When to apply the center of mass velocity formula?

[Hamiltonian]

Homework Statement

1. if two bodies of masses m are moving toward each other with a constant speed v and 2v find the speed of the CoM.
2. If two bodies of masses m are accelerating towards each other due to the force of gravity on each of them what is the speed of the CoM.

Relevant Equations

$$V_{cm} = \frac {m_{1}v_{1} + m_{2}v_{2}}{m_{1}+m_{2}}$$

In question 1. since there is no external force on the system of particles(and since it was initially at rest) shouldn't the $V_{cm}$ be zero?
But the correct answer applies the above stated formula for $V_{cm}$ and gets $V_{cm} = v/2$

and in question 2 again as there is no external force on the system the $V_{cm} = 0$ (as they will also collide at the CoM) but here how exactly can you apply the above formula(maybe in a differential eqn form as they are accelerating) to get $V_{cm} = 0$

in short I am confused as to when to apply $$V_{cm} = \frac {m_{1}v_{1} + m_{2}v_{2}}{m_{1}+m_{2}}$$

 

[haruspex]

since it was initially at rest

Who says?

how exactly can you apply the above formula(maybe in a differential eqn form as they are accelerating)

The formula has velocities, you want accelerations.
How do you get from a velocity to an acceleration?

 

[Hamiltonian]

Who says?

so the CoM of a system can move with a velocity only if it was initially moving (and $F_{ext} = 0$)
is this reasoning correct?

also, in the second scenario since the CoM wasn't initially moving and there is no external force hence it should remain at rest. but here if $$A_{cm} = \frac {m_{1}a_{1} + m_{2}a_{2}}{m_{1}+m_{2}}$$ won't you get a wrong answer as $A_{cm}$ should equal zero

 

[haruspex]

so the CoM of a system can move with a velocity only if it was initially moving (and Fext=0)
is this reasoning correct?

Yes.

wont you get a wrong answer as Acm should equal zero

So what do you need to prove regarding $m_1a_1+m_2a_2$ in this scenario?

 

[Hamiltonian]

Yes.

So what do you need to prove regarding $m_1a_1+m_2a_2$ in this scenario?

in this scenario since their masses are equal and their accelerations are in the opposite direction $m_1a_1+m_2a_2 = 0$
but if their masses were say $m$ and $2m$ then the CoM will have a non zero acceleration
(as $(2ma-ma)/(3m) = a/3)$ but again $F_{ext} = 0$ and hence shouldn't $A_{cm}$ equal zero?

 

[jbriggs444]

in this scenario since their masses are equal and their accelerations are in the opposite direction $m_1a_1+m_2a_2 = 0$

Yes.

but if their masses were say $m$ and $2m$ then the CoM will have a non zero acceleration
(as $(2ma-ma)/(3m) = a/3)$ but again $F_{ext} = 0$ and hence shouldn't $A_{cm}$ equal zero?

You've assumed here that the accelerations of the two masses are equal -- that both are given by the same variable named $a$. But is that assumption correct?

 

[Hamiltonian]

Yes.

You've assumed here that the accelerations of the two masses are equal -- that both are given by the same variable named $a$. But is that assumption correct?

I get my mistake now.
so the equation $$A_{cm} = \frac {m_{1}a_{1} + m_{2}a_{2}}{m_{1}+m_{2}}$$ is always valid and $$V_{cm} = \frac {m_{1}v_{1} + m_{2}v_{2}}{m_{1}+m_{2}}$$ can be used only if $F_{ext} = 0$ and the center of mass is initially in motion.

 

[jbriggs444]

I get my mistake now.
so the equation $$A_{cm} = \frac {m_{1}a_{1} + m_{2}a_{2}}{m_{1}+m_{2}}$$ is always valid

Yes, this is correct. This equation can be derived by differentiating the one below.

$$V_{cm} = \frac {m_{1}v_{1} + m_{2}v_{2}}{m_{1}+m_{2}}$$ can be used only if $F_{ext} = 0$ and the center of mass is initially in motion.

No. This equation holds always. It can be derived by differentiating the equation I provide here:$$X_{cm}=\frac{m_1x_1 + m_2x_2}{m_1 + m_2}$$This equation holds always. It is the definition of the position ($X$) of the center of mass.

Your error is the one that I pointed out, the fallacy of equivocation. You used one variable to denote two accelerations which were not necessarily equal.