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B Fdt vs. Fdx

  1. Jun 18, 2017 #1

    JLT

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    For collisions between particles, Fdt goes to zero while Fdx does not
    F = mdv/dt
    mv + Fdt = mv
    if you have two particles colliding
    mva + mvb +Fdt = mva'+mvb'
    in the above case, Fdt goes away as there are equal and opposite forces between the two particles during the collision, linear momentum is conserved

    but
    F = m(dv/dt)(dx/dx)=m(dx/dt)(dv/dx)=mv(dv/dx)
    Fdx = mvdv

    KE1 + Fdx = KE2
    Fdx does not go away during a collision.

    Why does Fdt go away during collisions while Fdx does not?
     
  2. jcsd
  3. Jun 18, 2017 #2

    Dale

    Staff: Mentor

    Where did this come from?

    I don't understand. Why are you adding forces on two different objects?
     
  4. Jun 18, 2017 #3

    JLT

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    forces between two colliding objects. you can set it up so that momentum is conserved, while energy is not conserved.
     
  5. Jun 18, 2017 #4

    Dale

    Staff: Mentor

    Both momentum and energy are conserved, but energy can be converted from KE to other forms.

    But I still don't get why you are adding the forces on two objects. If I have two forces acting on one system then I can add them and get the net force. But if you have two forces acting on two different systems then you don't add them in any meaningful way.
     
  6. Jun 19, 2017 #5

    JLT

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    Sorry, I should have put a summation sign on the forces:
    mVa + mVb + Fab dt + Fba dt = mVa' + mVb' → forces are equal and opposite, cancel out, and you get conservation of linear momentum

    0.5 mVa^2 + 0.5mVb^2 + Fab dx + Fba dx = 0.5 m Va'^2 + 0.5 m Vb'^2

    I think it is because part of dx does not restore itself after the collision, even though Fab=-Fba Fdx is a scalar quantity not a vector and the magnitudes do not cancel out.
     
  7. Jun 19, 2017 #6

    Dale

    Staff: Mentor

    Why are you adding Fab and Fba, they are acting on two different systems.
     
  8. Jun 21, 2017 #7

    JLT

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    This: https://en.wikipedia.org/wiki/Elastic_collision
    vs. this: https://en.wikipedia.org/wiki/Inelastic_collision#Perfectly_inelastic_collision

    Elastic has conserved energy, while plastic does not. Fdt vs. Fdx. Fdt goes to zero (equal and opposite pairs), Fdx does not go to zero even though it is still equal and opposite pairs??

    mava + mbvb + 0 = mava' + mbvb' → this works for both elastic and plastic collisions, Fdt does not care if it is elastic or plastic

    0.5mava^2 + 0.5mavb^2 + Fdx = 0.5mava'^2 + 0.5mavb'^2 → Fdx only goes to zero for elastic collisions??
     
  9. Jun 21, 2017 #8
    The formulas in blue are not meaningful. You are adding finite terms (like mv) with terms containing infinitesimals (Fdt).
    If you look at the momenta before and after collision, the difference is not Fdt but the integral of Fdt over the duration of collision.
    Fdt does not "go to zero" during the collision but the integral (over the duration of interaction) may if there are no forces external to the system under consideration.

    Same observation applies to the equations where you add Fdx to finite terms.
     
  10. Jun 21, 2017 #9

    JLT

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    mva + mvb +Fabdt+Fbadt = mva'+mvb'

    Fab = Fba → internal equal and opposite forces, so in the absence of external forces momentum is conserved.

    0.5mava^2 + 0.5mavb^2 + Fabdx+Fbadx = 0.5mava'^2 + 0.5mavb'^2

    Same scenario, Fab = Fba, only energy is not conserved if the collision is plastic (e < 1)

    mv does not care if the collision is an elastic collision (e = 1) or a plastic collision
     
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