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Conservation of Momentum

  1. Jan 18, 2012 #1
    Hello form, Vaironl here.

    I have some few very basic questions. Why might you ask? I came from a trip about a week ago and in my physics class we are discussing momentum, and elasticity.
    I asked my teacher for help but he really is a bit busy at the moment, and the semester/quarter will soon be done.

    I have this problem explained better on a image see it below.
    http://img703.imageshack.us/img703/9095/problemhm.jpg [Broken]

    I tried to solve this, in the following order:
    Find the total momentum before : 1kg * 2m/s + 2kg * -2m/s = 2kgm/s + -4kgm/s = -2kgm/s
    Find total momentum after (This is were I get stuck): 1kg * -1m/s + 2kg * VEL = -2kgm/s + 2kg

    I really don't know what to do, sorry to bother you guys with such basic questions
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 18, 2012 #2
    The question is not a bother, it's the purpose of the forum.
    Your analysis of the before collision momentum looks very good. The after part is almost correct. The terms to the left of the equals sign are good and on the right the first term is as well. There is a "+ 2kg", is that a typo?
     
  4. Jan 18, 2012 #3
    No sorry for that I thought that I should have left it blank indicating I din't know the speed, but I see I created greater confusion.
     
  5. Jan 18, 2012 #4
    I think the best way to approach the second problem is to first find out the velocity.

    Remember that conservation of momentum

    m1v1(initial) + m2(v2)initial = (m1) v1final + m2v2final

    You know every single variable except for v2final. Solve for that.
     
  6. Jan 18, 2012 #5
    That's ok.
    Remember that since momentum is conserved, the momentum before the collision is the same as that after the collision. You have correctly calculated the momentum before the collision, -2kgm/s, so that's what the total after collision momentum must be as well. Your unknown speed is the VEL in your second term. Can you solve for that?
     
  7. Jan 18, 2012 #6
    I believe now I got it I was confused because I was trying to use two separte variables (1kg*-1m/s)+(2kg * VEL) but I notice that I can just add the kg and find the vel right?

    (1kg+2kg)(VEL)
    -2kgm/s = 2kg * Vel

    Vel = [itex]\frac{-2kgm/s}{2kg}[/itex]
    = -1m/s
     
  8. Jan 18, 2012 #7
    The velocities of the two masses are not necessarily the same(most of the time they are not). Look at the equation 15tungalbert posted, the before collision momentum(left side) is equal to the after collision momentum(right side).
    You are given the velocity of the 1kg mass post collision. Your only unknown is the velocity of the 2kg mass, post collision.
     
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