Question

A moving body of mass $m$ makes a head on elastic collision with another body of mass $2m$ which is initially at rest. Find the fraction of Kinetic energy lost by the colliding particle after collision.

Answer 1

For a head on elastic collision, the kinetic energy of the system remains conserved. It means that the total kinetic energy of the system before collision is equal to the total kinetic energy of the system after collision. For calculating the fraction of kinetic energy lost by the colliding particle after collision, we will find the individual kinetic energy of the colliding body before and after the collision.

Formula used:
Conservation of total kinetic energy of the system during an elastic collision,
$\dfrac{1}{2}{{m}_{1}}{{\left( {{u}_{1}} \right)}^{2}}+\dfrac{1}{2}{{m}_{2}}{{\left( {{u}_{2}} \right)}^{2}}=\dfrac{1}{2}{{m}_{1}}{{\left( {{v}_{1}} \right)}^{2}}+\dfrac{1}{2}{{m}_{2}}{{\left( {{v}_{2}} \right)}^{2}}$

Complete step by step answer:
Collision, also known as impact, is the sudden and forceful coming together in direct contact of two bodies. A head-on, or direct, collision means that the point of impact is on the straight line connecting the centre of gravity of each of the body.
Two types of collisions can be Elastic collision and Non-elastic collision.
In elastic collision, the total kinetic energy of the two colliding bodies, that is our system, remains conserved. There is no net loss in kinetic energy of the system as a result of an elastic collision. The total system’s kinetic energy before the collision equals the total system’s kinetic energy after the collision.
We are given that a moving body of mass $m$ makes a head on elastic collision with another body of mass $2m$ which is initially at rest.

By applying conservation of total kinetic energy of the system during collision, we have,
$\dfrac{1}{2}{{m}_{1}}{{\left( {{u}_{1}} \right)}^{2}}+\dfrac{1}{2}{{m}_{2}}{{\left( {{u}_{2}} \right)}^{2}}=\dfrac{1}{2}{{m}_{1}}{{\left( {{v}_{1}} \right)}^{2}}+\dfrac{1}{2}{{m}_{2}}{{\left( {{v}_{2}} \right)}^{2}}$
Where,
${{m}_{1}}$ is the mass of first particle
${{m}_{2}}$ is the mass of second particle
${{u}_{1}}$ is the initial velocity of first particle
${{u}_{2}}$ is the initial velocity of second particle
${{v}_{1}}$ is the final velocity of first particle
${{v}_{2}}$ is the final velocity of second particle
Or,
${{v}_{1}}=\left( \dfrac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right){{u}_{1}}+\left( \dfrac{2{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)$
We have,
$\begin{aligned} & {{m}_{1}}=m \\\ & {{m}_{2}}=2m \\\ & {{u}_{2}}=0 \\\ \end{aligned}$
Therefore,
$\begin{aligned} & {{v}_{1}}=\left( \dfrac{m-2m}{m+2m} \right){{u}_{1}}+\left( \dfrac{2\times 2m\times 0}{m+2m} \right) \\\ & {{v}_{1}}=-\dfrac{{{u}_{1}}}{3} \\\ \end{aligned}$
Now,
Initial kinetic energy of the colliding body is given by,
${{\left( K.E \right)}_{i}}=\dfrac{1}{2}{{m}_{1}}{{\left( {{u}_{1}} \right)}^{2}}$
As,
${{m}_{1}}=m$
Therefore,
${{\left( K.E \right)}_{i}}=\dfrac{1}{2}m{{\left( {{u}_{1}} \right)}^{2}}$
Final kinetic energy of the colliding body is given by,
${{\left( K.E \right)}_{f}}=\dfrac{1}{2}{{m}_{1}}{{\left( {{v}_{1}} \right)}^{2}}$
As,
$\begin{aligned} & {{m}_{1}}=m \\\ & {{v}_{1}}=-\dfrac{{{u}_{1}}}{3} \\\ \end{aligned}$
Therefore,

& {{\left( K.E \right)}_{f}}=\dfrac{1}{2}m{{\left( \dfrac{-{{u}_{1}}}{3} \right)}^{2}} \\\ & {{\left( K.E \right)}_{f}}=\dfrac{1}{18}m{{\left( {{u}_{1}} \right)}^{2}} \\\ \end{aligned}$$ Change in Kinetic energy of the colliding body is given by, $\Delta K.E={{\left( K.E \right)}_{i}}-{{\left( K.E \right)}_{f}}$ We have, $${{\left( K.E \right)}_{i}}=\dfrac{1}{2}m{{\left( {{u}_{1}} \right)}^{2}}$$ $${{\left( K.E \right)}_{f}}=\dfrac{1}{18}m{{\left( {{u}_{1}} \right)}^{2}}$$ Therefore, $$\begin{aligned} & \Delta K.E=\dfrac{1}{2}m{{\left( {{u}_{1}} \right)}^{2}}-\dfrac{1}{18}m{{\left( {{u}_{1}} \right)}^{2}} \\\ & \Delta K.E=\dfrac{4}{9}m{{\left( {{u}_{1}} \right)}^{2}} \\\ \end{aligned}$$ Fractional change in Kinetic energy of the colliding body is given by, $$\begin{aligned} & \dfrac{\Delta K.E}{{{\left( K.E \right)}_{i}}}=\dfrac{\dfrac{4}{9}m{{\left( {{u}_{1}} \right)}^{2}}}{\dfrac{1}{2}m{{\left( {{u}_{1}} \right)}^{2}}}=\dfrac{4}{9}\times \dfrac{2}{1}=\dfrac{8}{9} \\\ & \dfrac{\Delta K.E}{{{\left( K.E \right)}_{i}}}=\dfrac{8}{9} \\\ \end{aligned}$$ The fraction of kinetic energy lost by the colliding particle after collision is $\dfrac{8}{9}$. **Note:** In a head on elastic collision, there is no net loss in the total kinetic energy of the system. But, the individual kinetic energies of the two bodies may vary before and after the collision. The fractional change in kinetic energy of a colliding body is expressed as the ratio of change in kinetic energy of the body before and after the collision to the initial kinetic energy of the body before the collision.