Question

A 52.0-g Super Ball traveling at 28.5 m/s bounces off a brick wall and rebounds at 15.5 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.45 ms, what is the magnitude of the average acceleration of the ball during this time interval

Answer 1

The magnitude of the average acceleration of the ball during given time interval is 12753.62 m/s².

Step-by-step explanation:

Given that,

Mass of ball = 52.0 g

Initial speed u= 28.5 m/s

Final speed v=-15.5 m/s

Time
`t = 3.45*10^(-3)\ s`

We need to calculate the acceleration

Using equation of motion

`v = u+at`

Where, v = final velocity

u = initial velocity

a = acceleration

t = time

Put the value into the formula

`-15.5=28.5+a*3.45*10^(-3)`

`a=(-15.5-28.5)/(3.45*10^(-3))`

`a=−12753.62\ m/s^2`

Hence, The magnitude of the average acceleration of the ball during given time interval is 12753.62 m/s².

Answer 2

Step-by-step explanation:

It is given that,

Mass of the ball, m = 52 g = 0.052 kg

Initial speed of the ball, u = 28.5 m/s

Final speed of the ball, v = -15.5 m/s (it rebounds)

If the ball is in contact with the wall for 3.45 ms, t = 0.00345 s

Acceleration,
`a=(v-u)/(t)`

`a=(-15.5-28.5)/(0.00345)`

`a=-12753.62\ m/s^2`

So, the average acceleration of the ball during this time interval is
`-12753.62\ m/s^2`. Hence, this is the required solution.