We know, that for a general problem regarding variable mass, using the formula $F=\frac{dP}{dt} = M\frac{dv}{dt}+ v\frac{dM}{dt}$, can be dangerously invalid, as this is generally incorrect.
According to Wikipedia, and some other texts, the correct formula would be the following :
$$F_{ext} \space\space+ \space v_{rel}\frac{dm}{dt} = m\frac{dv}{dt}$$
This equation is true for both cases – when mass is gained and when mass is lost. Over here, if the system is set up in such a way where $v_{rel} = -v$, then we can say that $F_{ext}=\frac{dmv}{dt}$ is true. However this completely depends on the relative velocity.
Now, here is where I have a doubt regarding this equation for decreasing mass in a body :
According to Wikipedia, for a decreasing mass :
$p_i = mv$
$p_f = (m-dm)(v+dv) \bf – udm$ (assuming the mass is released in opposite direction to $v$)
Then, $p_f-p_i = mdv – (v+u)dm$
$F_{ext} + (v+u)\frac{dm}{dt} = m\frac{dv}{dt}\space .\space\space$ Here $u-(-v) = u+v = v_{rel}$
So, here we got the correct equation, and if $u=-2v, $ then we get our $F=\frac{dmv}{dt}$ equation.
But, now let us consider, a similar system where mass is decreasing, but the decreasing mass is being ejected in the same direction of motion. How do I derive this?
My approach :
$p_i = mv$
$p_f=(m-dm)(v+dv) \bf +udm$
$p_f-p_i = mdv+udm-vdm = mdv + v_{rel}dm. \space\space$ Here $u-v = v_{rel}$
Hence we get :
$F=m\frac{dv}{dt} + v_{rel}\frac{dm}{dt}$
This is different from my original formula. However, I can rewrite this as :
$$F+ v_{rel}\frac{-dm}{dt}=m\frac{dv}{dt} $$
Hence, I can say that if the motion of $m$ and $dm$ is in the same direction, the equation for force remains the same, except for a small minus sign in front of $|v_{rel}|$ i.e. the mod of relative velocity, if the mass is decreasing, and a plus sign, if the mass is increasing.
Hence, if $m,dm$ travel in the same direction, then:
$$F_{ext} \space\space \pm |v_{rel}|\frac{dm}{dt} = m\frac{dv}{dt}$$
$-$ when mass decreases; $+$ when mass increases.
Is this derivation correct?
Best Answer
Well let's put your result through a test.
If you have an object where no other force is present, and it starts ejecting mass, it will accelerate towards the opposite direction, which is pretty intuitive. so
$$m\frac{dv}{dt} < 0$$
If we, in another scenario, have that same object but it's showered with particles which will get absorbed, it will accelerate towards the direction of the velocity of the particles. so in this case:
$$m\frac{dv}{dt} > 0$$
now, this statement:
so if there is an observer on the object emitting mass, it will see the particles getting away, so it already implies the change in sign if you will.
The sign that you are putting cancels out the effect that is intended.
Think of it like this.
car A has velocity +V, and car B has velocity -V. the person in car B sees that car A is moving with +2V, but A sees that car B is moving with velocity -2V.
So if mass is incoming and attaching, the observer on the object sees the shower of particles travelling towards the direction of motion of both, meaning that it sees a positive reference velocity in it's values. So this expression still holds:
$$F_{ext} + v_{rel}\frac{dm}{dt}=m\frac{dv}{dt}$$
One must take into account that relative velocity can posses negative values when writing the scalar form to solve it.
Also, the rate of change of mass is positive if the mass is increasing and negative if the mass is decreasing.
So to sum up:
If particles are moving at speeds U and the object is moving at speed V:
$$v_{rel}\frac{dm}{dt}>0$$
$$v_{rel}\frac{dm}{dt}<0$$