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}kg\,m/s $