Question

A hollow smooth uniform sphere $A$ of mass ${}^{\prime }{m}^{\prime }$ rolls without sliding on a smooth horizontal surface. It collides elastically and head on with another stationary smooth solid sphere $B$ of the same mass $m$ and same radius. The ratio of kinetic energy of ${}^{\prime }{B}^{\prime }$ to that of ${}^{\prime }{A}^{\prime }$ just after the collision is

• A. $\frac{3}{2}$
• B. $\frac{3}{4}$
• C. $\frac{4}{5}$
• D. $\frac{5}{3}$

Answer 1

The correct option is A $\frac{3}{2}$

Let linear/translational velocity of sphere $=V_o$
Angular velocity of sphere $=\omega =V_o/R$
Assume $V_1$ and $V_2$ are the velocities of sphere $A$ and $B$ respectively, just after collision.


No horizontal force on the system, so linear momentum will be conserved.
$P_i=P_f$
$m{V}_0+0=m{V}_1+m{V}_2$
$\Rightarrow V_o=V_1+V_2...\left(1\right)$

$e=\frac{\text{velocity of seperation}}{\text{velocity of approach}}=\frac{v_2-v_1}{u_1-u_2}$
$=\frac{V_2-V_1}{V_o-0}=1$
For elastic collision $e=1$
$\Rightarrow V_2-V_1=V_0...\left(ii\right)$
From Eq. (i) and (ii),
$V_2=V_o$ & $V_1=0$

Since there is no torque acting on either sphere during collsion, their angular velocities about respective centres remains the same. i.e ${\omega }_A=\omega \&{\omega }_B=0$
Thus, Kinetic energy of $A$ after collision $=\frac{1}{2}m{V}_1^2+\frac{1}{2}I{\omega }^2$
Here, $V_1=0$
$(KE{)}_A=\frac{1}{2}\times \left(\frac{2}{3}m{R}^2\right)\times {\left(\frac{V_o}{R}\right)}^2=\frac{m{V}_o^2}{3}$
Kinetic energy of $B$ after collision $(KE{)}_B=\frac{1}{2}m{V}_2^2=\frac{1}{2}m{V}_o^2$

$\Rightarrow \frac{(KE{)}_B}{(KE{)}_A}=\frac{\frac{1}{2}m{V}_o^2}{\frac{m{V}_o^2}{3}}=\frac{3}{2}$