Question

As shown in the figure a ball is moving with a velocity $u=6\hat{i}+\hat{j}$ and it collides with a vertical wall which is parallel to the vector $\hat{j}$. If the coefficient of restitution between the ball and the wall is $0.5$ and mass of the ball is $m=1\text{kg}$. Find the impulse that acts on the ball.

• A. $3\hat{j}$
• B. $\frac{3}{2}\hat{j}$
• C. $-9\hat{i}$
• D. $-\frac{9}{2}\hat{i}$

Answer 1

The correct option is C $-9\hat{i}$

Given, $u=6\hat{i}+\hat{j}$
As we know impulse will act along the line of impact.
So, component of velocity parallel to the wall remians unchanged while component of velocity perpendicular to the wall will be, given by
$e=\frac{\text{Velocity of seperation}}{\text{Velocity of approach}}$
$\Rightarrow e=\frac{(0-v_x)}{(6-0)}=0.5\Rightarrow v_x=-3$

Hence, velocity of ball after the collision
$\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}=-3\hat{i}+\hat{j}$
Thus, impulse $\overrightarrow{J}=\Delta P=(m\overrightarrow{v}-m\overrightarrow{u})$ along $\text{LOI}$
$\Rightarrow \overrightarrow{J}=\Delta P=m(-3\hat{i}-6\hat{j})$
$\Rightarrow \overrightarrow{J}=1\times (-9\hat{i})=-9\hat{i}$