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Finding the momentum after a collision

  1. Mar 8, 2014 #1
    1. The problem statement, all variables and given/known data
    VKUyVNK.gif
    A ball is attached to one end of a rod. A second ball hits the ball with velocity u. The collision is inelastic and both balls after the collision travel together with a new velocity v.
    The rod is attached to the ceiling at a point with no friction. The moment of inertia for the rod is known and the mass of the rod is known and its length is L. The mass of the two balls are each known.

    Find an expression for the momentum before and after the collision.


    2. Relevant equations
    [itex]p=mv[/itex]
    [itex]v_{cm}=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/itex]


    3. The attempt at a solution
    I know how to find the momentum for the balls by simply multiplying their speed with their mass.
    My problem is with the rod. I'm not sure how to implement the rod so that I can use this in the equation for momentum. I know that I'm supposed to use the velocity at the center of mass but how do I calculate the velocity of center of mass?

    [itex]v_{cm}=\frac{m_{ball1}v_{ball1}+m_{ball2}v_{ball2}+m_{rod}\cdot?}{m_{ball1}+m_{ball2}+m_{rod}}[/itex]

    You see, I'm not sure what velocity to use for the rod when calculating the velocity of the center of mass.
     
  2. jcsd
  3. Mar 8, 2014 #2
    This asks for the momentum before and after the collision. So what is (and what kind is) the momentum before the collision?
     
  4. Mar 8, 2014 #3

    The momentum before the collision would be the ball with the velocity u:
    p=mu

    Where m is the known mass of the ball.
     
  5. Mar 8, 2014 #4

    haruspex

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    What is going to be conserved during the collision (that you can make use of)? hint:
    it is not linear momentum.
     
  6. Mar 9, 2014 #5
    Linear momentum is conserved when net force acting is zero, here there is a force acting, guess whats that.
    And you may get close to what haruspex is asking!!
     
  7. Mar 9, 2014 #6

    Im pretty sure it is the angular momentum about the rods axis, right?
     
  8. Mar 9, 2014 #7
    Is it the angular momentum?
     
  9. Mar 9, 2014 #8

    haruspex

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    Yes, if you take moments about the right point. When the collision occurs, there is also an impulse at the pivot point of the rod. (Otherwise it would move.) In order to keep that unknown out of the equation, what point will you take moments about?
     
  10. Mar 9, 2014 #9
    I would take moment about the rods axis, the point where it is attached to the ceiling, right?
     
  11. Mar 9, 2014 #10

    haruspex

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    Yes.
     
  12. Mar 9, 2014 #11
    But I'm still confused as to how I find the momentum after the collision.
    I know that the equation for momentum is:
    [itex]p=mv[/itex]

    I know that for the balls it is simply their mass multiplied by their velocity which are all known variables. But I also have to take the rod's momentum into consideration. And while I know the mass of the rod, I have a hard time figuring out the velocity of the rod to insert into the momentum equation. I just can't see how I'm supposed to use the conservation of angular momentum in this situation.
     
  13. Mar 9, 2014 #12

    haruspex

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    First, what is the angular momentum of the incoming ball about the rod's axis before collision?
    After collision, the balls are moving, initially, at speed v. What is the angular momentum of the two balls about the rod's axis now?
    For the rod itself, you know its length and how fast the lower end is moving, so what is its angular velocity?
     
  14. Mar 9, 2014 #13
    The incoming ball's angular momentum about the rod's axis before collision is [itex]L=Mul[/itex]

    The two ball's angular momentum after collision is [itex]L=2Mvl[/itex]

    The rod moves at velocity v and its angular velocity is [itex]\omega=\frac{v}{l}[/itex]

    Therefore the momentum before the collision is just the incoming ball:
    [itex]p=Mul[/itex]

    And the momentum after the collision is both balls together and the rod:
    [itex]p=2Mvl+m\frac{v}{l}[/itex]

    Would that be correct?
     
  15. Mar 9, 2014 #14

    haruspex

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    Not quite. You don't multiply an angular velocity by a mass to get angular momentum. What equation do you need instead?
     
  16. Mar 9, 2014 #15
    Is it [itex]L=I\omega[/itex]?

    If my angular momentum is conserved, that means the angular momentum before and after are equal each other.

    The angular momentum before the collision was [itex]L=Mul[/itex] so that has to be equal to the angular momentum after the collision:
    [itex]Mul=I\frac{v}{l}[/itex]

    Then I can find an expression for v. But isn't this the same velocity as the two balls? Don't the balls and rod share the same velocity so I can write it as:
    [itex]p=2Mv+mv[/itex]?
     
  17. Mar 9, 2014 #16

    haruspex

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    Yes. What is the moment of inertia of a uniform rod of given mass and length about one end?
     
  18. Mar 9, 2014 #17
    That is [itex]I=\frac{1}{3}ml^2[/itex]

    So then I have [itex]L=\frac{1}{3}ml^2\frac{v}{l}[/itex]

    But how do I go from the angular momentum to the momentum? I need to incorporate it into [itex]p=mv[/itex] somehow but I haven't been able to figure it out.
     
  19. Mar 9, 2014 #18

    haruspex

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    Using angular momentum you can find v. Once you have v you can find the linear momentum. If the balls are moving at speed v, how fast is the mass centre of the rod moving?
     
  20. Mar 9, 2014 #19
    I found my v to:
    [itex]v=\frac{u}{2+\frac{m}{3M}}[/itex]

    The CM velocity would be:
    [itex]v_{CM}=\frac{2mv}{2m}=v[/itex]

    So the velocity of CM would be equal to the speed of the balls?
     
  21. Mar 9, 2014 #20

    haruspex

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    No. This is just geometry. One end of the rod is moving at speed v, the other is stationary, so how fast is its centre moving?
     
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