Question

A projectile of mass $m$ is thrown with velocity $v$ making an angle ${60}^{\circ }$ with the horizontal. Neglecting air resistance, the change in momentum from the departure A to its arrival at B, along the vertical direction is :

• A. $2mv$
• B. $\sqrt{3}mv$
• C. $mv$
• D. $\frac{mv}{\sqrt{3}}$

Answer 1

Answer: (B)

$\sqrt{3}mv$

As the figure drawn above shows that at points $A$ and $B$ the vertical component of velocity is $v\sin {60}^{\circ }$ but their direction are opposite.
Hence , change in momentum is given by:
$\Delta p=mv\sin {60}^{\circ }-(-mv\sin {60}^{\circ })=2mv\sin {60}^{\circ }$
$=2mv\frac{\sqrt{3}}{2}=\sqrt{3}mv$