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

[Shing]

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

$$P_t=Mv(t)+{m_0}-t{{dm}\over{dt}}$$

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))

Thanks for your reading!

 

[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)!

 

[Shing]

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

 

[ehild]

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

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

 

[rwisz]

I would approach this problem by first identifying the relevant equations and principles that can be applied. In this case, we can use the equation for momentum, which states that the momentum of an object is equal to its mass multiplied by its velocity. We can also use the principle of conservation of momentum, which states that the total momentum of a system remains constant unless acted upon by an external force.

In this problem, the initial momentum of the freight car is zero since it is at rest. As the sand flows out of the car, the mass of the car decreases, but the external force F continues to act on it, causing it to accelerate. We can use the equation for momentum to set up an equation for the final momentum of the car:

P_f = (M-m)v

where P_f is the final momentum, M is the initial mass of the car, m is the mass of sand that has flowed out, and v is the final velocity of the car.

We also know that the total mass of the system (car + sand) is constant, so we can set up an equation for the conservation of momentum:

P_i = P_f

where P_i is the initial momentum, which we know is zero.

Substituting in the equation for P_f, we get:

0 = (M-m)v

Solving for v, we get:

v = 0

This means that when all the sand has flowed out of the car, the car will come to a stop. This makes intuitive sense because as the mass of the car decreases, the external force F will have a greater effect on its acceleration, causing it to come to a stop.

In summary, the speed of the freight car when all the sand is gone is 0. If you are having trouble understanding the mathematical solution, I would recommend reviewing the principles of momentum and conservation of momentum, as well as practicing solving similar problems to build your understanding.