Find the Speed of a Freight Car When All the Sand is Gone

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Homework Help Overview

The problem involves a freight car of mass M containing a mass of sand m, where a constant horizontal force F is applied while sand flows out at a constant rate. The objective is to determine the speed of the freight car when all the sand has been expelled, starting from rest.

Discussion Character

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Approaches and Questions Raised

  • The original poster attempts to set up an equation involving momentum but struggles with the undefined nature of logarithmic terms at t=0. Some participants suggest a different approach to find the time when the mass is zero, using the relationship between mass flow rate and time.

Discussion Status

Participants are exploring various interpretations of the problem, with some providing guidance on the equations of motion and momentum. There is an ongoing exchange of ideas regarding the correct application of forces and mass conservation in the context of the problem.

Contextual Notes

There is a mention of the total mass being conserved and the implications of expelled mass on momentum, which raises questions about the assumptions made in the original setup.

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Homework Statement


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

The Attempt at a Solution


[tex]P_t=Mv(t)+{m_0}-t{{dm}\over{dt}}[/tex]

I was trying to solve [tex]m_0-t(dm/dt)=0[/tex]

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

but I can't solve it.

as it turns out to be

[tex]\int {{dt}\over{t}}=\int {{dm}\over{m_0}}[/tex]

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))

Thanks for your reading!
 
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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
[tex]F = \frac{dp}{dt} = m\frac{dv}{dt} + v\frac{dm}{dt}[/tex]

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)!
 
Thanks!
I got this:
[tex]v(t')=\frac{\dot{P}}{\dot{m}}[/tex]
is this correct?
 
Mindscrape said:
Let me give you another little bit of help
[tex]F = \frac{dp}{dt} = m\frac{dv}{dt} + v\frac{dm}{dt}[/tex]

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

[tex]F = \frac{dp}{dt} = m\frac{dv}{dt} + u\frac{dm}{dt}[/tex]

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

ehild
 

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