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!

Homework Help: Momentum question involving an elastic collision

  1. Mar 22, 2012 #1
    1. A 62 kg skater traveling at 2.8 m/s accidentally bumps elastically into a 76 kg skater moving at 1.6m/s in the same direction
    Find the velocity of each skater after the collision

    2. m1v1 + m1v2 = m1v3 + m2v4

  2. jcsd
  3. Mar 22, 2012 #2
    As you stated, momentum is conserved but so is one other thing. Can you figure out what?
  4. Mar 22, 2012 #3
    Kinetic energy is also conserved.

    This is all the work I have done so far.

    (62)(2.8) + (76)(1.6) = 62v1 + 76v2

    0.5(62)(2.8)^2 + 0.5(76)(1.6)^2 = 0.5(62)(v1)^2 + 0.5(76)(v2)^2

    295.2 = 62v1 + 76v2 and 340.32 = (31)(v1)^2 + (38)(v2)^2

    v2= (295.2 - 62v1)/76

    340.32 = (31)(v1)^2 + (38)[(295.2 - 62v1)/76]^2

    0 = (31)(v1)^2 + 357.778v1 + (25.289)(v1)^2 + 573.30947 - 340.32
    Last edited: Mar 22, 2012
  5. Mar 22, 2012 #4


    User Avatar
    Homework Helper

    It all looks good except the last line. In fact, just the term with v1 looks incorrect. Try checking the bracket expansion again?
  6. Mar 22, 2012 #5
    I fixed it.
    When I solve it I get -0.7366 and -5.6195 for v1 but neither of those are answers.
  7. Mar 22, 2012 #6


    User Avatar

    Staff: Mentor

    Did you substitute these back to verify that your answers correctly solve your equation?

    There is always the option of using an online calculator to check your answer, to give yourself confidence that you have the solution, e.g., http://www.quickmath.com/webMathema...+(31)(v1)^2+++(38)((295.2+-+62v1)/76)^2&v2=v1

    Note: Mathematica ignores square brackets, so you must change them to round parentheses ().
  8. Mar 23, 2012 #7


    User Avatar
    Homework Helper

    If you set up the equation correctly, one root has to be the same as the original velocity.

    You can follow a much simpler method of solving.

    Arrange the equations both for momentum and energy so that the initial and final velocity of one body is at one side and those of the other body on the other side:


    As v≠u for either body, you can divide the second equation with the first one, and obtain a simple first-order system of equations


Share this great discussion with others via Reddit, Google+, Twitter, or Facebook