I know if taking the integral of $F=ma$, then I can get $p=mv$. I'm weak in calculus, so I wondered how to do this exactly.

Is there anything wrong in my logic below?

\begin{align}\int F\left(t\right)\,{\rm d}t&=\int ma\,{\rm d}t \\ &=m\int a\left(t\right)\,{\rm d}t \end{align} where by definition, $\int F\,{\rm d}t=p$ and $\int a\,{\rm d}t=v$. Plug these in above, I get, $$p=mv$$

Equivalently, how to derive from $m_1a_1=m_1a_1$ to $m_2v_2=m_2v_2$?

Can I do

Since $$a_1=\frac{{\rm d}v_1}{{\rm d}t_1},\quad a_2=\frac{{\rm d}v_2}{{\rm d}t_2}\tag{1}$$ and $$m_1a_1=m_2a_2\tag{2}$$ Plug (1) into (2), then get $$ m_1\frac{{\rm d}v_1}{{\rm d}t_1}=m_2\frac{{\rm d}v_2}{{\rm d}t_2}\tag{3} $$ where ${\rm d}t_1={\rm d}t_2$. Then plug this into (3) to get, $$m_1{\rm d}v_1=m_2{\rm d}v_2$$ With $v_1={\rm d}v_1$ and $v_2={\rm d}v_2$, therefore $$m_1v_1=m_2v_2$$

  • $\begingroup$ The second law isn't $F = m a$, but $\sum F = \dot{p}$ $\endgroup$ – ja72 Oct 3 at 12:40

You are showing how force and momentum are related but are not showing momentum is conserved. I will show you a proof for 2 particles and you can generalize it for more.

Consider that $F_{12}$ and $F_{21}$ are the forces acting from 1 on 2 and vice versa. Then (using what you have derived that $F=\dot p$)

$$F = F_{12} + F_{21} + F_{ext} = \dot p$$

Where the last term is from any external forces. Then invoking the 1st law,if the frame is inertial that last term is zero.

Additionally, invoke the 3rd law st $F_{12} = -F_{21}$.

Then $\dot p = 0$ or integrating both sides $p = C $ where $C$ is some constant. So momentum is conserved.


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