Question

A small bob of mass $m$ is suspended by a massless string from a cart of the same mass $m$ as shown in the figure. The friction between the cart and horizontal ground is negligible. The bob is given a velocity $v_0$ in horizontal direction as shown. The maximum height attained by the bob is
(initially whole system (bob+string+cart) was at rest).

• A. $\frac{2v_0^2}{g}$
• B. $\frac{v_0^2}{g}$
• C. $\frac{v_0^2}{4g}$
• D. $\frac{v_0^2}{2g}$

Answer 1

The correct option is C $\frac{v_0^2}{4g}$

When ball reaches maximum height, let velocity of system (bob+string+cart) be $v$
By linear momentum conservation in horizontal direction,
for (bob+string+cart)
$m{v}_0=(m+m)v$
$v=\frac{v_0}{2}$
By mechanical energy conservation for (bob+string+cart)
$\frac{1}{2}m{v}_0^2+0+0=\frac{1}{2}(m+m)v^2+mgh+0$
$\frac{1}{2}m{v}_0^2-\frac{1}{2}\left(2m\right)\frac{v_0^2}{4}=mgh$
Solving it,
$h=\frac{v_0^2}{4g}$