What's the error in my solution (Freight car and hopper)

[yucheng]

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 a steady rate $b$ from a hopper at rest along the track. Find the speed when a mass of sand $m$ has been transferred.
(Kleppner, 2nd edition, problem 4.11)

Relevant Equations

N/A

Textbook solution:
$v$ is the instantaneous velocity,

$$P(t)=(M+b t) v$$
Then $$impulse = \Delta P = (M+b t) v = \int^{t}_{0} F dt'$$
Thus $$v=\frac{F t}{(M + bt)}$$
What I did instead was:

Let $M$ be the instantaneous mass, and $M_0$ be the initial mass, then $$M=M_{0} + b t$$
\begin{align*}
M \frac{d v}{d t}&=F \\
(M_{0} + b t) d v &= F d t \\
d v &= \frac{F}{(M_{0} + b t)} d t\\
\int^{v}_{0} d v' &= \int^{t}_{0} \frac{F}{(M_{0} + b t')} d t'\\
v &= \frac{F}{b}\ln{\frac{M_0 + b t}{M_0}} \\
\end{align*}

I would really appreciate it if you could point out where my conceptual error is. Do bear in mind that I have a huge tendency to just blindly manipulate equations :) I know, I'll have to overcome it. Thanks in advance!

 

[haruspex]

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 a steady rate $b$ from a hopper at rest along the track. Find the speed when a mass of sand $m$ has been transferred.
(Kleppner, 2nd edition, problem 4.11)
Relevant Equations:: N/A

Textbook solution:
$v$ is the instantaneous velocity,

$$P(t)=(M+b t) v$$
Then $$impulse = \Delta P = (M+b t) v = \int^{t}_{0} F dt'$$
Thus $$v=\frac{F t}{(M + bt)}$$
What I did instead was:

Let $M$ be the instantaneous mass, and $M_0$ be the initial mass, then $$M=M_{0} + b t$$
\begin{align*}
M \frac{d v}{d t}&=F \\
(M_{0} + b t) d v &= F d t \\
d v &= \frac{F}{(M_{0} + b t)} d t\\
\int^{v}_{0} d v' &= \int^{t}_{0} \frac{F}{(M_{0} + b t')} d t'\\
v &= \frac{F}{b}\ln{\frac{M_0 + b t}{M_0}} \\
\end{align*}

I would really appreciate it if you could point out where my conceptual error is. Do bear in mind that I have a huge tendency to just blindly manipulate equations :) I know, I'll have to overcome it. Thanks in advance!

Your method does not take into account that each element of added mass has to be brought to speed v.

 

[yucheng]

Your method does not take into account that each element of added mass has to be brought to speed v.

Honestly, I am very confused with the integrals. Would you explain more?

 

[Steve4Physics]

Your formula$$M \frac{d v}{d t}=F$$ is wrong. It only applies when M is constant

Use the product rule when finding $\frac {d(Mv)}{dt}$ as both M and v are functions of time.

 

[haruspex]

Your formula$$M \frac{d v}{d t}=F$$ is wrong. It only applies when M is constant

Use the product rule when finding $\frac {d(Mv)}{dt}$ as both M and v are functions of time.

Although that works here, I have a distaste for writing $\frac{dp}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}$, and I'm not alone.
At this level of physics at least, mass is neither created nor destroyed, so if $\dot m$ is nonzero we are not looking at a closed system. Mass is entering or leaving the system, and it could be taking momentum away with it or bringing momentum in.
You get away with it in this question because the sand had no horizontal velocity before being added to the load. A student applying the same equation to sand being dropped by a moving cart can go astray.

 

[haruspex]

Honestly, I am very confused with the integrals. Would you explain more?

Suppose the wagon were merely kept going at constant velocity, your equation says that would take no force, but the added sand will increase the momentum of the wagon+load system. As each grain of sand is added, it takes an impulse to get it up to speed.

 

[yucheng]

Suppose the wagon were merely kept going at constant velocity, your equation says that would take no force, but the added sand will increase the momentum of the wagon+load system. As each grain of sand is added, it takes an impulse to get it up to speed.

Hmm, looking back at my equation, it seems that it tells me for each instantaneous mass, the force causes an acceleration proportional to the mass. It is wrong because I am assuming that $M+\Delta m$, where $\Delta m$ is the mass of the newly added sand, already has velocity $v$, and then the equation holds true (since it is acceleration of the whole system, but since the sand added and the wagon have different speeds, they can't be considered as a homogeneous body?), while in fact it starts from rest. Is this what you meant?

 

[yucheng]

Your formula$$M \frac{d v}{d t}=F$$ is wrong. It only applies when M is constant

Use the product rule when finding $\frac {d(Mv)}{dt}$ as both M and v are functions of time.

$$M\frac{d v}{d t} + v \frac{d M}{d t} = F$$ looks impossible to solve!

Since $\frac{d M}{d t} = b$ is constant, $$M\frac{d v}{d t} + v b= F$$

 

[haruspex]

$$M\frac{d v}{d t} + v \frac{d M}{d t} = F$$ looks impossible to solve!

Since $\frac{d M}{d t} = b$ is constant, $$M\frac{d v}{d t} + v b= F$$

It's not as bad as it looks. You can write down M as a function of t.

 

[yucheng]

$$M\frac{d v}{d t} + v \frac{d M}{d t} = F$$ looks impossible to solve!

Since $\frac{d M}{d t} = b$ is constant, $$M\frac{d v}{d t} + v b= F$$

Let me try to continue.

$$(M+b t)\frac{d v}{d t} + v b= F$$

My first instinct is to multiply everything out...

$$(M+b t)d v + v b dt= F dt$$
$$(M+b t)d v = (F - v b) dt$$
$$\frac{1}{(F - v b)}d v = \frac{1}{(M+b t)} dt$$
...
$$v = \frac{Ft}{M+bt}$$

Okay, it works, thanks!

 

[kuruman]

Although that works here, I have a distaste for writing $\frac{dp}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}$, and I'm not alone.

I think that the equation itself is OK because it follows from the product rule of differentiation. Problems arise when folks misuse it. One must keep in mind that the generalized form of Newton's second law is $\vec F_{\text{Net}}=\frac{d\vec P}{dt}$ and that, to use it correctly, one must assemble the LHS first and then set it equal to the RHS. This avoids the pitfall of treating $m\dot v$ or $v\dot m$ as if they were external forces.

 

[haruspex]

Problems arise when folks misuse it.

Of course, but that happens so often.

There is no point in using that form, instead of F=ma, if we are only considering an external force on what is otherwise a closed system. So consider the case of two adjacent wagons traveling E at the same velocity. A worker on A is shovelling sand directly across (from the worker's perspective) to wagon B.
Wagon B is gaining momentum at steady velocity, so the arriving sand must constitute an external force on the system, even though it imparts no EW force where it lands.

Hardly surprising students get confused.

 

[kuruman]

There is no point in using that form, instead of F=ma, if we are only considering an external force on what is otherwise a closed system.

I concede. There is also no point in using a sledgehammer to kill a fly when a fly swatter will suffice.