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I read that if we want to perform conservation of energy taking one object as frame of reference, we need to assume that the other object has the effective mass and velocity.

And effective mass is given by:

M1 × M2 /( M1 + M2)

But I have the following problem: enter image description here

Can someone explain?

Edit: I guess I should add the source in case I misinterpreted it.

Question: enter image description here

Solution: enter image description here

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    $\begingroup$ I highly doubt your statement of "if we want to perform conservation of energy taking one object as frame of reference, we need to assume that the other object has the effective mass and velocity." $\endgroup$
    – Amin
    Dec 26, 2015 at 6:49
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    $\begingroup$ I agree with @Amin moreover kinetic energy is frame dependent, it need not be same in all frames, important thing is that it is conserved in that particular frame. $\endgroup$
    – Courage
    Dec 26, 2015 at 7:04
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    $\begingroup$ Where did you read this false statement? $\endgroup$
    – CuriousOne
    Dec 26, 2015 at 7:15
  • $\begingroup$ Please see the edit. $\endgroup$ Dec 26, 2015 at 8:49

1 Answer 1

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The definition of reduced mass $\mu$ in your formula can only be used under this condition

In this case it is when $m_1 \vec{v_1} + m_2\vec{ v_2}=0$

This condition is true when there is a central force,

In a central force

$$\vec{F_{12}}=-\vec{F_{21}}$$

$$m_1\vec{\dot{v_1}}=-m_2\vec{\dot{v_2}}$$

Or $m_1 \vec{v_1} + m_2\vec{ v_2}=0$ (Assuming initial conditions to be zero)

But the example shown in the first picture doesn't satisfy this condition, in this case

$m_1 \vec{v_1} + m_2\vec{ v_2}=m\vec{v_{\text{CM}}}\tag{1}$

Where $$m=m_1+m_2$$

there is a non-zero $v_{CM}$ unlike the central force case

The relative velocity is

$$v=v_2-v_1\tag{2}$$

Kinetic energy $$T=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$

Substituting $(1)$ and $(2)$ in $T$ and writing $\mu=\frac{m_1m_2}{m_1+m_2}$ you get

$$T=\frac{1}{2}m{v_{CM}}^2+\frac{1}{2}\mu v^2 \tag{3}$$

This is the correct formula for using the reduced mass

In your problem the total kinetic energy as in the first case is $T=17$,

$$v_{cm}=\frac{10}{3}$$

$$m=3$$ $$\mu=\frac{2}{3}$$ $$v=4-3=1$$

substituting these values in $(3)$ you get $T=17$

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  • $\begingroup$ I read the Wikipedia page but I could not understand this: 'In the computation one mass can be replaced by the reduced mass, if this is compensated by replacing the other mass by the sum of both masses. '. Could you explain that? $\endgroup$ Dec 26, 2015 at 13:33
  • $\begingroup$ And also could you clarify what you were saying about kinetic energy and being frame dependent. I did not quite get that. $\endgroup$ Dec 26, 2015 at 13:40
  • $\begingroup$ I think the statement in the article precisely means equation $(3)$, where 1 of the masses has been replaced by $\mu$ and the other by $m=m_1+m_2$, regarding K.E being frame dependent, in your equation using reduced mass you've done $\frac{1}{2}\mu (v_1+v_2)^2$ which indicates that your frame is fixed wrt one of the masses, this K.E may not be equal to to the K.E you obtained above. $\endgroup$
    – Courage
    Dec 26, 2015 at 13:50
  • $\begingroup$ I think instead of saying central force, you should say absence of external force $\endgroup$
    – Max Payne
    Mar 31, 2016 at 7:00

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