Elastic Collisions in 1-D

স্থিতিস্থাপক সংঘর্ষ -

In elastic collisions in 1D (pun intended), we can use 1D vectors or displacement vectors to track an object's motion. If one object $1$ has mass $m_1$ and moves with velocity $v_1$, while object $2$ has mass $m_2$ and moves with velocity $v_2$, then

$$m_1v_{1,initial}+m_2v_{2,initial}=m_1v_{2,final}+m_2v_{2,final}$$

by conservation of momentum. Since the collision is elastic, conservation of kinetic energy tells us

$$\frac12m_1v_{1,initial}^2+\frac12m_2v_{2,initial}^2=\frac12m_1v_{2,final}^2+\frac12m_2v_{2,final}^2$$

The only difference between 1D collisions and 2D or 3D collisions is that velocity can only be in one direction. That is, we don't need to use vectors to calculate momentums, just make sure you pay attention to the fact that velocity is a signed quantity in 1D.

It is important to understand 1D collisions because 2D and 3D collisions rely on 1D components. Therefore, 1D collisions are the foundation of higher dimension collisions.

An elastic collision is a collision in which kinetic energy is conserved. Suppose that we have $2$ objects of masses $m_1$ and $m_2$. They collide with initial velocities $v_{1i}$ and $v_{2i}$. We wish to calculate their velocities after the collision, denoted $v_{1f}$ and $v_{2f}$, respectively.

First, we consider the equation for the conservation of momentum: $$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$$ $$m_1(v_{1i}-v_{1f})=m_2(v_{2f}-v_{2i})$$

Next, we consider the equation for the conservation of kinetic energy: $$\frac{1}{2}m_1v^2_{1i}+\frac{1}{2}m_2v^2_{2i}=\frac{1}{2}m_1v^2_{1f}+\frac{1}{2}m_2v^2_{2f}$$ $$m_1(v^2_{1i}-v^2_{1f})=m_2(v^2_{2f}-v^2_{2i})$$

Dividing the second equation by the first equation gives us: $$v_{1i}+v_{1f}=v_{2f}+v_{2i}$$ $$v_{2f}-v_{1f}=v_{1i}-v_{2i}$$ This division is allowed because the masses are not zero, $v_{1i}-v_{1f}\ne 0$, and $v_{2f}-v_{2i}\ne 0$. These last two hold because we are looking for the situation after the collision, not before. Obviously, $v_{1f}=v_{1i}$ and $v_{2f}=v_{2i}$ would satisfy the equations, but that is not what we want.

Note that $v_{2f}-v_{1f}$ is the final velocity of object $2$ relative to object $1$, while $v_{1i}-v_{2i}$ is the initial velocity of object $1$ relative to object $2$. Therefore, we can say the following very useful result: In an elastic $1$-dimensional collision, the relative velocities switch signs.

This result is very powerful, because we no longer have to use any quadratic equations. Instead, we can solve the following linear system, equivalent to the original system: $$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$$ $$v_{2f}-v_{1f}=v_{1i}-v_{2i}$$ This is very useful in any context requiring fast calculation of the velocities, and it also helps with a general understanding of what occurs in an elastic collision.

There is really no point in trying to solve this system and memorize it, as far as I know, but it is very useful to know this result and use it to simplify the system of equations. But if you insist: $$v_{1f}=\frac{2m_2v_{2i}+(m_1-m_2)v_{1i}}{m_1+m_2}$$ $$v_{2f}=\frac{2m_1v_{1i}+(m_2-m_1)v_{2i}}{m_1+m_2}$$ Note that these solutions are completely symmetric, as they should be. But again, these are just for the sake of completeness and are not worth memorizing.

However, as I stated earlier, the important result is worth remembering: In an elastic $1$-dimensional collision, the relative velocities switch signs.

I will now show you a very easy way to remember this result. Since we are dealing with relative velocities, it makes sense to shift our reference frame. Let us look in the reference frame of object $1$ (of course, we could also choose object $2$, if we wished; the same reasoning will apply). In this reference frame, object $1$ stays stationary, while object $2$ approaches it, bounces off it, and moves away from it.

But wait! If object $1$ is always stationary, we can think of it as a wall! And when object $2$ collides with the wall, it will move away with the same speed that it approached, since the collision is elastic! Therefore, the speed at which object $2$ approaches object $1$ is the same speed at which object $2$ moves away from object $1$ after the collision! This means that the relative velocities must switch signs.