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Conservation of Momentum of a box of mass

  1. Apr 16, 2014 #1
    1. The problem statement, all variables and given/known data
    Box A of mass 1.20 kg is sliding to the right across a frictionless table at a speed of 2.52 m/s. Box A collides with Box B which has a mass of 2.56 kg, and Box A bounces straight back to the left with a speed of 0.665 m/s.
    A) What is the momentum of Box A before the collision?
    3.02 kg*m/s To the Right Correct Answer
    B) What is the momentum of Box A after the collision?
    0.798 kg*m/s Correct: To the Left Correct Answer
    C) What is the momentum of Box B after the collision? This is the one I need help with!

    2. Relevant equations
    Momentum(P)=mass(kg)*Velocity(kg*m/s)
    Momentum(P-before)=Momentum(P-after)

    3. The attempt at a solution
    A) P=m*V=(1.20 kg)(2.52 m/s)=3.02 kg*m/s To the right CORRECT
    B) P=m*V=(1.20 kg)(.665 m/s)=.798 kg*m/s To the Left CORRECT
    This is the part I am having trouble with
    C) Momentum(before)=Momentum(after); P=m*V
    PBeforeA+PBeforeB=PAfterA+PAfterB
    [(1.20 kg)(2.52 m/s)]+[(2.56 kg)(0 m/s)] = [(1.20 kg)(.665 m/s)]+[]
    [3.02 kg*m/s]+[0 kg*m/s] = [.798 kg*m/s]+[PAfterB]
    [(3.02 kg*m/s)/(.798 kg*m/s)] = PAfterB
    PAfterB = [(3.02 kg*m/s)/(.798 kg*m/s)]
    PAfterB = 3.78 kg*m/s : To the Right

    The answer is incorrect. I am assuming the [To the right] portion is correct because BoxA and BoxB collided; this means that they are asserting equal and opposite force on each other and because BoxA went to the Left, BoxB must go to the Right.

    This leaves the incorrect portion to be the Momentum of BoxB after the collision. I double checked my math, a numerous amount of time, so I don't know where I am going wrong.

    Help is appreciated!
     
  2. jcsd
  3. Apr 16, 2014 #2

    haruspex

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    Decide which direction, left or right, is positive. Say it's right. If I tell you box A has momentum +0.8kgm/s at some point, which way is it moving?
     
  4. Apr 16, 2014 #3
    So if I say the right is positive, then anytime a box is moving to the right-there would be a + sign and if it was moving to the left, a -sign.

    [+3.02 kg*m/s]+[0 kg*m/s] = [-.798 kg*m/s]+[+PAfterB]
    [(+3.02 kg*m/s)-(-.798 kg*m/s)] = +PAfterB
    +PAfterB = [(+3.02 kg*m/s)-(-.798 kg*m/s)]
    PAfterB = 3.818 kg*m/s To The Right

    Is this correct?
     
  5. Apr 16, 2014 #4

    haruspex

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    Looks right.
    (You also did some very strange things later on in the OP, like dividing instead of subtracting. But you got it right this time.)
     
  6. Apr 16, 2014 #5
    Thank you so much, that makes more sense!

    I caught the dividing error after I posted the question to the forum! C:
     
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