Question

A super-ball is to bounce elastically back and forth between two rigid walls at a distance d from each other. Neglecting gravity and assuming the velocity of super-ball to be $V_0$ horizontally, the average force (in large time interval) being exerted by the super-ball on one wall is:

• A. $\frac{1}{2}\frac{m{v}_0^2}{d}$
• B. $\frac{m{v}_0^2}{d}$
• C. $\frac{2m{v}_0^2}{d}$
• D. $\frac{4m{v}_0^2}{d}$

Answer 1

The correct option is B $\frac{m{v}_0^2}{d}$

Change in momentum when ball strikes one wall is:
$\Delta p=-m{v}_0-m\left(v_0\right)=-2m{v}_0$
The time it takes to come back and hit the wall again is :
$t=\frac{2d}{-v_0}$
Hence force applied is $\frac{\Delta p}{t}=\frac{m{v}_0^2}{d}$