Question

A projectile of mass $m$ is fired with velocity $v$ from a point $P$ at $\theta ={45}^o$. Neglecting air friction, the magnitude of change of momentum between the leaving pt P and arriving pt Q is:

• A. $\frac{mv}{\sqrt{2}}$
• B. $2mv$
• C. $\frac{1}{2}mv$
• D. $\sqrt{2}mv$

Answer 1

The correct option is D $\sqrt{2}mv$

Let the change in the momentum be $\Delta p$

and

$\Delta p=p_f-p_i$

$=m(v_f-v_i)=m\left[\left(\frac{v}{\sqrt{2}}\hat{i}-\frac{v}{\sqrt{2}}\hat{j}\right)-\left(\frac{v}{\sqrt{2}}\hat{i}+\frac{v}{\sqrt{2}}\hat{j}\right)\right]$

$=-\sqrt{2}mv\hat{j}$

Therefore,

$|\Delta p|=\sqrt{2}mv$