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Conservation of momentum and kinetic energy

  1. Jan 22, 2014 #1
    1. The problem statement, all variables and given/known data

    Two 0.3 Kg gliders collide elastically on a frictionless track. Prior to the collision , their total kinetic energy is 0.52 J and their total momentum is 0.12 Kg m/s
    along the track. Calculate the final velocities of the gliders.

    2. Relevant equations
    m1vi1 + m2vi2 = m1vf1 + m2vf2
    1/2m1vi1^2 + 1/2m2vi2^2 = 1/2m1vf1^2 + 1/2m2vf2^2



    3. The attempt at a solution

    tried subbing into conservation of momentum equation, and the conservation of energy formula, then subbing 1 into 2 to find final velocities, but its not working. Can someone tell me what im doing wrong or if the solution i am using is correct?

    m1vi1 + m2vi2 = m1vf1 + m2vf2
    -0.12 = (0.3)vf1 + (0.3)vf2
    -0.4 = vf1 + vf2
    -0.4 - vf2 = vf1

    1/2m1vi1^2 + 1/2m2vi2^2 = 1/2m1vf1^2 + 1/2m2vf2^2
    0.52 = (0.3)(vf1)^2 + (0.3)(vf2)^2
    √(0.52/0.3) = vf1 + vf2
    1.32= vf1 + vf2

    sub 1 into 2

    1.32 = -0.4 - vf2 + vf2

    This is where i am confused, because vf2-vf2 is obviously 0. I know somethings not right, and some assistance would be kindly appreciated :) The answers if needed are supposed to be 1.5 m/s
    and 1.1 m/s
     
  2. jcsd
  3. Jan 22, 2014 #2

    lightgrav

    User Avatar
    Homework Helper

    line 3 in your KE paragraph: √a²+b² is NOT a+b .
    you might square vf1 's formula , to plug into the KE line 2 ... eliminating it ... then solve for vf2.

    otherwise, try subtracting the center-of-mass velocity, to get relative velocities. the relative velocity after collision is negative of the relative velocity before collision (since elastic; otherwise it is -e times vrel , with 0<e<1).
     
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