A spherical ball of mass `m_(1)` collides head on with another ball of mass `m_(2)` at rest . The collision is elastic . The fraction of kinetic energy lost by `m_(1)` is `:`

Options

A. `(4m_1m_2)/((m_1+m_2)^2)`
B. `m_1/(m_1+m_2)`
C. `m_2/(m_1+m_2)`
D. `(m_1m_2)/((m_1+m_2)^2)`

Answer

A

Explanation

According to momentum conservation, we get `m_1v_(1i)=m_1v_(1f)+m_2v_(2f)`…(i)
where `v_(1i)` is the initial velocity of spherical ball of mass `m_1` before collision and `v_(1f)` and `v_(2f)` are the final velocities of the balls of masses `m_1 and m_2` after collision.
According to kinetic energy conservation, we get
`1/2m_1v_(1i)^(2)=1/2m_1v_(1f)^(2)+1/2m_2v_(2f)^(2)`
`m_1v_(1i)^(2)=m_1v_(1f)^(2)+m_2v_(2f)^(2)` ...(ii)
From Eqs. (i) and (ii) , it follows that
`m_1v_(1i)(v_(2f)-v_(1i))=m_1v_(1f)(v_(2f)-v_(1f))`
or `v_(2f)(v_(1i)-v_(1f))=v_(1i)^2-v_(1f)^2=(v_(1i)-v_(1f))(v_(1i)+v_(1f))`
`therefore v_(2f)=v_(1i)+v_(1f)`
Substituting this in Eq.(i), we get
`v_(1f)=((m_1-m_2))/(m_1+m_2)v_(1i)`...(iii)
The initial kinetic energy of the mass `m_1` is
`K_(1i)=1/2m_1v_(1i)^2`
The final kinetic energy of the mass `m_1` is
`K_(1f)=1/2m_1v_(1f)^2=1/2m_1((m_1-m_2)/(m_1+m_2))^2 v_(1i)^2`(Using (iii))
The fraction of kinetic energy lost by `m_1` is
`f=(K_(1i)-K_(1f))/K_(1i)=(1/2m_1v_(1i)^2 -1/2m_1((m_1-m_2)/(m_1+m_2))^2 v_(1i)^2)/(1/2m_1v_(1i)^2)`
`=1-((m_1+m_2)/(m_1+m_2))^2=(4m_1m_2)/((m_1+m_2)^2)`