Question

A spherical rigid ball is released from rest and starts rolling down an inclined plane from height $h\, =\, 7\,m$, 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

• A. 6 m/s
• B. 8 m/s
• C. 10 m/s
• D. 12 m/s

Answer 1

Answer: (C)

As collision is elastic and masses of colliding objects are equal, so here 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
$ω =$ angular speed, then by energy conservation, we have
$mgh=\frac{1}{2}mv^{2} \times\frac{1}{2}I\omega^{2}$
Here $\omega=\frac{v}{R} $ and $I=\frac{2}{5}$
Substituting and solving for velocity, We have
$v=\sqrt{\frac{10}{7}gh}$
$ =\sqrt{\frac{10\times10\times7}{7}}=10\, m/s$
$\therefore $ Velocity of block after collision = $10 \,m/s$.