Link to Figure: http://www.twitpic.com/10mkku A car is released from rest on a f

Question

Link to Figure: http://www.twitpic.com/10mkku A car is released from rest on a frictionless inclined plane (Figure 5.3). If a car suffers a nearly elastic collision it will coast back up the ramp a distance L_f before reversing direction. If the car has a mass of 0.3 kg, the ratio of height to width of the ramp is 12/75, the initial displacement is 1.95 m, and the change in momentum is 0.82 kg*m/s, how far will it coast back up the ramp before changing directions? Please help me this question! I dont understand how to approach it. I have been stuck on it for a while now. I keep getting 0.0090m but it is incorrect. Any help would be appreciated! Thank you!

Explanation / Answer

tan theta = 12/75 = 0.16

displacement = 1.95 m for final velocity = 0.82/0.3 = 2.733 m/s

when it is going back , the downward acceleration will be = g sin theta = 9.8* 0.1579 = 1.54742 m/s2

when there is an elastic collision , the velocity will remain the same

therfore distance = (v^2/2a) = (2.733^2/2*1.54742) = 2.14 m

it goes up a distance of 2.14 m

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