Two particles of masses `m_1 and m_2` moving with velocities `u_1 and u_2` , respectively and making an angle `theta` between them, collide with each other. After collision, the 1st particle travels in the initial direction of motion of the 2nd, and vice-verses. Find the velocities of the two particles after collision. Under what condition, would this collision be elastic ?

Explanation

Suppose `v_1, v_2` are the velocities of the two particles, respectively, after collision . The particles before and after collision move as shown in Fig.1.44 .It also shows the chosen directions of the x and the y-axis.

For momentum conservation along the x-axis, we get,
`m_1u_1 cos""(theta)/(2) +m_2u_2 cos ""(theta)/(2) =m_1v_1 cos ""(theta)/(2) +m_2 v_2 cos ""(theta)/(2) or, m_1u_1+m_2u_2=m_1v_1+m_2v_2` .......(1)
Similarly along the y-axis, we get,
`-m_1u_1sin""(theta)/2+m_2u_2sin ""theta/2=m_1v_1sin ""theta/2-m_2v_2 sin""theta/2 or, -m_1u_1+m_2u_2=m_1v_1-m_2v_2`.....(2)
Adding equations (1) and (2) ,
`2m_2u_2=2m_1v_1`
`or, v_1=(m_2)/(m_1)u_1`.......(3)
Subtracting equation (2) from (1) ,
`2m_1u_1=2m_2v_2`
`or, v_2=(m_1)/(m_2)u_1`.......(4)
The kinetic energy before collision is,
`K_1=1/2m_1u_1^2+1/2u_2^2`
and that after collision is
`K_2=1/2m_1v_1^2+1/2m_2v_2^2=1/2m_1((m_2)/(m_1)u_2)^2+1/2 m_2((m_1)/(m_2)u_1)^2`
=`1/2 (m_2)/(m_1) m_2u_2^2+1/2(m_1)/(m_2) m_1u_1^2`
Here `K_1ne K_2` , so the collision is inelastic, in general. As a special case , it would be an elastic collision if `K_1=K_2` . It is possible only when `m_1=m_2` , i.e., the two particles are of equal masses.