# Elastic Collisions

## Key Questions

• billiard balls
curling rocks
shuffleboard pucks
bowling ball and pins

Always take time to think because in real-life situations, most collisions are a combination of elastic and inelastic.

• ${m}_{1} , {m}_{2}$ be the two bodies
${u}_{1} , {v}_{1}$ be initial and final velocities of ${m}_{1}$
${u}_{2} , {v}_{2}$ be initial and final velocities of ${m}_{2}$

let the initial velocity of ${m}_{2}$ = o. So, ${u}_{2}$ = 0.

As the collision is elastic, Kinetic Energy and Momentum are conserved.

Initial Momentum is ${m}_{1} {u}_{1}$
Final Momentum is ${m}_{1} {v}_{1} + {m}_{2} {v}_{2}$

So, ${m}_{1} {u}_{1} = {m}_{1} {v}_{1} + {m}_{2} {v}_{2}$
$\Rightarrow$ ${m}_{1} {u}_{1} - {m}_{1} {v}_{1} = {m}_{2} {v}_{2}$
$\Rightarrow$ ${m}_{1} \left({u}_{1} - {v}_{1}\right) = {m}_{2} {v}_{2}$ $\to$ Equation'1'

Initial Kinetic Energy is $\frac{1}{2} {m}_{1} {u}_{1}^{2}$
Final Kinetic Energy is $\frac{1}{2} {m}_{1} {v}_{1}^{2} + \frac{1}{2} {m}_{2} {v}_{2}^{2}$

So, $\frac{1}{2} {m}_{1} {u}_{1}^{2} = \frac{1}{2} {m}_{1} {v}_{1}^{2} + \frac{1}{2} {m}_{2} {v}_{2}^{2}$
$\Rightarrow$ ${m}_{1} {u}_{1}^{2} = {m}_{1} {v}_{1}^{2} + {m}_{2} {v}_{2}^{2}$
$\Rightarrow$ ${m}_{1} {u}_{1}^{2} - {m}_{1} {v}_{1}^{2} = {m}_{2} {v}_{2}^{2}$
$\Rightarrow$ ${m}_{1} \left({u}_{1}^{2} - {v}_{1}^{2}\right) = {m}_{2} {v}_{2}^{2}$
$\Rightarrow$ ${m}_{1} \left({u}_{1} - {v}_{1}\right) \left({u}_{1} + {v}_{1}\right) = {m}_{2} {v}_{2}^{2}$

From Equation '1', ${m}_{1} \left({u}_{1} - {v}_{1}\right) = {m}_{2} {v}_{2}$

$\Rightarrow$ ${m}_{2} {v}_{2} \left({u}_{1} + {v}_{1}\right) = {m}_{2} {v}_{2}^{2}$
$\Rightarrow$ ${u}_{1} + {v}_{1} = {v}_{2}$

Now, ${v}_{2} = {u}_{1} + {v}_{1}$

From Equation '1', ${m}_{1} \left({u}_{1} - {v}_{1}\right) = {m}_{2} {v}_{2}$

So, ${m}_{1} \left({u}_{1} - {v}_{1}\right) = {m}_{2} \left({u}_{1} + {v}_{1}\right)$
$\Rightarrow$ ${m}_{1} {u}_{1} - {m}_{1} {v}_{1} = {m}_{2} {u}_{1} + {m}_{2} {v}_{1}$
$\Rightarrow$ ${v}_{1} = \frac{{u}_{1} \left({m}_{1} - {m}_{2}\right)}{{m}_{1} + {m}_{2}}$

now, solve the equation '1' using ${v}_{1} = {v}_{2} - {u}_{1}$ to get ${v}_{2}$

I think, ${v}_{2} = \frac{2 {m}_{1} {u}_{2}}{{m}_{1} + {m}_{2}}$

Final momentum of of ${m}_{1}$ is ${m}_{1} {v}_{1}$ and Final momentum of ${m}_{2}$ is ${m}_{2} {v}_{2}$

:D

PS : Sorry if the math is clumsy, I'm new here.

Elastic collision is the collision where there occurs no loss in net kinetic energy as the result of collision.

#### Explanation:

Total Kinetic energy before the collision= Total kinetic energy after the collision

For example,

Bouncing back of a ball from the floor is an example of elastic collision.

Some other examples are:-
$\implies$collision between atoms
$\implies$collision of billiard balls
$\implies$balls in the Newton's cradle... etc.