A uniform solid sphere A of mass m is rolling without sliding on a smooth horizontal surface. It collides elastically and head-on with another stationary hollow sphere B of the same mass and radius. Assuming friction to be absent everywhere, the ratio of the kinetic energy of B to that of A just after the collision is

Question

A uniform solid sphere A of mass m is rolling without sliding on a smooth horizontal surface. It collides elastically and head-on with another stationary hollow sphere B of the same mass and radius. Assuming friction to be absent everywhere, the ratio of the kinetic energy of B to that of A just after the collision is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-uhgrpxqdxymp.png" /> (A) 5:2 (B) 1:1 (C) 2:3 (D) 3:2

Tardigrade Admin · Accepted Answer

Since the two bodies have the same mass and collide head-on elastically, the linear velocities get exchanged.
Hence, just after the collision

B

will move with velocity

v_{0}

and

A

becomes stationary but continues to rotate at the same initial angular velocity

\frac{v _{0}}{R}

.
So after the collision, the kinetic energies of

A

and

B

are

K_{B} = \frac{1}{2} m v_{0}^{2}

K_{A} = \frac{1}{2} I ω^{2} = \frac{1}{2} (\frac{2}{5} m R^{2}) \cdot (\frac{v _{0}}{R})^{2}

\Rightarrow \frac{K _{B}}{K _{A}} = \frac{5}{2}

$5 : 2$

$1 : 1$

$2 : 3$

$3 : 2$