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

Here, `m_(1)=1.2kg, u_(1)=20cm//s, m_(2)=1.2kg, u_(2)=0`
If `upsilon_(1)` and `upsilon_(2)` are velocities of the two blocks after collision, then accordin to the principle of conservation of momentum, `m_(1)u_(1)+m_(2)u_(2)=m_(1)upsilon_(1)+m_(2)upsilon_(2)`
`1.2xx20+0=1.2upsilon_(1)+1.2upsilon_(2) ` `:. upsilon_(1)+upsilon_(2)=20(cm//s)` ...(i)
velocity of approach `=u_(1)-u_(2)=20cm//s,` velocity of separation `=upsilon_(2)-upsilon_(1)`
By definition, `e=(upsilon_(2)-upsilon_(1))/(u_(1)-u_(2))`
`(3)/(5)=(upsilon_(2)-upsilon_(1))/(20) :. upsilon_(2)-upsilon_(1)=(20xx3)/(5)=12` ...(ii)
From (i) and (ii), `upsilon_(1)=4cm//s, upsilon_(2)=16cm//s`
Loss in K.E. `=(1)/(2)m_(1)u_(1)^(2)-(1)/(2)(m_(1))upsilon_(1)^(2)-(1)/(2)m_(2)upsilon_(2)^(2)=(1)/(2)xx1.2((20)/(100))^(2)-(1)/(2)(1.2)((4)/(100))^(2)-(1)/(2)xx1.2((16)/(100))^(2)`
`=2.4xx10^(-2)-0.096xx10^(-2)-1.536xx10^(-2)`