Question #24797

A 50g ball travelling at 25.0m/s is bounced off a brick wall and rebounds at 22.0m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.50ms, what is the magnitude of the average acceleration of the ball during this time interval?

Expert's answer

The change that the velocityhas undergone

dv = 22.0 m/s - (-25.0 m/s) = 47.0 m/s

It has taken place during time interval dt = 3.50 ms,

so the average acceleration is

a = dv/dt = (47.0 m/s) / (3.50 ms) = 13428.6 m/s^2

Using the mass of the ball and the 2nd Newton's Law, one can also estimate the average force of interaction between the ball and the wall.

Answer: 13428.6 m/s^2

dv = 22.0 m/s - (-25.0 m/s) = 47.0 m/s

It has taken place during time interval dt = 3.50 ms,

so the average acceleration is

a = dv/dt = (47.0 m/s) / (3.50 ms) = 13428.6 m/s^2

Using the mass of the ball and the 2nd Newton's Law, one can also estimate the average force of interaction between the ball and the wall.

Answer: 13428.6 m/s^2

## Comments

kathy23.02.13, 02:27thanks so much

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