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?


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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.