Acceleration in an elastic collision

  • Thread starter Saw
  • Start date
  • #1
Saw
Gold Member
529
9
Incidentally in another thread I made this reasoning to calculate the velocity increase in an elastic collision:

In any collision:

[tex]
\frac{{m_1 }}{{m_2 }} = \frac{{a_2 }}{{a_1 }} = \frac{{\frac{{\Delta v_2 }}{t}}}{{\frac{{\Delta v_1 }}{t}}} = \frac{{\Delta v_2 }}{{\Delta v_1 }}
[/tex]

If the system formed by the two bodies is closed (no external force) and the collision is perfectly elastic (no internal dissipation of energy), then the relative speed of the two bodies does not change. For example, in the frame of m1, m2 was approaching at a certain v and after the collision it recedes in the opposite direction. This means that:

[tex]
\Delta v_1 + \Delta v_2 = 2v_{rel}
[/tex]

After some algebra:

[tex]
\begin{array}{l}
\frac{{m_1 }}{{m_2 }} + 1 = \frac{{\Delta v_2 }}{{\Delta v_1 }} + 1 = \frac{{m_1 + m_2 }}{{m_2 }} = \frac{{\Delta v_2 + \Delta v_1 }}{{\Delta v_1 }} = \frac{{2v_{rel} }}{{\Delta v_1 }} \to \Delta v_1 = 2v_{rel} \frac{{m_2 }}{{m_1 + m_2 }} \\
\frac{{m_2 }}{{m_1 }} + 1 = \frac{{\Delta v_1 }}{{\Delta v_2 }} + 1 = \frac{{m_2 + m_1 }}{{m_1 }} = \frac{{\Delta v_1 + \Delta v_2 }}{{\Delta v_2 }} = \frac{{2v_{rel} }}{{\Delta v_2 }} \to \Delta v_2 = 2v_{rel} \frac{{m_1 }}{{m_1 + m_2 }} \\
\end{array}
[/tex]
My questions:

(a) I haven't seen this approach in the texts I use. Is this correct? Am I missing anything?

(b) If correct, is it also valid for acceleration, not only for velocity increase? It seems to me so, since the interaction (the collision) will take the time it takes, whatever it is, but in any case the same time for both bodies and just after it, after the bodies start separating, there is no interaction any more, so no more acceleration...
 

Answers and Replies

  • #2
Saw
Gold Member
529
9
It seems the question was unnoticed but I am really curious about it. Any comment? Thanks.
 
  • #3
Meir Achuz
Science Advisor
Homework Helper
Gold Member
3,529
112
There is nothing wrong, but I don't see anything new.
 
  • #4
Saw
Gold Member
529
9
There is nothing wrong, but I don't see anything new.
Thanks. No, really, I was not thinking I was inventing anything. Just wanted to check it was all right since I had not seen the "2 times relative velocity" term written in the sources I handle. Do you know where I can find something alike?
 
  • #5
Saw
Gold Member
529
9
After some investigation, I've seen that the principle that the relative velocity does not change is derived by combining (i) the equation for conservation of total momentum (which in turn stems from the idea that the velocity centre of mass of the system remains the same) with (ii) the equation for conservation of total kinetic energy, for elastic collisions. Logically, in an inelastic collision, the final relative velocity will be lower, not higher.

I have then this doubt: if one plays with the formula for conservation of total momentum alone so as to try to find the relation between initial and final relative velocity, could you conclude that there are two unique valid solutions for that equation, namely, final relative velocity is equal or lower than the initial one? In other words, is that principle somehow contained in the equation for conservation of total momentum? If so, can it be mathematically shown?
 
  • #6
649
2
If you use the conservation of momentum alone, you will be able to get an arbitrary final relative velocity, since it is possible for the momentum to be conserved for any final relative velocity.

For an incoming total momentum 0 and equal masses, any pair of outgoing velocities {-v,v}, and thus any relative velocity in (-infinity,infinity), will be allowed by the momentum conservation. This is a one-parameter family of solutions (when space is one-dimensional), so the solution is determined by imposing one more equation, the energy conservation equation.

Torquil
 
  • #7
Saw
Gold Member
529
9
If you use the conservation of momentum alone, you will be able to get an arbitrary final relative velocity, since it is possible for the momentum to be conserved for any final relative velocity.

For an incoming total momentum 0 and equal masses, any pair of outgoing velocities {-v,v}, and thus any relative velocity in (-infinity,infinity), will be allowed by the momentum conservation. This is a one-parameter family of solutions (when space is one-dimensional), so the solution is determined by imposing one more equation, the energy conservation equation.
Ok, I see. Could we thus state the following?

- The conservation of momentum equation (just like the principle that the velocity of the centre of mass remains the same, just like Newton's Third Law m1a1 = - m2a2) only indicates how the total acceleration (in absolute numbers) is distributed (in inverse proportion to masses), but says nothing about the magnitude of such total acceleration.

- The conservation of energy principle sets an upper limit to that total acceleration (= twice the relative velocity).

- Assuming the system is closed, the composition of the materials involved in the collision (their elasticity degree = how well they transmit, in particular, kinetic energy) determines the specific magnitude, within the said limit, of the total acceleration: from 1 * relative velocity (in a perfectly inelastic collision) to 2 * relative velocity (perfectly elastic).
 
  • #8
649
2
Ok, I see. Could we thus state the following?

- The conservation of momentum equation (just like the principle that the velocity of the centre of mass remains the same, just like Newton's Third Law m1a1 = - m2a2) only indicates how the total acceleration (in absolute numbers) is distributed (in inverse proportion to masses), but says nothing about the magnitude of such total acceleration.
If an elastic collision between N particles takes place during a time interval T, then momentum conservation says that

m_1*a_1 + m_2*a_2 + m_3*a_3 + ... + m_N*a_N = 0

where m_i is the mass of the i'th particle, and a_i is the average acceleration during the time T. So as you say, it gives a linear relation among these, and doesn't determine their exact values.

- The conservation of energy principle sets an upper limit to that total acceleration (= twice the relative velocity).
Conservation of energy demands that

0.5*(m_1*u_1^2 + m_2*u_2^2 + ... + m_N*u_N^2 ) = 0.5*(m_1*v_1^2 + m_2*v_2^2 + ... + m_N*v_N^2)

where u are incoming velocities and v are outgoing velocities. This equation defines a sphere in N-dimensional v-space, with a radius given in terms of the incoming velocities. So as you say, no particle can have an average acceleration that will result in a |velocity| larger than the radius of this sphere.
 
  • #9
Saw
Gold Member
529
9
Thanks. That geometrical view is really very illustrative. Is there any text/site you can point at where that can be visualized?

On the other hand, I've thought of these formulas that maybe you could check:

Naming e = coefficient of restitution = final relative velocity / initial relative velocity

(ranging from 1 in perfectly elastic collisions to 0 in perfectly inelastic collisions),

I suppose one could state this rule of thumb:

∆vm = (1+e) vrel • M / (m+M)

∆vM = (1+e) vrel • m / (m+M)

Also, could this "coefficient of restitution of the interaction" (e) be calculated following an analogy with the "effective stiffness constant of two interacting springs": e = k1 · k2 /(k1 + k2)?
 

Related Threads on Acceleration in an elastic collision

  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
1
Views
3K
Replies
1
Views
1K
  • Last Post
Replies
8
Views
624
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
1K
Replies
17
Views
5K
  • Last Post
Replies
9
Views
6K
Top