Momentum and its relation to Force

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

F⃗=dp⃗dt\vec{F} = \frac{d\vec{p}}{dt}

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

F⃗=mdv⃗dt+v⃗dmdt\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,

F⃗=ma⃗\vec{F} = m\vec{a}

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

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

F⃗=ma⃗\vec{F} = m\vec{a}

F⃗=mdv⃗dt\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)

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

F⃗=dp⃗dt\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)

F⃗=dp⃗dt=d(mv⃗)dt=mdv⃗dt+v⃗dmdt\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 mdv⃗dt=d(mv⃗)dtm\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 F⃗=dp⃗dt\vec{F} = \frac{d\vec{p}}{dt} before he showed Newton's Second Law. Whats the proof?

Note by Saad Haider
7 years, 8 months ago

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