Variable Mass system : Sand falling into a Freight car

[Su6had1p]

Homework Statement

An empty freight car of mass M starts from rest under an applied
force F. At the same time, sand begins to run into the car at steady
rate b from a hopper at rest along the track.
Find the speed when a mass of sand m has been transferred.

Relevant Equations

Below is my approach which doesnot seem to agree with the given solution, please check and correct my solution.
let us start with mass considerations -
$$M(t) = M_{c} + m(t)$$
where $M(t)$ is the mass of the system, $m(t)$ mass of the falling sand both of which are time varying, $M_{c}$ is the mass of the freight car, which is constant with time.
taking the time derivative on both sides we get,
$$\frac{dM}{dt} = \frac{dm}{dt}$$
also, since it is given that the rate of mass falling is $b$
we can solve the following differential equation, keeping in mind $M(0) = M_{C}$
$$\frac{dM}{dt} = b$$
$$M = M_{C} +bt$$
initially the freight car is moving at some speed say $v$
now let's calculate the change in momentum
$$p(t) = M v$$
$$p(t+ dt) = (M + dm)(v+dv)$$
$$dp = p(t+ dt) - p(t) = Mdv+vdm$$
$$\frac{dp}{dt} = M\frac{dv}{dt} + v\frac{dm}{dt}$$
from earlier relation,
$$\frac{dp}{dt} = M\frac{dv}{dt} + v\frac{dM}{dt}$$
Since the net external force is $F$, and rate of mass is $b$
$$ F = M\frac{dv}{dt} + bv$$
rearranging and solving,
$$\int \frac{dv}{F-bv} = \int \frac{dt}{M_{C}+bt} $$
$$\frac{ln(F-bv)}{-b} = \frac{ln(M_c+bt)}{b} +C$$
It's given that at $t=0, v=0$
$$C = \frac{ln(F+M_{C})}{-b}$$
the particular solution therefore becomes,
$$ln(F-bv) = ln(F+M_{C})-ln(M_{C}+bt)$$
exponentiating both sides,
$$F-bv = \frac{F+M_{C}}{M_{C}+bt}$$
the final part of the equation asks speed when $m$ amount of mass is transferred. which means $M=m$
$$m = M_{C}+bt$$
$$F-bv = \frac{F+m-bt}{m}$$

My solution doesnot match with the given solution (Its from Kleppner and Kolenkow 2ed) which is
$$v=\frac{Ft}{M+bt}$$
where $M$ is the initial mass of the system i guess.

I dont understand what step i am doing wrong, the given solution proceeds with impulse momentum theorem which is fine but i would like to solve it in this way since this is the approach that organically came to my mind.

 

[kuruman]

You may wish to read this article that discusses how to approach variable mass problems and the pitfalls to watch for. There is a step-by-step method to follow and illustrated with examples. As you will see, the solution to your posted problem is simpler than you presented. The key point is to have a clear idea in your mind what the system of interest is.

 

[TSny]

I dont understand what step i am doing wrong, the given solution proceeds with impulse momentum theorem which is fine but i would like to solve it in this way since this is the approach that organically came to my mind.

I think you are ok up through the following line:

$$\frac{ln(F-bv)}{-b} = \frac{ln(M_c+bt)}{b} +C$$

But your result for $C$ is not correct:

It's given that at $t=0, v=0$
$$C = \frac{ln(F+M_{C})}{-b}$$

Check your work.

Your method is fine, but it's a bit lengthy. Using the impulse-momentum theorem yields the answer quickly (without calculus).

My solution doesnot match with the given solution (Its from Kleppner and Kolenkow 2ed) which is
$$v=\frac{Ft}{M+bt}$$
where $M$ is the initial mass of the system i guess.

I'm not sure what you're taking to be the system. But, $M$ is given to be the mass of the empty car.

 

[Su6had1p]

But your result for C is not correct:

Yeah, I just checked somehow i forgot that
$$log(a) + log(b) = log(ab) ≠ log(a+b)$$

 

[Su6had1p]

@TSny @kuruman thanks for your help.

 

[PeroK]

@TSny @kuruman thanks for your help.

Just to emphasise the point that in Newtonian physics, mass cannot be created or destroyed. Newton's second law, therefore, applies to an object of fixed mass:
$$\vec F = \frac {d\vec p}{dt} = m\frac{d \vec v}{dt}$$If the mass of an object is changing, then mass must be moving into or out of the object - and the state of motion of that mass must be taken into account. It's true that in certain cases you can naively assume that Newton's second law applies to an object of variable mass and the mathematics works out. But, in general, it does not.

If the mass of an object is changing, then the definition of the object is changing. And you need to be careful using Newton's second law.

 

[kuruman]

If the mass of an object is changing, then the definition of the object is changing.

I grappled with this point when I was putting together the article mentioned in post #2. Eventually I decided to reserve the word "object" solely in the context of fixed mass. However, a "system" in the context of Newton's laws may consist of many "objects". Thus, to deal with variable mass problems, I thought it would be best to separate objects, which always have fixed mass, from systems, which may have variable mass. All these need to be identified before one writes the appropriate equation(s). Of course, this separation is subject to the constraint that the sum of all objects' masses must be equal to the sum of all systems' masses.

In this problem the objects are: (a) the empty freight car and (b) all the sand that could conceivably fall in it. The systems are: (a) the car plus the sand in it at a given moment in time (system of interest), (b) the amount $dm$ of sand that interacts with the car at that moment and (c) the remainder of the sand that is yet to fall in the car.

Noting that the sand system $dm$ does not transfer horizontal momentum to the system of interest, we can write Newton's law $$
\begin{align} & \frac{dP}{dt}=F \implies P=F~\Delta t \nonumber \\& V(t)=\frac{P}{M}=\frac{F~\Delta t}{M(t)}.\nonumber \end{align}$$Surprise, surprise! The change in momentum from $t=0$ to time $t$ is the impulse delivered by the external force. The manner in which the sand is added is immaterial as long as it drops in vertically.