Question

A Particle of mass ${m_1}$ and velocity $u$ collides elastically (in one dimension) with a stationary particle of mass ${m_2}$. Derive the expression for the velocities of ${m_1}$ and ${m_2}$ after the collision. Prove that bodies of identical masses exchange their velocities after head on collision.

Answer 1

Hint: In this question we use the concept of principle of conservation of momentum and the principle of conservation of kinetic energy to find the expression between the velocities of two particles after an elastic collision.

Complete solution:
Let us assume that the velocity after collision of particle of mass ${m_1}$ is ${v_1}$and the velocity of the stationary particle of mass ${m_2}$ is ${v_2}$. The initial velocity of the particle is given as $u$.
As the collision is an elastic collision, we can write from the law of conservation of momentum that,
${m_1}u = {m_1}{v_1} + {m_2}{v_2}$
$ \Rightarrow {m_2}{v_2} = {m_1}\left( {u - {v_1}} \right)......\left( 1 \right)$
And from the law of conservation of kinetic energy we can write that,
$\dfrac{1}{2}{m_1}{u^2} = \dfrac{1}{2}{m_1}{v_1}^2 + \dfrac{1}{2}{m_2}{v_2}^2$
$ \Rightarrow {m_2}{v_2}^2 = {m_1}\left( {{u^2} - {v_1}^2} \right)......\left( 2 \right)$
Now we divide equation (2) by equation (1)
$ \Rightarrow \dfrac{{{m_2}{v_2}^2}}{{{m_2}{v_2}}} = \dfrac{{{m_1}\left( {{u^2} - {v_1}^2} \right)}}{{{m_1}\left( {u - {v_1}} \right)}}$
After simplification we get,
$ \Rightarrow {v_2} = u + {v_1}$
Hence the expression of velocities of particle of mass ${m_1}$ and ${m_2}$
${v_2} = u + {v_1}$
So after a head on elastic collision between two bodies the kinetic energy and momentum will be constant.
Let the initial velocity of the two bodies are ${u_1},{u_2}$ respectively and the velocity after collision are ${v_1},{v_2}$ respectively. The mass of the two bodies are identical, let it be $m$.
As the momentum is constant,
$m{u_1} + m{u_2} = m{v_1} + m{v_2}$
$ \Rightarrow {u_1} + {u_2} = {v_1} + {v_2}$
Now the kinetic energy is constant, hence $\dfrac{1}{2}m{u_1}^2 + \dfrac{1}{2}m{u_2}^2 = \dfrac{1}{2}m{v_1}^2 + \dfrac{1}{2}m{v_2}^2 \Rightarrow {u_1}^2 + {u_2}^2 = {v_1}^2 + {v_2}^2$
Solving these equations, we get ${v_2} = {u_1}$ and ${v_1} = {u_2}$.
Hence, the velocity exchanges after collision.

Note: In an inelastic collision the energy is conserved and the momentum of the collision is conserved. As we know that the Law of conservation of momentum states that the momentum before a collision is equal to the momentum after the collision if no no external force acts on the system. Such as two elastic balls collide horizontally then the momentum is conserved between the two balls. When the first ball stops the second ball gains momentum and moves, hence the momentum is conserved. Law of conservation of energy states that within an isolated system there are no changes in energy. It changes from one form of energy to another.