Question

A projectile ofmass m is thrown with a velocity v making an angle 60${}^o$ with the horizontal, neglecting air resistance, the change in momentum from the departure A to its arrival at B, along the vertical directions:

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

Answer 1

The correct option is B $\sqrt{3}$mv

Momentum of the body in horizontal direction remains constant throughout the motion as projectile, Initial momentum of the body in vertical direction (upward)
$=p_i=mvsin{60}^o=mv\frac{\sqrt{3}}{2}$
Final momentum in vertical direction (down ward) $=p_f=-mvsin{60}^o=-mv\frac{\sqrt{3}}{2}$
$\therefore$ Change in momentum $=p_i-(-p_f)$
$=2mv\frac{\sqrt{3}}{2}=\sqrt{3}mv$