What has the theory of relativity got to do with car crashes? Well it all has to do with kinetic energy. Einstein said if two objects have the same relative velocity between them it doesn’t matter if one is at rest or both are moving, any interaction between them would be identical in both cases.
Initial kinetic energy
Some believe that the kinetic energy dissipated in a crash involving two cars would be different in different situations even if the relative velocity between the cars was the same. The argument they make is that a car traveling twice as fast as an identical car would have four times the kinetic energy due to this formula – kinetic energy = ½ mv².
Citing this formula they say a car hitting a stationary car has twice the combined kinetic energy of two cars each traveling at half that speed towards each other and colliding. So even though the relative velocity is the same, they claim the extent of the damage would be less in the second case. This is a matter of serious concern, so who is right? You might not be surprised to find out that Einstein is, and here is the explanation why.
Two types of collision
We can understand this subject more clearly if we imagine two scenarios, or two types of collisions, elastic and inelastic. A perfectly elastic collision happens when after the collision both objects rebound with the same relative velocity they had at the beginning, though of course in opposite directions. An inelastic collision is when both objects stay in contact with each other after the collision.
Now let us look at both types of collision in both cases, starting with the cars that are moving towards each other with the same speed, and involved in an inelastic collision. The law of the conservation of momentum states that the combined mass times velocity of the two vehicles will be the same before and after impact. Remember that velocity is a vector and thus has direction as well as magnitude. So let’s say the car moving from right to left has a negative velocity.
The momentum before the crash = mv + m(-v) = mv – mv = 0 .
According to the law, the momentum after the crash must also equal zero. This is what we would expect: two identical cars hitting each other at the same speed head on and remaining in contact would come to an abrupt halt. Therefore all the kinetic energy of the cars has been converted into other forms of energy or damage shall we say.
That energy loss = ½mv² + ½mv² = mv² .
When only one car is moving but the relative velocity is the same we have the following equation.
The momentum before the crash = m(2v) + 0 = 2mv .
As the momentum after the crash must be the same as before and both cars remain in contact, they must now both be moving but with only half the original speed of the previously moving car.
The kinetic energy difference or loss = m(2v)²/2 – 2mv²/2 = 4mv²/2 – mv² = mv² .
Notice that this is the same energy loss or damage as in the case where both cars are moving. This is because at the moment of impact some of the previous kinetic energy remains, and is not converted.
At the other extreme, in an elastic collision, no kinetic energy is lost; it is only transferred. As in a Newton’s cradle when one of the end ball bearings hits the others, the opposite one is expelled with almost the same speed. In this case, it’s not the kinetic energy we look at, but the forces involved, which of course would be transmitted to the inelastic passengers. We can compare these forces by looking at the individual change in momentum of each car. The overall momentum remains the same which in the case where both cars are moving is given by the equation – momentum = mv + m(-v) = m(-v) + mv = 0 .
As both cars have had their velocities reversed after the collision, they have both undergone a change in momentum of 2vm or 2mv. Force is directly proportional to the change in momentum and is inversely proportional to the time taken; the latter would simply depend on the material of the cars.
In the other case involving a stationary and a moving car the total momentum must also be the same before and after the crash, so momentum = 2vm + 0 = 0 + 2vm = 2mv.
Here of course, all the momentum has been transferred to the previously stationary car and therefore again the individual change in momentum for both cars also equals 2mv.
So in conclusion, as the damage or force is the same when the relative velocity is the same, whether the collision is elastic or inelastic, we can reasonably infer that in an ordinary car crash, which would contain elements of both types of collision, the outcome would also be the same whether both cars are in motion or not.
The examples given assume that no other objects are involved in the collisions. To simplify the equations they were head-on collisions where frictional forces from the ground were regarded as small compared to the forces between the vehicles. The calculations show that it is the relative velocity that is important even in collisions involving moderate speeds.
The results are not surprising because a term has been wrongly used here – you may have guessed that the term is “stationary”. When we say something is stationary we mean it is not moving in relation to the ground, but as we know the ground is not stationary. At the equator, the ground is moving at 1000 mph around the center of the earth. The earth is moving at 66,000 mph around the sun, the sun is moving around the center of our galaxy, and the galaxies are flying apart at tremendous speeds. There is no fixed point in space to measure absolute speeds from, and therefore all motion is relative.