Derive an expression for kinetic energy in pure rolling.

Explanation

In perfectly inclastic collision, the loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, heat. light etc. Let `KE_(i)` be the total kinetic energy before collision and `KE_(f)` be the total kinetic energy after collision.
Total kinetic energy before collision,
`KE_(i)=(1)/(2)m_(1)u_(1)^(2)+(1)/(2)m_(2)u_(2)^(2)` ...(1)
Total kinetic energy after collision.
`KE_(f)=(1)/(2)(m_(1)+m_(2))v^(2)` ...(2)
Loss of KE, `DeltaQ=KE_(f)-KE_(i)=(1)/(2)(m_(1)+m_(2))v^(2)-(1)/(2)m_(1)u_(1)^(2)-(1)/(2)m_(2)u_(2)^(2)` ...(3)
Substituting equation `v = (m_(1)u_(1)+m_(2)u_(2))/((m_(1)+m_(2)))` in equation (3), and on simplifying (expand v by using the algebra `(a +b)^(2) = a^(2) +b^(2) + 2ab` , we get
Loss of KE, `DeltaQ=(1)/(2)((m_(1)m_(2))/(m_(1)+m_(2)))(u_(1)-u_(2))^(2)`