Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The difference between kinematics and dynamics that Grisha pointed to in his answer comes up in a similar but slightly more complicated case:

Consider (in 1D) two point-like particles connected by a spring ("the molecule"), the right of them being hit by a third particle ("the bullet") coming from the right with initial velocity $v_0$. Let the molecule be in rest initially. Let all three particles have equal mass $m$.

After the collision took place the bullet will move to the right with velocity $v_b$, the molecule will move to the left with velocity $v_m$ and oscillate with frequency $\omega$ and amplitude $\alpha$.

The total energy of the molecule and its total momentum after the collision are then (I hope I didn't make any serious errors):

$E = \frac{m}{2}(v_1^2 + v_2^2) = m (v_m^2 + \alpha^2\omega^2)$

$p = 2mv_m$

The conservation laws yield:

$v_m = \frac{1}{3}(v_0 + \sqrt{v_0^2 - 3\alpha^2\omega^2})$

$v_b = v_0 - 2v_m$

Since we know that $v_m \geq \frac{1}{2}v_0$ we get

$0 \leq \alpha \leq \frac{v_0}{2\omega}$

I.e., the range in which the amplitude $\alpha$ can vary depends on the spring constant $k$ (which is associated with the frequency by $\omega \propto \sqrt{k/m}$): if the spring constant tends to $\infty$ the amplitude must tend to 0 and the molecule must behave approximately like a rigid rod of mass $2m$, as would have been expected.

In any case, the amplitude $\alpha$ is not determined by the conservation laws, only its range of possible values. (This is analoguous to the case of two colliding point-like particles, see answers to Where do particles go after collision?)

But: At least I would see it as unphysical if the molecule would not oscillate after the collision for finite spring constants $k>0$, which is nevertheless allowed by the conservation laws. I cannot imagine an "inner structure" or a micro-process taking place during the collision which would produce this behaviour. What results could inverse scattering experiments yield?

How am I to think about this mind-boggling puzzle?

[Postscript] In the case of two colliding point-like particles it is easy to specify the dynamics in a most natural way, at least in the case when the two collide with diametrical velocities: "for symmetry reasons" their velocities after the collision shall be in the same/opposite direction. (For real point-like particles without inner structure that's the only law/rule that makes sense.)

But now the puzzle continues: Which "natural" law could determine the solution of the molecule-bullet-problem?

share|improve this question
    
The question is equivalent to "What sets the value of $v_b$?", right? I remember working out this problem in intro mechanics, but there the assumption was that $\alpha=0$ (we model the spring as always being in equilibrium), and this then set the value of $v_b$. –  j.c. Nov 12 '10 at 19:35
    
Your first formula for E looks off, why the \dots over the velocities? –  Thomas Themel Nov 13 '10 at 10:37
    
Sorry, removed them! –  Hans Stricker Nov 13 '10 at 10:57
add comment

3 Answers

up vote 1 down vote accepted

I'm not entirely sure that this is what you're looking for, but here's a Lagrangian take on the system. I choose coordinates so that $x_1(0)=0,x_2(0)=0$, so

$$L=m/2(\dot x_1^2 + \dot x_2^2) - k/2 (x_1-x_2)^2$$

It's easier to solve if we go to coordinates $$X=1/2(x_1+x_2), \Delta x = 1/2 (x_1-x_2)$$ so that $$L = m(\dot X^2 + \Delta x^2) - 2k\Delta x^2$$. This gives us the following Euler-Lagrange equations -

$$\frac{d}{dt}\left(2m\dot X\right) = 0$$

$$\frac{d}{dt}\left(2m\dot\Delta X\right) = -4k\Delta x$$

As you'd expect, the center-of-mass motion is constant and the second equation describes an oscillating $\Delta X$. What determines the amplitude of your oscillation here is the initial condition - the bullet strikes one end of the molecule and transfers momentum $p_0$there, so that you get initial conditions $$x_1=0,x_2=0,\dot x_1 = p_0/m, \dot x_2 = 0$$. If we parametrize the oscillation as $\Delta x = \alpha \sin(\omega t + \phi)$, at $t=0$ we get

$$\Delta x(0) = \alpha \sin(\phi) = 0, \Delta \dot x = \alpha \omega \cos(\phi) = p_0/2m $$

which gives us

$$\phi=0, \alpha = \frac{p_0}{2\omega m}$$.

So of course $\alpha$ is completely determined in this picture.

Comparing with the model in your question, I would say you mistreat the collision by assuming that the bullet collides with the entire molecule as a rigid unit, while in fact the collision is between the bullet and one atom, both of equal mass, so that $v_b$ will be zero, $v_m$ becomes exactly $v_0/2$ and you can calculate your $\alpha$ from there.

I'm wondering a little about the validity of my approach when going to infinite $k$ since that SHOULD change the collision process to be 'with the entire system' and I'd like to see a kind of $k$ dependence in $v_b$, but I can't see how to properly treat this at the moment.

share|improve this answer
    
You name it. Neither I "can see how to properly treat this at the moment". –  Hans Stricker Nov 13 '10 at 14:19
    
My intuition says that there must be a continuous function that mediates between the two extremal cases: (1) infinite $k$ $\rightarrow$ bullet interacts with a rigid unit and will be reflected accordingly, (2) $k=0$ $\rightarrow$ bullet interacts with an atom of the same mass and comes to rest. I believe there should be a difference for the collision process depending on how strongly the collision partner of the bullet is interacting with its molecule partner. –  Hans Stricker Nov 13 '10 at 15:40
add comment

I believe, Thomas' answer is correct, even when I didn't think of the bullet colliding with the entire molecule as a rigid unit, literally.

The critical point is, why the target takes the momentum of the bullet instantly and completely (stopping the latter instantly), even when antagonized by the spring.

That's because the force of the spring is finite, while the force-by-collision is infinite. This state of affairs changes only - but drastically then -, when the force of the spring - i.e. $k$ - is infinite, too.

Or when we consider a smooth potential between the bullet and the target like $V(\Delta x) \propto \beta\text{e}^{-\beta \Delta x^2}$ or - more "realistically" something like $V(\Delta x) \propto \frac{1}{\beta \Delta x}$ - with $\beta \rightarrow \infty$ giving "hard sphere collision".

share|improve this answer
add comment

There is only a paradox if you assume an infinitely rigid spring AND and infinitely elastic collision. For a collision that occurs on a timescale much smaller than the natural oscillation period of the molecule, the incident atom stops dead. For a collision that is spread out over several oscillations of the molecule, the atom recoils as though it had struck a target of mass 2m. The problem is solved by trivial methods so long as you can assume that the ratio between the oscillation period and the contact time is very large or very small.

(Of course, this isn't how atoms and molecules work at all. It's really a billiard ball and spring problem, and we've just used the "atom/molecule" terminology as a convenient shorthand.)

share|improve this answer
    
Is this thread on real molecules or or some hypothetical classical model? –  Georg May 9 '11 at 9:22
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.