A 53.5-g Super Ball traveling at 27.5 m/s bounces off a brick wall and rebounds at 21.5 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 4.05 ms, what is the magnitude of the average acceleration of the ball during this time interval

Explanation:

Acceleration is the rate of change of velocity of a body. It is expressed as;

acceleration = change in velocity/time

acceleration a = v-u/t

Given

v = 27.5m/s

u = -21.5m/s (rebounds in the opposite direction)

time t = 4.05ms = 0.00405secs

Required

acceleration

Substitute the given values into the formula

a = 27.5-(-21.5)/0.00405

a = 27.5+21.5/0.00405

a = 49/0.00405

a = 12,098.77m/s²

Hence  the magnitude of the average acceleration of the ball during this time interval is 12,098.77m/s²

we need to know the time interval during which the force acts. impulse of force = momentum change