Question

A ball of mass m is projected at an angle of ${45}^o$ with the horizontal. Taking air resistance negligible, the total change in momentum when it strikes the ground is

• A. $\frac{mv}{\sqrt{2}}$
• B. 2 mv
• C. $\sqrt{2}mv$
• D. $0$

Answer 1

Answer: (C)

$\sqrt{2}mv$

Change in momentum $=mvsin\theta -(-mvsin\theta )$
$=2mvsin\theta$
Given, $\theta =\frac{\pi }{4}$
Change in momentum $=\sqrt{2}mv$