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. $0{\text{Kg ms}}^{-1}$
• B. $4.2{\text{Kg ms}}^{-1}$
• C. $2.1{\text{Kg ms}}^{-1}$
• D. $1.4{\text{Kg ms}}^{-1}$

Answer 1

The correct option is B $4.2{\text{Kg ms}}^{-1}$

Velocity of ball just before striking the ground is,

$v_1=\sqrt{2gh}=\sqrt{2\times 10\times 10}=10\sqrt{2}\text{m/s}$

Let the upward direction be the $(+ve)-y$ axis

$\overrightarrow{v_1}=-10\sqrt{2}\hat{j}$

If it reaches the same height after collision, the collision is perfectly elastic, and the speed remains same, only the direction gets reversed.

$\overrightarrow{v_2}=10\sqrt{2}\hat{j}$

From Impulse- Momentum theorem we have:

$|J|=|\Delta P|$

$|J|=m|10\sqrt{2}\hat{j}-(-10\sqrt{2}\hat{j})|$

$|J|=0.15\left[2\right(10\sqrt{2}\left)\right]=3\sqrt{2}\text{Kg m/s}$

$|J|=4.2\text{Kg m/s}$

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

Why this Question? Note: Impulse-Momentum theorem states that, $|J|=|\Delta P|$