Question

A ball of mass $m$ is dropped onto a floor from a certain height. The collision is perfectly elastic and the ball rebounds to the same height and again falls. Find the average force exerted by the ball on the floor during a long time interval.

Answer 1

Let height be 'h' from where the ball of mass m is dropped onto a floor.

Before collision

$v_1=\sqrt{2gh}$

$v_1=0$

After collision

$v_2=-\sqrt{2gh}$

$v_2=0$

Now, the force is

$F=ma$

$F=m\times \frac{d(v_2-v_1)}{dt}$

$F=m\times \frac{2\sqrt{2gh}}{t}....\left(I\right)$

Now, from equation of motion

$v=u+gt$

$\sqrt{2gh}=gt$

$t=\sqrt{\frac{2h}{g}}$

Now, the total time

$T=2t$

$T=2\sqrt{\frac{2h}{g}}$

Now, put the value of t in equation (I)

$F=m\times \frac{2\sqrt{2gh}}{t}$

$F=m\times \frac{2\sqrt{2gh}}{2\sqrt{\frac{2h}{g}}}$

$F=mg$

Hence, the force is exerted by the ball on the floo r$mg$