Question

A sphere of mass $M=2\text{kg}$ moving with a velocity of $2\sqrt{2}\text{m/s}$ collides with another sphere of mass $m=1\text{kg}$ initially at rest at an angle of ${45}^{\circ }$ with the horizontal. If the coefficient of restitution of the collision is 0.5, what will be the final velocity(in $\text{m/s}$) of the sphere having mass $m$ ?

• A. $\sqrt{2}\hat{i}+\sqrt{2}\hat{j}$
• B. $-\sqrt{2}\hat{i}-\sqrt{2}\hat{j}$
• C. $-\sqrt{2}\hat{i}+\sqrt{2}\hat{j}$
• D. $\sqrt{2}\hat{i}-\sqrt{2}\hat{j}$

Answer 1

The correct option is D $\sqrt{2}\hat{i}-\sqrt{2}\hat{j}$

(L.O.I = Line of impact)
Before collision

The sphere of mass $m=1\text{kg}$ is at rest.
The component of velocity of the sphere $M$ along the line of impact,
$u_1=u\cos {45}^{\circ }=2\sqrt{2}\times \frac{1}{\sqrt{2}}=2\text{m/s}$
The component of velocity of the sphere $M$ perpendicular to the line of impact,
$u_1^{\prime }=u\sin {45}^{\circ }=2\sqrt{2}\times \frac{1}{\sqrt{2}}=2\text{m/s}$

After collision (along L.O.I)

Applying the P.C.L.M along L.O.I
$p_i=p_f$
$M{u}_1+mu=M{v}_1+m{v}_2$
$\Rightarrow$ $2\times 2+1\times 0=2v_1+v_2$
$\Rightarrow 2v_1+v_2=4...\left(i\right)$
$u=0$ As sphere of mass $m$ is at rest initially.

Coefficient of restitution $e=\frac{v_2-v_1}{u_1-u_2}$
$\Rightarrow 0.5=\frac{v_2-v_1}{u_1-0}\Rightarrow 0.5\times u_1=v_2-v_1\Rightarrow 0.5\times 2=v_2-v_1$
$\Rightarrow v_2-v_1=1....\left(ii\right)$

Using equation $\left(i\right)$ and $\left(ii\right)$ we get,
$v_2=2\text{m/s}$.

After collision (perpendicular to L.O.I.)

No component of impulse acts perpendicular to L.O.I. Therefore $v^{\prime }=u_1$ and $v^{\prime \prime }=0$ as component of initial velocity perpendicular to the line of impact remains unchanged.


$\therefore v_{net}$ of sphere of mass $m=v_2$


$\therefore$ The final velocity vector of the second sphere is $v_2\cos {45}^{\circ }\hat{i}-v_2\sin {45}^{\circ }\hat{j}=\sqrt{2}\hat{i}-\sqrt{2}\hat{j}\text{m/s}$