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## Homework Statement

[This problem is part of a preparation for modifying the concepts involved in special relativity.]

Two cars collide and lock together. They are each a mass of 800 kg and were traveling at a steady 20 m/s in opposite directions prior to collision.

a) What's the kinetic energy before and after collision?

b) Analyze a similar collision where one of the cars initially approaches the other (which is at rest) at velocity 40 m/s.

c) Are these two situations equivalent?

d) What can we conclude about kinetic energy?

## Homework Equations

[/B]

Energy conservation, momentum conservation

## The Attempt at a Solution

[/B]

Okay, I'll call the 'ground frame' the frame where the cars are moving toward each other, each with the same speed. And I'll call the 'object frame' the frame of the rightmost object, i.e. the frame where that object is at rest. Quantities in the object frame will be primed.

First, the ground frame analysis. The total KE prior to collision is twice the KE of each car, so

$$KE_i = 2(\frac{mv^2}{2}) = mv^2$$

The object after collision has KE given by

$$KE_f = \frac{2mv_{f}^{2}}{2} = 0$$

From conservation of momentum, I know that v_f = 0. (So the cars lock together, making a composite object of twice the mass of an individual car. This object can't be moving, because the system started with no momentum, so it has to end with no momentum.)

Now for the object frame analysis. (I'll be expressing speeds here as multiples of the speed in the ground frame.) In this frame, the KE prior to collision is

$$KE'_i = \frac{m(2v)^2}{2}$$

i.e. so the rightmost object is not moving, while the leftmost object zooms rightward at 2v m/s (in our problem this is the 40 m/s).

Since changing reference frame cannot change the fact that the cars lock together, the KE after collision will be

$$KE'_f = \frac{(2m)v'_f}{2}$$

Again I can use conservation of momentum to figure out what v'

_{f}is. (This is the velocity of the post-collision composite object in the object frame.)

The initial momentum in the object frame was

$$p_i = m(2v)^2 = 4mv^2$$

Since momentum is conserved, this must also be the final momentum of the post-collision composite object:

$$p_f = (2m)v'_f^2 = 4mv^2$$

(not sure where that box came from or how to get rid of it, sorry)

Therefore

$$v'_f = v\sqrt(2)$$

Not sure how useful this last result is. I do notice that this since v'

_{i }= 2v

_{i}, I therefore have that

$$v'_f = v'_i \frac{\sqrt(2)}{2}$$

So that's what I have. I'm not sure what the lesson is here. I hope it's not just that KE of a system depends on the reference frame we use to observe the system, because that's obvious just from the fact that velocity is relative.