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!

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
    and
    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.

    Cheers
     
  6. Sep 27, 2009 #5
    Thankyou so much! This was such a help to say the least
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: The collision between two pucks
Loading...