Question

A body of mass 2 kg is projected from the ground with a velocity $20m{s}^{-1}$ at an angle $30{}^{\circ }$ with the vertical. If $t_1$ is the time in seconds at which the body is projected and $t_2$ is the time in seconds at which it reaches the ground, the change in momentum in kg ms-1 during the time $(t_2{t}_1)$ is:

• A. 40
• B. 40 $\sqrt{3}$
• C. 50 $\sqrt{3}$
• D. 60

Answer 1

Answer: (B)

Momentum of the body in the horizontal direction remains unchanged throughout the motion as projectile. Initial moment of the body in vertical upward direction. $p_1=mv\cos {30}^o=\frac{\sqrt{3}}{2}mv$ Final momentum of body in downward direction, $p_2=mv\cos {30}^o=\frac{\sqrt{3}}{2}mv$ $\therefore$ Change in momentum, $\Delta p=p_2-(-p_1)$ $=p_2+p_1$ $=\frac{\sqrt{3}}{2}mv+\frac{\sqrt{3}}{2}mv$ $=\sqrt{3}mv$ Given, m = 2 kg, v = 20 m/ s $\therefore$ $\Delta p=\sqrt{3}\times 2\times 20$ $=40\sqrt{3}kgm/s$