A spherical rigid ball is released from rest and starts rolling down an inclined plane from height $h=7 \mathrm{m}$, as shown in the figure. It hits a block at rest on the horizontal plane (assume elastic collision). If the mass of both the ball and the block is $m$ and the ball is rolling without sliding, then the speed of the block after collision is close to

As collision is elastic and masses of colliding objects are equal, so there is exchange of total energy of moving mass to stationary mass. 

Hence, velocity of block $=$ velocity of sphere just before collision. 

For sphere, if $v=$ velocity of translation of centre of mass at the bottom of plane and $\omega =$ angular speed, then by energy conservation, we have

$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

Here, $\omega =\frac{v}{R}$ and $I=\frac{2}{5}m{R}^2$

Substituting and solving for velocity, we have 

$v=\sqrt{\frac{10}{7}gh}=\sqrt{\frac{10\times 10\times 7}{7}}=10 \mathrm{m}/\mathrm{s}$

$\therefore$ Velocity of block after collision $=10 \mathrm{m}/\mathrm{s}$.