0
$\begingroup$

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 $$m_1v_1=m_2v_2$$

$\endgroup$
  • $\begingroup$ The second law isn't $F = m a$, but $\sum F = \dot{p}$ $\endgroup$ – ja72 Oct 3 at 12:40
0
$\begingroup$

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.

$\endgroup$

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.