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What is the velocity of the center of mass of the system?

  1. Mar 26, 2013 #1
    a(n) proton (mass 1.67x10^-27 kg) of kenetic energy 1.58x10^-12 J is moving toward a(n) proton (mass 1.67x10^-27) at rest. what is the velocity of the center of mass of the system

    I used the equation k=1/2mv^2 -> sqrt(2k/m)=v
    sqrt(2*1.58x10^-12 /1.67x10^-27)=4.34x10^7
  2. jcsd
  3. Mar 27, 2013 #2
    What would be the instantaneous position of the center of mass provided you know the positions of the two protons?

    What is the relation between position (vector) and velocity?
  4. Mar 27, 2013 #3
    would the position be zero.
  5. Mar 27, 2013 #4
    The velocity of the center of mass is zero only in its own frame of reference. Stick to the frame of reference where the velocity of the proton is as given in the problem.

    Okay, what formula gives the position of center of mass in this frame of reference?
  6. Mar 28, 2013 #5
    Xcm=(x*y........)/total mass
  7. Mar 28, 2013 #6
    Yes that is correct (I assume y is the mass of an object there).

    Now if you differentiate that equation with respect to time, what would you get?
  8. Mar 28, 2013 #7
    Which one is suppose to be time?
  9. Mar 28, 2013 #8
    I think you got me wrong,

    Your initial equation is
    [itex]X_{CM}=\frac{\Sigma m_ix_i}{\Sigma m_i}[/itex] right?

    Where [itex]x_i [/itex] and [itex] m_i[/itex] are the position vector and mass of the [itex]i^{th}[/itex] object.

    Now differentiate wrt time i.e. find [itex]\frac{dX_{CM}}{dt}[/itex].

    P.S. This helps you derive the relation between velocity of CM and velocity of the constituent particles.
    Last edited: Mar 28, 2013
  10. Mar 28, 2013 #9
    The derivative would be m/M
  11. Mar 28, 2013 #10
    Let's go back to the basics.

    First of all what does dx/dt represent?

    Secondly when you differentiate kx (where k is a constant, and x is a position vector) with respect to time, what do you obtain?
  12. Mar 28, 2013 #11
    dx/dt is the instantaneous rate of change of x with respect to t.

    kx would be k
  13. Mar 28, 2013 #12

    Doc Al

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    d(kx)/dx = k, but what about d(kx)/dt?
  14. Mar 28, 2013 #13
    would it be -1/t^2
  15. Mar 28, 2013 #14

    Doc Al

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    Staff: Mentor

    Much simpler than all that. Just use the chain rule:
    d(y)/dt = d(y)/dx * dx/dt
  16. Mar 28, 2013 #15
    Think about it like this:

    If kx were to change by a small amount into k(x+dx) in the time interval dt, what would be the rate of change of kx with respect to time?
  17. Mar 28, 2013 #16
    I know how to take the derivative but im confused, is the d suppose to be coefficient or number where it can be ignored?
  18. Mar 29, 2013 #17
    No. d by itself doesn't have any meaning. dx means an infinitesimal change in the quantity x. It's like sinx. Sin by itself doesn't have any meaning.
  19. Mar 29, 2013 #18
    so derivative is k/t?
  20. Mar 29, 2013 #19


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    Science Advisor

    I can see no reason to use the derivative at all. Since the two protons have the same mass, the center of mass will be exactly half way between them and so its "velocity" will be half the velocity of the moving proton.
  21. Mar 29, 2013 #20
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