Question

Two smooth spheres made of identical material having masses $m$ and $2m$ undergo an oblique impact as shown in figure. The initial velocities of the masses are also shown. The impact force is along the line joining their centres. The coefficient of restitution is 5/9. The velocities of the masses after the impact are:

• A. $\frac{10}{3}\hat{i}+8j,\frac{5}{3}\hat{i}+\frac{4}{3}j,$
• B. $\frac{5}{3}\hat{i}-8j,\frac{-5}{3}\hat{i}+4j,$
• C. $\frac{10}{3}\hat{i}-8\hat{j},\frac{-5}{3}\hat{i}+4\hat{j},$
• D. None of these

Answer 1

The correct option is C $\frac{10}{3}\hat{i}-8\hat{j},\frac{-5}{3}\hat{i}+4\hat{j},$

Velocity components before collision


Line of impact is along x-axis. Hence, velocities perpendicular to the x-axis remain unchanged.
Hence y component velocity after collision
For $A$:
$v_{Ay}=-10\sin {53}^{\circ }\hat{j}=-10\times \frac{4}{5}\hat{j}=-8\hat{j}$
For $B$:
$v_{By}=5\sin {53}^{\circ }=5\times \frac{4}{5}\hat{j}=4\hat{j}$

x - component of velocities after collision.
By momentum conservation (because no external force is present)
$2m\left(5\cos {53}^{\circ }\right)-m\times \left(10\cos {53}^{\circ }\right)=2m{v}_{Bx}+m{v}_{Ax}$
$\Rightarrow 2v_{Bx}+v_{Ax}=0$ ...(I)

As we know, $e=\frac{\text{velocity of seperation}}{\text{velocity of approach}}$ in horizontal direction
i.e $e=\frac{v_{Ax}-v_{Bx}}{5\cos {53}^{\circ }+10\cos {53}^{\circ }}=\frac{5}{9}$
or $v_{Ax}-v_{Bx}=5$ ...(II)

From eq. (I) and (II) $v_{Ax}=\frac{10}{3}{v}_{Bx}=\frac{-5}{3}$
Hence, after collision,
Velocity of sphere $A,v_A=\frac{10}{3}\hat{i}-8\hat{j}$
Velocity of sphere $B,v_B=\frac{-5}{3}\hat{i}+4\hat{j}$