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If the rate of change in velocity in a particle (of mass m) caused due to a force F is dv/dt, then

F = m dv/dt

It may be argued that this is how we define force. But my question is:

Can there be any kind of force, which is so strange that no matter how we write the formula for the force, we will find that F = m dv/dt fails in at least some cases?

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closed as unclear what you're asking by Qmechanic Feb 8 '18 at 10:23

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  • $\begingroup$ I close this post as unclear what you're asking partly since title question and main body question seem different. To reopen this post (v1), consider to align them. $\endgroup$ – Qmechanic Feb 8 '18 at 10:24
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If $F = \frac{d(mv)}{dt}$ fails in just one example, then you have disproven Newton's 2nd law by falsification. Then you will get the next Nobel price.

The thing is that Newton's 2nd law is not provable. But after infinitely many experiments, no one has ever been able to disprove it with normal size scales and speeds. We therefore trust it to be always true. We can't prove that a pen will fall when we let go - but we are still pretty sure it will, because it always does.

Newton's 2nd law is therefore called a law of nature, and it is one of the most established ones in physics, I would say.

So the answer to your question is without a doubt, although I can't prove it: No.

As the other answers mention, though, note that the mass $m$ is included in the derivative in the complete description of Newton's 2nd law, since it was originally formulated by I. Newton as the change in momentum:

$$F=\frac{dp}{dt}=\frac{d(mv)}{dt}$$

So, we are talking about a change in momentum and not only a change in velocity. And this is what my explanation above coves.

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    $\begingroup$ It's the Nobel Prize, not the Noble price. (I can't edit this answer to correct two characters.) Someone won the Nobel Prize in Physics for exactly that almost 100 years ago, for "for his services to theoretical physics, and especially for his discovery of the law of the photoelectric effect". $\endgroup$ – David Hammen Feb 8 '18 at 10:26
  • $\begingroup$ On a less sarcastic note, while scientific conjectures are easily disprovable by one observation (plus confirmation), no scientific theory is provable. $\endgroup$ – David Hammen Feb 8 '18 at 10:29
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The formula you mentioned is for a special case where the mass doesn't change. The general case is, $$\vec{F}=\frac{d \vec{p}}{dt}=\vec{v}\frac{dm}{dt} + m\frac{d \vec{v}}{dt}$$ where $\vec{p}$ is the momentum of the object.

For example, a missile loses mass by burning fuel, so your formula will not suffice to describe its motion.

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This formular holds only true for time-constant masses. The original formular is $$\vec{F} = \dot{\vec{p}}$$ which, taking the derivative, leads with to $$\vec{F} = m\dot{v}+\dot{m}v$$ So $$F=m\dot{v}$$ is only true if the second term is zero, so if $\dot{m}$ is zero.

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  • $\begingroup$ Thank you for pointing this out. But if mass is constant, then is F = m dv/dt always true? $\endgroup$ – Avinash Feb 8 '18 at 9:22
  • $\begingroup$ $\vec{F}=\dot{\vec{p}}$ is always true. But for example the momentum of a massless particle is not $p=mv$, but $p=\frac{h}{\lambda}$. So, if the mass is constant and non-zero, then it is true. $\endgroup$ – Densch Feb 8 '18 at 9:28

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