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Impulse, Momentum, and Force

  1. Mar 19, 2012 #1
    1. The problem statement, all variables and given/known data

    2000kg stationary cannon
    25kg cannon ball fired horizontally @ 250m/s
    Assume no friction acting on the cannon
    ***What is the velocity of the cannon after firing the ball?

    2. Relevant equations
    PT = PT’
    ^initial momentum = final momentum
    m1v1 + m2v2 = m1v1’ + m2v2’

    3. The attempt at a solution

    m1v1 + m2v2 = m1v1’ + m2v2’
    (2000kg)(0m/s) + (25kg)(0m/s) = (2000kg)(v1’) + (25kg)(250m/s)
    0 = (2000kg)(v1’) + 6250kg•m/s
    v1’ = (–6250kg•m/s)/(2000kg)
    v1’ = –3.125m/s
    After rounding:
    v1’ = –3.1m/s
    Therefore, the velocity of the cannon after firing the cannonball is 3.1m/s[backward]


    I think I got it.. if someone could confirm my answer for me that would be awesome!
    thank you forum!
     
  2. jcsd
  3. Mar 19, 2012 #2
    That's right. Also, I don't know how you're writing the problem on your own paper, but it's good practice never to put the numbers in until the last second. So for example,
    $$
    m_1v_1+m_2v_2=m_1v_1'+m_2v_2'\Rightarrow0=m_1v_1'+m_2v_2'\\
    \Rightarrow m_1v_1'=-m_2v_2'\Rightarrow v_1'=-\frac{m_2v_2'}{m_1}
    $$
    Then you finally plug in the numbers and your solution falls out. This way it's easier to keep track of dimensions and to rearrange a result to get another result in a more complicated problem.
     
  4. Mar 19, 2012 #3
    yeah my book actually doesn't do it that way but I'm definitely going to write it like that from now on. Thanks again SadScholar, you rock!
    if you're bored I seem to always have topics I need help with >.>
    https://www.physicsforums.com/showthread.php?t=588460
     
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