Question

A sphere of mass $10\text{kg}$ is moving on a horizontal plane with a velocity $(10\hat{i}+\hat{j})\text{m/s}$ where it collides with a vertical wall which is parallel to $\hat{j}$. If the coefficient of restitution is $0.5$, find the impulse (in SI units) acted on the sphere.

• A. $150\hat{i}-\hat{j}$
• B. $-15\hat{i}$
• C. $-150\hat{i}$
• D. $55\hat{i}+\hat{j}$

Answer 1

The correct option is C $-150\hat{i}$

Given, velocity of the sphere $\overrightarrow{v}=10\hat{i}+\hat{j}$.
The situation is as shown in the below figure.


Momentum of the individual particle will be conserved perpendicular to $\text{LOI}$. Here along $\hat{j}$ (along the vertical wall) momentum of individual particle will be conserved.

As, $e=\frac{\text{Velocity of seperation}}{\text{Velocity of approach}}$ along $\text{LOI}$
$\Rightarrow 0.5=\frac{0-v_x}{10-0}$
$\Rightarrow v_x=-5\hat{i}$ where, $v_x$ is the speed of sphere after collision, along the $x-$axis or along the $\text{LOI}$.

Hence, velocity of sphere after collision is
$\overrightarrow{v}=-5\hat{i}+\hat{j}$
So, impulse on sphere is given by
$\overrightarrow{J}=△P=m\overrightarrow{v}-m\overrightarrow{u}$
$\Rightarrow \overrightarrow{J}=m\left[\right(-5\hat{i}+\hat{j})-(10\hat{i}+\hat{j}\left)\right]=10\times [-15\hat{i}]$
$\Rightarrow \overrightarrow{J}=-150\hat{i}$