Question

In the figure shown, a ring $A$ is initially rolling without sliding with a velocity $v$ on a horizontal surface of the body $B$ (of same mass as $A$). All surfaces are smooth. $B$ has no initial velocity. What will be the maximum height reached by $A$ on $B$?

• A. $\frac{3v^2}{4g}$
• B. $\frac{v^2}{4g}$
• C. $\frac{v^2}{2g}$
• D. $\frac{v^2}{3g}$

Answer 1

The correct option is B $\frac{v^2}{4g}$


initial energy $\left(E_i\right)=\frac{1}{2}m{v}^2+\frac{1}{2}\left(M{R}^2\right){\omega }^2$

at fianl position,
$E_f=mgh+\frac{1}{2}\left(M{R}^2\right){\omega }^2+\frac{1}{2}(m+m)v^{\prime 2}$ ....(1)

using momentum conservation,
$mv=\left(2m\right)v^{\prime }{v}^{\prime }=\frac{v}{2}$
on putting on eq.(2)

$E_f=mgh+\frac{1}{2}\left(M{R}^2\right){\omega }^2+\frac{1}{2}\left(2m\right)\frac{v}{2}$

So, $E_i=E_f$
$\frac{1}{2}m{v}^2+\frac{1}{(}M{R}^2){\omega }^2$= $mgh+\frac{1}{2}\left(M{R}^2\right){\omega }^2+\frac{1}{2}\left(2m\right)\frac{v}{2}$
on solving further,
$h=\frac{v^2}{4g}$