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Q&A

If you're stopped and about to be hit from behind, should you brake or release the brake?

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The title is a framing for a theoretical question; I'm not asking for practical advice. A friend was recently in this situation and my attempts to apply what I remember of a couple semesters of college physics were inconclusive.

Suppose you are in a stopped car. You notice another car behind you moving too quickly to stop before hitting you. Your goal is to minimize your injury. Strictly from a physics perspective of these two cars, are you better off braking hard or releasing the brake so you'll be pushed? My gut feeling (assuming no one's in front of you) is that you should do the latter, but my friend and I got into a conversation about elastic and inelastic collisions and the effects of friction and of the masses of the two vehicles (if not similar), and now neither one of us knows how to answer the question with physics rather than with gut feels.

A comment pointed out that the speed of the moving car is important. Assume the stopped car is in city traffic (say, stopped at a traffic signal) and the car bearing down on it is going about 25mph.

In this collision, some force will be absorbed by the car being hit, some will be absorbed by the car doing the hitting, some will go into forward momentum (the front car being pushed), and some will be lost to friction. Some of the force that goes into the car being hit will be transmitted to the occupants and some to the car itself (crumple zones etc). How should I think about these forces and their distribution in the two scenarios (braking and not braking)? And how do the relative masses of the vehicles affect the outcome? A car hitting a car seems different from a motorcycle hitting a semi (or vice versa), but when does a less-extreme difference still matter?

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What's the closing speed? (3 comments)

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Basic answer

It depends on what you care about.

If you're primary concern is to minimize injury to the people in your car, then hold the brake firmly in ordinary low speed cases. This minimizes the motion of your car, which is what jerks around the passengers and causes injury. Actually the first thing I would do in this situation is to make sure my head is solidly against the head rest of the seat. That will reduce it flopping around and causing whiplash.

If the other car is going really fast, then the strategy is different. Now you have to consider the car getting crushed, crushing you along with it. In that case, letting your car roll will reduce the amount it gets crushed a little bit. Think of the limiting case where a concrete wall is coming at you. The collision impulse (force × time) is less the more your car moves.

On balance in ordinary circumstances, I'd probably make sure my head is against the headrest, then let the car roll as much as possible to minimize damage to the car. The impact on my body will be slightly higher, but it is coming from the back, well supported by the seat, and I can brace myself to prepare for it. This is, of course, assuming there is open space in front of you that you can safely enter.

In the above case, it would even be good to deliberately drive the car forward accelerating as hard as you can. That minimizes the closing speed between you and the other car, and increases the distance the other car travels before impact. If the other car brakes at the last moment, then it will be slowing down. Leaving it more room to slow down will reduce the impact. Think of the limiting case where the other car is going 10 MPH. If you can speed up to 10 MPH before it hits you, then there won't be any impact at all.

Clearing up some physics

In this collision, some force will be absorbed by the car being hit, some will be absorbed by the car doing the hitting, some will go into forward momentum (the front car being pushed), and some will be lost to friction.

This is confusing force, impulse, momentum, and energy.

An inelastic collision with friction to the ground is complicated. It's not all about forces, and they aren't "absorbed". In fact, I'd think of the forces as the result of other things.

In the end, energy and momentum will be conserved, although in the braking case some of the momentum will be imparted onto the planet, so it does effectively vanish from your point of view.

Let's walk thru an intuitive scenario of what happens during the collision. We'll start with the no-brakes case, since that's easier to analyze.

As the cars touch, there will be an equal but opposite force on both cars. This is true regardless of the weight of each car, and how much each car deforms or changes speed. This is the one thing in this whole mess you can count on. The force on your car will be pushing it forwards, while exactly the same size force will be pushing the other car backwards.

The relative masses of the two cars dictate how much each of them changes speed. The heavier car will speed up (if the front car) or slow down (if the rear car) more slowly. This change in speed is called acceleration. The occupants of the cars feel a force due to this acceleration. Higher acceleration of your body means a higher force is applied to it by the seat or other parts of the car. So while the force on each car is the same, the force on occupants is not the same. The guy driving the loaded cement truck crashing into your car won't be bothered by the collision as much as you will be.

In the end, the total momentum, which is mass × velocity, will be the same as before the collision. This is the law of conservation of momentum. For example, let's say your car weighs 700 kg (1500 pounds), and the other car 1000 kg (2200 pounds). Before the collision your car is going 0 m/s, and the other car 10 m/s (22 MPH). The total momentum (all in the forward direction) is:

    (700 kg)(0 m/s) + (1000 kg)(10 m/s) = 10,000 kg m/s

The end result will be the same.

But what about crumple zones? Those are for absorbing energy. We can't escape the momentum balance above, but note that there is still one degree of freedom unaccounted for. You can pick any velocity you want for one car, but that then dictates the final velocity of the other car.

In practice, the actual result depends on how "bouncy" the collision is. If the two cars rebounded from each other like two bouncy balls, the back car may actually end up going backwards, with the front car going quite fast forward due to the momentum balance above. A less bouncy collision is better for the occupants of both cars. For example, if both cars ended up being stuck together at the end, they would both be going 5.88 m/s (13 MPH) forward.

Now look at the resulting kinetic energy, which is mass times the square of speed, divided by 2. In the above example, the initial energy is:

    (700 kg)(0 m/s)2/2 + (1000 kg)(10 m/s)2/2 = 50,000 kg m2/s2 = 50 kJ

But the final energy in the not-bouncy case is:

    (700 kg)(5.88 m/s)2/2 + (1000 kg)(5.88 m/s)2/2 = 29.4 kJ

Energy needs to be conserved too, so where did the missing 20.6 kJ go? It was dissipated by the deforming bodies of the two cars. Put another way, it took energy to transform the sleek shape of both cars into the resulting twisted metal. This is the point of crumple zones. They are structural areas designed to be deformed on collision, thereby absorbing energy. They make collisions less bouncy, reducing the final kinetic energy, which reduces the speed change, which reduces the forces felt by occupants.

But wait, what's this "absorb energy" hand waving? Isn't energy supposed to be conserved too? Yes it is. Deforming the metal doesn't just make energy disappear. It transforms it into a form less likely to hurt the occupants. In this case, that form is heat. The missing 20.6 kJ ultimately went to heating the metal that got twisted. To put this in perspective, that's like running a 2 kW space heater for 10 seconds. That's enough to notice the warmth if you touched the deformed parts right after the collision.

There is a lot more that could be said, but this is getting long already so I'll quit here.

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Clear explanation, thanks (1 comment)

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