Effective mass of two objects 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:

Can someone explain?
Edit: I guess I should add the source in case I misinterpreted it.
Question:

Solution:

 A: 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 $T=17$
