Question

A ball falls from 20 m height on floor and rebounds to 5 m. Time contact is 0.02 sec. Find acceleration during impact.

Answer 1

According to Newtons second law of motion, force acting on a body is equal to the rate of change of momentum during impact

$F=\frac{\Delta p}{\Delta t}$

Also, $F=ma$

Therefore,

$ma=\frac{p_2-p_1}{\Delta t}$

or

$a=\frac{m{v}_2-(-m{v}_1)}{m\Delta t}$

or

$a=\frac{v_2+v_1}{\Delta t}$

Therefore,

$a=\frac{\sqrt{2\times 10\times 20}+\sqrt{2\times 10\times 2}}{0.02}$

or

$a=\frac{20+10}{0.02}=1500m/s^2$