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Momentum is the time derivative of force. $F = \frac{dp}{dt}$.
But isn't impulse $J=Ft$, $F=\frac{J}{t}$?
Can impulse be the time derivative of force too, or is it just defined as an integral?

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Usually, the term impulse means the difference in momentum:

$$\vec J=\Delta \vec p$$

Force is the time derivative of momentum:

$$\vec F=\frac{d\vec p}{dt} \Leftrightarrow d\vec p=\vec F \;dt$$

In a scenario where you consider a finite amount of transfered momentum, or when $d\vec p\approx \Delta \vec p$, then you can rewrite as:

$$\vec J=\vec F \;\Delta t$$

$\Delta t$ is often just denoted $t$ to mean the time duration. So there you have it. In the case of non-constant force, you would how to add up all the (maybe infinitely many) small impulses, which gives the usual integral version of the formula:

$$\vec J=\vec {J_1}+\vec {J_2}+\vec {J_3}+\cdots=\vec {F_1} \;\Delta t_1+\vec {F_2}\;\Delta t_2+\vec {F_3} \;\Delta t_3+\cdots\quad\Leftrightarrow\\ \vec J=\int \vec F \;d t$$

Impulse and momentum are closely related, but not exactly the same. Rather, one is the difference in the other.

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You have written wrong formula for impulse.it is not just time there instead it is change in time(delta t). So, now if you look over the final formula that is F=J/delta t(not just t) it is same as F=dp/dt as, J =dp. So you got nothing new like you observed in your question . You got same equation.impulse is just a fancy word for change in momentum(generally big changes not delta p) as it is used often.

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  • $\begingroup$ Your formula should not be written that way. The correct formula is $J=\int_{t_1}^{t_2} F.dt$ $\endgroup$ – harshit54 Dec 25 '18 at 18:06

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