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Marble collisions, momentum, velocity

  1. Nov 18, 2007 #1
    1. The problem statement, all variables and given/known data

    Marble A, mass 5g moves at a speed of 20 cm/s. It collides with a second marble B, mass 10 g, moving at 10 cm/s in the same direction. After the collision, marble A continues with a speed of 8 cm/s in the same direction. A) Calculate the momentum of each marble before the collision. B) Calculate the momentum of each marble after the collision. C) What is the speed of marble B after the collision?

    2. Relevant equations

    F=m(Vf-Vi)/t


    3. The attempt at a solution

    I converted all the numbers first

    Mass A= .005kg
    Vi A = .2 m/s
    Mass B= .1 m/s
    Vi B = .2 m/s
    V after = .8 m/s

    A) Marble A:
    (.005kg)(02m/s) = .001 kg*m/s

    Marble B:
    (.01kg)(.1m/s)=.001 kg*m/s

    B) (.001+.001)=(.005kg+.01kg)V
    v=1.33 m/s

    What did I do wrong in B? I'm pretty sure A is right.
     
  2. jcsd
  3. Nov 18, 2007 #2
    Just skimming over,

    Total Momentum After = Total Momentum Before

    P2' = P2

    m2v2' = m2v2
    (.01 Kg)(.08 m/s) = (.01 Kg)v2

    algebra to isolate v2


    do you see what i did? Momentum after must equal momentum before, you know the momentum before the collision was m2v2, it has to equal momentum after, which is m2v2' your only unknown is v2' which you can solve using your data.. i did this usuing your definitions for the mass and velocities, i didn't check those.
     
    Last edited: Nov 18, 2007
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