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ParthKohli

  • one year ago

How to derive \({\bf \vec{F}} = \dfrac{m(v_f - v_i)}{t}\) or \({\rm \vec{F}} \propto \dfrac{m(v_f - v_i)}{t}\), for that matter? Note, you do not have to use \({\bf \vec{F}} = m{\bf \vec{a}}\)

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  1. ParthKohli
    • one year ago
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    @amistre64

  2. DLS
    • one year ago
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    \[\LARGE F=\frac{dp}{dt}\] Can we use this? :P

  3. ParthKohli
    • one year ago
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    Good idea! ^_^ No.

  4. amistre64
    • one year ago
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    force is equal to change in momentum divided by change in time

  5. DLS
    • one year ago
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    -_- @ParthKohli

  6. ParthKohli
    • one year ago
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    @amistre64 Isn't that just what DLS said...

  7. amistre64
    • one year ago
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    i was just seeing if i remembered how to interpret the question correctly :)

  8. ParthKohli
    • one year ago
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    \[F = \dfrac{\Delta p}{\Delta t} = \dfrac{\Delta mv}{\Delta t}\]Is the above true? How did you derive it?

  9. amistre64
    • one year ago
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    Ft = p = mv2-mv1

  10. amistre64
    • one year ago
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    \[\vec F\Delta t=\Delta \vec p = m(\vec v_f-\vec v_i)\]

  11. amistre64
    • one year ago
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    not sure what you mean by how you derive it.

  12. ParthKohli
    • one year ago
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    Does not seem so convincing. :-|

  13. amistre64
    • one year ago
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    do we have to first prove that momentum is a product of mass and velocity? i never know when i have to reinvent the wheel on these things

  14. ParthKohli
    • one year ago
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    No, not that one. lol, got it

  15. amistre64
    • one year ago
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    changing an objects momentum requires applying a force over a given interval of time, right?\[\vec F\Delta t=\Delta \vec p\]

  16. ParthKohli
    • one year ago
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    I'd rather say...\[\vec F \Delta t \propto \Delta \vec p\]

  17. amistre64
    • one year ago
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    that style went out with Descartes :)

  18. ParthKohli
    • one year ago
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    I feel like a newbie here!

  19. ParthKohli
    • one year ago
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    Can you give me a link to the derivation? I got how you mean \(\Delta \vec p =m(\vec v _f - \vec v_i)\) and there's only a little work left... but I didn't get how \(\vec F \Delta t = \Delta \vec p \).

  20. DLS
    • one year ago
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    You are indirectly asking for the proof of F=ma which only Isaac Newton knows.

  21. DLS
    • one year ago
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    I think @shubhamsrg can derive F=ma. Its a piece of cake for him.

  22. ParthKohli
    • one year ago
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    Of course, of course!

  23. amistre64
    • one year ago
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    an object remains at a constant velocity (change in velocity is acceleration) unless acted upon by a force. therefore in order to change an objects momentum, you would need to apply a constant force

  24. amistre64
    • one year ago
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    over a given time interval, the mass "acquires" a new velocity, and therefore a change in momentum. does that sound about right?

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