Question

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

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

Answer 1

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


As we know, Change in momentum $\Delta P=P_f-P_i$

$\therefore \Delta P=m(v_f-v_i)......\left(1\right)$

Given, the velocity of the projectile is $v$ which makes an angle $\theta$ with the horizontal.

Let velocity at point $\text{P}$ and $\text{Q}$ be $v_i$ and $v_f$ respectively.

$v_i=v_x\hat{i}+v_y\hat{j}$

$=v\cos {45}^{\circ }\hat{i}+v\sin {45}^{\circ }\hat{j}$

$=\frac{v}{\sqrt{2}}\hat{i}+\frac{v}{\sqrt{2}}\hat{j}$

Similarly,

$v_f=v_x\hat{i}+v_y\hat{j}$

$=v\cos {45}^{\circ }\hat{i}+v\sin {45}^{\circ }(-\hat{j})$

$=\frac{v}{\sqrt{2}}\hat{i}-\frac{v}{\sqrt{2}}\hat{j}$

From equation $\left(1\right)$, we get,

$\Delta P=m[(\frac{v}{\sqrt{2}}\hat{i}-\frac{v}{\sqrt{2}}\hat{j})-(\frac{v}{\sqrt{2}}\hat{i}+\frac{v}{\sqrt{2}}\hat{i})]$

$\Rightarrow \Delta P=-2m\frac{v}{\sqrt{2}}\hat{j}=-\sqrt{2}mv\hat{j}$

$\left|\begin{array}{c}\Delta P\end{array}\right|=\sqrt{2}mv$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(A\right)$ is the correct answer.