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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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The notation $|\vec x|$ usually means the magnitude $\sum_i x_i^2$ for a real vector. –  nullUser Aug 13 '12 at 14:28
    
It's not clear what the words "Relative to the particle moving at $u$" mean. Your first equation does not hold when $|u|$ becomes large and the $v_i$ don't follow suit somehow. –  Christian Blatter Aug 13 '12 at 15:12
    
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 $\vec{u}$ –  Magpie Aug 14 '12 at 18:57
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1 Answer

up vote 1 down vote accepted

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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Wait, you lost me at the first step; I'm not seeing what you did there, at all. Can you clarify that for me? –  Magpie Aug 13 '12 at 16:25
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@Magpie: I tried to clarify. –  enzotib Aug 13 '12 at 16:48
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@enzotib Relative velocity of an object A wrt to a reference object B is obtained by subtracting the absolute velocity of the reference object B from the absolute velocity of the object A, while their "absolute" velocities are wrt some common reference frame. –  Shuhao Cao Aug 13 '12 at 17:40
    
@ShuhaoCao: cannot understand if you think I made a mistake or it is just a clarification. –  enzotib Aug 13 '12 at 17:43
    
@enzotib Sorry I at'ed the wrong person..., should have at'ed OP. –  Shuhao Cao Aug 13 '12 at 17:55
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