#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.
