Question

A body of mass 2 kg is projected from the ground with a velocity $20m{s}^{-1}$ at an angle ${30}^o$ with the vertical. If $t_1$ is the time in second at which the body is projected and $t_2$ is the time in second at which it reaches the ground, the change in momentum in $kgm{s}^{-1}$ during the time $(t_2-t_1)$ is

• A. $40\sqrt{2}$
• B. $40\sqrt{3}$
• C. $25\sqrt{3}$
• D. $45$

Answer 1

Answer: (B)

$40\sqrt{3}$

Initial moment of the body in vertical upward direction
$p_1=mvcos{30}^o=\frac{\sqrt{3}}{2}mv$
Final momentum of body in downward direction
$p_2=mvcos{30}^o=\frac{\sqrt{3}}{2}mv$
Change in momentum
$\Delta p=p_2-(-p_1)$
$=p_2+p_1$
$=\frac{\sqrt{3}}{2}mv+\frac{\sqrt{3}}{2}mv$
$=\sqrt{3}mv$
$=\sqrt{3}\times 2\times 20$
$=40\sqrt{3}$