Question

A particle having mass $0.5kg$ is projected under gravity with a speed of $98m/sec$ at an angle of ${60}^{\circ }$.The magnitude of the change in momentum $($in N sec$)$ of particle after $10seconds$ is :

• A. $0.5$
• B. $49$
• C. $98$
• D. $490$

Answer 1

The correct option is B $49$

There is no change in horizontal velocity , hence no change in momentum in horizontal direction.The vertical velocity at $t=10sec$ is
$v=98\times \sin {60}^{\circ }-\left(9.8\right)\times 10=113.13m/sec$
So change in momentum in vertical direction is
$(0.5\times 98\times \frac{\sqrt{3}}{2})-\left[\right(0.5\times 13.13\left)\right]$
$=42.434+6.56=48.997=49$