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Is there ever a situation where the distinction between $F = m \frac{dv}{dt}$ and $F = \frac{dp}{dt}$ is important? I can't think of a situation where one is true and not the other (assuming only conservation of momentum).

Edit: Obviously it is important to take a changing mass into account (e.g. for a rocket) when you're considering a full time evolution, i.e. $F(t) = m(t) \frac{dv}{dt}$ (or in the relativistic case, perhaps something like $F(t) = m(t) \frac{d}{dt} \left( \frac{p}{m} \right)$ with $m$ the rest-mass). And perhaps there is a nontrivial relationship between the rate of change of mass, and the forces being exerted (again, e.g. with a rocket --- where the mass loss is tied to the propulsion). What is not clear is that there should ever be a $F = v\frac{dm}{dt}$ term.

Edit 2: My understanding of the solution:
There should not be a $dm/dt$ term, as pointed out by @garyp. The change in momentum expression is, however, more accurate because $p \neq mv$ in general (e.g. in relativistic cases, or when considering massless systems). It would seem that either one must take the caveat that $dp/dt$ cannot be used for mass-varying systems, or take the much less conceptual or aesthetically pleasing expression that $F = m \frac{d (\gamma v)}{dt}$ (which still only applies to classical systems).

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There is never a $F=v\, dm/dt$ term if $v$ is the velocity of the system. My answer below tries (poorly, it seems) to make that point. Try this link. Very explicit, I think. –  garyp May 22 at 11:56
Addendum: Newton's Second Law applies only to closed (constant mass) systems. –  garyp May 22 at 12:03

5 Answers 5

up vote 10 down vote accepted

In classical mechanics Newton's second law applies only to constant mass systems. In those cases there's no difference between $F=ma$ and $F=\mathrm{d}p/\mathrm{d}t$. However, in special relativity the latter is valid, but the former is not. A relativistic definition of momentum is required: $p=\gamma m v$.

Some details

[rework of my original answer] Some of the responses given so far answer the OP, but have weaknesses that might lead to misconceptions about Newton 2. I'll try to address the issue.

Some of the answers so far are not entirely correct if by $p$ is meant $mv$ of the rocket. Newton's second law is valid only for constant mass systems. $F=\mathrm{d}p/\mathrm{d}t$ leads to the rocket equation by accident if the propellent is exhausted in a direction opposite to the direction of motion. If you are very careful about what you mean by $p$ a correct analysis can be made, but $p=m_\mathrm{rocket}v_\mathrm{rocket}$ does not work.

This Wikipedia entry. is the clearest statement of that fact that I've found.

To see why, consider a system comprising the rocket plus its remaining fuel and remaining propellant, which is what I think the other responders intend. (I may be wrong, but if so, they should clarify what exactly is their system.)

Imagine the system moving at constant velocity, and having two thrusters pointing perpendicular to the direction of motion, and directly opposite each other, and producing identical constant thrust by expelling exhaust gas. The thrust force of the two are equal and opposite. So the net force on the rocket is zero. Then $F_{net}=0$ and $\mathrm{d}v/\mathrm{d}t = 0$, and $v \neq 0$ and $\mathrm{d}m/\mathrm{d}t \neq 0$. Blindly applying $F=\mathrm{d}p/\mathrm{d}t$ leads to $0=v\,\mathrm{d}m/\mathrm{d}t$, a contradiction which can only be resolved if the mass is unchanging.

Comments on some comments

The system is losing momentum by losing mass, but it's velocity is not changing: there is no net force on the system. Momentum is conserved in the closed system consisting of the rocket and the exhausted propellant. The system consisting of the rocket plus the fuel and propellant not yet exhausted (fuel remaining in the tank) is an open system. Conservation of momentum does not apply to open systems.

Careful analysis of a variable mass system lead to $$ \vec{F}_\mathrm{ext}=\vec{u}\frac{\mathrm{d}m}{\mathrm{d}t} + m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}$$ where $\vec{u}$ is the velocity of the mass leaving the system relative to the velocity of the system, and $\vec{F}_\mathrm{ext}$ is the external force on the system. For a rocket $\vec{F}_\mathrm{ext}=0$, and $$ 0=\vec{u}\frac{\mathrm{d}m}{\mathrm{d}t} + m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}$$ This is not the same as $$ 0=\vec{v}\frac{\mathrm{d}m}{\mathrm{d}t} + m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}$$ where $\vec{v}$ is the velocity of the rocket.

Trying to write $F = m\,\mathrm{d}v/\mathrm{d}t + v\,\mathrm{d}m/\mathrm{d}t$ for a variable mass system is not correct. here's a nice discussion whose first sentences are "In mechanics, a variable-mass system is a collection of matter whose mass varies with time. Newton's second law of motion cannot directly be applied to such a system because it is valid for constant mass systems only."

Aeronautical engineers know all this. It's only physicists who are confused. I checked some books. Symon's and John R. Taylor's classical mechanics texts get it right, as does Halliday, Resnick, and Walker.

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Er...no. You have to compute the thrust properly for each engine and then form the vector sum of the forces just as you have to form vector sum of all other kinds of forces. –  dmckee May 21 at 4:39
@garyp Isn't your example violating conservation of momentum? After loss of mass, the momentum in the direction of travel has gone down, which means velocity must increase. so, instead of 0=v.dm/dt, I think we get 0=v.dm/dt+m.dv/dt, which makes sense –  tpb261 May 21 at 6:07
@dmckee Yes, and in my example there are two thrusters operating, and they are pointed exactly opposite, so that the net force is zero. –  garyp May 21 at 11:30
@zhermes A better conclusion is that $\vec F = d\vec p/dt$ for a closed system. If you want to consider an open system, you have to be careful. –  rob May 21 at 15:50
@garyp "there is no net force on the rocket, and its speed does not change". Confused. Considering the closed system of the rocket and the masses that are flying off (in opposite directions and perpendicular to direction of motion of rocket), then 1. there is no change in momentum in horizontal direction - This is clear. 2. In the direction of motion, if velocity hasn't changed, then there is clearly a reduction of momentum (it becomes, (M-2*m)*v from M*v). So, to compensate, v must increase. I am considering the closed system, so where is the flaw? How is v remaining constant for you? –  tpb261 May 22 at 3:47

Yes, and a rocket is a good example. In $$F=m\left( \frac{dv}{dt} \right)$$ you assume mass is constant. If mass is variable, like a rocket burning off fuel, you need to account for it, $$F=\frac{dp}{dt}=\frac{d}{dt}\left( mv \right)\; =\; m\frac{dv}{dt}\; +\; v\frac{dm}{dt}$$ where the m is decreasing. If you fiddle with this you might get the Rocket Equation. Start by examining the motion with m replaced by $$\left( m+\Delta m \right)$$ where the delta m can be negative, and use some ideas from conservation of momentum.

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It seems like the $dm/dt$ term will expressly not lead to a force. Consider a 'rocket' as two masses connected by a string; cutting the string (decreasing the 'rocket's mass) will obviously not accelerate it. –  zhermes May 21 at 2:59
In the case of rocket motion the force most often analyzed is that of propulsion, wherein you are not passively cutting a string but flinging the reaction mass backward as hard as you can manage. Both terms are necessary to understand the dynamics of the system and trying to use $F = m\dot{v}$ is fatal. –  dmckee May 21 at 3:06
It is probably worth looking for the meaning of the two terms. The m dv/dt term is obviously ma with instantaneous mass of the rocket. But what is vdm? Momentum of the fuel spent in dt? That is flung away from the rocket? I think this is why rocket equation derivations start with a momentum equation - so that these are defined in the beginning. Hmmm. –  C. Towne Springer May 21 at 5:12
@C.TowneSpringer This is why I wrote a regrettably long and unpopular answer of my own. You do answer the OP's question, but the missing details leave lingering questions. –  garyp May 21 at 13:13
@garyp I taught enough high school physics and math that I try to only answer if I think I understand an OP's question and intent, and I don't have to go to my books. I went further down the rocket equation route then erased when it got complicated. I risk making errors so I keep it simple. That and an old habit of asking questions or posing problems at the end. I liked yours though. –  C. Towne Springer May 21 at 14:04

In electrodynamics, the Lorentz force on a moving charge $q$ $$ \vec F = q \left( \vec E + \vec v \times\vec B \right) $$ can be rewritten as the force per unit volume $dV$ on charge density distribution, $\rho$, with current density $\vec J$: $$ \vec F = dV\ \vec f = dV \left( \rho\vec E + \vec J \times \vec B \right). $$ You can use Maxwell's source-term equations $$ \begin{align*} \vec\nabla\cdot\vec E &= \frac{\rho}{\epsilon_0} & \vec\nabla\times\vec B - {\epsilon_0\mu_0}\frac{\partial\vec E}{\partial t} = {\mu_0 \vec J} \end{align*} $$ to eliminate the charge and current distributions and write this force density $\vec f$ solely in terms of the fields. It takes some doing, but you wind up with $$ \vec f = \vec\nabla\cdot\mathbf{T} - \epsilon_0\mu_0 \frac{\partial \vec S}{\partial t} $$ where $\vec S$ is the Poynting vector and $\mathbf T$ is the Maxwell stress tensor. This shows that you may have momentum entering and leaving the fields in a region of space, even when there isn't any mass around to accelerate.

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For over 200 years the equations of motion enunciated by Newton were believed to describe nature correctly, and the first time that an error in these laws was discovered, the way to correct it was also discovered. Both the error and its correction were discovered by Einstein in 1905.

Newton's Second Law, which we have expressed by the equation
was stated with the tacit assumption that m is a constant, but we now know that this not true, and that the mass of a body increases with velocity.

So, your two equations differ to greater extent for the situations in which a body movies with greater velocity.

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Just a little bit of addition:

Newton's 2nd Law is F = ma = m dv/dt

This is correct for constant mass and in non-relativistic context.

However since mass can be assumed constant in Newton's original formulation is equivalent to:

F = dp/dt

And this form is correct for both variable mass AND in a relativistic context (of course with the appropriate interpreations)

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I think we've already established that $F = dp/dt$ does not apply for variable mass, while $F = ma$ does. –  zhermes May 23 at 2:13
Ok but this half-correct also, for example relativistic mass is variable mass (depending on velocity), so variable mass depending on velocity for example is indeed covered, but i agree not all cases of variable mass are covered –  Nikos M. May 23 at 10:26

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