Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This question already has an answer here:

I was just wondering... I believe that if a car travelling 50 miles per hour crashes into a wall, the result should be the same as crashing to another car also travelling 50 miles per hour (but in the other direction of course)

Is this true? Why is that?

share|improve this question

marked as duplicate by Qmechanic Oct 26 '13 at 0:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3 Answers 3

up vote 3 down vote accepted

This is true if the wall is so strong that it does not move and is not damaged in the course of the crash, so this can be only approximately true. So why is this (approximately) true? Because, on the one hand, the consequences of a crash are determined by the accelerations in the course of the crash, on the other hand, due to the symmetry, accelerations will be approximately the same in the crash of two identical vehicles as in a crash of a vehicle against a very strong wall (in both cases, no parts of the first vehicle can travel beyond the plane of symmetry of the two vehicles or beyond the nearest surface of the wall). However, some differences are still possible, as parts of the vehicle can rebound in a different way in these two cases (the collision can be (in)elastic to a different extent in these two cases).

share|improve this answer

From a simplistic POV assuming complete preservation of kinetic energy & momentum (no sound, no whizzing parts flying everywhere), both scenarios are essentially the same if the cars are of the same mass.

When the car hits a wall head-on with velocity v, it will bounce back with velocity -v; similarly if a two cars of same mass collide head-on they will exchange velocities in the opposite directions since both the cars are traveling at same speed, they essentially bounce back at the same speed = -v

.. Of course its better to have one car crash into a wall than 2 cars crash hurting ~twice the people

share|improve this answer

There's something no one has covered yet. Let's assume full theoretical perfectness, the two objects really are completely symmetric in how they approach each other. In the case of a car, this even means that one car will have the driver's seat on the right side and one will have it on the left.

Crashing into a wall has one component that the train-on-train collision does not - coefficient of friction. Take the direction of movement to be the x-axis for one of the trains, which is perpendicular to the yz-plane of collision. Nothing crosses the yz-plane. If there was shrapnel from one train, it would collide perfectly with the identical shrapnel from the other plane (suspending disbelief for a moment). Something is still different, which is the dampening in the yz directions. In both cases of a strong wall and a train-on-train collision, there is highly inelastic physics in the x-direction, however, in the case of train-on-train, there is no energy dissipation in the yz-plane whatsoever.

Imagine that I throw a tennis ball into the air in the general direction of a perfect mirror clone of myself. The ball bounces off the ball that my mirror image throws. The ball gains no rotational momentum whatsoever. It bounces back toward me off the imaginary yz-plane discussed above. If I threw the same ball at a wall, it would:

  • slow down more
  • gain rotational momentum
  • land closer to the center of the system

For the case of a train wreck these factors probably aren't a huge deal, but for an entirely physics-based discussion, the zy-plane coefficient of friction is the one factor that separates the perfectly ideal situations.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.