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Unformatted text preview: Therefore, the final momentum must be north of northeast, which is path B. Exercise 12. We begin by calculating the momentum of the ball before and after the collision with the wall. We choose the direction toward the wall to be the positive direction. ( 29 ( 29 ( 29 ( 29 0.2 kg 16 m/s 3.2 kg m/s 0.2 kg 8 m/s 1.6 kg m/s i i f f p mv p mv = = = × = =-= -× We now calculate the change in the momentum, being careful of the plus and minus signs. We know that the impulse is equal to the change in momentum. Therefore, we can use the given time to calculate the average force acting on the ball. ( 29 1.6 3.2 kg m/s 4.8 kg m/s 4.8 kg m/s 120 N 0.04 s f i av av p p p I F t p p F t ∆ =-= --× = -× = ∆ = ∆ ∆-× = = = -∆ Therefore, the average force has a magnitude of 240 N (and acts away from the wall)....
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This document was uploaded on 10/29/2011 for the course PHSX 103 at MSU Bozeman.
- Fall '02