A ball is released from rest at the top of a ramp. The ramp is on wheels, and is

Question

A ball is released from rest at the top of a ramp. The ramp is on wheels, and is free to roll either direction. How fast, and in what direction, will the ramp be moving the instant the ball rolls off the bottom of the ramp? The ramp has a mass of 2.5 kilograms; the ball has a mass of 90 g, and the height of the ramp is 1 meter. Assume there is no friction or air resistance.  Also, assume that once it leaves the ramp it is essentially at 0 m from the ground, so it has no gravitational potential energy.

Explanation / Answer

E = mgh + .5mv^2

0 = (.09)(9.8)(1) - (.5)(2.5+.09)v^2

Because the ball will have 0 height at the bottom of the ramp, all of the energy must be converted into kinetic energy. Also, the ramp and ball will have the same momentum at the point when the ball falls off the ramp because the ball and the ramp exert an equal and opposite force on each other in order to cause them to move in opposite directions. So:

KE = .5mv^2 = 8.82

m = 2.59

v = 2.61

(2.59)(2.61) = (.09)v1 + (2.5)(v2)

because (.09)v1 = (2.5)v2, we can set them equal to each other and simply divide the left side by 2 to get one of the values and plug in for the other value.

(2.59)(2.61) = (2)(.09)v1

v1 = 37.555

v2 = 1.35198

The ramp will be moving in the -x direction with a speed of -1.35198 m/s

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