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Conservation of energy and momentum?

  1. Jul 25, 2009 #1
    1. The problem statement, all variables and given/known data

    2 boys of identical masses are standing on 2 identical trolleys A and B(facing each other) which are at rest on a frictionless horizontal surface.The boy on trolley A then throws a ball of mass m horizontally with velocity V with respect to the earth and the boy o trolley B catches it.If the mass of a trolley with a boy is M,the respective final velocities of trolleys A and B are

    1) -mV/M and -mV/M+m
    2)-mV/M+m and mV/M+m
    3) -mV/M and mV/M+m
    4) -mV/M-m and mV/M+m
    5) -V and V

    2. Relevant equations

    3. The attempt at a solution
    momentum before = after
    mV = (M+m ) VB - MVA

    conservation of energy
    1/2mV2 = 1/2 MVA2 + 1/2(M+m)VB2
    and VA = [tex]\sqrt{[mV^{2}-(M+m)VB^{2}/M]}[/tex]
    which just looks so wrong,i'm too afraid to continue.

    Hope someone can help.

  2. jcsd
  3. Jul 25, 2009 #2


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    Homework Helper

    Hi leena19! :smile:

    First, energy is not conserved in collisions unless the exam question somehow tells you it is!

    In this case, the question actually tells you that these are perfectly inelastic collisions (because both masses either start or finish with the same velocity), and so energy certainly isn't conserved. :wink:
    Yes, but that's one equation for all three masses.

    You have two collisions here, so do one equation for each collision separately! :smile:
  4. Jul 25, 2009 #3
    So momentum before throwing ball = momentum just after throwing the ball ?
    0 = mV - MVA
    mV = MVA
    VA= mV/M ?

    Momentum before catching the ball= momentum after catching it ?
    mV - MVA = MVB+ mVB
    0 = (M+m)VB
    VB=0 which is obviously wrong
    though i'm not sure where i'm going wrong :(

    The answer is (3),but I still don't know how to get it .
    Last edited: Jul 25, 2009
  5. Jul 25, 2009 #4
    I get the speeds ,but the directions are still wrong?
    mV = (m+M)VB
    VB=mV/(m+M) ?
  6. Jul 26, 2009 #5


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    Staff Emeritus
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    Gold Member

    This equation seems to assume that the ball is thrown in the positive direction, and that therefore the trolley recoils in the negative direction, and the negative sign is put in explicitly. However, if you do this, it means that your VA has a different meaning from the VA in the answer key. Yours is the magnitude of the velocity of trolley A only, with no information about the direction (this is contained in the minus sign). Theanswer key assumes VA indicates both the magnitude and the direction of the velocity (i.e. its sign is intrinsic). You can do this by NOT assuming anything about the direction (sign) of VA and merely writing:

    sum of final momenta = 0

    mV + MVA = 0

    In this way, the fact that VA is negative shows up naturally in the math, in the final answer.
  7. Jul 26, 2009 #6
    Thank you very much!
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