Question

A woman bowls a bowling ball with initial speed v. The ball rolls up a steep ramp, comes to a stop, and then rolls back to the woman's position. Assuming that the force of friction on the ball is small but significant, with what speed is the ball traveling when it returns to the woman?
a speed slightly less than v

a speed slightly greater than v

a speed exactly equal to v

a speed close to zero

Answer 1

The ball returns to the woman with a speed slightly less than v

Explanation:

We can solve this problem by using the law of conservation of energy. In fact, in absence of friction, the total mechanical energy of the ball would be conserved:

$E=U+K= const.$

where

U is the gravitational potential energy

K is the kinetic energy of the ball

In this problem, the ball is thrown up along the ramp. At the beginning (bottom) of the ramp, the ball has only kinetic energy, given by

$K=\frac{1}{2}mv^2$

where m is the mass of the ball and v its initial speed. As the ball goes up, its kinetic energy decreases (since v decreases) and its potential energy U increases (as the height, h, increases). When the ball stops, all the kinetic energy has been converted into potential energy. However, when the ball travels back down along the slope, the potential energy is converted back into kinetic energy (as the ball gains speed again): when it reaches again the bottom of the ramp, all the potential energy has been re-converted back into kinetic energy, which would be again

$K=\frac{1}{2}mv^2$

Because the mechanical energy must be conserved: so, the final speed of the ball in absence of friction would be the same.

However, in this situation we have friction acting along the slope. This means that the force of friction does some work on the ball during its motion, and therefore part of the total mechanical energy, E, is "lost" (converted into thermal energy due to friction) during the motion. This means that when the ball reaches the bottom again, its final kinetic energy will be less than its initial kinetic energy:

$K'$

And this also implies that its final speed will be less than its initial speed:

$v'$

so the correct answer is

a speed slightly less than v

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