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For an elastic collision of a mass \[m_{1}\] moving with a velocity \[v\] with a mass \[m_{2}\] moving with speed \[v_{2}\] after collision I'm getting the expression for v2 as \[v_{2}=\frac{2m_{1}v}{m_{1}+m_{2}}=\frac{2v}{(1+\frac{m_{2}}{m_{1}})}\] Is this correct?
I do not completely understand your question, which seems to be missing some information. We need initial velocities of both objects to find the final velocities. Denoting initial velocities u1, u2, and final velocities v1 and v2, then we get from conservation of momentum: \(m_1 u_1+m_2 u_2 = m_1 v_1+m_2 v_2\) ........(1) Equation (1) applies whether collision is elastic or not. Also, sign is important, so velocities are positive in a given direction, assuming the collision is rectilinear. then, for an elastic collision, there is no loss of energy, so equating kinetic energies and multiplying the equation by a factor of two (to get rid of the 1/2 on both sides): \(m_1 u_1^2+m_2 u_2^2 = m_1 v_1^2+m_2 v_2^2\) ........(2) These two equations will enable you to solve for v1 and v2, given u1 and u2, or vice versa.
Sorry I meant that \[m_{2}\] is at rest and the velocity \[v_{2}\] is it's velocity after collision \[m_{1}v^2=m_{1}v_{1}^2+m_{2}v_{2}^2\] eliminating v1 using equation of kinetic energy im getting \[v_{2}=\frac{2m_{1}v}{m_{1}+m_{2}}=\frac{2v}{(1+\frac{m_{2}}{m_{1}})}\]

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Yes, your expression is correct. From momentum: \(v1=-(m2~v2-m1~v)/m1\) substitute into the energy equation, \(m1~v^2=m1~(-(m2~v2-m1~v)/m1)^2+m2~v2^2\) and solving for v2 gives: \(\Large v_2=\frac{2~m_1~v}{m_1+m_2}\) (sorry, was too lazy to put subscripts in the intermediate steps!)

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