Question

A particle of mass $m$ strikes another particle of the same mass of rest. Find the angle between the velocities of particles after the collision, if the collision is elastic.

Answer 1

Elastic collision:

Case 1: When line of impact is along the velocity

1. An elastic collision is a type of collision in which there is no net loss in kinetic energy in the system as a result of the collision.
2. Both the momentum and the kinetic energy are well-conserved quantities in such types of collisions.
3. It can be both, either one-dimensional or two-dimensional.
4. In the real world, it is nearly impossible for a perfectly elastic collision to occur because there is always bound to be some energy conversion. However small may be.
5. When a ball at a billiard sports table hits another ball, it is an example of elastic collision.
6. When you throw a ball on the ground and it bounces back to you, there is no net change in the kinetic energy of the ball, and hence, it is an elastic collision.
7. As per given in the question, the collision is elastic and both the masses are the same.
8. Hence, we know according to the law of conservation of momentum and also they are symmetric, the first mass will immediately transfer momentum to the second mass and hence velocity will be the exactly same as in the previous case.
9. Hence angle made by the velocities will be Zero.

Thus, the angle between the velocities of particles after the collision will be zero.

Case 2:When line of impact is not along the velocity

In this case, if the line of impact is not along the velocity, then both the particles will move in different directions with the different angles with the horizontal.

Let $m$is the mass of both the particles and $\overrightarrow{p}$ is the momentum of the first particle, while the second particle is at rest.

Now, after the collision, the first mass will move with momentum $\overrightarrow{p_1}$while making an angle $\alpha$with the horizontal.

Similarly, the second particle will move with momentum $\overrightarrow{p_2}$ while making an angle $\beta$ with the horizontal.

Now we will calculate the resultant of both the momentum vectors.

$\left|p_{res}\right|=\sqrt{{p^2}_1+{p^2}_2+2p_1{p}_2.\cos \left(\alpha +\beta \right)}=p_f$= final momentum after the collision.

Now equalizing the initial momentum and final momentum.

$$\begin{gathered} p=\sqrt{{p^2}_1+{p^2}_2+2p_1{p}_2.\cos \left(\alpha +\beta \right)} \\ {p}^2={p^2}_1+{p^2}_2+2p_1{p}_2.\cos \left(\alpha +\beta \right).......\left(1\right) \end{gathered}$$

Now, we will conserve kinetic energy.

We know, Kinetic energy = $\frac{p^2}{2m}$

So initial kinetic energy, $K{E}_i=\frac{p^2}{2m}+0=\frac{p^2}{2m}$

Final kinetic energy, $K{E}_f=\frac{{p^2}_1}{2m}+\frac{{p^2}_2}{2m}$

Now, for elastic collision, initial kinetic energy = final kinetic energy $\Rightarrow \frac{p^2}{2m}=\frac{{p^2}_1}{2m}+\frac{{p^2}_2}{2m}$

$\Rightarrow p^2={p^2}_1+{p^2}_2.....\left(2\right)$

Solving $\left(1\right)and\left(2\right)$, we get

$$\begin{gathered} {p^2}_1+{p^2}_2={p^2}_1+{p^2}_2+2p_1{p}_2.\cos \left(\alpha +\beta \right) \\ \Rightarrow 0=2p_1{p}_2.\cos \left(\alpha +\beta \right)(p_{1}and{p}_{2}can\text{'}tbezero) \\ \cos \left(\alpha +\beta \right)=0 \\ \left(\alpha +\beta \right)=90° \end{gathered}$$

Now, we know velocity is also in the direction of momentum.

Thus, the angle between the velocities of particles after the collision will be $\alpha +\beta =90°$