m_1, m_2 be the two bodies
u_1, v_1 be initial and final velocities of m_1
u_2,v_2 be initial and final velocities of m_2
let the initial velocity of m_2 = o. So, u_2 = 0.
As the collision is elastic, Kinetic Energy and Momentum are conserved.
Initial Momentum is m_1u_1
Final Momentum is m_1v_1+m_2v_2
So, m_1u_1 = m_1v_1+m_2v_2
rArr m_1u_1-m_1v_1=m_2v_2
rArr m_1(u_1-v_1)=m_2v_2 -> Equation'1'
Initial Kinetic Energy is 1/2m_1u_1^2
Final Kinetic Energy is 1/2m_1v_1^2+1/2m_2v_2^2
So, 1/2m_1u_1^2 = 1/2m_1v_1^2+1/2m_2v_2^2
rArr m_1u_1^2 = m_1v_1^2+m_2v_2^2
rArr m_1u_1^2-m_1v_1^2 = m_2v_2^2
rArr m_1(u_1^2-v_1^2) = m_2v_2^2
rArr m_1(u_1-v_1)(u_1+v_1) = m_2v_2^2
From Equation '1', m_1(u_1-v_1)=m_2v_2
rArr m_2v_2(u_1+v_1)=m_2v_2^2
rArr u_1+v_1=v_2
Now, v_2=u_1+v_1
From Equation '1', m_1(u_1-v_1)=m_2v_2
So, m_1(u_1-v_1)=m_2(u_1+v_1)
rArr m_1u_1-m_1v_1=m_2u_1+m_2v_1
rArr v_1= (u_1(m_1-m_2))/(m_1+m_2)
now, solve the equation '1' using v_1=v_2-u_1 to get v_2
I think, v_2=(2m_1u_2)/(m_1+m_2)
Final momentum of of m_1 is m_1v_1 and Final momentum of m_2 is m_2v_2
:D
PS : Sorry if the math is clumsy, I'm new here.
