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In this problem there are three particles, with velocities $\vec{v_1}, \vec{v_2}$ and $\vec{u}$.

Relative to the particle moving at $u$, the velocities $v_1$ and $v_2$ are of equal magnitude and are perpendicular. Accordingly, show: $\left|\vec{u}-\frac{1}{2}(\vec{v_1}+\vec{v_2})\right|^2=\left|\frac{1}{2}(\vec{v_1}-\vec{v_2})\right|^2$

I couldn't work out whether to put this into components or not. I only got it to work for $v_1\not=v_2$

I have taken $\left|\vec{u}-\frac{1}{2}(\vec{v_1}+\vec{v_2})\right|$

To mean $\left(\vec{u}-\frac{1}{2}(\vec{v_1}+\vec{v_2})\right)$

Then I expand out the LHS that to get:

$|\vec{u}|^2-\vec{u}\cdot\vec{v_1}-\vec{u}\cdot\vec{v_2}+\frac{1}{4}(\vec{v_1}+\vec{v_2})^2$

Then I am left with something different to the RHS unless I make some certain assumptions.

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    I'm not sure either but I interpreted it to mean that the motion of $\vec{v_1}$ and $\vec{v_2}$ is with respect to $\vec{u}$2012-08-14

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The relative velocities are defined as $\vec{v}_1-\vec{u}$ and $\vec{v}_2-\vec{u}$.

If we start from

$\left|\vec{u}-\frac{1}{2}(\vec{v_1}+\vec{v_2})\right|^2=\left|\frac{1}{2}(\vec{v_1}-\vec{v_2})\right|^2$

and separate $\vec{u}=\frac{1}{2}\vec{u}+\frac{1}{2}\vec{u}$ on the LHS and add $\frac{1}{2}\vec{u}-\frac{1}{2}\vec{u}=\vec{0}$ on the RHS, we have, grouping the terms opportunely,

$\left|-\frac{1}{2}(\vec{v_1}-\vec{u})-\frac{1}{2}(\vec{v_2}-\vec{u})\right|^2=\left|\frac{1}{2}(\vec{v_1}-\vec{u})-\frac{1}{2}(\vec{v_2}-\vec{u})\right|^2$

if we now evaluate the square of the modulus as $|\vec{a}+\vec{b}|^2=\vec{a}^2+\vec{b}^2+2\vec{a}\cdot\vec{b}$, we have

$\frac{1}{4}(\vec{v_1}-\vec{u})^2+\frac{1}{4}(\vec{v_2}-\vec{u})^2=\frac{1}{4}(\vec{v_1}-\vec{u})^2+\frac{1}{4}(\vec{v_2}-\vec{u})^2$ where we have taken into account $(\vec{v}_1-\vec{u})\cdot(\vec{v}_2-\vec{u})=0$, as per hypothesis.

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    I feel a bit better now. This is really 'just' algebra once you know how to deal with the frame of reference bit, I suppose.2012-08-13