Question

In the figure shown, a ring A is initially rolling without sliding with a velocity v on the horizontal surface of the body B (of same mass as A). All surfaces are smooth. B has no initial velocity. The maximum height reached by A on B is $\frac{v^2}{xg}$. Find $x$.

Answer 1

Given that all surfaces are smooth.
Therefore, intial and final Rotational Kinetic energy do not change.
As there is no external force, momentum is conserved.
At the highest point reached by ring, its horizontal velocity would be zero relative to that wedge.
Initial momentum = Final momentum.
mv = (m+m)v' $\Rightarrow v^{\prime }=\frac{v}{2}$
Applying Energy equation,
$\frac{1}{2}m{v}^2=mgh+\frac{1}{2}(m+m)(\frac{v}{2}{)}^2$
$\Rightarrow h=\frac{v^2}{4g}$
Therefore, $x=4$.