# Momentum and its relation to Force

Sir Isaac Newton First proposed that $\vec{F} = \frac{d\vec{p}}{dt}$ before he proposed that $\vec{F} = m\vec{a}$ from the first formula it is easy to prove the second.

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

$\vec{F} = \frac{d(m\vec{v})}{dt}$

$\vec{F} = m\frac{d\vec{v}}{dt}+\vec{v}\frac{dm}{dt}$

From the definition of acceleration and assuming mass to be a constant,

$\vec{F} = m\vec{a}$

However first of all the books always prove that $\vec{F} = m\vec{a}$ as because acceleration is directly proportional to force and inversely proportional to mass. From $\vec{F}=m\vec{a}$ they then prove that $\vec{F} = \frac{d\vec{p}}{dt}$

I have a few problems with this. The book starts by saying the following.

$\vec{F} = m\vec{a}$

$\vec{F} = m\frac{d\vec{v}}{dt}$

They say since mass is a constant it can be moved into the derivative. (Remember the fact that mass is a constant)

$\vec{F} = \frac{d(m\vec{v})}{dt}$

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

From here they conclude that hence we can use the product rule and find force for varying mass (e.g. rockets using up fuel)

$\vec{F} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt}$

this doesn't make sense as in the proof they said mass is constant and that is why u can have $m\frac{d\vec{v}}{dt} = \frac{d(m\vec{v})}{dt}$. If mass was varying we can't do that. So can someone give me the right proof.

Secondly and lastly, how did Newton show $\vec{F} = \frac{d\vec{p}}{dt}$ before he showed Newton's Second Law. Whats the proof?

6 years, 8 months ago

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