Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Frequently I see the expression $$F = \frac{dp}{dt} = \frac{d}{dt}(mv) = \frac{dm}{dt}v + ma,$$ which can be applied to variable mass systems.

But I'm wondering if this derivation is correct, because the second law of Newton is formulated for point masses.

Furthermore if i change the inertial frame of reference, only $v$ on the right side of the formula $F = \frac{dm}{dt}v+ma$ will change, meaning that $F$ would be dependent of the frame of reference, which (according to me) can't be true.

I realize there exists a formula for varying mass systems, that looks quite familiar to this one, but isn't exactly the same, because the $v$ on the right side is there the relative velocity of mass expulsed/accreted. The derivation of that formula is also rather different from this one.

So my question is: is this formula, that I frequently encounter in syllabi and books, correct? Where lies my mistake.

share|cite|improve this question

You'll probably find the wikipedia article useful:

It is formulated so that $F+v_{rel} \frac{dm}{dt} = ma$, where $v_{rel}$ is the relative velocity of the mass being ejected to the center of mass of the body. This takes care of your question about reference frames, because $v$ will be the same in all frames. The term gets moved to the left side of the equation because $-v$ describes the velocity of the center of mass relative to the ejected matter.

share|cite|improve this answer

You can only apply Newton's second law to closed systems. But, since you are applying second law to a open system, you are getting contradictory results. The correct procedure for solving variable mass system, is by calculating the change in momentum and then equating it to

    Force = (change in momentum)/small time interval in which change occurred.

Here is a article on it that you can find useful, apart from the wikipedia article. See, this website.

share|cite|improve this answer

So my question is: is this formula, that I frequently encounter in syllabi and books, correct? Where lies my mistake.

The procedure $$\mathbf F = \frac{d\mathbf p}{dt} = \frac{d}{dt}(m\mathbf v) = \frac{dm}{dt}\mathbf v + m\mathbf a,$$

is based on the erroneous idea that the equation

$$ \mathbf F = \frac{d\mathbf p}{dt} $$

is valid for systems with variable mass. It keeps around probably also because of some relativity texts saying $\mathbf F = d\mathbf p/dt$ is more general than $\mathbf F = m\mathbf a$ because the former remains valid in relativity.

But in non-relativistic mechanics (what Newton himself was concerned with) this equation is valid only when the system does not lose or gain parts - which is quite a common case.

In the more rare case the system of interest (imaginary region in space) does acquire or lose material parts, the equation $$ \mathbf F = \frac{d\mathbf p}{dt} $$ is no longer valid. Another equation may be derived from this one applied to the whole supersystem (system + incoming/leaving parts). This can be done because the supersystem has constant mass. The new equation has the form

$$ \mathbf F + \mathbf F_{parts} = m\frac{d}{dt}\left(\frac{\mathbf p}{m}\right) $$ where $\mathbf F_{parts}$ is force on the system due to incoming/leaving parts and both $m$ and $\mathbf p$ are generally time-dependent. As you can see, both left-hand and right-hand side differs from the simple equation we began with.

share|cite|improve this answer
+1. It might be worth mentioning that using $\mathbf F = \frac{d \mathbf p}{dt}$ would mean that force becomes a frame-dependent quantity in a variable mass system. Most people who work with variable mass systems see this as anathema and use $\mathbf F = m\mathbf a$ instead. – David Hammen Oct 23 '14 at 20:02
Indeed, only the meaning of $\mathbf a$ may be a little obscure in some cases, like when mass is removed from the body while both external and parts' force vanishes - the center of mass may accelerate, but clearly $\mathbf a$ has to be zero. This is because $\mathbf a$ is not acceleration of the center of mass anymore (this is valid only for constant mass cases), but is defined as $\mathbf p/m$. – Ján Lalinský Oct 23 '14 at 20:16
  1. Whether a mass can be considered a point or not depends on the scale at which it is studied. The earth can be considered to be a point mass when we are studying its motion around the sun but not so when we are studying its own rotation. Newton's laws are applied to systems of many particles.

  2. Newton's second law says that the rate of change of momentum of a system is proportional to the applied force. We choose units in such a manner that the constant of proportionality is 1. With this definition, the equation $md\vec{v}/dt + \vec{v} dm/dt = \vec{F}$ makes sense with $\vec{v}$ being the velocity of the body in the same frame of reference in which other vectors are measured.

share|cite|improve this answer

protected by Qmechanic Oct 23 '14 at 18:13

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.