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Totally elastic and momentum collision equation help! Thanks

  1. Nov 10, 2011 #1
    1. The problem statement, all variables and given/known data

    A 12kg wad of clay is hurled horizontally at a 100g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50m before coming to rest. If the coefficient of friction between the bloack and the surface is .650, what is the SPEED of the clay immediately before impact?

    2. Relevant equations

    This is a perfectly inelastic collision correct?

    My book tells me in the answer section to use the following formula, but Im not sure where it came from or why to use this. Am I to use this for equations related to inelastic collisions and friction involved?

    1/2(m1+m2)V2^2 = friction(m1+m2)gL

    How did it come up with that???? I guess I am just having trouble figuring out how I was suppose to come up with that formula.

    Any clarification?



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 10, 2011 #2

    Doc Al

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    Staff: Mentor

    Yes.

    It's telling you that the kinetic energy the system had after the collision must equal the work done by friction. Note that V2 is the speed of the 'clay+block' after the collision. You'll use that to figure out the speed of the clay before the collision.
     
  4. Nov 10, 2011 #3
    Do you think that above equation is to be used before impact or after impact?

    Now i notice that another poster beat me.
     
  5. Nov 10, 2011 #4
    Ok great.

    When looking at this problem... What is the first and second things I should be trying to figure out. Mainly, when looking at this problem specifically, what would I be looking to do first?

    I want to systematically break this equation down from start!
     
  6. Nov 10, 2011 #5
    I really am not sure... ?
     
  7. Nov 10, 2011 #6

    Doc Al

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    Break the problem into two stages:
    (1) The collision itself. What's conserved?
    (2) The movement against friction after the collision.

    The equation you have deals with part #2.

    What equation can you come up with to deal with #1?
     
  8. Nov 10, 2011 #7
    I thought for step one, I would use:

    Vf = m1V1i + m2V2i / (m1 + m2)

    ?
     
  9. Nov 10, 2011 #8

    Doc Al

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    Good. (Careful with parentheses!)

    What does Vf correspond to?

    What about V1i and V2i?
     
  10. Nov 10, 2011 #9
    Vf would be velocity of the system?

    v1i would be for the clay and v2i would be wood ???
     
  11. Nov 10, 2011 #10

    Doc Al

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    Yes, immediately after the collision.

    Right.
     
  12. Nov 10, 2011 #11
    Ok, so first thing I do when looking at this is figure out the Vf of the system after the collision.

    With the formula I supplied for the first step Vf = m1V1i + m2V2i / (m1 + m2) , how do I go from there, to here: 1/2(m1+m2)V2^2 = friction(m1+m2)gL
     
  13. Nov 10, 2011 #12
    Doc? Can you explain the above reply to me if you can, or anyone else? :)
     
  14. Nov 11, 2011 #13

    Doc Al

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    Right. You'll do that by solving the second part first, since you have all the needed data. (It's often the case that its best to work a problem 'backwards'.)

    You'll solve the second step first.

    How will you relate the two steps? Hint: How does Vf in the first formula relate to V2 in the second? (Generally it's a good idea to use the same symbols for the same quantities. Otherwise things get confusing.)

    In your first formula, Vf = (m1V1i + m2V2i) / (m1 + m2), what is V2?
     
  15. Nov 11, 2011 #14
    Oh ok Doc, im starting to understand it a little better. Just having trouble coming to a conclusion on what exact formula to use. Like what stpes I take to get the right formula....

    You ask what is V2 ? Well, thats the Velocity of the wood, which is zero, correct? What significance is that may I ask?

    Again, MUCH thanks for the help!

    EDIT: I guess what im missing is how I would come to the formula to get the speed of the system, as detailed here: 1/2(m1+m2)V2^2 = friction(m1+m2)gL

    I have never seen that before!
     
    Last edited: Nov 11, 2011
  16. Nov 11, 2011 #15

    Doc Al

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    Yes, in your conservation of momentum equation (which describes the collision), V2 stands for the initial speed of the block, which is zero. You'll need that equation to solve for the initial speed of the bullet. (You'll use this equation last.)

    That's just an energy equation stating that the change in KE of the system after the collision is equal to the work done by friction. What does V2 stand for here?
     
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