Question

a block of mass 2kg moving at 2m/s collides head on with another block of equal mass kept at rest . if the loss of kinetic energy is half of max possible loss of kinetic energy of system,if the coefficient of restitution is
$\alpha \div \sqrt{2.}$
then find
$\alpha$

Answer 1

(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).