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)

40 $\sqrt{3}$

Answer 2

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 $\therefore$ $\Delta p=\sqrt{3}\times 2\times 20$ $=40\sqrt{3}kg\,m/s$