Sand flowing out of the car

1. Aug 20, 2013

Saitama

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 the 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 $t=0$.

2. Relevant equations

3. The attempt at a solution
Let $\frac{dm}{dt}=\lambda$.
At t=0, momentum of the system, P(0)=0.
At some later time t, $P(t)=(M+m-\lambda t)v$ where v is the velocity of car at time t.

When all the sand falls out, $m=\lambda t \Rightarrow t=m/\lambda$. Hence $P(m/\lambda)=Mv$.
$$F=\frac{P(t)-P(0)}{t}=\frac{P(m/\lambda)}{m/\lambda}=\frac{M \lambda v}{m}$$
Therefore, the speed of car is
$$v=\frac{Fm}{M\lambda}$$
Does this look correct?

Any help is appreciated. Thanks!

2. Aug 20, 2013

TSny

Your expression for $P(t)$ only includes the momentum of the cart and the sand still left in the cart. But the sand that falls out also has horizontal momentum.

3. Aug 20, 2013

Saitama

Is $P(t)=(M+m-\lambda t)v+\lambda tv= (M+m)v$?

4. Aug 20, 2013

TSny

No. The particles of sand that fall out earlier have less horizontal velocity than the particles that fall out later.

5. Aug 20, 2013

Saitama

Do you mean I should evaluate P at time t and t+dt?

6. Aug 20, 2013

TSny

That approach should work.

7. Aug 20, 2013

Saitama

$P(t)=(M+m-\lambda t)v+\lambda tv'$

How do I write the expression for P(t+dt)?
The mass that has already fallen off ($\lambda t$) has still the same horizontal velocity but what about the mass which will fall off at t+dt? What about the mass which is still in the car? I need a few hints to make the equations.

8. Aug 20, 2013

TSny

You are only interested in the change in momentum of the entire system between t and t + dt. The last term in your expression above apparently represents the momentum of the sand that has already fallen out before time t. Is the momentum of that sand going to change between t and t + dt? If not, you can ignore it since it will not contribute to the change in momentum of the system between t and t + dt.

So, ignoring the sand that has fallen out before time t, the momentum of the system at time t is

$P(t)=(M+m-\lambda t)v(t)$

How would you express the momentum at time t + dt ignoring the sand that fell out before time t but including the sand that falls out between t and t + dt?

9. Aug 20, 2013

Saitama

After time t, the mass of sand remaining in car is $M+m-\lambda t$. At time (t+dt), mass $\lambda dt$ falls off with velocity v(t) i.e
$P(t+dt)=(M+m-\lambda (t+dt))v(t+dt)+\lambda (dt) v(t)$

Correct?

10. Aug 20, 2013

TSny

Yes.

11. Aug 20, 2013

Saitama

Erm...I end up with a really dirty expression for change in momentum.

$$\Delta P=P(t+dt)-P(t)$$
$$=M(v(t+dt)-v(t))+m(v(t+dt)-v(t))+\lambda(v(t)dt-tv(t)-tv(t+dt)-(dt)v(t+dt))$$

12. Aug 20, 2013

TSny

I believe you have one sign error in the parentheses multiplying the λ.

Can you express v(t+dt) in terms of v(t) and the acceleration a(t)? (to first-order accuracy in dt)

Then see if you can expand the equation for ΔP keeping terms only up to and including first order in dt.

13. Aug 20, 2013

Saitama

I cannot find the sign error.
Taylor series?
$v(t+dt)=v(t)+a(t)dt$. Looks good?
Do you mean this:
$\Delta P=P(t+dt)-P(t)=P(t)+P'(t)dt-P(t)=P'(t)dt$.

14. Aug 20, 2013

TSny

Check the sign of the 2nd term inside the parentheses multiplying λ.

Yes, looks good.

Yes, that's good. By expanding the expression for ΔP, I meant to expand out the right hand side of the equation for ΔP after substituting $v(t+dt)=v(t)+a(t)dt$.

15. Aug 20, 2013

Saitama

Honestly, I still cannot see it. :uhh:

I will write down the expressions again.
$P(t)=(M+m-\lambda t)v(t)$
$P(t+dt)=(M+m-\lambda(t+dt))v(t+dt)+\lambda v(t)(dt)$
See, there is only one term in $\lambda$ which has a plus sign.

I think you are right about the sign. When I evaluate $\Delta P$, I get a term $v(t+dt)+v(t)$ which is undesired. I mean it hinders in simplification.

I will reply to other thread later when I return back home. Thank you for your time TSny! :)

16. Aug 21, 2013

haruspex

I don't see how that happens. It doesn't when I do it.

17. Aug 21, 2013

ehild

Just call the mass at time t as M(t). From t to t+Δt, the mass changes by -λΔt. The velocity changes from v(t) to v(t)+Δv.

The momentum of the car and load at time t is P(t)=M(t)v. Δt later, the mass of the car is M(t)-λΔt, its velocity is v+Δv, and the mass of the fallen-out sand is λΔt, its (average) velocity is something between v and v+Δv. Let it be v+Δv'

The change of momentum during t and t+Δt is

(M(t)-λΔt)(v+Δv)+λΔt(v+Δv')-Mv=FΔt.

Expanding and simplifying, MΔv+λΔt(Δv'-Δv)=FΔt
The second term on the left-hand side can be ignored, So you get the equation

M(t)Δv=FΔt -->(M+m-λt)dv/dt=F

A linear de, solve.

ehild

18. Aug 21, 2013

Saitama

I finally found the sign error.

I plugged in v(t+dt)-v(t)=a(t)dt and simplified the expression further. This is what I got,
$$\Delta P=(M+m)a(t)dt+\lambda(a(t)(dt)^2-ta(t)(dt))$$
I think I have to neglect the term with (dt)^2.
Substituting $\Delta P$ with $P'(t)dt=(M+m-\lambda t)a(t)dt \Rightarrow F=(M+m-\lambda t)dv/dt$ which is the same D.E ehild has shown.

Thanks for the help TSny! :)

Why we deal with average velocity? I am sorry if I miss something obvious, I am not too much familiar with variable mass systems.

19. Aug 21, 2013

arildno

20. Aug 21, 2013

ehild

We wrote the equation for conservation of momentum for the time period Δt. The sand that leaves the car at the beginning of the Δt period of time has the velocity v, the amount leaving at the end of that period has v+Δv. The momentum of the spilt sand is included to the momentum at t+Δt, but it does not have an exactly defined velocity, that is why I used v' for the average velocity for the fallen-out sand.
Anyway, the term proportional to the product ΔtΔv is a second-order small quantity, it can be ignored.

ehild