Question

The projectile motion of a particle of mass $5\text{g}$ is shown in the figure. The initial velocity of the particle is $5\sqrt{2}{\text{ms}}^{-1}$ and the air resistance is assumed to be negligible. The magnitude of the change in momentum of the particles between the points $A$ and $B$ is $x\times {10}^{-2}{\text{kg-ms}}^{-1}$. The value of $x$, to the nearest integer, is _____.

Answer 1

The change in momentum of the particle between points $A$ and $B$ is given by,

$\Delta \overrightarrow{p}=\overrightarrow{p_B}-\overrightarrow{p_A}=m({\overrightarrow{v}}_B-{\overrightarrow{v}}_A)$

$\Rightarrow \Delta \overrightarrow{p}=m\left[\right(5\sqrt{2}\cos {45}^{\circ }\hat{i}-5\sqrt{2}\sin {45}^{\circ }\hat{j})-(5\sqrt{2}\cos {45}^{\circ }\hat{i}+5\sqrt{2}\sin {45}^{\circ }\hat{j}\left)\right]$

$\Rightarrow \Delta \overrightarrow{p}=-2\times 5\times {10}^{-3}\times 5\sqrt{2}\times \frac{1}{\sqrt{2}}\hat{j}$

$\Rightarrow \Delta \overrightarrow{p}=(-0.05\hat{j}){\text{kg-ms}}^{-1}=(-5\times {10}^{-2}\hat{j}){\text{kg-ms}}^{-1}$

So, $\Delta p=5\times {10}^{-2}{\text{kg-ms}}^{-1}=x\times {10}^{-2}{\text{kg-ms}}^{-1}$

Hence, $x=5$