Question

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-
(1) 1 J
(2) 10 J
(3) 0.1J
(4) 0.5 J​

Answer 1

Answer:

Option (1) 1 J

Explanation:

h = 1 m

m = 0.1 kg

Velocity of the ball when it is about to hit the block

From the third equation of motion

$v=\sqrt{2gh}$

$v=\sqrt{2\times 10\times 1}$

$\implies v=\sqrt{20}$

The kinetic energy of the ball

$K=\frac{1}{2}mv^2$

$K=\frac{1}{2}\times 0.1\times (\sqrt{20})^2$

$K=\frac{1}{2}\times 0.1\times 20$

$K=1$ Joule

Initially the block is at rest on the table

Total kinetic energy = 1 Joule

Since the collision is elastic, the total kinetic energy after hitting, imparted to the block will be 1 Joule

Hence option (1) is correct.

Hope this helps.