$ \newcommand{\quantity}[2]{ #1 \;\mathrm{#2}} $ $ \newcommand{\units}[1]{\mathrm{#1}}$

Elastic collisions

You can derive a pair of equations to describe the velocities of objects with masses of $m$ and $M$ and initial velocities of $u_{a}$ and $u_{b}$ involved in an elastic collision.

In elastic collisions, both momentum and kinetic energy are conserved. The conservation of momentum is:

Total momentum before = Total momentum after

$$p_{a\,initial}+p_{b\,initial}=p_{a\,final}+p_{b\,final}$$ $$mu_{a}+Mu_{b}=mv_{a}+Mv_{b} \tag{1}$$

And as kinetic energy is

Total kinetic energy before = Total kinetic energy after

$$E_{a\,initial}+E_{b\,initial}=E_{a\,final}+E_{b\,final}$$ $$m{u_{a}}^{2}+M{u_{b}}^{2}=m{v_{a}}^{2}+M{v_{b}}^{2} \tag{2}$$

Taking the conservation of momentum, we can rearrange it to bring all the $m$ and $M$ to the same side and then factorise each side:

$$m\left(u_{a}-v_{a}\right)=M\left(v_{b}-u_{b}\right) \tag{3}$$

And doing the same for the conservation of energy:

$$m\left({u_{a}}^{2}-{v_{a}}^{2}\right)=M\left({v_{b}}^{2}-{u_{b}}^{2}\right) \tag{4}$$

As we have the terms $\left({u_{a}}^{2}-{v_{a}}^{2}\right)$ and $\left({v_{b}}^{2}-{u_{b}}^{2}\right)$ we can use the difference of two squares:

$$a^{2}-b^{2}=\left(a+b\right)\times\left(a-b\right)$$

To produce:

$$m\left(u_{a}+v_{a}\right)\times\left(u_{a}-v_{a}\right)=M\left(v_{b}+u_{b}\right)\times\left(v_{b}-u_{b}\right) \tag{5}$$

Dividing equation 5 by equation 3, the we can cancel out the mass terms as well as a lot of the $v$ and $u$ terms.

$$\require{cancel} \frac{\cancel{m}\left(u_{a}+v_{a}\right)\times\left(\cancel{u_{a}-v_{a}}\right)} {\cancel{m}\left(\cancel{u_{a}-v_{a}}\right)} =\frac{\cancel{M}\left(v_{b}+u_{b}\right)\times\left(\cancel{v_{b}-u_{b}}\right)} {\cancel{M}\left(\cancel{v_{b}-u_{b}}\right)}$$

Which leaves us with:

$$u_{a}+v_{a}=v_{b}+u_{b}$$

Or

$$v_{a}=v_{b}+u_{b}-u_{a} \tag{6}$$

This is an interesting equation as it tells us that in an elastic collision, the relative speed of the two objects as they approach, $u_{a}-u_{b}$ is equal the relative speed of the two objects after the separate, $v_{b}-v_{a}$, i.e.:

the speed of approach = the speed of separation

We are now able to substitute equation 6 into the conservation of momentum equation, eq 1:

$$mu_{a}+Mu_{b}=m\left(v_{b}+u_{b}-u_{a}\right)+Mv_{b}$$

The bracket can be multiplied out and all the terms containing a $u$ term brought to the same side.

\begin{align} mu_{a}+Mu_{b}&=m\left(v_{b}+u_{b}-u_{a}\right)+Mv_{b}\\ \\ mu_{a}+Mu_{b}&=mv_{b}+mu_{b}-mu_{a}+Mv_{b}\\ \\ mu_{a}+Mu_{b}-mu_{b}+mu_{a}&=mv_{b}+Mv_{b}\\ \end{align}

The right hand side can now be factored and the left can be simplified.

$$2mu_{a}+u_{b}\left(M-m\right)=v_{b}\left(m+M\right)$$

Therefore:

$$v_{b}=\frac{2mu_{a}}{\left(m+M\right)}+\frac{u_{b}\left(M-m\right)}{\left(m+M\right)} \tag{7}$$

If one of the objects is initially stationary when the two collide, then the equation simplifies to:

$$v_{b}=\frac{2mu_{a}}{\left(m+M\right)} \tag{8}$$

To find the final velocity of the other object we can either use the relative speeds of approach and separation, or we can substitute equation 7 into equation 6:

$$v_{a}=\frac{2mu_{a}}{\left(m+M\right)}+\frac{u_{b}\left(M-m\right)}{\left(m+M\right)}+u_{b}-u_{a}$$

We can now factor for $u_{1}$ and $u_{2}$ to get:

$$v_{a}=\left(\frac{2m}{m+M}-1\right)u_{a}+\left(\frac{M-m}{m+M}+1\right)u_{b}$$

We need to get rid of the 1s and bring both terms in the brackets over a common denominator, which is $\left(m+M\right)$ as,

$$1=\frac{\left(m+M\right)}{\left(m+M\right)}$$

The equation becomes:

$$v_{a}=\left(\frac{2m-\left(m+M\right)}{m+M}\right)u_{a}+\left(\frac{M-m+\left(m+M\right)}{m+M}\right)u_{b}$$

Which simplifies to:

$$v_{a}=\left(\frac{m-M}{m+M}\right)u_{a}+\left(\frac{2M}{m+M}\right)u_{b} \tag{9}$$

And, if the one of the objects is stationary in the collision, the equation becomes:

$$v_{a}=\left(\frac{m-M}{m+M}\right)u_{a} \tag{10}$$

The two equations, 8 and 10 also help explain some of the examples of momentum that you have studied, for example, the Newton’s cradle or the collision between two objects with the same mass on an air track, where the initial velocity of one of them is zero.

$$v_{b}=\frac{2mu_{a}}{\left(m+M\right)}\: \mathrm{and}\: v_{a}=\left(\frac{m-M}{m+M}\right)u_{a}$$

In equation 8 we can see that if the masses are the same, they will all cancel out and therefore $v_{b}=u_{a}$ so the second object will continue with the same velocity as the first. And in the second of the two equations, $m_{a}-m_{b}=0 $ so $v_{a}$ is equal to zero, i.e. it will stop in the collision, which is exactly what we observe.