Question

Two equal spheres of mass $m$ are in contact on a smooth horizontal table. A third identical sphere impinges symmetrically on them and is reduced to rest. Find the loss of K.E

• A. $\frac{1}{5}m{u}^2$
• B. $\frac{1}{6}m{u}^2$
• C. $\frac{2}{3}m{u}^2$
• D. $\frac{3}{5}m{u}^2$

Answer 1

The correct option is B

$\frac{1}{6}m{u}^2$

Let $u$ = velocity of sphere $A$ before impact. As the sphere are identical, the triangle $ABC$ formed by joining their centres is equilateral. The spheres $B$ & $C$ will move in direction $AB$ and $AC$ after impact making an angle of ${30}^{\circ }$ with the original lines of motion of ball $A$.

Let $v$ = speed of the other balls after impact

Momentum Conservation:

$mu$ = $mvcos{30}^{\circ }+mvcos{30}^{\circ }$

$u$ = $v\sqrt{3}$ ........(i)

for an oblique collision, we have to take components along normal i.e., along $AB$ for balls $A$ and $B$

$v_B-v_A$ = $-e(u_B-u_A)$ $\Rightarrow$ $v-0$ = $-e(0-ucos{30}^{\circ })$

$v$ = $eucos{30}^{\circ }$

combining (i) and (ii), we get $e$ = $\frac{2}{3}$.

Loss in KE = $\frac{1}{2}m{u}^2-2\left(\frac{1}{2}m{v}^2\right)$ = $\frac{1}{2}m{u}^2-m{\left(\frac{u}{\sqrt{3}}\right)}^2$ = $\frac{1}{6}m{u}^2$