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Homework Help: Momentum and relative distance

  1. Jun 17, 2012 #1
    1. The problem statement, all variables and given/known data
    The attached figure below shows object A (mass,m), which is placed on a table, and wagon B (mass,M), which is in contact with the table. The top of the table and the top of the wagon are at the same height. A is made to slide on the table so that it transfers to the top of B with speed v0. At that instant, A begins sliding on the top of B, and B begins to move on the floor. Shortly afterwards, A comes to rest with respect to B, and A and B then travel with a constant speed. The coefficient of kinetic friction between A and the top of B is μ'. B moves smoothly on the floor. The size of A is negligible.

    What is the distance from the left side of B to the point where A came to rest with respect to B?

    2. Relevant equations
    Non-conservation energy due to friction between A and B;
    ƩE2-ƩE1=Work ...(1)

    Momentum conserve(non elastic collision)
    ƩP1=ƩP2 ..(2)

    SA/B=SA-SB ..(3)

    3. The attempt at a solution

    I) When A slides on B surface, I find SA by using (1)

    ∴SA=[itex]\frac{v^{2}_{0}}{2gμ'}[/itex] ...(4)
    II) A and B move together after collision
    From (2), I got the velocity(v') after collision

    Using (1) again to find displacement of B (B moves by friction of A and B)
    Thus, -[itex]\frac{1}{2}[/itex](m+M)([itex]\frac{mv^{2}_{0}}{m+M}[/itex])^2 = -μ'mgSB

    and SA/B= SA-SB =[itex]\frac{Mv^{2}_{0}}{2μ'g(M+m)}[/itex]

    Is it right?? Who can tell me that my procedure to solve this problem is correct??

    help is appreciate :))
    Thanks a lot

    Attached Files:

  2. jcsd
  3. Jun 18, 2012 #2


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    Can you explain why you set the final kinetic energy of block A equal to 0? Relative to the earth, block A never comes to rest. Seems to me that you need to stick to one frame of reference. So, if you say that the initial velocity of A is vo, then you are using the earth frame of reference. If you stay in the earth frame of reference, the final velocity of A will not be zero.

    Also, I don't understand why you're lumping the masses together when finding the distance B moves. Shouldn't you just use the mass of B alone for M in [itex]\frac{1}{2}[/itex]Mv'2?

    However, when I work the problem I get the same final answer as you do! So, maybe your method is ok. But I don't follow it.
  4. Jun 18, 2012 #3

    I think I misunderstood with this problem because I think velocity of A = 0 relative to the Earth so I set final kinetic energy of A = 0. Because of this I understand that 2 masses (A and B) stick together too...

    What's the equation did you set???
  5. Jun 18, 2012 #4


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    In the earth frame block A has a final velocity of

    v' = [itex]\frac{mvo}{m+M}[/itex]

    So, it seems to me that in your first equation under "The attempt of a solution" your zero should be replaced by the kinetic energy of block A when it has its final velocity. Does that seem right to you?

    When setting up the similar equation for block B, I think you should use just the mass M of block B rather than (m+M) when setting up the final kinetic energy of block B: [itex]\frac{1}{2}[/itex]Mv'2. Does that also seem right?

    If I make these changes and work through the algebra, I get different answers than you did for Sa and Sb. But I get the same answer as you did for the difference: Sa - Sb! :bugeye: I find that kind of amazing.
  6. Jun 18, 2012 #5
    In block B: Did you set equation of energy like this



    and from your mentioned in #2, I get SA=v[itex]^{2}_{0}[/itex](mM+M2)/2μg(m+M)2

    but when I find SA/B I can't eliminate M2 term....
  7. Jun 18, 2012 #6


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    I get (2mM + M2) rather than (mM + M2) in the numerator of the expression for SA. I agree with your expression for SB.
  8. Jun 19, 2012 #7
    now I get SA like you get :) This is from my wrong calculation

    but when find SA/B, I can't manage the answer as you can :((
  9. Jun 19, 2012 #8
    Oh!!! now I can do it!!!

    Thank you :)
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