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Analyzing Inelastic Collision Data from a graph

  1. Apr 13, 2014 #1
    1. The problem statement, all variables and given/known data

    A head on collision between mass and mass B occurs as described in the F vs s graph below.

    Complete the table in the data section

    Find [itex]\vec{x}[/itex]A
    Find [itex]\vec{x}[/itex]B

    see attached page for clarification

    2. Relevant equations
    P=m*v
    [itex]\Delta[/itex]P=F*t
    Ek=[itex]\stackrel{1}{2}[/itex]mv2
    [itex]\vec{d}[/itex] = [itex]\vec{v}[/itex]t
    W=F*[itex]\vec{d}[/itex]


    3. The attempt at a solution

    I've attempted to use the impulse and the momentum formula to apply to the graph but the numbers just isn't matching up I am unsure of how the 0 and 75 was arrived at just by observing the graph nor can I figure out how velocity 1 was 24 vs v2 at 0.

    This is algebraic physics not calculus.
     

    Attached Files:

    Last edited: Apr 14, 2014
  2. jcsd
  3. Apr 13, 2014 #2

    Andrew Mason

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    I don't quite understand your graph. There are two forces: the force on A and the force on B. There is one line joining several data points (F, s). Which parts of the graph relate to the force on A and which relate to the force on B?

    AM
     
  4. Apr 13, 2014 #3

    Simon Bridge

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    The impulse is the area under the F vs t graph.
    You graph is F vs s (is s displacement?) ... what equations have force and displacement together?

    Physics without calculus??!
    Oh and I see the "lab" did not involve any actual practical work.

    It looks a bit like sudoku: you can't just fill it in left to right. The force-displacement graph can be used to tell you the work, and so the change in energy. You need the relations between work, energy, and force, and displacement to understand it.

    The graph looks like the force-displacement for one object from the point of contact.
    It approaches the target under acceleration of 8N force pointing the other way, comes to rest, then bounces off.
    The force during the recoil is less than during the approach.
    Is this correct?

    These look to be pretty big objects though.
     
    Last edited: Apr 13, 2014
  5. Apr 14, 2014 #4
    Well this is a collision type problem, which means conservation of momentum occurs, so I'd assume you are correct but i'm not so sure, myself. I mean I have the impulse and Kinetic energy recorded per area of contact with the force turning points. But I am not sure how to factor in the time thing or what xa means, i'm assuming it's displacement of object A and object B but I have no clue how to find that out from the graph or the data given. especially with time involved, I dont know where 1s is or where 0s was.

    I do not know which relates to which, what I do know is that this is a inelastic collision so im assuming that that is the total force/displacement of the two objects together.
     
    Last edited: Apr 14, 2014
  6. Apr 14, 2014 #5

    Andrew Mason

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    ??? The motion of the centre of mass of the two objects cannot change if the only forces are due to the collision. So if the graph is supposed to represent the force on the two bodies stuck together after the collision, the force should be 0.

    Can you give us the whole problem as it was presented to you, before you wrote anything on the graph? Can you then give us all the data that you have and tell us where it comes from?

    AM
     
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