Comment by ScottBurson | Hacker News Reader

Comment by ScottBurson | Hacker News Reader

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Walter's answer is good, but here's another. When the two cars collide, it's possible to imagine that they are exact mirror images of each other and hit exactly head-on, so that a large sheet of paper hung vertically at exactly the collision point would not be torn. Of course we know that wouldn't literally happen in the real world, but it is possible. This thought experiment demonstrates that the collision is equivalent to hitting a brick wall at 50mph, not 100.

Alternatively, imagine one car is parked (in neutral, with its brake off) and the other car hits it at 100. The center of mass of the two cars is moving at 50 both before and after the collision (conservation of momentum); after it, the cars will be moving at that average speed. The impact will again be equivalent to the original scenario.

The original claim probably results from a conflation of these two scenarios.

ETA: So what student drivers should be told is that hitting another car head-on is like hitting a brick wall at the same speed. For this to be exactly true, the momenta (mass * velocity) of the two vehicles have to be equal and opposite, but to communicate the general idea, I don't think we have to go into that.

Please help me understand where my intuition (or maybe my assumptions/simplifications) is wrong.

Assume two perfectly elastic cars. When they collide at 50mph, each car will bounce backwards at 50mph (due to conservation of momentum), representing a change in velocity of 100mph — identical to the brick wall case.

I feel that introducing deformation or other energy dissipation to the equation kind of takes it out of the “high school physics” realm, right? What else am I missing?

Edit: ah I see, the car will bounce off the brick wall at 100mph as well, resulting in a 200mph change in velocity. I guess you could explain it then that the effect of the impact is felt entirely in one car in the brick wall case, and it’s spread out over two cars in the head-on case?

> When they collide at 50mph, each car will bounce backwards at 50mph

This is the incorrect part. They would both go to zero velocity/momentum.

Momentum is a vector quantity, so has a direction and magnitude. Two identical cars with the same speed going opposite directions would have the same magnitude of momentum, but opposite sign. After colliding, their sum would be zero.

If you watch billiards you would see kinda the same thing going on.

Edit: completely messed this up. Other comments are more correct

Your explanation is not correct.

In a perfectly elastic collision, both kinetic energy and momentum are conserved. In a perfectly inelastic collision, kinetic energy is not conserved (because it is converted to heat), but momentum is always conserved.

So lets say you have 2 objects of the same mass traveling toward each other at the same speed. In a perfectly elastic collision, the balls objects will "bounce" off each other, going back in the opposite directions. In that case momentum is conserved (as you note, it's a vectored metric, so before and after the the total momentum of the system is 0), but so is kinetic energy, because you still have 2 masses traveling at the same speeds (think about if you have a Newton's cradle and pull both end balls up and drop them at the same time - they'll both bounce back).

In a perfectly inelastic collision, both masses will essentially crush and come to a complete stop where they collide. Again, momentum is conserved (it's still 0 before and after the collision) but kinetic energy is not conserved because it's all converted to heat of the 2 objects.

Oh man did I screw that up. You are right of course, just kind of skipped over the “elastic” part and completely messed up the billiards example.

Thanks for the correction.

(Also, how does a “throwaway” account get that much karma??)

> I feel that introducing deformation or other energy dissipation to the equation kind of takes it out of the “high school physics” realm, right?

I don't think so. In my high school physics class I learned about both fully elastic and fully inelastic collisions. The math basically works out the same in the fully inelastic collision case, which I did here, https://news.ycombinator.com/item?id=40628932

The elasticity doesn't matter for the equivalence of the two scenarios (head-on collision at 50 vs. brick wall at 50). We've assumed, albeit implicitly, that in the head-on case, the cars have equal and opposite momentum. Whether the collision is perfectly elastic, perfectly inelastic, or somewhere in between, a car will experience the same forces in the two scenarios (assuming, of course, that the elasticity is equal in the two cases).

The bounce in mechanics is defined by the coefficient of restitution, which is how elastic something is. Technically it's related to the energy lost in the collision.

Perfectly elastic objects bounce apart with mirrored velocities. No energy is lost.

Perfectly inelastic objects just stop. All of the energy is dissipated through noise, heat, and deformation.

Momentum is conserved in both.

For drivers ed, cars are almost perfectly inelastic.

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