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## Homework Statement

Prove that, for any 1D elastic collision between two particles: as viewed from the centre of mass (or zero momentum) frame of reference, the velocity of each particle after the collision has the same magnitude but opposite sign to its velocity before the collision.

**2. The attempt at a solution**

For i=1,2, let [itex]m_i, u_i, v_i[/itex] be the mass, lab velocity before collision and lab velocity after the collision respectively.

The velocity of the centre of mass is:

[itex]V_{CM}=\frac{m_1u_1+m_2u_2}{m_1+m_2}=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/itex] (1)

Since the total momentum in the CM frame is zero, we have

[itex]m_1(u_1-V_{CM})+m_2(u_2-V_{CM})=0[/itex] (2)

[itex]m_1(v_1-V_{CM})+m_2(v_2-V_{CM})=0[/itex] (3)

Conservation of kinetic energy:

[itex]m_1u_1^2+m_2u_2^2=m_1v_1^2+m_2v_2^2[/itex] (4)

I need to show that

(i) [itex](u_1-V_{CM})=-(v_1-V_{CM})[/itex]

(ii) [itex](u_2-V_{CM})=-(v_2-V_{CM})[/itex]

I've been playing around with equations (1)-(4) (don't see much point copying my scribbles up on here) but can't seem to come up with (i) or (ii). Can anyone point me in the right direction to solving these equations? I'd very much appreciate it :)

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