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

Options

A. `5//9`
B. `1//9`
C. `8//9`
D. `4//9`

Answer

C

Explanation

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`