Deriving force from momentum using d(mv)/dt

[Mohmmad Maaitah]

Homework Statement

How did we Derive force from momentum

Relevant Equations

F=ma
P=mv
F=dp/dt

How did the d(mv)/dt become the other two?
Can someone explain how do we derive for new formulas in physics?

 

[malawi_glenn]

Product rule of differentiation

 

[Mohmmad Maaitah]

Product rule of differentiation

Product rule of differentiation

Thank you I just missed it.

 

[haruspex]

You need to be very careful using this $dp=m.dv+v.dm$ form.
Consider e.g. a water tank mass m speed v leaking water at rate $\rho$. If we blindly apply that formula we get $\dot p= m\dot v+v\dot m=ma-v\rho$. Since there appear to be no external forces we might conclude $ma=v\rho$, so the tank is accelerating.
The flaw is that the leaking water carries momentum away with it, which should be represented as an external force $-v\rho$. Plugging that in gives $a=0$.

My advice: don’t use it.

 

[Frabjous]

You need to be very careful using this $dp=m.dv+v.dm$ form.
Consider e.g. a water tank mass m speed v leaking water at rate $\rho$. If we blindly apply that formula we get $\dot p= m\dot v+v\dot m=ma-v\rho$. Since there appear to be no external forces we might conclude $ma=v\rho$, so the tank is accelerating.
The flaw is that the leaking water carries momentum away with it, which should be represented as an external force $-v\rho$. Plugging that in gives $a=0$.

My advice: don’t use it.

So why do rockets work? I do not understand the point you are trying to make.

 

[haruspex]

So why do rockets work? I do not understand the point you are trying to make.

You can solve rocketry questions using conservation of momentum quite easily without using that equation explicitly. In time $\delta t$ a mass of fuel $\mu \delta t$ is ejected at speed $u$ relative to the craft of remaining mass $m$. $m\delta v=u\mu\delta t$. This leads to $v=u\ln(\frac{m_0}{m-\mu t})$ as desired.

You can argue that I am effectively using the equation, but quoting the equation and then trying to apply it has pitfalls, such as the one I outlined. In the rocketry context, what exactly is the system which m represents?

 

[erobz]

@Frabjous

Just for clarity, correcting post #11 using the proper way to capture the accumulation of momentum in the ejecta from @anuttarasammyak:

$$\sum F_{ext} = \frac{d}{dt} \left( p_{\rm{rocket}} + p_{\rm{fuel}} + p_{\rm{ejecta}} \right)$$

Mass of the rocket is constant $M_r$

The mass of fuel lost from the rocket is gained by the ejecta $\frac{dM_e}{dt} = -\frac{dM_f}{dt}$

The velocity of the ejecta in the rest frame is $v_{e/O} = v_r - v_{e/r}$ and is constant after being ejected, but the momentum of the ejecta is accumulating, hence:

$$\sum F_{ext} = \frac{d}{dt} \left( M_r v_r + M_f v_r + \int \frac{dM_e}{dt}( v_r - v_{e/r}) ~dt \right)$$

Applying the derivative:

$$\sum F_{ext} = \overbrace{\left( M_r\frac{dv_r}{dt} \right)}^{\rm {rocket}} +\overbrace{\left(M_f\frac{dv_r}{dt} + \frac{dM_f}{dt}v_r\right)}^{\rm{fuel}}+ \overbrace{ \frac{dM_e}{dt}( v_r - v_{e/r})}^{\rm{ejecta}} $$

$$\sum F_{ext} = \left( M_r + M_f \right) \frac{dv_r}{dt}+ \cancel{ \left( \frac{dM_f}{dt}v_r + \frac{dM_e}{dt}v_r \right)}^0 -\frac{dM_e}{dt} v_{e/r}$$

The Rocket Equation:

$$ \sum F_{ext} = \left( M_r + M_f \right) \frac{dv_r}{dt} + \frac{dM_f}{dt} v_{e/r}$$

The issue comes up often about taking $F = \frac{dp}{dt} = \frac{d}{dt} \big( mv \big)$.

We (several posters) took the time not too long ago to figure out how to apply it for "The Rocket Equation". I think this is what @haruspex means by "you have to be careful". It wasn't immediately obvious how to get it to work out.

The whole exchange:
https://www.physicsforums.com/threa...oblem-confused-about-newtons-2nd-law.1050482/