Question

A body of mass $1kg$ is projected at an angle ${30}^{\circ }$ with horizontal on a level ground at a speed $50m/s.$ The magnitude of change in momentum of the body during its flight is
$(g=10m/s^2)$

• A. $50kgm{s}^{-1}$
• B. $100kgm{s}^{-1}$
• C. $25kgm{s}^{-1}$
• D. Zero

Answer 1

The correct option is A $50kgm{s}^{-1}$


We know that in a horizontal plane, Initial velocity of projectile = Final velocity of projectile
$\overrightarrow{p_i}=mu\cos \theta \hat{i}+mu\sin \theta \hat{j}\overrightarrow{p_f}=mu\cos \theta \hat{i}-mu\sin \theta \hat{j}|\overrightarrow{\Delta p}|=|\overrightarrow{p_f}-\overrightarrow{p_i}|=2mu\sin \theta$
$=2\times 1\times 50\times \frac{1}{2}=50kgm{s}^{-1}$
Alternate Solution:
In the horizontal direction, momentum $p$ doesn’t change.
$\Delta p_x=0$
In vertical direction,
$\Delta p_y=mu\sin \theta -(-mu\sin \theta )=2mu\sin \theta$
$=2\times 1\times 50\times \frac{1}{2}=50kgm{s}^{-1}$