Question

Collision derivation problem. A car is released from rest on a frictionless inclined plane (Figure 5.3). EXAMPLES: Calculate the momentum pi at the end of the plane in terms of the measured quantities x, y, L, and m. Assume is very small so that h/L is approximately equal to y/x. (Hint: use conservation of energy and the fact that K=1 2mv2=p2 2m.)
[Answer: .]
If a car suffers a nearly elastic collision it will coast back up the ramp a distance Lf before reversing direction. What is the momentum pf immediately following the collision? The general expression for the change in momentum suffered in a collision is is = - . What is p (the magnitude of ) in terms of x, y, Li, Lf, and m?
[Answer: .]
This is the expression you should use in the experiment. Make sure you understand how to derive these equations.
QUESTION:
If the car has a mass of 0.3 kg, the ratio of height to width of the ramp is 11/110, the initial displacement is 1.8 m, and the change in momentum is 0.62 kg*m/s, how far will it coast back up the ramp before changing directions?

Answer 1

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. If this helps then kindly rate 5-stars

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 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

solution:

tan? = 13/90  ---> ? = 8.219o

Use energy to get the speed at bottom:

PE = KE = mgh = m g (Lsin?)

     = 0.3 kg * 9.8 * (2.05 * sin 8.219)

     = 0.8616 J

v = (2 * KE / m)1/2

   = 2.397 m/s

Momentum at bottom:  p = mv

= 0.719 kg m/s towards the left

Logically, change in momentum is 0.80 kg m/s to the right due to the force acting on the block

So momentum after

= - 0.719 + 0.80 = 0.0801 kg m/s

Speed after = 0.27 m/s

PE gained = KE = 1/2 * 0.272 * 0.3 kg

               = 0.0109 J

?Lsin? = PE / mg

           = 0.00372 m

? L = 0.026 m

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