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Speed of 2 Balls of Clay

  1. Nov 15, 2009 #1
    I am a little lost on where to begin with this problem. In my mind, it has something to do with a 90 degree angle, so it would require the use of sin or cos. But I may be wrong. Any help would be appreciated. It is due tonight.

    Problem: A 20 g ball of clay traveling east at 4.5 m/s collides and sticks together with a 60 g ball of clay traveling north at 8.5 m/s. What is the speed of the resulting ball of clay?
     
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  3. Nov 15, 2009 #2

    cepheid

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    We rely on the fact that certain physical quantities are conserved during collisions in order to be able to predict what the outcome will be. Can you think of a physical quantity that is always conserved during a collision of two masses (in the absence of any external forces acting on those masses?)
     
  4. Nov 15, 2009 #3
    The mass of the object?
     
  5. Nov 15, 2009 #4

    cepheid

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    Well, yes, in most of the problems you will consider, total mass is conserved, but that is not the relevant quantity here. Are you familiar with the quantity momentum, and the fact that the momentum of a system is conserved (so long as there is no net external force on that system)?
     
  6. Nov 15, 2009 #5
    Not very familiar, but I know about it and use it some. How is that relevant for this?
     
  7. Nov 15, 2009 #6

    cepheid

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    Well, momentum is always conserved in collisions, which is very useful, because it tells you that:

    total momentum before collision = total momentum after collision.

    Of course, momentum is a vector, which means it has a direction. But you can consider momentum to be conserved separately for each component (east and north) of that vector.
     
  8. Nov 15, 2009 #7
    Okay. Does that mean that their individual momentum's added together will equal the momentum once they have collided and combined?
     
  9. Nov 15, 2009 #8

    cepheid

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    Yes, added together *as vectors* (taking direction into account). Do you know how to add vectors? It is easier in this problem because the initial momentum of one of the masses has only an eastward component, and the other mass's momentum is entirely northward.
     
  10. Nov 15, 2009 #9
    I have a little. Doesn't it involve the use of sin? And since it is exactly east and north, it would be 90 degrees?
     
  11. Nov 15, 2009 #10

    cepheid

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    For this one, the eastward component of the momentum before and after are equal. So that gives you the eastward component of the speed of the combined clay mass afterwards.

    The northward component before and after are equal. That relation gives you the northward component of the speed of the combined clay mass after the collision.

    You add together these two component vectors to get the resultant (total) velocity vector (magnitude and direction). You add vectors by placing them tip to tail, and then drawing the resultant vector from the tail of the first to the tip of the second. There will be some angle involved (the angle of the total velocity vector relative to the horizontal), but it won't be the 90 degree angle, it will be one of the other two angles in the right angled triangle. The resultant is the hypotenuse. Based on basic trigonometry you can calculate it's length and the angle in question.
     
  12. Nov 15, 2009 #11
    I know how to draw them, but how will that help me solve this problem? And I am still lost on how to get the momentum. Do I use the equation J = MfVf - MiVi? (lowercase letters in the equation are meant to be subscripts)
     
  13. Nov 15, 2009 #12

    cepheid

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    Total momentum before equals momentum after, right?

    What's going on before the collision? You have two balls of clay, each with different masses going at different velocities. So the total momentum before is the sum of the momenta of the individual balls. I'm sure that using p = mv, you can write down an expression for that. Of course, it's easier if you break this up into equations for eastward and northward momentum separately, so that you'll actually have two momentum equations.

    What's going on after the collision? You have single mass of clay going at a single velocity, which can also be broken up into eastward and northward components. Again, not that hard to write down the expressions for the total (eastward or northward) momenta if you know how momentum is defined.
     
  14. Nov 15, 2009 #13
    So if I do what you said and break them into separate momentum's, I get:

    P east = (.02 kg)( 4.5 m/s) = .09
    P north = (.06 kg)(8.5 m/s) = .51

    Is that right? And what do I do with them if they are correct? Add them like vectors?
     
  15. Nov 15, 2009 #14

    cepheid

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    You don't quite add them up yet. These are the eastward and northward components of the momentum *before* the collision. The whole point is that these are equal to the eastward and northward components of the momentum *after* the collision, because momentum is *conserved* (which means that the total momentum doesn't change in the collision). So, you use the fact that these have to be equal to the eastward and northward components of momentum after the collision to calculate the eastward and northward components of the velocity of the combined mass after the collision. *Then* you add those velocity components together to get the resultant (or total) velocity.
     
  16. Nov 15, 2009 #15
    I'm confused... If these components are equal before and after the collision, then wouldn't I just add these up and get the same thing?
     
  17. Nov 15, 2009 #16

    cepheid

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    The same total momentum? Yes.

    The same total velocity? NO.

    Here is an outline of the math. Let's call the mass of the eastward ball m1 and the mass of the northward ball m2. The combined mass after the collision is M = m1 + m2.

    So we have:

    total momentum before collision = total momentum after collision​

    In algebraic symbols (boldface means vectors):

    pbefore = pafter

    p1 + p2 = pf

    m1v1 + m2v2 = Mvf

    where the subscript f is for "final" momentum. But we want to break this equation up into two equations, one for each of the horizontal and vertical components of velocity. I'm going to call the east-west direction the "x-axis" and the north-south direction the "y-axis." Then the velocity of ball 1 is entirely in the +x direction and the velocity of ball 2 is entirely in the +y direction (which is really convenient, because it means one of the initial momenta will be zero in each equation, as shown below):

    Conservation of momentum in x-direction:

    m1v1 + m2(0) = Mvf,x
    Conservation of momentum in y-direction:

    m1(0) + m2v2 = Mvf,y

    Once you solve each of these equations for the x and y components of the final velocity, vf,x and vf,y, you can add these components together to get the total final velocity.
     
  18. Nov 15, 2009 #17
    But doesn't the zero in your equations make the one part cancel out and leave me with what I had before?
     
  19. Nov 16, 2009 #18

    cepheid

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    No, it doesn't, and that would have been apparent if you had worked through the algebra starting from those two equations I gave you (it is not actually hard). I don't know if you figured it out before your assignment was due, but notice that those equations give you:

    [tex] v_{f,x} = \frac{m_1}{M}v_1 = \frac{m_1}{m_1 + m_2}v_1 [/tex]

    [tex] v_{f,y} = \frac{m_2}{M}v_2 = \frac{m_2}{m_1 + m_2}v_2 [/tex] ​

    Momentum may be conserved, but velocity is not, in general. By the way, momentum conservation is a central idea in physics, and many collision problems in introductory physics are similar to this one, so it would be a good idea for you to get a handle on it.
     
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