## ParthKohli 2 years 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 $${\bf \vec{F}} = m{\bf \vec{a}}$$

1. ParthKohli

@amistre64

2. DLS

$\LARGE F=\frac{dp}{dt}$ Can we use this? :P

3. ParthKohli

Good idea! ^_^ No.

4. amistre64

force is equal to change in momentum divided by change in time

5. DLS

-_- @ParthKohli

6. ParthKohli

@amistre64 Isn't that just what DLS said...

7. amistre64

i was just seeing if i remembered how to interpret the question correctly :)

8. ParthKohli

$F = \dfrac{\Delta p}{\Delta t} = \dfrac{\Delta mv}{\Delta t}$Is the above true? How did you derive it?

9. amistre64

Ft = p = mv2-mv1

10. amistre64

$\vec F\Delta t=\Delta \vec p = m(\vec v_f-\vec v_i)$

11. amistre64

not sure what you mean by how you derive it.

12. ParthKohli

Does not seem so convincing. :-|

13. amistre64

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

No, not that one. lol, got it

15. amistre64

changing an objects momentum requires applying a force over a given interval of time, right?$\vec F\Delta t=\Delta \vec p$

16. ParthKohli

I'd rather say...$\vec F \Delta t \propto \Delta \vec p$

17. amistre64

that style went out with Descartes :)

18. ParthKohli

I feel like a newbie here!

19. ParthKohli

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

You are indirectly asking for the proof of F=ma which only Isaac Newton knows.

21. DLS

I think @shubhamsrg can derive F=ma. Its a piece of cake for him.

22. ParthKohli

Of course, of course!

23. amistre64

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

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