You might wonder why all the molecules in a gas in thermal equilibrium don't have exactly the same speed. After all, when two molecules collide, doesn't the faster one always lose energy and the slower one gain energy? And if so, wouldn't repeated collisions eventually bring all the molecules to some common speed? Describe an example of a billiard-ball collision in which this is not the case: The faster ball gains energy and the slower ball loses energy. Include numbers, and be sure that your collision conserves both energy and momentum.

We are given that,

Consider two billiard ball of equal masses m, one moving with velocity $v\left(\cos \right(\theta )i+\sin (\theta \left)j\right)$and the

Other moving with velocity $u\left(\cos \right(\varphi )i+\sin (\varphi \left)j\right)$

Also assume $v>u$

So initial momentum of the system

$m\left(\cos \right(\theta )\hat{i}+\sin (\theta )\hat{j}+\cos (\varphi )\hat{i}+\sin (\varphi \left)\hat{j}\right)=\mathrm{\pi }$

Now when these balls collide

Let final velocities of the balls be

$v\text{'}\left(\cos \right(\theta \text{'})i+\sin (\theta \text{'}\left)j\right)andu\text{'}\left(\cos \right(\varphi \text{'})i+\sin (\varphi \text{'}\left)j\right)$ {$v\text{'}isfinalvelocities$}

Hence final momentum

$m\left(\cos \right(\theta \text{'})i+\sin (\theta \text{'})j+\cos (\varphi \text{'})i+\sin (\varphi \text{'}\left)j\right)=\mathrm{\pi }$

From conservation of momentum

$\left(v\right(\cos \left(\theta \right)i+\sin \left(\theta \right)j)+u(\cos \left(\varphi \right)i+\sin \left(\varphi \right)j\left)\right)=v\text{'}\left(\cos \right(\theta \text{'})i+\sin (\theta \text{'}\left)j\right)+u\text{'}\left(\cos \right(\varphi \text{'})i+\sin (\varphi \text{'}\left)j\right)$

Also, for perfectly elastic collision

Coefficient of restitution $=1$

Then

$[v\text{'}(\cos (\theta \text{'})i+\sin (\theta \text{'})j)-u\text{'}(\cos (\varphi \text{'})i+\sin (\varphi \text{'})j\left)\right]=u\left(\cos \right(\varphi )i+\sin (\varphi \left)j\right)-v\left(\cos \right(\theta )i+\sin (\theta \left)j\right)$

Hence we get the following equations

$v\cos \left(\theta \right)+u\cos \left(\varphi \right)=v\text{'}\cos (\theta \text{'})+u\text{'}\cos (\varphi \text{'})$

$v\sin \left(\theta \right)+u\sin \left(\varphi \right)=v\text{'}\sin (\theta \text{'})+u\text{'}\sin (\varphi \text{'})$

$v\text{'}\cos (\theta \text{'})-u\text{'}\cos (\varphi \text{'})=u\cos \left(\varphi \right)-v\cos \left(\theta \right)$

$v\text{'}\sin (\theta \text{'})-u\text{'}\sin (\varphi \text{'})=u\sin \left(\varphi \right)-v\sin \left(\theta \right)$

$2v\text{'}\cos (\theta \text{'})=2u\cos \left(\varphi \right)$

$2v\text{'}\sin (\theta \text{'})=2u\sin \left(\varphi \right)$

So we get $v\text{'}=u$

Hence the velocities are exchanged for billiard ball with equal mass

Hence for equal masses one cannot say that the faster ball will gain energy and the slower ball will Not but for unequal masses the faster ball can gain speed and the slower one can lose speed Depending on the mass.