Question

A ball of mass $2\text{kg}$, travelling with a velocity $2\hat{i}+3\hat{j}+4\hat{k}\text{m/s}$ receives an impulse of $4\hat{i}+2\hat{j}\text{N-s}$. What is the velocity of the ball immediately after the impulse is imparted?

• A. $4\hat{i}+4\hat{j}+4\hat{k}\text{m/s}$
• B. $2\hat{i}+2\hat{j}+2\hat{k}\text{m/s}$
• C. $6\hat{i}+5\hat{j}+4\hat{k}\text{m/s}$
• D. $6\hat{i}+5\hat{j}\text{m/s}$

Answer 1

The correct option is A $4\hat{i}+4\hat{j}+4\hat{k}\text{m/s}$

Given, initial velocity $\overrightarrow{v_i}=(2\hat{i}+3\hat{j}+4\hat{k})\text{m/s}$ and $m=2\text{kg}$
Impulse imparted
$\overrightarrow{J}=(4\hat{i}+2\hat{j})\text{N.s}$

Applying $\overrightarrow{J}=m\Delta \overrightarrow{v}$
$\overrightarrow{J}=m(\overrightarrow{v_f}-\overrightarrow{v_i})...\left(i\right)$
where $\overrightarrow{v_f}$ is the velocity of ball just after receiving the impulse

Putting the values in Eq. ($i$)
$(4\hat{i}+2\hat{j})=2[\overrightarrow{v_f}-\overrightarrow{v_i}]$
$(2\hat{i}+\hat{j})=\overrightarrow{v_f}-\overrightarrow{v_i}$
$\Rightarrow \overrightarrow{v_f}=\overrightarrow{v_i}+(2\hat{i}+\hat{j})$
$\Rightarrow \overrightarrow{v_f}=(2\hat{i}+3\hat{j}+4\hat{k})+(2\hat{i}+\hat{j})$
$\therefore \overrightarrow{v_f}=(4\hat{i}+4\hat{j}+4\hat{k})\text{m/s}$