sol14 - Therefore, the final momentum must be north of...

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PHYSICS 103 Chapter 6 -- HW #14 Questions 20. The product of force and time that yields the largest impulse will produce the largest change in momentum. Using the data given, we obtain 6 N × 3 s = 18 N s and 4 N × 5 s = 20 N s. Therefore, the second pair produces the larger change in momentum. 28. We begin by calculating the initial momenta of the two particles. For the left-hand mass we have mV to the right and for the right-hand mass we have 3 mV to the left. Therefore the initial momentum of the system is 2 mV to the left. Because the momentum of the system must be to the left after the collision, the conjoined pair will be moving to the left. 38. The magnitude of the momentum of the eastbound object is mV and the magnitude of the momentum of the northbound object is 2 mV . The vector sum of these two vectors points more north than east.
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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.

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