1. Jul 18, 2010

metalrose

1. The problem statement, all variables and given/known data

A freight car of mass M contains a mass of sand m'. At t = 0 a
constant horizontal force F is applied in the direction of rolling and at
the same time a port in the bottom is opened to let the sand flow out at
constant rate dm/dt. Find the speed of the freight car when all the sand
is gone. Assume the freight car is at rest at t = o.

2. Relevant equations

F = d(P)/dt

3. The attempt at a solution

I applied the above equation to the system of "freight car".

Which is,

F = d(P)/dt, where P is the momentum of the freight car at any instant of time.

Now P = mv, where m and v are the total mass and velocity of the freight car at any instant. Now, d(P)/dt should be

v(dm/dt) + m(dv/dt) .................. (right?)

But I looked up the solution, and over there it is instead given,

d(P)/dt = d(mv)/dt = m(t) x (dv/dt)

where m(t) is the mass of the freight car at time 't'.

So essentially, in the solution, m has been taken out of differentiation. My question is, why is it so?

What is wrong with my differential equation, d(P)/dt = d(mv)/dt = v(dm/dt) + m(dv/dt) ?

Thanks

2. Jul 18, 2010

ehild

Nothing is wrong, it is correct.

ehild

3. Jul 18, 2010

ehild

No law is violated, the car and mass is not a closed system, the sand flows out. Read the text of the problem.

ehild

4. Jul 18, 2010

metalrose

@ehild

Thanks for the reply, but I didn't get you in your second reply above. All I want to know is, that why has the first term on the right hand side in the following equation been dropped :

d(P)/dt = v(dm/dt) + m(dv/dt)

i.e., why is v(dm/dt) 0 in the formal solutions I looked up?

Thanks

5. Jul 18, 2010

ehild

My second post was the answer to somebody else who has deleted his post meanwhile.

You are right, dm/dt can not be ignored, proceed, build up a differential equation for v(t).

The "official" solution are entirely wrong sometimes. Do not worry.

ehild

6. Jul 18, 2010

metalrose

@ehild

Thank you. But are you totally sure that the solutions are wrong and I am right? This is a question from Kleppner and Kloenkow's An Introduction To mechanics and many univ.'s use it as a text and I found the solutions to this question by University fo Chicago, and most probably , the solutions were given by some prof. So are you absolutely sure that the sol. is incorrect?

7. Jul 18, 2010

ehild

Do not believe anybody if that what he says contradicts to the laws of Physics and Mathematics. F=m dv/dt is valid only for constant mass. F=dp/dt is more general, valid for variable mass, and it is consistent with Newton's original statement.
The equation of motion for rockets is also derived from it, and the rockets go up, don't they? See this:

http://hyperphysics.phy-astr.gsu.edu/HBASE/rocket.html

I would like to see how you proceed with that differential equation.

(That d(P)/dt = d(mv)/dt = m(t) x (dv/dt) in the solution can be just a typo. Look how it goes further.
By the way, I used to teach at a big university and I know, how many mistakes can be present in the solutions we give out. I always asked my students to report any errors they found in the books and notes or tests.)

ehild

8. Jul 18, 2010

metalrose

@ehild

Thanks a ton. And yes, I have already proceeded with that differential equation and have got the answer out : v(when all sand runs out) = F.(dm/dt)ln[(M+m)/M]

And the solution doesn't have a typo, the answer doesn't match, and the answer proceeds only on the basis of F = m(t) x dv/dt in there.

I know , that the official solution went against the laws, but I thought that maybe I am missing out some subtle point in the question that makes the first term drop out.

But as you say, there isn't anything I'm missing out, I'll leave it at that and trust you!!

Thanks a lot again..!!

Cheers..!!!

9. Jul 18, 2010

Swap

you just find accln. in terms of dm/dt. and a= F/m. since m is changing u have to express it in terms of M, m and dm/dt. treat dm/dt as a constant (as given) and then integrate both sides. the answer will be in natural log form.btw ur answer is bit wronf. instead of dm/dt it should be dt/dm.

10. Jul 18, 2010

ehild

Do not leave it so without checking the solution. I got a different result. Show your work in detail.

ehild

11. Jul 19, 2010

metalrose

@ ehild and swap,

I worked out the problem again and now I am getting a totally different answer!!!
Here is how I worked out the entire solution. Please tell me where I have gone wrong:

F= d(P)/dt = d(mv)/dt = v(dm/dt) + m(dv/dt)

Let's denote dm/dt by 'k', and dm/dt will have a negative sign because the sand is flowing out, so basically the dm/dt in the above equation is equal to -k .

then,

F = -vk + (M + m - kt)dv/dt

--> F + vk = (M + m - kt)dv/dt

--> dt/(M + m - kt) = dv/(F + vk)

Integrating both sides,

-1/k x ln(M + m - kt)/(M + m - kt0) = 1/k x ln(F + vk)/(F + v0k)

t0 and v0 are both equal to 0. also the time 't' at which all sand runs out is t=m/k

Putting these in the above, we get,

(M + m)/M = (F + vk)/F

--> v = Fm/Mk = Fm.dt/M.dm

12. Jul 19, 2010

Swap

ya this is the correct answer!!!

13. Jul 19, 2010

hikaru1221

I think there is actually a problem with this equation, and the solution in the textbook is right. There should be caution when taking differentiation of the term mv. The general form of the Newton's 2nd law's equation applying to these kinds of problems is:

$$m\frac{d\vec{v}}{dt}=\vec{F}+\vec{u}\frac{dm}{dt}$$

where $$\vec{u}$$ is the velocity of the separate part relative to the system.

Let's go from the start, instead of applying directly that equation. Consider the 2 moments t and t+dt, and the system considered during this period is the car + sand on it + the amount of sand spilling out.
1/ At t:
_ The mass of the system is m
_ The velocity of the system is v
_ The momentum is p = mv
2/ At t+dt:
_ The system now consists of 2 parts: (car+sand on it) & (sand spilling out)
_ The velocity and the mass of the 1st part are v+dv and m+dm (dm<0 of course)
_ The horizontal velocity and the mass of the 2nd part are v and -dm.
Note that the 2nd part, at the time of spilling out of the car, maintains its horizontal velocity component.
_ The momentum of the system: p+dp = p(part 1) + p(part 2) = (m+dm)(v+dv) - vdm = mv + mdv.
Therefore: Fdt = dp = mdv, as the solution in the book gives.

The problem with the differentiation method (d(mv) = mdv+vdm) is that it assumes every part in the system behaves exactly like each other, while in this problem, the 2 parts don't.

EDIT: To elaborate on this, let's consider the system consisting of only the car and the sand on it. Then this system is defined only during the time period from t to t+dt (when we move to t+2dt, it is another system). Because the sand of mass -dm which has just spilled out gains no change in horizontal velocity, it doesn't exert forces on the system. Moreover, in this period, the system does not lose any mass - this is an important point. Therefore: Fdt=mdv.

By the way, if the sand is not spilling out of the car but into the car, then the equation d(mv)=mdv+vdm is valid

Just my own thought, so please correct me if I'm wrong, ehild and swap

Last edited: Jul 19, 2010
14. Jul 19, 2010

metalrose

@ ehild and swap

Please comment on hikaru1221 's answer..!!! I knew there is something subtle in there. I guess hikaru is right, though I'll have to re-read his answer....but please comment ehild and swap!!!

15. Jul 20, 2010

ehild

Apologies, Hikaru is right, and your book is right, I looked after and this is the way such problems (rocket) are solved. Now I remember that I bumped into Metalrose's problem with dp/dt=F, and came to the conclusion that it can not be applied in the normal way in general, but I had forgotten about it. Sorry to having misled you, Metalrose.

The mass can not be considered as a simple time-dependent variable as it is conserved in the universe.

Think that a spaceship travels with constant velocity in empty space and a piece gets lose. Will this event change the velocity of either the ship or the lost piece?

If using Δp=Δm*v+m*Δv=0 it involves Δv=-Δm/m *v which is not true.

We have to use the conservation of momentum instead: vi is the initial velocity of the ship, vf is the final one, and vp is the velocity of the lost piece. m*vi=(m-Δm)vf + Δm*vp. If vp=vi, vf is also equal to vi.

In case of an external force, the change of the momentum is equal FΔt, where F is the external force on the whole system.

Assume that a piece Δm loses contact with a bigger stone of mass m when it falls down. The change of the momentum in Δt time is

[(m-Δm)vf+Δm*vp]-m*vi=mgΔt--->

m*vf-Δm*vf+Δm*vp-m*ví=m(vf-vi)+Δm(vp-vf)=mgΔt.

If vp=vf, mΔv=mgΔt follows, m is the original mass, and this leads to dv/dt=g, so it is correct, both parts move with the same acceleration g and gain the same amount of velocity in time Δt.

In your problem, F is an external force, independent on mass, but should act on the whole system, even on the lost piece in order that mΔv=F*Δt hold.
But the pulling force does not act on the split sand. It leaves with the initial velocity. So we have the equation

m(vf-vi)+Δm(vi-vf)=FΔt--->Δv*(m-Δm)=FΔt.

It would be the same result when we treated the problem as an explosion which takes zero time and after that an acceleration of the remainder mass.

One can say that Δm can be ignored with respect to m, and m-Δm can be replaced by m(t) instead of m(t-Δt). Well, I do not like it too much but it is true enough if the change of the mass is very small with respect to the mass.

So your book is right. As Maths is true, so dp/dt=mdv/dt+dm/dt*v, dp/dt = F is not true in general. It is better to use the conservation of momentum. The force is a very problematic concept.

Care has to be taken when defining a system and force acting on the different parts of the system.

In the original problem, the sand flows out of the car, but why? If there is a hole, the amount of mass above it will fall out, but the other amount should reach there. That means that some amount of sand has some velocity with respect to the car. If the sand moves with respect to the car, there must be friction. The force F acts on the car, and normally, a normal force acts on the sand. But there is no normal force at the trap to accelerate the sand in the same way as the car is accelerated. It is friction there that keeps the sand moving together with the car. But the sand slips to the trap, and falls out, so its velocity is lower than that of the car. But the friction from the sand acts on the car, too, so the car will accelerate not by F, but F-Ffr. The Ffr term is related to the mass of the sand and changing in time. This is quite a complex problem and not defined enough.

Well, I have to look after a bit more...

ehild

16. Jul 20, 2010

hikaru1221

I don't think that's the reason to omit Δm. Even though Δm<<m, it won't make sense if we omit Δm just because of that, since this means m=const. Instead, we omit ΔmΔv, which is way too small compared to mv or vΔm or mΔv.

:surprised
I think math is true and either does physics. F=dp/dt is still true in general cases. It's just that we have to apply the right math. Math says, if p=mv, then p is a multi-variable function depending on both variables m and v, which means both m and v must associate with p. Physics interprets that as m and v are 2 quantities describing the whole body (system), and this whole body has a momentum of p. However, in the system of this problem in particular, m and v can NOT describe the whole system, but just a part of it.

To clarify this: $$Fdt = dp_{total} = dp_{car+sand}+dp_{sand-out}$$
Now we have: $$p_{car+sand}=mv$$ as m and v are good enough to describe the system (car+sand) at time t (only at time t, not 2t, or 3t). Therefore: $$dp_{car+sand}=mdv+vdm$$. However, as during dt (and only from t to t+dt), dm=0, so $$dp_{car+sand}=mdv$$
Moreover: $$dp_{sand-out}=0$$. We come back to the expected result.

Strictly speaking, friction should be taken into account, and the problem is unsolvable or very complicated. But I think the problem doesn't demand that much. Instead, we can model the problem as when the (-dm) amount of sand starts to spill out, it experiences free fall, i.e. it is independent from the car from that moment. Therefore, in the next period dt, the force only has effect on the car + remaining sand, which leads directly to the equation: Fdt = mdv.
I haven't thought of a better way to model the problem.

Last edited: Jul 20, 2010
17. Jul 20, 2010