Answer to Question #141519 in Mechanics | Relativity for Xavier

Question

A ball of mass 0.05kg strikes a smooth wall normally 4 times in 2 seconds with a velocity of 10m/s. Each time the ball rebounds with the same speed of 10 m/s. Calculate the average force on the wall. Draw a sketch showing how the momentum varies with time over the 2 seconds.

Answer 1

Since the force is equal to the ratio of pulse change to time change, then

$F=\frac{\Delta p}{\Delta t}=\frac{m\times \Delta v}{\Delta t}$ ;

Since the period is equal to the ratio of time to the number of oscillations, $T =\frac {t} {N} =\frac {2} {4} = 0.5$ s.

For the half of the period there is a change in speed from 10 to -10 m/s, then

$F=\frac{0.05\times (10+10)}{0.5\times0.5}=4$ N.