A block of mass 1.2 kg moving at a speed of 20 cm /s collides head on with a similar block kept at rest. The coefficient of restitution is 3/5. Find the loss of kinetic energy during the collision.

Explanation

Suppose the first block moves at a speed `v_1` and the second at `v_2` after the collision. Since the collisionis head on the two blocks moe along the origiN/Al direction of motion of the block.
by conservation of linear momentum
`(1.2kg)(20cm/s)=(1.2kg)v_1+(1.2kg)v_2`
or, `v_1+v_2=20cm/s`
the velocity of separation is `v_2-v_1` and the velocity of approach is 20 cm/s. As the coeffficient of restitutioin is 3/5 we have,
`v_2-v_1=(3/5)xx20cm/s=12cm/s`........ii
By i and ii
`v_1=4cm/s and v_2=16cm/s`
The loss in kinetic energy is
`1/2(1.2kg)[20cm/s)^2-(4cm/s)^2-(16cm/s^2]`
`=(0.6kg)[0.04m^2/s^2-0.0016 m^2/s^2-0.0256m^2/s^2]`
`=(0.6kg)(0.0128m^2/s^2)=7.7xx10^-3J`.