Body `A` of mass `4m` 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 colliding body `A` is :

Options

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

Answer

B

Explanation

From law of conservation of momentum will be
`m_1u_1=m_1v_1+m_2v_2`
`m_2v_2=m_1u_1-m_1v_1`
From the law of conservation of KE. we have
`1/2m_1u_1^2=1/2m_1v_1^2+1/2m_2v_2^2`
Rewriting (1) as
`m_2v_2=m_1(u_1-v_1)`
Rewriting (2) as
`m_2v_2^2=m_1(u_1^2-v_1^2)`
Dividing (4) with (3)
`v_2=u_1+v_1`
eliminating `v_2` from (1) and (5) we get
`(m_1u_1-m_1v_1)/(m_2)=u_1+v_1`
`((m_1-m_2)/(m_1+m_2))u_1=v_1`
Fraction of K.E of `m_1` carried by `m_2` is
`1-[(m_1-m_2)/(m_1+m_2)]^2=(4m_1m_2)/((m_1+m_2)^2)`
This is also equal to the fractional transfer after colliding body.
Fractional transfer of KE of colliding body
`(DeltaKE)/(KE)=(4(m_1m_2))/((m_1+m_2)^2)`
`=(4(4m)2m)/((4m+2m)^2)`
`=(32m^2)/(36m^2)=8/9`
This fractional transfer is equal to the fraction lost by A.