Question

A body having momentum $P$ is moving in positive $x$ direction. If the direction of velocity is changed by angle ${120}^{\circ }$ from the original direction, find the magnitude of the impulse on the body when it changes its direction.

• A. $\sqrt{2}P$
• B. $\frac{P}{2}$
• C. $\sqrt{3}P$
• D. $\frac{\sqrt{3}P}{2}$

Answer 1

The correct option is C $\sqrt{3}P$

Let initial momentum be $P_i$.

Direction of momentum is the same as the direction of velocity.
Direction of initial momentum is +ve $x$ direction.
$\overrightarrow{P_i}=P\hat{i}$
and $\overrightarrow{P_f}=P\cos 120\hat{i}+P\sin 120\hat{j}$
$=\frac{-P}{2}\hat{i}+\frac{\sqrt{3}P}{2}\hat{j}$
Impulse = change in momentum
$\overrightarrow{J}=\overrightarrow{P_f}-\overrightarrow{P_i}$
$=\frac{-P}{2}\hat{i}+\frac{\sqrt{3}P}{2}\hat{j}-P\hat{i}$
$=\frac{-3P}{2}\hat{i}+\frac{\sqrt{3}P}{2}\hat{j}$
$\therefore \left|J\right|=\sqrt{{\left(\frac{3}{2}P\right)}^2+{\left(\frac{\sqrt{3}}{2}P\right)}^2}$
$=\sqrt{3}P$