Question

Show that after direct elastic collision of two bodies of equal mass, their velocities are interchanged. ​

Answer 1

Given :

Mass of two bodies are equal

To Find :

The mass of two bodies interchanged.

Solution :

Let the initial velocity of the bodies of mass $m_{1}$ and $m_{2}$ be $u_{1}$ and $u_{2}$, and their final velocity (after collision) be $v_{1}$ and $v_{2}$ respectively.

From the law of conservation of momentum,

         $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

or,      $m_{1}$ (u$_{1}$ - v$_{1}$)      =  m$_{2}$ (v$_{2}$ - u$_{2}$)     --------------> equation (1)

In an elastic collision, the total kinetic energy of the particles will also be conserved.

       $\frac{1}{2} m_1u_1^{2} + \frac{1}{2} m_2u_2^{2}$  =   $\frac{1}{2} m_1v_1^{2} + \frac{1}{2} m_2v_2^{2}$

or,    $m_{1}$ ($u_1^{2} - v_1^{2}$)      =   m$_{2}$ ($v_2^{2} - u_2^{2}$)     --------------> equation (2)

Now, (2) + (1) gives,

$v_{2}$ = u$_{1}$ + v$_{1}$ - u$_{2}$ ------(3) and  v$_{1}$ = v$_{2}$ - u$_{1}$ + u$_{2}$ ---------(4)

Substituting the value of $v_{2}$ in equation (1) we get,

v$_{1}$ = $\frac{m_1 - m_2}{m_1 + m_2} u_1 + \frac{2m_2}{m_1 + m_2} u_2$

Similarly, taking value of v$_{1}$from (4) and substituting it in equation (1), we get,

 $v_{2}$ = $\frac{m_2 - m_1}{m_1 + m_2} u_2 + \frac{2m_1}{m_1 + m_2} u_1$

Let $m_1 = m_2$ = m

v$_{1}$ = $\frac{m - m}{m + m} u_1 + \frac{2m}{m + m} u_2$

v$_{1}$ = 0 +  u$_{2}$

∴  v$_{1}$  =   u$_{2}$

$v_{2}$ = $\frac{m - m}{m + m} u_2 + \frac{2m}{m + m} u_1$

∴   v$_{2}$  =  u$_{1}$

Hence proved that after direct elastic collision of two bodies of equal mass, their velocities are interchanged. ​