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Conservation of Momentum Help

  1. Oct 17, 2013 #1
    We just started learning Conservation of Momentum, and I feel completely lost. This is the following question I was asked:

    1. The problem statement, all variables and given/known data
    A 2.00-kg object moving at 6.00 m/s collides with a 4.00-kg object that is initially at rest. After the collision, the 2.00-kg object moves backward at 1.00 m/s. This is a one-dimensional collision, and no external forces are acting.

    a) Calculate the velocity of the 4.00-kg object after the collision.

    My attempt was to simply use the Momentum equation:

    2. Relevant equations & The attempt at a solution

    mA1vA1 + mB1vB1 = mA2vA2 + mB2vB2

    So:
    vB2 = (mA1vA1 + mB1vB1 - mA2vA2)/mB2

    Resulting in the velocity to be 3.5m/s.

    I was just looking for someone to confirm that answer, because I am unsure if I did it right.

    In question b) it asks:
    Calculate the velocity of the centre of mass of the two objects (i) before the collision and (ii) after the collision.
    What does that mean?

    Thank you for helping.
     
    Last edited: Oct 17, 2013
  2. jcsd
  3. Oct 17, 2013 #2
    Hi JamesRV

    Welcome to PF!!!

    The result of part a) looks right .

    For part b) use Vcm = (m1v1+m2v2)/(m1+m2) .

    Apply this before and after collision .What do you get ?

    Please take care of the signs .
     
    Last edited: Oct 17, 2013
  4. Oct 17, 2013 #3
    Thank you Tanya!

    I used the equation you gave me, it makes logic sense taking the total momentum and dividing by the mass.
    I got 2m/s for both cases, which seems right!
     
  5. Oct 17, 2013 #4
    Yes..Well done :thumbs:

    So what does this tell you about collision ?
     
  6. Oct 17, 2013 #5
    Doesn't it just mean that the momentum is conserved? I thought momentum was always conserved, is it not?
     
  7. Oct 17, 2013 #6
    Yes... And the velocity of Center of Mass remains unaffected in a collision.
     
  8. Oct 17, 2013 #7
    Okay, thanks again!
     
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