A sphere of mass 0.1 kg is attached to a cord of 1m length. Starting from the height of its point of suspension this sphere hits a block of same mass at rest on a frictionless table, If the impact is elastic, then the kinetic energy of the block after the collision is

The correct answer is 0.5 J.

1. Understanding the Initial State and Motion of the Sphere:
The sphere of mass $m=0.1\text{kg}$ is released from the height of its point of suspension, which means the cord of length $L=1\text{m}$ is initially horizontal. As the sphere swings down to the lowest point, it converts its potential energy into kinetic energy.

The velocity $v$ of the sphere at the bottom of the swing can be determined using the acceleration due to gravity ($g=10{\text{m/s}}^2$):
$v=\sqrt{gL}$
Substituting the given values:
$v=\sqrt{10\times 1}=\sqrt{10}\text{m/s}$

2. Kinetic Energy of the Sphere Before Collision:
The kinetic energy ($K_{\text{sphere}}$) of the sphere just before hitting the block is:
$K_{\text{sphere}}=\frac{1}{2}m{v}^2$
Substituting the mass and velocity:
$K_{\text{sphere}}=\frac{1}{2}\times 0.1\times {\left(\sqrt{10}\right)}^2=\frac{1}{2}\times 0.1\times 10=0.5\text{J}$

3. Collision Dynamics:
When the sphere collides elastically with a block of the same mass ($m=0.1\text{kg}$) which is initially at rest:
For a head-on elastic collision between two equal masses, the velocities are completely interchanged. This means the sphere comes to rest, and the block moves forward with the exact velocity (and therefore the exact kinetic energy) that the sphere had immediately before the impact.

Consequently, the kinetic energy of the block after the collision is:
$K_{\text{block}}=K_{\text{sphere}}=0.5\text{J}$