If the car has a mass of 0.3 kg, the ratio of height to width of the ramp is 13/

Question

If the car has a mass of 0.3 kg, the ratio of height to width of the ramp is 13/90, the initial displacement is 2.05 m, and the change in momentum is 0.80 kg*m/s, how far will it coast back up the ramp before changing directions?

_________ m
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. Use the follow equations:If the car has a mass of 0.3 kg, the ratio of height to width of the ramp is 13/90, the initial displacement is 2.05 m, and the change in momentum is 0.80 kg*m/s, how far will it coast back up the ramp before changing directions?

Explanation / Answer

m = 0.3 kg, y/x = 13/90, Li = 2.05 m, p = 0.80 kg*m/s, find Lf p = m*sqrt(2gy/x) * [sqrt(Li) + sqrt(Lf)] sqrt(Li) + sqrt(Lf) = p/[m*sqrt(2gy/x)] sqrt(Lf) = p/[m*sqrt(2gy/x)] - sqrt(Li) so Lf = {p/[m*sqrt(2gy/x)] - sqrt(Li)}^2 = 0.0234 m

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