Question

A ball of mass $2\text{kg}$ is dropped from a height of $20\text{m}$. After striking the ground, the ball bounces back with half of its striking speed. Find the impulse imparted by the normal reaction from the ground. Take $g=10{\text{m/s}}^2$.

• A. $40\text{N-s}$
• B. $80\text{N-s}$
• C. $0\text{N-s}$
• D. $60\text{N-s}$

Answer 1

The correct option is D $60\text{N-s}$

Ball is dropped from $20\text{m}$ height, therefore it reaches the ground with speed $v_0$.
$|v_0|=\sqrt{2gh}=\sqrt{2\times 10\times 20}$
$=-20\text{m/s}$
(consider upward direction as positive)

Let $v_1$ be the speed at which it bounces back
Given $v_1=\frac{v_0}{2}=10\text{m/s}$

Then, Impulse $J=m\Delta v$
$\Rightarrow J=m(v_1-v_0)=2\times (10-(-20\left)\right)$
$\therefore J=60\text{N-s}$