Question

Two identical particles of same mass, having velocities opposite to each other, equal in
magnitude, collide head on. During collision 50% of kinetic energy is lost. Coefficient of restitution is:
(1) $\dfrac{1}{{\sqrt 2 }}$
(2) $\dfrac{1}{2}$
(3) $\dfrac{2}{3}$
(4) $\dfrac{1}{4}$

Answer 1

We know that the coefficient of restitution is given by the ratio of the relative velocity of the given particles before the collision to the relative velocity of particles after collision. We will use the given statement to write the relation between initial and final kinetic energies.

Complete step by step answer:
Assume:
The mass of two identical particles is m.
The initial velocity of the first mass is ${V_1} = V$.
The initial velocity of the second mass is ${V_2} = - V$.
We can write the expression for the initial kinetic energy of the first particle as below:
${K_1} = \dfrac{1}{2}mV_1^2$
We will substitute V for ${V_1}$ in the above expression to get the initial kinetic energy of the first
particle.

{K_1} = \dfrac{1}{2}m{\left( V \right)^2}\\\ = \dfrac{1}{2}m{V^2} \end{array}$$ We can write the expression for the initial kinetic energy of the second particle as below: $${K_1} = \dfrac{1}{2}mV_2^2$$ We will substitute $$ - V$$ for $${V_2}$$ in the above expression to get the initial kinetic energy of the second particle. $$\begin{array}{c} {K_1} = \dfrac{1}{2}m{\left( { - V} \right)^2}\\\ = \dfrac{1}{2}m{V^2} \end{array}$$ We can write the expression for the final kinetic energy of both the particles as below: $$\begin{array}{c} {K_3} = \dfrac{1}{2}m{U^2}\\\ = {K_4} \end{array}$$ Here U is the final velocity of both the particles, $${K_3}$$ is the final kinetic energy of the first particle and $${K_4}$$ is the final kinetic energy of the second particle. It is given that 50% of kinetic energy is lost when the given identical particles collide with each other that means the summation of the final kinetic energy of both the particles is equal to the half of their initial kinetic energy of so we can write: $${K_1} + {K_2} = \dfrac{1}{2}\left( {{K_3} + {K_4}} \right)$$ We will substitute $$\dfrac{1}{2}m{V^2}$$ for $${K_1}$$, $${K_2}$$ and $$\dfrac{1}{2}m{U^2}$$ for $${K_3}$$,$${K_4}$$ in the above expression. $ \dfrac{1}{2}m{V^2} + \dfrac{1}{2}m{V^2} = \dfrac{1}{2}\left( {\dfrac{1}{2}m{U^2} + \implies \dfrac{1}{2}m{U^2}} \right)\\\ \implies {V^2} = \dfrac{1}{2}{U^2}\\\ \implies V = \dfrac{1}{{\sqrt 2 }}U $ We can write the expression for the restitution coefficient as below: $$C = \dfrac{{{V_1} - {V_2}}}{{U - \left( { - U} \right)}}$$ We will substitute V for $${V_1}$$, $$ - V$$ for $${V_2}$$ in the above expression. $ C = \dfrac{{V - \left( { - V} \right)}}{{U - \left( { - U} \right)}}\\\ \implies C = \dfrac{V}{U} $ We will substitute $$\dfrac{1}{{\sqrt 2 }}U$$ for V in the above expression to get the value of the coefficient of restitution. $$\begin{array}{c} C = \dfrac{{\left( {\dfrac{1}{{\sqrt 2 }}U} \right)}}{U}\\\ = \dfrac{1}{{\sqrt 2 }} \end{array}$$ **So, the correct answer is “Option 1”.** **Note:** After colliding, the velocity of both the particles becomes the same but opposite in direction, but due to the square of velocity in kinetic energy expression final kinetic energy of both the particles became the same.