Trolley with mass of $m_0=1 \ kg$ is moving without friction on the railway track. It is raining so there is a constant mass flow of water $\Phi_m=0.1\ kg/s$. Constant force $F=0.1 \ N$ is accelerating the trolly horizontally.

What is the velocity at time $t$ if the trolly is stationary initially ?

I tried two different aproaches and got different results. I graphed the both functions and noticed that both were similar at $t=0$.

1. Newton's law

$$F=m(t)a$$ $$F=(m+\Phi t) \frac {dv}{dt}$$ $$\int dv=F \int\frac{dt}{m+\Phi t}$$

..integrated 0 to v; and 0 to t

$$v=\frac F\Phi \ln(m+\Phi t)$$

2. Momentum

$$(m+\Phi t)v - 0 = \int Fdt$$ as $F=const.$

$$v=\frac{Ft}{m+\Phi t}$$

Am I missing some concept behind differential equations?

  • $\begingroup$ One way of deciding which of the two is right would be to look at the limits $t\to \infty$ and $t\to 0$ :) $\endgroup$ – Danu Aug 30 '14 at 15:31
  • $\begingroup$ You've not applied your initial condition in the first attempt: it should be $v=\frac F\Phi \ln(1+\frac{\Phi t}{m})$ $\endgroup$ – Holographer Aug 30 '14 at 15:50
  • 1
    $\begingroup$ As you think about this: it is important to understand that Newton's second law is valid only for systems whose mass is constant. It is not valid for systems whose mass is changing. $\endgroup$ – garyp Aug 30 '14 at 16:38
  • $\begingroup$ Yes, it seems I've missed that fact. $\endgroup$ – xan Aug 30 '14 at 16:40
  • $\begingroup$ I should mention that the restriction to constant mass systems is frequently overlooked, and confusion is common. Most textbooks mention it explicitly, but it's easy to gloss over. I particularly like this very clear statement on Wikipedia. $\endgroup$ – garyp Aug 30 '14 at 17:14

Newton's 2nd law in differential form (ignoring vectors) is

$$F_\text{net}=\frac{dp}{dt}=\frac{d}{dt}\left(p_\text{train} + p_\text{water in trolley} + p_\text{rain just hitting trolley}\right). \tag{1}$$

You must take into account the change in momentum of the rain that occurs when it falls into the trolley and accelerates up to the speed of the train. This is in addition to knowing that the force exerted on the train acts on a body whose mass is increasing. (There are subtleties associated with this wording; see comments below.) It seems in method 1 you're not accounting for everything.

Now, Eqn. 1 can be re-written as

$$dp = F_\text{net}\,dt$$

then you can integrate both sides to get the general form of Newton's second law in integral form. After that, take into account the momentum of the rain + train at both the initial and final times. Luckily $p_\text{i,rain}=0$, so this simplifies some things.

| cite | improve this answer | |
  • 1
    $\begingroup$ Careful how you express this. Newton's second law is only valid for constant mass systems. This applies to both forms of the law. Applying the product rule as you suggest leads to $F=m\frac{dv}{dt}$. $F=ma$ is not the simplified form, it is the correct non-relativistic form. $F=\frac{dp}{dt}$ is useful in special relativity. $\endgroup$ – garyp Aug 30 '14 at 16:41
  • $\begingroup$ Applying the product rule leads to $F=m\frac{dv}{dt}+v\frac{dm}{dt}$. It's valid for systems with changing masses. This is used, for example, in the context of rockets. $\endgroup$ – BMS Aug 31 '14 at 4:39
  • 1
    $\begingroup$ I'm afraid that that equation is not valid. It usually does not work in the context of rockets (there's a case in which it works accidentally). See this SE answer and the Wikipedia link referenced in that answer. $\endgroup$ – garyp Aug 31 '14 at 11:31
  • $\begingroup$ So you're saying that one can apply $F=dp/dt$ as long as the system I'm considering consists of the train and the rain that eventually falls into the train. I can buy that; definitely more subtle than I realized. $\endgroup$ – BMS Aug 31 '14 at 16:44

You should be careful, since you need to take account of the force that the rain exerts on the trolley, or the momentum of the rain. Your second approach does this rather nicely, (with the assumption that the rain falls vertically, and hence doesn't contribute to the initial momentum).

In the first approach, you could redo it to add the force that the system must exert on the rain that is just landing in the trolley, and hence (by NIIL) the extra resistance that provides.

| cite | improve this answer | |
  • $\begingroup$ That's a very intuitive way to go about it, thank you! $\endgroup$ – xan Aug 30 '14 at 16:44

It looks like in this problem the rain moves at whatever $v$ the train is currently at, giving an infinite amount of energy as $lim_{t\rightarrow \infty}$. However it doesn't look like your using energy anyway, just be careful if you do.

As BMS said, use the product rule. This gives you $F= {\Phi}{v(t)}+\frac{dv}{dt}{(m+\Phi)}$ then subtract $\Phi v(t)$ to the other side and then move things around and integrate over $t$ to get $$\frac{1}{m+\Phi}\int{dt}=\int\frac{dv}{F-\Phi v(t)}$$ I ended up with $$v=\frac{F}{\Phi}\left(1-exp\left[\frac{-\Phi t}{(m+\Phi)}\right]\right)$$ Not too pretty with the $\Phi$ in 3 places but it satisfies $v(0)=0$ and has a convergent final velocity which seems to appeal to my intuition.It could be a useful exercise to solve this problem but assuming instead that the rain always falls vertically.

| cite | improve this answer | |
  • $\begingroup$ Well acctually in the question it was stated that the rain is falling vertically, I forgot to mention it. $\endgroup$ – xan Aug 30 '14 at 23:19
  • $\begingroup$ The right hand side of your first inline equation is incorrect. It shoud be $m+\Phi t$. $\endgroup$ – Bernhard Aug 31 '14 at 6:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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