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momentum = mass * velocity

Differentiating both sides leads to

force = mass * acceleration

since the mass doesn't participate in the differentiation as it is constant.

Is this a sound definition, or a pointless use of differentiation? Or are there systems (like rockets) where the mass is also a variable wrt time?

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closed as unclear what you're asking by ja72, tpg2114, Waffle's Crazy Peanut, Dimensio1n0, Qmechanic Nov 9 '13 at 12:02

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

Of course it is sound--how is it any less so than acceleration being the rate of change of velocity (you've shown why the two are equivalent yourself!). And of course rockets have variable mass, but you've written a formula that works instantaneously; just integrate if mass changes. What are you actually asking? –  Rex Kerr May 6 '12 at 15:01
See this; I believe it's quite close. –  Noein May 6 '12 at 15:25
While differentiating with constant mass you get the second, but that is not always the case. –  ja72 Nov 29 '12 at 4:36

4 Answers 4

The prescription $\mathbf{p} = m\mathbf{v}$ only holds in non-relativistic contexts, while $F = \frac{\mathrm{d} \mathbf{p}}{\mathrm{d} t}$ is true in all contexts.

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Ooh that too... –  Manishearth May 6 '12 at 15:36

In the simplest case, the system consists of a single object acted on by a constant external force. Since it is only the object's velocity that can change, not its mass, the momentum transferred is

$$Δp = mΔv ,$$

which with the help of a = F/m and the constant-acceleration equation a = Δv/Δt becomes

$$Δp = maΔt$$

$$= FΔt .$$

Thus the rate of transfer of momentum, i.e. the number of kg·m/s absorbed per second, is simply the external force,

relationship between the force on an object and the rate of change of its momentum; valid only if the force is constant.

This is just a restatement of Newton's second law, and in fact Newton originally stated it this way.

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$\newcommand{\d}[1]{\frac{\mathrm d #1}{\mathrm dt}}$

Well, yes. Rockets work on this very principle. By throwing out some mass in the form of exhaust gases in the opposite direction, the rocket gains momentum (gain since it is thrown out in the opposite direction). This can be equated to a force.

If mass changes, $$F=\d p=m\d v+v\d m=ma+v\d m$$

Note that in empty space, the rocket has no net force on it. So we get $$0=F=m\d v+v\d m=ma+v\d m\implies ma=-v\d m$$

This lets the rocket accelerate without a force. If there was a gravitational force or something, we get: $$\frac{Gm_1m_2}{r^2}=m\d v+v\d m=ma+v\d m$$

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If a piece of a rocket is detached, it does not mean there is a force acting on the the rocked, does it?. –  Vladimir Kalitvianski May 6 '12 at 15:25
@VladimirKalitvianski: No, it means that the rocket accelerates. I'll clarify that, one sec. Thanks! –  Manishearth May 6 '12 at 15:31

In addition to what everyone else said above, the $F = \frac{dp}{dt}$ relation also makes the link between Newton's third law and conservation of momentum completely obvious, even moreso since there are cases where Newton's third law doesn't really apply, but conservation of momentum is always true.

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