

Car Collisions and Gallilean Relativity
Name: Dan H.
Status: educator
Age: 50s
Location: N/A
Country: N/A
Date: 10/27/2003
Question:
I have never read a clear explanation of this apparent contradiction, and hope
you can shed some light. Simple (Galilean) relativity tells us that 2 cars approaching at
50 mph are equivalent, in either's frames of reference, to one car at rest and the other
approaching at 100 mph. But the total energy available, should they collide, is not
equivalent. This is seen by comparing 2(0.5mv^2) when each of the cars is seen at 50 mph,
to 0.5mv^2 with either at 100 mph. There seems to be more energy in the latter case by a
factor of 2.
So what is the answer to the question: Which is more dangerous? Two cars colliding at 50
mph, or one car, moving at 100 mph, colliding with another car at rest. Both cars, of
course, have the same mass.
Replies:
The two situations are equivalent, at least in principle. Notice that after a collision
with one car initially at rest, the two cars, locked together, are moving at a speed of
v/2, by conservation of momentum:
mv = 2mv/2
where m is the mass of either car and v is the initial speed of the moving car. They, of
course, slow down rapidly due to friction with the ground. Note that the final kinetic
energy of the two cars, locked together, before friction with the ground slows them down,
is
0.5 (2m)(v/2)^2 = mv^2/4, just half the kinetic energy of the initial car. This means that
the energy dissipated in the collision internally to the cars is also mv^2/4.
If the two cars are each moving at speed v/2, their total kinetic energy, which is then all
dissipated internally is
2*0.5*m(v/2)^2 = mv^2/4,
exactly the same!
This is utilized in all modern high energy particle accelerators. If a high energy proton
collides with a stationary proton, most of the energy goes into the kinetic energy of the
particles moving mostly in the direction of the initial proton to conserve momentum.
Their kinetic energy can be much more than half the initial kinetic energy in a relativistic
collision. Therefore we now use colliders to collide two protons (or other particles)
head on so all the energy goes into probing the inner structure of the particles and is
not "wasted" in the common downstream motion of the particles.
Best, Dick Plano...
Dan,
Which is more dangerous actually depends more on friction than amount of energy. If the
collision occurs on ice, there is essentially no difference. In the 5050 collision, there
is no motion after the collision. Kinetic energy is zero. In the 0100 collision, the
combined cars end up moving together at about 50mph. The "extra" energy is in the final
kinetic energy. The drivers feel the same things in either case. In the 5050 collision,
all kinetic energy is used. In the 0100 collision, only half is used.
If the collision occurs on concrete, the 0100 collision is more dangerous. The extra
resistive force due to friction causes both cars to experience greater force. Consider
hitting a punching bag free to move or a punching bag braced against a wall.In both
collisions, all kinetic energy is used. The friction force depends on how the ground is
moving. Since the ground moves differently in different reference frames, you get
different outcomes in different frames.
Dr. Ken Mellendorf
Physics Professor
Illinois Central College
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