Question

A particle having a mass $0.5\, kg$ is projected under gravity with a speed of $98 \,m / s$ 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 $10 s$ 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) $
$=98(0.866-1)=-98 \times 0.134 m / s$
Vertical momentum of the ball after $10 s =m v$
$=0.5 \times(-98 \times 0.134) kg - m / s$
Initial vertical momentum of the ball
$=m v \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 $
$=49 \,kg - m / s$