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I found the folowing formula for 1d elastic colition:

$$v_1 +u_1 =v_2+u_2 $$

(v,u means the velocity before and after the collision respectively)

I tried to derivate it from momentum and energy conservations but didn't see how it works. Moreover that, when I tried to see if it works it seems to work for mass ratio of 1,2 and infinity (wall colision).

A problem I saw was that obviously when there is no collision the law dous not apply. Can anyone help me pinpoint what I miss?

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Conservation of momentum and kinetic energy: $$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$ $$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$$

can be rewritten to: $$m_1\left(v_1 - u_1\right) = -m_2\left(v_2 - u_2\right)$$ $$m_1\left(v_1^2 - u_1^2\right) = -m_2\left(v_2^2 - u_2^2\right)$$

Substituting: $v_i^2 - u_j^2 = \left(v_i+u_j\right)\left(v_i-u_j\right)$ yields: $$ m_1\left(v_1 - u_1\right) = - m_2\left(v_2 - u_2\right)$$ $$m_1\left(v_1+u_1\right)\left(v_1-u_1\right) = -m_2\left(v_2+u_2\right)\left(v_2-u_2\right)$$

Dividing the latter by the former equation yields: $$ v_1 + u_1 = v_2 + u_2$$ which is the equation you are looking for

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  • $\begingroup$ Thanks for the great answer $\endgroup$
    – os20
    Commented Apr 23, 2019 at 20:39
  • $\begingroup$ Can you please explain why it dous not work when there is no collision? $\endgroup$
    – os20
    Commented Apr 23, 2019 at 20:40
  • $\begingroup$ @os20 - this works just fine without collision; all this is is a statement about conservation of momentum and kinetic energy. In case there is no collision, momentum and kinetic energy are still conserved but there is simply no change in velocities. Setting $v_1=u_1$ and $v_2=u_2$ still works in the equations but you simply get a trivial answer of $0=0$ which is true. $\endgroup$
    – nluigi
    Commented Apr 23, 2019 at 20:44
  • $\begingroup$ $$2V_1=2v_2$$ is what i get after substituting. And if in the no collision case there is no reason for that to be true. $\endgroup$
    – os20
    Commented Apr 23, 2019 at 20:48
  • $\begingroup$ @os20 - I see what you mean... my previous comment was regarding the original system of equations at the start of my answer from which (i hope) it is clear that even for no collisions the conservation laws still apply. The reason why you get an undefined answer when using the derived equation is because in the derivation we divide one equation by the other, but by doing so we have to assume that the equation we divide by doesn't evaluate to 0. When there is no collision, i.e. $v_1=u_1$ and $v_2=u_2$ it evaluates exactly to 0 and so the operation and the resulting equation is undefined. $\endgroup$
    – nluigi
    Commented Apr 23, 2019 at 20:56
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here's the proof.

Let $$KE_1 = \frac{1}{2}m_1(v_1^2 - u_1^2)$$

Let $$KE_2 = \frac{1}{2} m_2(v_2^2 - u_2^2)$$

Since it is an elastic collision, KE must be conserved. $$KE_1 - KE_2 = 0$$

$$KE_1 = KE_2$$

$$\frac{1}{2}m_1(v_1^2 - u_1^2) = \frac{1}{2}m_2(v_2^2 - u_2^2)$$

$$m_1 = m_2.\frac{v_2^2 - u_2^2}{v_1^2 - u_1^2}$$

By conservation of momentum,

$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$

Substituting for $$m_1$$,

$$m_2.\frac{v_2^2 - u_2^2}{v_1^2 - u_1^2}u_1 + m_2u_2 = m_2.\frac{v_2^2 - u_2^2}{v_1^2 - u_1^2}v_1 + m_2v_2$$

Cancel $$m_2$$,

$$\frac{v_2^2 - u_2^2}{v_1^2 - u_1^2}u_1 + u_2 = \frac{v_2^2 - u_2^2}{v_1^2 - u_1^2}v_1 + v_2$$

$$\frac{v_1-u_1}{u_2-v_2}.\frac{(v_2 + u_2)(v_2 - u_2)}{(v_1 + u_1)(v_1 - u_1)} = 1$$

$$\frac{(-1)(u_2 + v_2)}{v_1 + u_1} = 1$$

$$(-1)(u_2 + v_2) = v_1 + u_1$$

$$|u_2 + v_2| = |v_1 + u_1|$$

Let me know if you need to clarify any part of the proof.

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  • $\begingroup$ Thanks! It's very clear. A bit different than the other answer but i ges you can prove a correct thing in many ways ... Now i need to understand the minus sign difference but what is a minus sign between physicists???;-) $\endgroup$
    – os20
    Commented Apr 23, 2019 at 21:07
  • $\begingroup$ No, you don't need a modulus if you take care of vector directions. $\endgroup$
    – Bill N
    Commented Apr 23, 2019 at 21:10
  • $\begingroup$ @d_g, you are not dealing with 'absolute' kinetic energies but with changes in kinetic energies. In that case to have conservation of kinetic energy you need to have $\Delta {KE}_1 + \Delta {KE}_2 = 0$ which fixes your problem with the minus sign. $\endgroup$
    – nluigi
    Commented Apr 23, 2019 at 21:13
  • $\begingroup$ @nluigi right, that works. I think this proof doesn't take care of vector directions, which is why I get that minus sign. $\endgroup$
    – d_g
    Commented Apr 24, 2019 at 5:36

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