# A Mass flow car

1. Aug 17, 2010

### Shing

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 et the sand flow out at tconstant rate dm/dt. Find the speed of the freight car when all the sand is gone. Assume the car is at rest t=0

3. The attempt at a solution
$$P_t=Mv(t)+{m_0}-t{{dm}\over{dt}}$$

I was trying to solve $$m_0-t(dm/dt)=0$$

such that I know the t when all the sand is gone

but I can't solve it.

as it turns out to be

$$\int {{dt}\over{t}}=\int {{dm}\over{m_0}}$$

it doesn't make sense at all when t=o! (undefined for In0)

also, the equation of the momentum still has two unknowns for one equation, (even I take dp/dt, I still not sure about d^2/dt(x))

2. Aug 17, 2010

### Mindscrape

You're solving for the time when the mass runs out in the wrong way. Maybe once you get that you'll be back on track. You know that the mass is draining at a constant speed

dm/dt=-k(kg/s)

gives

m(t)=m0-kt

want to know when mass is 0

m(t)=0 => m0=kt

so t_(mass=0)=m0/k

check dimensions to see if they are okay kg/(kg/s) = s (check!)

What is P_t supposed to represent?

Let me give you another little bit of help
$$F = \frac{dp}{dt} = m\frac{dv}{dt} + v\frac{dm}{dt}$$

you know dm/dt, so now you'll have a differential equation for v that you should be able solve. Of course, your final solution will be an analytic one given that you don't know the rate or the force, but you do know they are constant and so you don't have to worry about F(t) or k(t)!

3. Aug 20, 2010

### Shing

Thanks!
I got this:
$$v(t')=\frac{\dot{P}}{\dot{m}}$$
is this correct?

4. Aug 20, 2010

### ehild

Take care, the total mass is conserved, F=dp/dt refers to the whole system of mass, the spilt sand included. The cart changes momentum due to expelled mass only when that mass has some velocity u relative to the cart. The equation

$$F = \frac{dp}{dt} = m\frac{dv}{dt} + u\frac{dm}{dt}$$

is valid, and u=0 in this case, so F=m(t)a.

ehild