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ParthKohli
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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 \({\bf \vec{F}} = m{\bf \vec{a}}\)
 one year ago
 one year ago
ParthKohli Group Title
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 \({\bf \vec{F}} = m{\bf \vec{a}}\)
 one year ago
 one year ago

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ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
@amistre64
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
\[\LARGE F=\frac{dp}{dt}\] Can we use this? :P
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Good idea! ^_^ No.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
force is equal to change in momentum divided by change in time
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
_ @ParthKohli
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
@amistre64 Isn't that just what DLS said...
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
i was just seeing if i remembered how to interpret the question correctly :)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
\[F = \dfrac{\Delta p}{\Delta t} = \dfrac{\Delta mv}{\Delta t}\]Is the above true? How did you derive it?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
Ft = p = mv2mv1
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
\[\vec F\Delta t=\Delta \vec p = m(\vec v_f\vec v_i)\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
not sure what you mean by how you derive it.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Does not seem so convincing. :
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
No, not that one. lol, got it
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
changing an objects momentum requires applying a force over a given interval of time, right?\[\vec F\Delta t=\Delta \vec p\]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
I'd rather say...\[\vec F \Delta t \propto \Delta \vec p\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
that style went out with Descartes :)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
I feel like a newbie here!
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
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 \(\vec F \Delta t = \Delta \vec p \).
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
You are indirectly asking for the proof of F=ma which only Isaac Newton knows.
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
I think @shubhamsrg can derive F=ma. Its a piece of cake for him.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Of course, of course!
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
over a given time interval, the mass "acquires" a new velocity, and therefore a change in momentum. does that sound about right?
 one year ago
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