1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Internal and center of mass energies

  1. Apr 27, 2014 #1
    1. The problem statement, all variables and given/known data

    An object with kinetic energy K explodes into two pieces, each of which moves with twice the speed of the original object.

    Compare the internal and center-of-mass energies after the explosion.

    3. The attempt at a solution

    Let K1 be the kinetic energy of the main object before the explosion. k2 and k3 be the kinetic energies after explosion.

    K1 = k2 + k3

    Let K1 = 0.5m1v12 = K


    0.5m1v12 = 0.5m2v22 + 0.5m3v32

    and since each pieces has twice the velocity of the original piece befor explosion

    0.5m1v2 = 0.5m2(2v)2 + 0.5m3(2v)2

    ∴ 0.5m1v2 = 0.5(4v2)(m2+m3)
  2. jcsd
  3. Apr 27, 2014 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    I am not following what you are doing. You seem to be equating the original kinetic energy to the sum of the final kinetic energies of the two parts? Why would k1 = k2 + k3?

    Think of the problem this way: In the frame of reference of the centre of mass does the body have any energy before the explosion? How about after?

  4. Apr 27, 2014 #3
    I equate them because the total energy is the sum of the KE of each particle.

    However, I may be wrong.

    I think it's good to ask what internal kinetic energy is and what center of mass kinetic enegry is and why kint + kcm = Ktotal
  5. Apr 27, 2014 #4
    What's the definition of the frame of reference of the center of mass?
  6. Apr 27, 2014 #5


    User Avatar
    Homework Helper

    The problem asks the center-of-mass energy and the internal energy after the explosion. I do not see any of them in your post.

    I see, you assumed conservation of energy. It is not true in an explosion.

  7. Apr 27, 2014 #6

    Where can we go from here?

    Ktotal = [itex]\sum[/itex]0.5mivcm2 + [itex]\sum[/itex]0.5mivirel2 + [itex]\sum[/itex]mivirelvcm

    Edit: why is a particle relative velocity to COM zero?
    Last edited: Apr 27, 2014
  8. Apr 27, 2014 #7
    kint = 0.5m1vcm = m1vcm2
    there are 2 pieces so 2m1vcm^2
    kint/kcm = 4 but the answer is 3.

    I spent 4 hrs on this. It's not going anywhere and very agonizing/
  9. Apr 27, 2014 #8


    User Avatar
    Homework Helper

    The CoM will travel with the same velocity as before the explosion, as the forces are internal. So the center-of mass KE is the same as the initial KE:

    KECoM=1/2 m1v12

    If m2 and m3 are the masses after the explosion, and m2+m3=m1, the total KE is

    KE(total)=1/2 (m2v12+m3v32=2v12m1

    The internal energy is the difference between the total energy and the energy of the translation of the CoM:

    KE(internal)= KE(total)-KE(CoM)

    So what is the ratio KEi/KE(CoM)?

  10. Apr 27, 2014 #9
    I have managed to solve it by drawing it out and using geometry. However, one question persist when I attempt to derive K= Kint + Kcm.

    Why is [itex]\sum[/itex] mivirel2 = 0?
  11. Apr 27, 2014 #10


    User Avatar
    Homework Helper

    Are vi rel the relative velocities with respect to the CoM? How did you get that ∑ mivirel2 = 0? It is certainly wrong. What are the relative velocities? The pieces move in opposite directions, and have different masses. The velocity of one piece is v2=2v1, its relative velocity with respect to the CoM is v2rel=v1.
    The other piece has velocity v3=-2v1. Its relative velocity is -2v1-v1=-3v1.

    Last edited: Apr 27, 2014
  12. Apr 27, 2014 #11

    My initial premise is

    K = [itex]\sum[/itex]0.5 mivi2

    but vi = (vcm + virel) (it would be great if someone can give me an intuitive sense of why vi = (vcm + virel)

    then we have K = [itex]\sum[/itex]0.5mi(virel+vcm).(vcm+virel)

    = [itex]\sum[/itex]0.5mivcm2 + [itex]\sum[/itex]mivcm . virel2 + [itex]\sum[/itex] mivirel2

    and then the book states [itex]\sum[/itex]mivcm.virel = 0 because virel = 0 since it is the particle's velocity relative to the COM.
    Unsure what this means.
  13. Apr 27, 2014 #12


    User Avatar
    Homework Helper

    I edited my previous post, read it.

    Your formula is wrong. There is no square in the second term.

    Recall how the CoM is defined.
    The velocity of the CoM is VCoM=(∑mivi)/∑mi.

    Virel=Vi-VCoM. Virel is not zero, but ∑miVirel=0

    Last edited: Apr 27, 2014
  14. Apr 27, 2014 #13
    The square was a typo. It's hard to see what is going on with all the brackets when using tex.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted