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. Prove that $e=2/3$ and find the loss in KE.

Answer 1

Let $u$ be the velocity of sphere $A$ before impact. As the sphere are identical, the triangle ABC formed by joining their centres is equilibrium. The sphere $B$ and $C$ will move in direction $AB$ and $AC$ after impact making an angle of ${30}^o$ with the original line of motion os sphere $A$.
Let $\nu$ be the speed of the other spheres after impact.
From momentum conservation,
$mu=m\nu cos{30}^o+m\nu cos{30}^o$
$u=\nu \sqrt{3}$ (i)
From Newton's experimental law, for an oblique, we have to take components along the normal, i.e. along $AB$ for sphere $A$ and $B$. Hence,
${\nu }_B-{\nu }_A=-e(u_B-u_B)$
$⟹\nu -0=-e(0-ucos{30}^o)$
$\nu =eucos{30}^o$ (ii)
Combining Eqs. (i) and (ii), we get $e=2/3$
Loss in KE$=\frac{1}{2}m{u}^2-2\left(\frac{1}{2}m{\nu }^2\right)$
$\frac{1}{2}m{u}^2-m{\left(\frac{u}{\sqrt{3}}\right)}^2=\frac{1}{6}m{u}^2$