Question

A projectile of mass $m$ is fired with velocity $v$ from a point as shown in figure, neglecting air resistance, what is the change in momentum when leaving $P$ and arriving at $Q$ ?

• A. $0$
• B. $mv$
• C. $4mv$
• D. $\frac{2mv}{\sqrt{3}}$

Answer 1

The correct option is B $mv$

Horizontal component will be same but vertical component will be completely -ve of initial so change in momentum $=mvsin\theta -(-mvsin\theta )=2mvsin\theta =2mv\ast sin{30}^o=2mv\frac{1}{2}=mv$