A block of mass 2kg moving at 2m/s collides head on with another block of equal mass kept at rest. (A) find the max possible lost in kinetic energy due to collision. (B) if the actusl lost in kinetic energy is half of this max value, find the value of e?

(a) maximum kinetic energy is lost in inelastic collision when the two masses move together with same velocity `v2` after collision.By conservation of momentum mv1=(m+m)v2 v2=1 m/s. Initail KE=1/2 2 2^2=4 J. Final KE=1/2 4 1^2= 2 J. Loss in KE= 4-2=2 J (b) Loss in KE=1 J. Final KE=4-1=3 J. 3=1/2 m(v1)^2 + 1/2 m(v2)^2, 3= (v1)^2 + (v2)^2 also , 4=mv1 +mv2, v1+v2 =2, v1=2 -v2(put this value in the KE equation and solve) v1=1 +(1/2)^1/2, v2= 1 - (1/2)^1/2, velocity of separation= v1-v2 =2^1/2, velocity of approach =2 e=(2^1/2) / 2 =1/(2^1/2).

Sir but inelastic collisions is not given anywhere..

First of all please read about elastic and inelastic collision. Then you will know that the maximum kinetic energy is lost during perfectly inelastic collision in which the two bodies move with the same speed after collision. The kinetic energy is conserved in the elastic collision and the maximum loss is in perfectly inelastic collision.As in the question it is given to find the maximum possible loss in KE ,so we will consider the perfectly inelastic collision.

Approved

ok thanku sir