3
$\begingroup$

I am getting stuck in a really easy problem in Statistical Mechanics that involves elastic collisions, it is really very shameful that I am getting stuck at such a simple thing, but from:

$$\|\vec{v_1}\|^2 +\|\vec{v_2}\|^2 = \|\vec{u_1}\|^2 +\|\vec{u_2}\|^2$$ and $$\vec{v_1}+\vec{v_2} = \vec{u_1} + \vec{u_2}$$

How can I get $$\|\vec{v_2}-\vec{v_1}\|=\|\vec{u_2}-\vec{u_1}\|$$

I tried completing the square in the first equation like:

$$\vec{v_1}\cdot\vec{v_1} +\vec{v_2}\cdot\vec{v_2} -2\vec{v_1}\cdot\vec{v_2}= (\vec{v_2}-\vec{v_1})\cdot(\vec{v_2}-\vec{v_1})=\|\vec{v_2}-\vec{v_1}\|^2= \vec{u_1}\cdot\vec{u_1} +\vec{u_2}\cdot\vec{u_2} -2\vec{v_1}\cdot\vec{v_2}$$

and then using the second equation to get:

$$=\vec{u_1}\cdot\vec{u_1} +\vec{u_2}\cdot\vec{u_2} -2\vec{v_1}\cdot(\vec{u_1}+\vec{u_2}-\vec{v_1})$$

but I cannot seem to be able to simplify this to $$\vec{u_1}\cdot\vec{u_1} +\vec{u_2}\cdot\vec{u_2} -2\vec{u_1}\cdot\vec{u_2} = \|\vec{u_2}-\vec{u_1}\|^2$$

Can someone help me with this? I am sure it is quite simple, but since I am stuck I am losing way too much time on this.

Improve this question
$\endgroup$
3
  • $\begingroup$ The algebraic solution to this problem is very far from simple. If you ever have the option, use the solution that involves the center of mass reference frame. $\endgroup$
    – David White
    Commented Mar 19, 2021 at 2:50
  • $\begingroup$ Hum?@DavidWhite Could you elaborate on that ? You mean that if I try to solve this problem in the reference frame of the center of mass I can diminish the number of variables and solve directly, whithout resorting to that squares trick that I had forgot ? $\endgroup$
    – Felipe Dilho
    Commented Mar 27, 2021 at 7:29
  • $\begingroup$ Yes. The center of mass reference frame makes this problem much easier. $\endgroup$
    – David White
    Commented Mar 27, 2021 at 16:05

2 Answers 2

Reset to default
6
$\begingroup$

I'm typing in mobile so I'm ignoring all the vector signs. The problem is to show that $$a^2 + b^2 = c^2 + d^2$$ and $$a + b = c + d$$ Gives you $$|a - b| = |c-d|$$ From the second eq you get by squaring both sides $$a^2 + b^2 + 2ab = c^2 + d^2 + 2cd$$ Using the first equation you then get $$ab = cd$$ Now subtract $2ab= 2cd$ on both sides of the first equation and you get $$(a-b)^2 = (c-d)^2$$ Which is the required answer. To get the vector equivalent just replace the regular product with the dot product.

Improve this answer
$\endgroup$
1
  • $\begingroup$ It is really a talent to write all that latex on mobile $\endgroup$
    – Brian
    Commented Mar 19, 2021 at 20:44
6
$\begingroup$

Use the reduced mass identity to write the total energy as $$ \frac 12 m_1 |{\bf v}_1|^2+ \frac 12 m_2 |{\bf v}_2|^2\\= \frac 12 \frac{m_1m_2}{m_1+m_2} |{\bf v}_1-{\bf v}_2|^2+ \frac 12 (m_1+m_2)|{\bf v}_{CofM}|^2. $$ Then, if energy is conserved (definition of elastic) we must have that $|{\bf v}_1-{\bf v}_2|$ is the same before and after.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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