Inelastically colliding cars

5 December 2020

Until today I thought that, ignoring any (very important in practice) crunch zone effects, two cars heading towards each other with speed $\frac v 2$ each will collide with the same force as a car heading towards a solid wall at speed $v$. It turns out this is not true at all.

Making things worse, there’s an intuitive-sounding explanation for the difference between the two cases that turns out not to be true at all. If you try applying high-school physics to this problem, in the two-car scenario, the total kinetic energy seems to be $\frac 1 2 (2m)\left(\frac v 2\right)^2 = \frac 1 4 mv^2$, but in the one-car scenario, it’s $\frac 1 2 mv^2$, twice as much ($m$ is the mass of one car). If we assume the collision is perfectly inelastic, the kinetic energy is completely converted into heat and general destruction, so we could think the car-and-wall case is twice as bad.


The reason this explanation is wrong is it’s not explicit about the frame of reference in which energy is computed, so it gets away with making illegal transformations. Let’s look at the two-car example first, and let’s assume all cars are point particles and in a vacuum, having accelerated towards each other with ion engines or something.

Maybe the most natural frame of reference to consider is the midpoint between the two cars (which is inertial so nothing weird happens). In this referential, the collision looks like this: the cars head towards each other with a total energy of $\frac 1 4 mv^2$, until they reach the midpoint, collide inelastically, and end up with a kinetic energy of 0 and a thermal energy of $\frac 1 4 mv^2$.

We can also almost consider one car’s frame of reference, but not exactly. Initially there’s no problem, because the car is undergoing no acceleration, so the frame is inertial, but when they collide, the car undergoes (infinitely fast) deceleration, which makes the frame not inertial at all, breaking all conservation laws. What we can do is take a frame of references that initially moves at $\frac v 2$ with the car, but keeps moving when the cars collide.

In this frame of reference, the initial kinetic energy is $\frac 1 2 mv^2$, twice as much as before, but there’s also final kinetic energy of $\frac 1 2 (2m)\left(\frac v 2\right)^2 = \frac 1 4 mv^2$ as both cars are now stopped in the midpoint frame, but moving at $\frac v 2$ in our initially-following-a-car frame.

This gives us a net $\frac 1 4 mv^2$ energy dissipated too, so all is well.

The end result here is that it doesn’t matter which car is moving (in fact, the question of which car is moving doesn’t make sense), which is not surprising at all because of Galilean invariance.

Total kinetic energy

So what’s different about the car-and-wall case, if not movement? Let’s model the wall as another sphere-or-point-particle with mass $M$ much larger than $m$. For instance, we could take $M$ to be the mass of the Earth, assuming the wall is attached to the ground with a completely rigid link, and the Earth itself is an undeformable sphere. (The mass of the Earth is about 1022 times the mass of a car).

Let’s think in the frame of reference that is initially centered on the wall, and doesn’t move when the collision happen. This frame is also inertial.

What happens after the collision? Not just kinetic energy is conserved in an inertial frame, momentum is too. Since this time, unlike in the car example, we have nonzero total momentum, we’ll need to end up with the same nonzero total momentum. In other words, in an inertial frame where only the car is initially moving, we cannot assume that the car and wall will have zero speed after the collision.

In practice, this means the car will transfer some momentum to the Earth itself through the wall. The amount of momentum transferred will obviously have no practical impact on the Earth’s motion, but it does play a role in the energy dissipation calculations.

By conservation of momentum, we have $mv = (m+M)v_f$, or $v_f = \frac{mv}{m+M}$, where $v_f$ is the speed of the crashed-car-and-wall.

What’s the final kinetic energy here? It’s $\frac 1 2 (m+M)v_f^2=\frac {m^2v^2} {2(m+M)}$. Since we had $\frac 1 2 mv^2$ coming in, the kinetic energy dissipated into car crushing is:

But since $m$ is by assumption very small compared to $M$, this simplifies to $\frac 1 2v^2\frac{mM}{M} = \frac 1 2 mv^2$, which is indeed the twice as much as the car-versus-car case.


This final expression also explains where the difference comes from: the problem isn’t relative motion, but mass of the collider. It’s in fact intuitively obvious (well, to me at least) that, all other things being equal, it’s better to hit a very small object than a very large one. Taking $M$ to be the mass of the Earth (effectively infinity) makes sense to model a wall, so this type of collision will be very bad. We could instead take something like $M=20m$, corresponding to a large laden truck hitting a car, and end up with a similar figure (the total energy would be 20/21≈95% of the wall case, still almost twice as bad as hitting a car). On the other hand, as the collider gets smaller, the collision will be relatively less energetic: hitting a very fast grain of sand won’t do much harm.