Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is

According to conservation of momentum,4mu1=4mv1+2mv2 ⇒2u1−v1=v2……..(i)From conservation of energy,12(4m)u12=12(4m)v12+12(2m)v22 ⇒2u12−v12=v22…… (ii)   From (i) and (ii)  2u12−v12=4u1−v12Now, fraction of loss in kinetic energy for mass 4m,ΔKKi=Ki−KfKi=1/2(4m)u12−12(4m)v1212(4m)u12Substituting (iii) in (iv), we get ΔKKi=89