Question

A 50.0 g Super ball travelling at 25.0 m/s bounces off a brick wall and rebounds at 22.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.50 ms, what is the magnitude of the average acceleration of the ball during this time interval?

Answer 1

Here let us consider that the ball initially traveling along the negative x-axis and after bounce, it travels along the positive x-axis. Therefore the initial velocity of the ball before bounce is v_i=$-25i( \tfrac{m}{s} )$ and after bounce the final velocity is

v_f=$22i( \tfrac{m}{s} )$ .

The time interval of contact of the ball with wall is $\Delta t=3.5ms=0.0035s.$

Now as we know the average acceleration is given by

$$a= \tfrac{\Delta v}{\Delta t}=\tfrac{\Delta v_{f}-\Delta v_{i}}{\Delta t}$$

Therefore the magnitude of the average acceleration:

$$a=1.34*10^{4} ( \tfrac{m}{s^{2}} )$$