Impulse-momentum: During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. If the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball?

Question

Quizplus · Accepted Answer

Using the impulse-momentum theorem, we have:

$$	ext{Impulse} = 	ext{Change in momentum}$$

$$	ext{Force} 	imes 	ext{Time} = \Delta 	ext{p} = 	ext{p}_{	ext{f}} - 	ext{p}_{	ext{i}}$$

$$	ext{Force} = \frac{\Delta 	ext{p}}{	ext{Time}} = \frac{	ext{p}_{	ext{f}} - 	ext{p}_{	ext{i}}}{	ext{Time}}$$

where $\Delta 	ext{p}$ is the change in momentum and $	ext{Time}$ is the time the ball was in contact with the wall.

We can calculate the change in momentum as:

$$\Delta 	ext{p} = 	ext{p}_{	ext{f}} - 	ext{p}_{	ext{i}} = 	ext{m}(	ext{v}_{	ext{f}} - 	ext{v}_{	ext{i}})$$

Plugging in the given values:

$$\Delta 	ext{p} = 0.200 	ext{ kg} 	imes (12.0 	ext{ m/s} - (-20.0 	ext{ m/s})) = 6.40 	ext{ kg m/s}$$

And the time given is $	ext{Time} = 60.0 	ext{ ms} = 0.060 	ext{ s}$.

Substituting these into the equation for force:

$$	ext{Force} = \frac{\Delta 	ext{p}}{	ext{Time}} = \frac{6.40 	ext{ kg m/s}}{0.060 	ext{ s}} = 107 	ext{ N}$$

Therefore, the magnitude of the average force applied to the ball is 107 N. The answer is (B).

Impulse-Momentum: During a Collision with a Wall, the Velocity of a 0.200-Kg