Extension - Elastic collisions

Elastic collisions

You can derive a pair of equations to describe the velocities of objects with masses $m$ and $M$ and initial velocities $u_{a}$ 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 the conservation of 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 $M$ terms to the same side, and 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:

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

Now divide equation 5 by equation 3. On the left, $m$ and the whole factor $\left(u_a-v_a\right)$ both appear in the numerator and denominator, so they cancel; the same happens with $M$ and $\left(v_b-u_b\right)$ on the right:

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

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 to the relative speed of the two objects as they 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, equation 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 $u$ 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 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, this 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 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 $u_{a}$ and $u_{b}$ 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 of $\left(m+M\right)$, using:

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

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 one of the objects is stationary in the collision, this becomes:

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

Equations 8 and 10 also help explain some of the examples of momentum you've studied, such as a Newton's cradle, or the collision between two objects of the same mass on an air track, where the initial velocity of one of them is zero:

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

In equation 8, if the masses are the same, they cancel out entirely, leaving $v_{b}=u_{a}$: the second object continues with exactly the same velocity the first one had. And in equation 10, if $m-M=0$, then $v_{a}=0$: the first object simply stops in the collision, which is exactly what's observed.