Super Basic Collision Problem - Preparation for SR

  • #1

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 = 2vi, 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.
 

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
16,171
8,204
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:
It was good to here, but that's not momentum.
 
  • #3
It was good to here, but that's not momentum.
Ha ha, dumb mistake. Sorry. Yes, so the initial momentum in the object frame is

$$p'_i = m(2v) = 2mv$$

Since momentum is conserved, this must also be the final momentum, so

$$p'_f = 2mv = (2m)v'_f$$

from which it follows that

$$v'_f = v$$

And since v'i = 2v, so v = (1/2)v'i, so

$$v'_f = \frac{v'_i}{2}$$

which seems to make good physical sense, since the object that is moving is now twice as massive and so should be going only half as fast.
 
  • #4
950
418
In your expression for KEf’ vf’ should be squared. You had the momentum error. You fixed that. You need to fix this error too. Once you correctly calculate the change in kinetic energy in each case the answer to c and d will become obvious
 
  • Like
Likes Ghost Repeater and PeroK
  • #5
In your expression for KEf’ vf’ should be squared. You had the momentum error. You fixed that. You need to fix this error too. Once you correctly calculate the change in kinetic energy in each case the answer to c and d will become obvious
Thanks for pointing our my calculation error there.

So here's a summary of what I've got so far.

Ground Frame:

$$KE_i = 2 (\frac{1}{2}mv_i^2) = mv_i^2$$

(I changed notation here a bit, using ##v_i## to represent initial speed in the ground frame.)

$$KE_f = \frac{1}{2}(2m)v_f^2 = 0$$ (since conservation of momentum tells me that ##v_f = 0##)

Total kinetic energy change is

$$\Delta(KE) = KE_f = KE_i = 0 - mv_i^2 = -mv_i^2$$

Object Frame:

$$KE'_i = \frac{1}{2}m(2v_i)^2$$

$$ KE'_f = \frac{1}{2}(2m)v'_f^2 $$

(Sorry, don't know where these damn boxes are coming from.)

$$p'_i = m(2v_i) = 2mv_i $$

$$p'_f = 2mv'_f = 2mv_i$$ (due to momentum conservation)

From this last it follows that ##v_i = v'_f##

Total kinetic energy change is

$$\Delta(KE') = KE'_f = KE'_i = mv'_f^2 - 2mv_i^2 = mv_i^2 - 2mv_i^2 = -mv_i^2 = \Delta(KE)$$

So what I conclude about kinetic energy is that

$$\Delta(KE') = \Delta(KE)$$

I.e. the kinetic energy changes by the same amount in each frame?

I am really struggling to see the lesson I'm supposed to be learning here.
 
  • #6
950
418
I am really struggling to see the lesson I'm supposed to be learning here.
Oh, don’t be so negative! You get it.
 
  • #7
Oh, don’t be so negative! You get it.
I may be overthinking it, since I know the problem is supposed to be preparation for modifying these Newtonian concepts for SR later on.

I just have a 'so what?' feeling about this 'result', perhaps because I am not familiar enough with relativity. Is it surprising that the change in KE should be the same in both frames? I can't think of any good reason why it shouldn't. This is an inelastic collision, so kinetic energy shouldn't be conserved. But I don't see any reason why simply looking at the exact same situation, just in a different coordinate system, should make any difference at all to how much kinetic energy the system loses.

Is the point perhaps that even though the KE varies from one coordinate system to another, the change in KE does not?
 
  • Like
Likes jbriggs444 and PeroK
  • #8
950
418
Is the point perhaps that even though the KE varies from one coordinate system to another, the change in KE does not?
Seems like a good insight to me. Everything you already know seems obvious, so if that was already your understanding, you won’t get too excited about it. However, for a lot of people it is fairly alarming that the momentum and the kinetic energy vary with reference frame. It’s reassuring to see that the usefulness of the concepts of energy and momentum aren’t broken.
 
  • #9
950
418
You can go a little further and relate the change in momentum and energy from reference frame to reference frame to the apparent momentum and energy of the system as a whole.
 
  • #10
591
195
For an inelastic collision of this nature, the equal and opposite impulse(s) experienced by the colliding masses is/are given by the following simple equation:

Δp = μΔv where μ is the reduced mass [m1*m2/(m1 + m2)] of the colliding pair and Δv is their relative velocity.

The KE lost is given by an equally simple equation:

ΔKE = 1/2 μ (Δv)^2

Since both impulse and change in KE depend only on relative velocity, the two situations you describe are equivalent.
 
  • #11
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,535
6,431
$$ KE'_f = \frac{1}{2}(2m)v'_f^2 $$

(Sorry, don't know where these damn boxes are coming from.)
The LaTeX processing takes fright at the sequence v'_f^2. You can remove the ', the _ or the ^ and it is happy. Strange.
v'^2_f is fine: ##v'^2_f##
 

Related Threads on Super Basic Collision Problem - Preparation for SR

  • Last Post
Replies
21
Views
3K
  • Last Post
Replies
6
Views
825
Replies
3
Views
921
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
9
Views
954
  • Last Post
Replies
3
Views
949
  • Last Post
Replies
1
Views
1K
Replies
1
Views
7K
Replies
7
Views
2K
Top