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# find momentum absorbed by wall after struck by ball

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A tennis ball with a mass of 0.50 kg is thrown at a wall at a speed of 20 m/s. The ball hits the wall perpendicularly and bounces off in an elastic collision.

a) How much momentum does the wall absorb when struck by the tennis ball?

b) A tennis serving machine shoots tennis balls at the wall, as described above, at a rate of eight tennis balls per second. What is the average force on the wall?

https://brainmass.com/physics/velocity/momentum-absorbed-wall-after-struck-ball-307826

## SOLUTION This solution is FREE courtesy of BrainMass!

Because it is an elastic collision, the initial speed must be equal to the final speed, that is, the final speed is 20 m/s. In term of velocity, the final velocity is -20 m/s.

To calculate the momentum the wall absorbs when struck by the tennis balls, we can calculate the momentum lost by the ball.

∆p=p_f-p_f
=mv_f-mv_i
=0.50 kg×(-20m/s)-0.50 kg×20m/s
=-20 kg m/s

Since the ball lost 20 kg m/s of momentum, the wall absorbs 20 kg m/s of momentum.

∆p=F∆t=m∆v
F∆t=20 kg m/s
F=(20 kg m/s)/∆t

The tennis serving machine shoots 8 balls/second, so one ball is shot every 1/8 second = 0.125 second.

F=(20 kg m/s)/∆t
=(20 kg m/s)/(0.125 s)
=160 N

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!