# Why is force described as rate of change of momentum?

momentum = mass * velocity

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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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. – Mussri 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

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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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

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

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