Collisions

We can use the law of the conservation of momentum to, among rather a lot of other things, analyze collisions. Of course, we can use the many exciting techniques of vector theory along with this law to analyze collisions in $ n$ dimensions for any $ n$ (well, any integer $ n$), but in order to avoid catastrophe the Physics C curriculum is restricted to one and two dimensions, though the one-dimensional case is simplistic enough that it’s used sparingly on the test, and then almost exclusively for multiple-choice questions. It is important to remember, though, that these principles are extensible.

The basis for all of these is found in two equations:

  • Momentum is defined as $ \mathbf p=m\mathbf v$
  • Momentum is conserved; that is, $ \mathbf p_1 = \mathbf p_2$ iff $ \sum\mathbf F_{\textrm{external}}=0$

Anyway, there are several things to know or find out when considering collisions: initial speed, direction, and mass for each particle involved in the collision (we’re still dealing exclusively with point particles for the moment) as well as the final speed, direction of each particle (depending on how they collide, we may also need to consider the final mass of each). Given a certain amount of this information, we can set up an equation or systems thereof and solve it/them for the remaining value(s).

Perhaps the simplest case is that of a particle of mass $ m$ moving with initial velocity $ \mathbf v_{1i}$ in one dimension towards a stationary particle also of mass $ m$. What will be the final velocities of each? The moving particle has initial momentum $\displaystyle \mathbf p_{1i}=m\mathbf v_{1i}$ while the stationary particle has initial momentum $\displaystyle \mathbf p_{2i}=m(0)=0$

We add these momenta to find the total initial momentum of the system: $\displaystyle \mathbf p_i=m\mathbf v_{1i}$ and recall that the law of the conservation of momentum requires, since there is no external net force, that the final momentum be identical. To determine the final velocity or velocities, we do need to know how the particles behave when they collide.

First let’s suppose that they stick together (an inelastic collision). Now we have a new “particle” with mass $ 2m$, so all we are missing is the final velocity, which we’ll call $ \mathbf v_f$. The final momentum will thus be $ \mathbf p_f=2m\mathbf v_f$. We set the initial momentum equal to the final momentum: $\displaystyle m\mathbf v_{1i} = 2m\mathbf v_f$ and solve for $ \mathbf v_f$, getting $\displaystyle \mathbf v_f=\tfrac 12 \mathbf v_{1i}$

On the other hand, if they remain separate—which we call an elastic collision —, similar steps will show that the first particle now becomes stationary and the second takes on velocity $ \mathbf v_{2f}=\mathbf v_{1i}$.

We can readily extend this to a case with a case where the first particle has $ \mathbf v_{1i}$ and $ m_1$ while the second particle has $ \mathbf v_{2i}$ and $ m_2$ rather than being stationary. We’ll again deal first with the case where the particles stick together; then, $\displaystyle \sum\mathbf p_i = m_1\mathbf v_{1i} + m_2\mathbf v_{2i}$ and $\displaystyle \sum\mathbf p_f = (m_1+m_2)\mathbf v_f$ so $\displaystyle m_1\mathbf v_{1i} + m_2\mathbf v_{2i} = (m_1+m_2)\mathbf v_f$ therefore, $\displaystyle \mathbf v_f = \frac{m_1\mathbf v_{1i}+m_2\mathbf v_{2i}}{m_1+m_2}$

Now if the first particle stays stationary and the second begins moving, we have the same $ \sum\mathbf p_i$, while $\displaystyle \sum\mathbf p_f = m_2\mathbf v_{2f}$ so $\displaystyle m_1\mathbf v_{1i} + m_2\mathbf v_{2i} = m_2\mathbf v_{2f}$ giving $\displaystyle \mathbf v_{2f} = \frac{m_1\mathbf v_{1i} + m_2\mathbf v_{2i}}{m_2}$

That takes care of the nearly trivial one-dimensional case. Now, we will tackle the substantially more interesting two-dimensional situation. This gives us the opportunity to use our very favorite math technique, trigonometry. I know you’re as excited as I am.

Diagram for the rest of the lesson

Given the diagram, we’ll again suppose first that the collision is perfectly inelastic. We need to resolve the vectors $ \mathbf v_1$ and $ \mathbf v_2$ into their $ x$ and $ y$ components in order to find $ \sum\mathbf p_i$: $\displaystyle \mathbf v_{1x} = \Vert\mathbf v_1 \Vert\cos\theta_1\hat{\mathbf i}$ and $\displaystyle \mathbf v_{1y} = \Vert\mathbf v_1 \Vert\sin\theta_1\hat{\mathbf j}$ while similarly $\displaystyle \mathbf v_{2x} = \Vert\mathbf v_2 \Vert\cos\theta_2\hat{\mathbf i}$ and $\displaystyle \mathbf v_{2y} = \Vert\mathbf v_2 \Vert\sin\theta_2\hat{\mathbf j}$

Now we can add and multiply by $ m_1$ and $ m_2$ to get $\displaystyle \mathbf p_{ix} = \left({m_1\Vert\mathbf v_1 \Vert\cos\theta_1 + m_2\Vert\mathbf v_2 \Vert\cos\theta_2}\right)\hat{\mathbf i}$ and correspondingly, $\displaystyle \mathbf p_{iy} = \left({m_1\Vert\mathbf v_1 \Vert\sin\theta_1 + m_2\Vert\mathbf v_2 \Vert\sin\theta_2}\right)\hat{\mathbf j}$

The law of the conservation of momentum states not only that $ \mathbf p_i = \mathbf p_f$ but also that this equation is true in each component. So, given that the objects stick together, we know the final momentum in each direction: $\displaystyle \mathbf p_{fx} = \left({m_1+m_2}\right)\Vert\mathbf v_f\Vert\cos\theta_f\hat{\mathbf i}$ and $\displaystyle \mathbf p_{fy} = \left({m_1+m_2}\right)\Vert\mathbf v_f\Vert\sin\theta_f\hat{\mathbf j}$

We begin the end by equating $ p_{ix}$ with $ p_{fx}$: $\displaystyle \mathbf p_{ix} = \left({m_1\Vert\mathbf v_1 \Vert\cos\theta_1 + m......hbf i} = \left({m_1+m_2}\right)\Vert\mathbf v_f\Vert\cos\theta_f\hat{\mathbf i}$ which can readily be solved for $ \Vert\mathbf v_f\Vert\cos\theta_f$: $\displaystyle \Vert\mathbf v_f\Vert\cos\theta_f = \frac{m_1\Vert\mathbf v_1 \Vert\cos\theta_1 + m_2\Vert\mathbf v_2 \Vert\cos\theta_2}{m_1+m_2}$ but of course that’s meaningless since it’s really two variables rather than the one we’d hoped for. Still, we equate $ p_{iy}=p_{fy}$ and find that $\displaystyle \Vert\mathbf v_f\Vert\sin\theta_f = \frac{m_1\Vert\mathbf v_1 \Vert\sin\theta_1 + m_2\Vert\mathbf v_2 \Vert\sin\theta_2}{m_1+m_2}$

That leaves us with two equations and two unknowns. We actually can’t do any more unless we’re given more information in the statement of the question; typically, we’d be given either the final speed of the angle at which the (newly combined) object flies off. Given either, we can find the other.

Now, iff the collision is elastic (that is, if the objects do not stick together), we can follow the process as far as we did, but also apply the fact that kinetic energy is conserved in the collision. That is, $\displaystyle \tfrac{1}{2}m_1\Vert\mathbf v_1\Vert^2+\tfrac{1}{2}m_2\Vert\mathbf v_2\Vert^2=\tfrac{1}{2}\left({m_1+m_2}\right)\Vert\mathbf v_f\Vert^2$ and now we have precisely enough information to solve for everything we need.

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