Question

A particle of mass $2\text{kg}$ moving with a velocity of $3\text{m/s}$ is acted upon by a force which changes its direction of motion by an angle of ${90}^{\circ }$ without changing its speed. What is the magnitude of impulse experienced by the particle?

• A. $6\text{N-s}$
• B. $2\text{N-s}$
• C. $3\sqrt{2}\text{N-s}$
• D. $6\sqrt{2}\text{N-s}$

Answer 1

The correct option is D $6\sqrt{2}\text{N-s}$

Given $m=2\text{kg}$ is the mass of the particle.
Let initial momentum $\overrightarrow{p_1}=m{v}_1\hat{i}$
$\overrightarrow{p_1}=2\times 3\hat{i}=6\hat{i}$
Final momentum $\overrightarrow{p_2}=m{v}_2\hat{j}$
$\overrightarrow{p_2}=2\times 3\hat{j}=6\hat{j}$
From impulse-momentum theorem:
$\overrightarrow{J}=m\Delta \overrightarrow{v}=\Delta \overrightarrow{p}$
$=\overrightarrow{p_2}-\overrightarrow{p_1}$
$=6\hat{j}-6\hat{i}$

Thus, magnitude of impulse experienced by the particle is:
$\therefore |\overrightarrow{J}|=\sqrt{6^2+6^2}=6\sqrt{2}\text{N-s}$