Is change in momentum over time equal to average force? Hi guys I’m trying to get my head round some basic physics I understand the change in velocity x mass = change in momentum
Using this does change in momentum/ time = Mass x acceleration as velocity/time = Acceleration
So my understanding from this is will we get an average force? As F=ma ?
 A: Alright, let us analyse this question.
What we know is that p = mv (p being momentum, m being mass and v being velocity)
So, $$\frac{p}{t} = \frac{mv}{t}$$
Which is, $$\frac{p}{t} = m(\frac{v}{t})$$
Now, $\frac{v}{t} = acceleration$
$$\frac{p}{t} = ma$$
As you said, F = ma $$\frac{p}{t} = F$$
Which is $$\Delta p = F$$
So yes, change in momentum is equal to the average force ( $\Delta p = F$ )
A: The changes (differences) you mention should really be differentials, so replace $\Delta p=\Delta (mv)$ with:
$$\mathrm{d}p=\mathrm{d}(mv)$$
which for constant mass $m$ becomes:
$$\mathrm{d}p=m\mathrm{d}v$$
Similarly:
$$a=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}$$
becomes:
$$a(t)=m\frac{\mathrm{d} v}{\mathrm{d }t}$$
Replacing the $\Delta$ with $\mathrm{d}$ means we shift from averages to instantaneous changes, used in Calculus to compute ever changing quantities like $a(t)$.  Only if quantities like $v$ are independent of time $t$ can differences $\Delta v$ be used accurately.
Read more on the differences between $\Delta$ with $\mathrm{d}$ on my relevant webinar.
