Ans.

The particle of mass \( m_1 \) collides head on with the particles of mass \( m_2 \).

The velocities of particles \( m_1 \) and \( m_2 \) before collision are \( u_1 \) and \( u_2=0 \) respectively.

After collision their respective velocities are \( v_1 \) and \( v_2 \).

Since the collision is perfectly elastic,

So according to the conservation of momentum we get

\( \displaystyle{m_1\cdot{u_1}+m_2\cdot{0}=m_1\cdot{v_1}+m_2\cdot{v_2}} \)

or,\( \displaystyle{m_1\cdot{u_1}=m_1\cdot{v_1}+m_2\cdot{v_2}} \)

or, \( \displaystyle{m_1(u_1-v_1)=m_2v_2}\tag{1} \)

And according to the conservation of energy,

\( \displaystyle{\frac{1}{2}m_1{u_1}^2+\frac{1}{2}m_2{u_2}^2=\frac{1}{2}m_1{v_1}^2+\frac{1}{2}m_2{v_2}^2} \)

or, \( \displaystyle{\frac{1}{2}m_1{u_1}^2=\frac{1}{2}m_1{v_1}^2+\frac{1}{2}m_2{v_2}^2} \)

or, \( \displaystyle{m_1(u_1^2-v_1^2)=m_2v_2^2}\tag{2} \)

Dividing equation (2) by equation (1), we get

\( u_1+v_1=v_2 \)

or, \( v_1=v_2-u_1\tag{3} \)

From equations (1) & (3), we get

\( m_1u_1=m_1(v_2-u_1)+m_2v_2 \)

or, \( m_1u_1=(m_1+m_2)v_2-m_1u_1 \)

therefore, \( \displaystyle{v_2=\frac{2m_1}{m_1+m_2}u_1} \)