Doubt related to the velocity of the center of mass

  • #1

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

  • #2
haruspex
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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?
 
  • #3
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 $$A_{cm} = \frac {m_{1}a_{1} + m_{2}a_{2}}{m_{1}+m_{2}}$$ wont you get a wrong answer as ##A_{cm}## should equal zero o_O
 
  • #4
haruspex
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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?
 
  • #5
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 shouldnt ##A_{cm}## equal zero?
 
  • #6
jbriggs444
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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 shouldnt ##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?
 
  • #7
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.
 
  • #8
jbriggs444
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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.
 
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