vf = mv/(m+m/4) = mv/(5m/4) = (4/5)v = (4/5)5 km/hr = 4 km/hr. Units and numbers check. 8. Given: mass fish 1 (m1) = 5 kg, mass fish 2 (m2) = 1 kg. Velocity of fish 1 (v1) = 1m/s toward fish 2 that is swimming toward it with a speed (v2) great enough to stop fish 1 when it swallows fish 2. We are asked to find fish 2’s speed. Answer: We know that pi, the initial momentum of the fish1 fish 2 system, must be zero because the the final momentum, pf, is 0. So m1v1 + m2v2 = 0 , or m1v1 = -m2v2, where negative sign means the momenta are in opposite directions (toward each other in this case). Solving for: v2 = - m1/m2v1 = 0 = -5 kg/1 kg 1 m/s = -5 m/s. Units and numbers check. Owner chapter 6 solutions

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