Question

A ball of mass $0.15\text{kg}$ is dropped from a height $10\text{m}$, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is nearly -

$(g=10{\text{m/s}}^2)$

• A. $1.4\text{kg-m/s}$
• B. $0\text{kg-m/s}$
• C. $4.2\text{kg-m/s}$
• D. $2.1\text{kg-m/s}$

Answer 1

The correct option is C $4.2\text{kg-m/s}$

Velocity of the ball just before hitting the ground,

$v_i=\sqrt{2gh}=\sqrt{2\times 10\times 10}=14.14\text{m/s}$

As the ball reaches the same height after bouncing back, from law of conservation of energy, the velocity just after hitting the ground is,

$v_f=-14.14\text{m/s}$

So, impulse,

$L=\Delta p=m(v_f-v_i)=0.15(-14.14-14.14)$

$\Rightarrow L=-4.242\text{kg-m/s}\approx -4.2\text{kg-m/s}$

Hence, option $\left(C\right)$ is the correct answer.