Question

In the figure shown a ring A is 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. What will be the maximum height (from initial position) 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 $\frac{v^2}{4g}$

It is given that all the surfaces are smooth and there is no external force acting on the system. Hence, momentum is conserved. So,

Initial momentum=final momentum

$mv=(m+m)v^{\prime }$

$v^{\prime }=\frac{v}{2}$

At highest point, horizontal velocity of ring relative to wedge$=0$

Applying conservation of energy equation,

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

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

$h=\frac{v^2}{4g}$ maximum height reached is $\frac{v^2}{4g}$