Question

A particle having a mass $0.5kg$ is projected under gravity with a speed of $98m/s$ at an angle of $60^{\circ }$. The magnitude of the change in momentum in $kg\cdot m/s$ of the particle after

• A. 490
• B. 98
• C. 49
• D. 0.5

Answer 1

Answer: (C)

Vertical velocity after $10s$ is
$u=\left(98\sin 60^{\circ }\right)-9.8\times 10$
$=98\times \frac{\sqrt{3}}{2}-98=98\left(\frac{\sqrt{3}}{2}-1\right)$
$=98(0.866-1)=-98\times 0.134m/s$
Vertical momentum of the ball after $10s=mv$
$=0.5\times (-98\times 0.134)kg-m/s$
Initial vertical momentum of the ball
$=mv\sin 60^{\circ }$
$=0.5\times 98\times \frac{\sqrt{3}}{2}=0.5\times 98\times 0.866$
Change in vertical momentum
$=0.5\times 98\times 0.866-(-0.5\times 98\times 0.134)$
$=0.5\times 98[0.866+0.134]$
$=0.5\times 98$
$=49kg-m/s$