Question

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 64 ms, what was the magnitude of the average force (in N) applied to the ball?

Answer 1

The magnitude of the average force applied to the ball during its collision with a wall is 100 N, calculated using the change in momentum and the time interval.

Step-by-step explanation:

To calculate the magnitude of the average force applied to the ball during the collision, we use the impulse-momentum theorem, which states that the impulse applied to an object is equal to the change in the object's momentum. The impulse can be calculated as the product of the average force and the time interval during which the force is applied.

The change in momentum (Δp) is given by the difference in the final momentum (pf) and the initial momentum (pi). The final velocity is 12.0 m/s away from the wall, so its direction is opposite the initial velocity which was 20.0 m/s toward the wall. Thus, the final momentum is 0.200 kg × 12.0 m/s and the initial momentum is 0.200 kg × (-20.0 m/s).

The change in momentum is:
Δp = pf - pi = (0.200 kg × 12.0 m/s) - (0.200 kg × (-20.0 m/s)) = 2.4 kg·m/s + 4 kg·m/s = 6.4 kg·m/s

Time interval (t) = 64 ms = 0.064 s (converting milliseconds to seconds)

Using the formula for impulse (Δp = F_avg × t), we get:
F_avg = Δp / t = 6.4 kg·m/s / 0.064 s = 100 N

Therefore, the average force applied to the ball during its collision with the wall is 100 N.