Question

A projectile of mass $m$ is fired with a velocity $v$ from point $P$ at an angle ${45}^{\circ }$. Neglecting air resistance, the magnitude of the change in momentum leaving the point $P$ and arriving at $Q$ is

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

Answer 1

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

Let $u$ be the initial velocity.
The horizontal component of the velocity does not change during the journey.
The vertical component changes from $\left(v\sin 45\right)to(-v\sin 45)$.

$\mid Impulse\mid =\mid \Delta \overrightarrow{p}\mid =\mid \overrightarrow{F}\cdot \Delta t\mid$

Here $\Delta T=\frac{2vsin{45}^o}{g}=\frac{\sqrt{2}v}{g}$

and $\mid \overrightarrow{F}\mid =mg$

$\therefore \mid \Delta \overrightarrow{p}\mid =mg\times \frac{\sqrt{2}v}{g}=\sqrt{2}mv$.