Question

A small ball (uniform solid sphere) of mass $m$ is released from the top of a wedge of the same mass $m$. The wedge is free to move on a smooth horizontal surface. The ball rolls without sliding on the wedge. The required height of the wedge are mentioned in the figure.The total kinetic energy of the ball just before it falls on the ground is:

• A. $2mgh$
• B. $\frac{31}{18}mgh$
• C. $\frac{13}{18}mgh$
• D. none of these

Answer 1

Answer: (D)

none of these

Initially, when the ball rolls down a height $h$,
$mgh=\frac{7}{10}m{v}^2+\frac{1}{2}m{v}^2$

or, $mgh=\frac{6}{5}m{v}^2$

or, $v=\sqrt{\frac{5gh}{6}}$

Now, total kinetic energy just before the ball falls down $=\frac{7}{10}m{v}^2+mgh$ $=\frac{19}{12}mgh$