Question

A ball of mass 50g is dropped from a height h=10m. It rebounds losing 75 percent of its total mechanical energy. If it remains in contact with the ground for $△t=0.01sec$, find the impulse of the impact force.

Answer 1

Velocity before collision=$v$.

$v=\sqrt{2gh}=10\sqrt{2}$

Velocity after collision is $u$.

$\frac{1}{2}m{u}^2=0.75\frac{1}{2}m{v}^{2}u=\frac{\sqrt{3}}{2}v=0.867v$

Change in momentom=$mv-(-mu)$

or impulse imparted=$mv+m\left(0.867v\right)F△t=1.867mvF=\frac{1.867\times 50\times {10}^{-3}\times 10\sqrt{2}}{0.01}=132N$