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Homework Help: The collision between two pucks

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Puck A has a mass of 0.294 kg and is moving along the x axis with a velocity of 5.55 m/s. It makes a collision with puck B, which has a mass of 0.588 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the final speed of
    Puck A
    Puck B

    2. Relevant equations
    I have spent an hour on this problem. I have two different eqns one for x and one for y, respectively:
    m1v0 = m1v1 cos 65 + m2v2cos 37
    0 = m1v1 sin 65 - m2v2 sin 37

    3. The attempt at a solution
    I solved for v2 from the eqn for y and tried to plug that value into the eqn for x. This did not work and Im running out of ideas and patience. Please help!!! There is also a pic for this problem..

    Attached Files:

  2. jcsd
  3. Sep 27, 2009 #2
    I think you accidentally messed up you arithmetic when solving the question. Your X and Y equations are correct so you've set up your question correctly. i think you have just made a mistake when solving because the substitution of V2' of you Y equation into your X equation is the correct thing to do.

    I got V1' = 3.41 m/s; V2' = 2.57 m/s; And these numbers make sense in my mind
  4. Sep 27, 2009 #3
    Your numbers worked out perfectly. I dont understand where i am going wrong, because i am not getting your numbers. can you do a step by step please? id really appreciate it.
  5. Sep 27, 2009 #4
    X-component: m1v1 = m1v1' cos 65 + m2v2'cos 37
    Y-Component: 0 = m1v1' sin 65 - m2v2' sin 37

    Y: 0 = m1v1' sin 65 - m2v2' sin 37
    m1v1' sin 65 = m2v2' sin 37
    m1v1' sin65/(m2sin37) = v2' -> Plug in numbers
    (0.294 kg)*(v1')*(sin65)/[(0.588 kg)(sin37)] = v2'
    0.75298*v1' = v2' -> substitute into the X

    m1v1 = m1v1' cos 65 + m2v2'cos 37
    (0.294 kg)(5.55 m/s) = (0.294 kg)v1'(cos65) + (0.588 kg)(0.75298v1')(cos37)
    1.6317 kg*m/s = 0.12425*v1' kg + 0.3536*v1' kg
    1.6317 kg*m/s = 0.4778*v1' kg
    v1' = 3.41 m/s -> substiture back into Y

    0.75298*v1' = v2'
    0.75298*(3.41 m/s) = v2'
    v2' = 2.57 m/s

    So this was my solution... a lot messier on the computer but i hope you can find where you and i differed in our solutions and this can help you avoid the same mistake in the future.

  6. Sep 27, 2009 #5
    Thankyou so much! This was such a help to say the least
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