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  4. 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 />

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Q. 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

NTA AbhyasNTA Abhyas 2020System of Particles and Rotational Motion

Solution:

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
$K_{B}=\frac{1}{2}mv_{0}^{2}$
$K_{\mathrm{A}}=\frac{1}{2} I \omega^2=\frac{1}{2}\left(\frac{2}{5} m R^2\right) \cdot\left(\frac{v_0}{R}\right)^2$
$\Rightarrow \frac{K_{B}}{K_{A}}=\frac{5}{2}$