Body A of mass 4 m moving with speed u collides with another body B of mass 2m, 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

Body A of mass 4 m moving with speed u collides with another body B of

1.	Body A of mass 4 m moving with speed u collides with another body B of mass 2m, 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
Answer» `5//9` `1//9` `8//9` `4//9` Solution :ACCORDING to conservation of momentum, `4 m u_1 = 4 mv_1 + 2 mv_2` `implies 2(u_1 - v_1) = v_1 "" ….(i)` From conservation of energy, `1/2(4m)u_1^2 = 1/2 (4m)v_1^2 + 1/2 (2m)v_2^2` `implies 2(u_1^2 - v_1^2) = v_2^2 "" ……(ii)` From (i) and (ii) `2(u_1^2 - v_1^2) = 4(u_1 - v_1)^2` `3v_1 = u_1 "" .......(iii)` Now, FRACTION of loss in kinetic energy for MASS 4 m, `(DeltaK)/(K_i) = (K_i - K_f)/(K_i) = (1//2 (4m) u_1^2 - 1/2 (4m)v_1^2)/(1/2 (4m)u_1^2) ""........(IV)` Substituting (iii) in (iv), we get `(DeltaK)/(K_i) = 8/9`