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Find the height the block rises to after an elastic collision

  1. Dec 13, 2016 #1
    1. The problem statement, all variables and given/known data

    Two blocks are free to slide along a frictionless wooden track ABC as shown below. The block of mass m1 = 5.09 kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m2 = 9.50 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

    Answer in m

    figure: p9-26.gif

    2. Relevant equations
    We'll use law of energy conservation: KEi+P.Ei=K.Ef+P.E
    +
    Vf=(m1-m2/m1+m2)Vi

    3. The attempt at a solution
    K.Ei+P.Ei=K.Ef+P.Ef
    m1gh=1/2 m1 V1i2

    So we get V1i=9.90 m/s.

    Substituting in V1f=(m1-m2/m1+m2)Vi, w get:
    V1f=-3.3m/s

    1)I want to know if what I did above is correct.

    2) My second question is

    I'll use the law of energy conservation again so we can find the maximum height:
    The second body will acquire velocity so there is kinetic energy after collision:

    1/2m1 v12+1/2 m2 V22=m1gh(new height)
    Is this one correct?
     
    Last edited: Dec 13, 2016
  2. jcsd
  3. Dec 13, 2016 #2
    Updated.
     
  4. Dec 13, 2016 #3
    Did not get this part.
    it is given that ##V_{1i} = 0##.
     
  5. Dec 13, 2016 #4
    This is velocity of mass 1 right before it collides with body 2.
     
  6. Dec 13, 2016 #5
    Yes that is correct. (if the calculations are correct).

    You did not clarify what v12 and v22 means here so I assume :-
    v12 - velocity of m1 after collision.
    v22 - velocity of m2 after collision.

    You will get the same height this way (because 1/2m1 v12+1/2 m2 V22 = initial energy of the body m1). Which is incorrect since some energy is transferred this way.
    You should not consider the kinetic energy of second body in your equation, i.e 1/2 m2 V22 should not come.

    Please use subscript and superscript in writing variables, it is hard to understand.
    Capture.PNG
     
  7. Dec 14, 2016 #6
    Sorry for this ambiguous statement. I meant to say that some energy is transferred to m2, so height of m1 will be less than initial.
     
  8. Dec 14, 2016 #7

    haruspex

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    The question looks a bit flawed to me. If the magnets are strong enough to prevent collision, m2 will start to move when m1 is still some way short of it. From the diagram, it will still be on the arc. This means momentum will not be conserved, since the normal force has a horizontal component and it will be more than merely enough to support m1.
    Glossing over that....
    Please use parentheses correctly. That cannot be what you mean.
    Please check that. I make it a bit less.
     
    Last edited: Dec 14, 2016
  9. Dec 14, 2016 #8

    PeroK

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    If you know that equation, can see you a good way to solve the problem without calculating any velocities?
     
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