Super Basic Collision Problem - Preparation for SR

In summary, the kinetic energy of a system changes depending on the reference frame we use to observe it. In the ground frame, the total kinetic energy is negative, while in the object frame the total kinetic energy is positive.
  • #1
Ghost Repeater
32
5

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.
 
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  • #2
Ghost Repeater said:
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
PeroK said:
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
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
 
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  • #5
Cutter Ketch said:
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
Ghost Repeater said:
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
Cutter Ketch said:
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?
 
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  • #8
Ghost Repeater said:
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
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
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
Ghost Repeater said:
$$ KE'_f = \frac{1}{2}(2m)v'_f^2 $$

(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##
 

1. What is a "Super Basic Collision Problem"?

A "Super Basic Collision Problem" is a type of physics problem that involves two objects colliding with each other. These types of problems often ask for the final velocity or the amount of energy lost during the collision.

2. What is SR in the context of this preparation?

SR stands for "Special Relativity" and refers to the theory of how objects behave at high speeds, specifically near the speed of light. In this context, SR is used to prepare for solving collision problems involving objects moving at high speeds.

3. What are some key principles to keep in mind when solving a Super Basic Collision Problem?

Some key principles to keep in mind include conservation of momentum and conservation of energy. These principles state that the total momentum and energy of a closed system (such as two colliding objects) remains constant before and after the collision. Additionally, the type of collision (elastic or inelastic) and the masses and velocities of the objects also play a role in solving the problem.

4. How can I prepare for solving Super Basic Collision Problems in SR?

To prepare for solving these types of problems, it is important to understand the concepts of momentum and energy, as well as how to apply the equations for special relativity. Practice problems and working through examples can also help improve problem-solving skills.

5. Are there any real-world applications for Super Basic Collision Problems in SR?

Yes, Super Basic Collision Problems in SR have real-world applications in fields such as astrophysics, particle physics, and engineering. These concepts can be used to understand and predict the behavior of objects colliding in high-speed environments, such as collisions between subatomic particles or spacecrafts.

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