Question

A ball of mass $0.2kg$ strikes an obstacle and moves at ${60}^0$ to its original direction. If its speed also changes from $20m/s$ to $10m/s$, the magnitude of the impulse received by the ball is

• A. $2\sqrt{7}Ns$
• B. $2\sqrt{3}Ns$
• C. $2\sqrt{5}Ns$
• D. $3\sqrt{2}Ns$

Answer 1

The correct option is B $2\sqrt{3}Ns$

$\text{Step 1: Momentum calculation}$ $\text{{Ref. Fig.}}$

Lets assume that initially ball is moving in positive $x$ direction.

$\therefore$ $\text{Initial momentum}$ $\overrightarrow{P_i}=m\overrightarrow{u_1}$

$=0.2\times 20\hat{i}kgm/s$

$=4\hat{i}kgm/s$

$\text{Final momentum}$ $\overrightarrow{P_f}=m\overrightarrow{u_2}$

$\overrightarrow{u_2}=10\cos {60}^0\hat{i}+10\sin 60\hat{j}m/s$

$=5\hat{i}+5\sqrt{3}\hat{j}m/s$

$\therefore P_f=0.2(5\hat{i}+5\sqrt{3}\hat{j})kgm/s$

$=(\hat{i}+\sqrt{3}\hat{j})kgm/s$

$\text{Step 2: Impulse calculation}$

Impulse is given by

$\overrightarrow{J}=\Delta \overrightarrow{P}=\overrightarrow{P_f}-\overrightarrow{P_i}$

$=(\hat{i}+\sqrt{3}\hat{j})-4\hat{i}Ns$

$\overrightarrow{J}=-3\hat{j}+\sqrt{3}\hat{j}Ns$

$\text{Magnitude of impulse}$ $|\overrightarrow{J}|=\sqrt{(-3{)}^2+(\sqrt{3}{)}^2}Ns$

$=\sqrt{12}$ $=2\sqrt{3}Ns$

Hence magnitude of impulse received by the ball is $2\sqrt{3}N.s.$

Option $B$ is correct.