Question

Derive an expression for the velocity of the two masses $m_1$ and $m_2$ with speeds $u_1$ and $u_2$ undergoing elastic collision in one direction.

Answer 1

Let two bodies of masses $m_1$ and $m_2$ moving with velocities $u_1$ and $u_2$ along the same straight line.

And consider the two bodies collide and after collision $v_1$ and $v_2$ be the velocities of two masses.

Before collision

Momentum of mass $m_1=m_1{u}_1$

Momentum of mass $m_2=m_2{u}_2$

Total momentum before collision

$p_1=m_1{u}_1+m_2{u}_2$

Kinetic energy of mass $m_1=\frac{1}{2}{m}_1{u_1}^2$

Kinetic energy of mass $m_2=\frac{1}{2}{m}_2{u_2}^2$

Thus,

Total kinetic energy before collision is

$K.E=\frac{1}{2}{m}_1{u_1}^2+\frac{1}{2}{m}_2{u_2}^2$

After collision

Momentum of mass $m_1=m_1{v}_1$

Momentum of mass $m_2=m_2{v}_2$

Total momentum before collision

$P_f=m_1{v}_1+m_2{v}_2$

Kinetic energy of mass $m_1=\frac{1}{2}{m}_1{v_1}^2$

Kinetic energy of mass $m_2=\frac{1}{2}{m}_2{v_2}^2$

Total kinetic energy after collision

$K_f=\frac{1}{2}{m}_1{v_1}^2+\frac{1}{2}{m}_2{v_2}^2$

So, according to the law of conservation of momentum

$\begin{array}{c}{m}_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\\ \Rightarrow m_1\left(u_1-v_1\right)=m_2\left(v_2-u_2\right)\end{array}$ $............\left(1\right)$

And according to the law of conservation of kinetic energy

$\begin{array}{c}\frac{1}{2}{m}_1{u_1}^2+\frac{1}{2}{m}_2{u_2}^2=\frac{1}{2}{m}_1{v_1}^2+\frac{1}{2}{m}_2{v_2}^2\\ \Rightarrow m_1\left({u_1}^2-{v_1}^2\right)=m_2\left({v_2}^2-{v_1}^2\right)\end{array}$ $.............\left(2\right)$

Now dividing the equation

$\begin{array}{c}\left(u_1+v_1\right)=\left(v_2+u_2\right)\\ \Rightarrow u_1-u_2=v_2-v_1\end{array}$

Therefore, this is, relative velocity of approach is equal to relative velocity of separation.