Locomotive throwing sand in a car

[Saitama]

Homework Statement

A sand-spraying locomotive sprays sand horizontally into a freight car situated ahead of it. The locomotive and freight car are not attached. The engineer in the locomotive maintains his speed so that the distance to the freight car is constant. The sand is transferred at a rate dm/dt = 10 kg/s with a velocity of 5 m/s relative to the locomotive. The car starts from rest with an initial mass of 2000 kg. Find its speed after 100 s.

Homework Equations

The Attempt at a Solution

I am actually not sure how to start because of the given constraint that distance between the locomotive and car stays constant.

Do I have to begin with evaluating P(t) and P(t+dt)? If so, what should be my system? Car and the incoming sand?

Any help is appreciated. Thanks!

 

[milesyoung]

Have a look at the first equation here:
http://en.wikipedia.org/wiki/Variable-mass_system

Considering the freight car to be a variable-mass system with no net external force applied, it should be simple to use.

 

[voko]

I am actually not sure how to start because of the given constraint that distance between the locomotive and car stays constant.

This basically means the mass rate and the velocity of the incoming sand are constant with regard to the car, and you can forget about the locomotive.

 

[Saitama]

This basically means the mass rate and the velocity of the incoming sand are constant with regard to the car, and you can forget about the locomotive.

I initially thought that I have to find the distance between the locomotive and car as a function of time, differentiate it and set the derivative to zero as the distance stays constant. Is this a wrong approach?

I still don't understand how to make the equations. Let the mass of car (without sand) be M. At time t, let the mass be M(t) and velocity of car be v(t).
$P(t)=M(t)v(t)+\text{momentum of incoming sand}$
But what do I write in place of "momentum of incoming sand"?

 

[voko]

Consider what happens during some brief time $\Delta t$.

 

[Saitama]

Consider what happens during some brief time $\Delta t$.

A mass of $\lambda \Delta t$ ($\lambda=dm/dt$) is added to the car.

 

[voko]

What happens with the car's momentum, it being the product of mass with velocity?

 

[Saitama]

What happens with the car's momentum, it being the product of mass with velocity?

Let the velocity of car change to $v(t+\Delta t)$. The momentum is $P(t+\Delta t)=(M(t)+\lambda \Delta t)v(t+\Delta t)$. Correct?

 

[voko]

Correct, but not too useful. How about expressing that in terms of $\Delta M$ and $\Delta v$?

 

[Saitama]

Correct, but not too useful. How about expressing that in terms of $\Delta M$ and $\Delta v$?

$P(t+\Delta t)=(M(t)+\Delta m)v(t+\Delta t)$
I don't see how to introduce $\Delta v$ here.

 

[voko]

$v(t + \Delta t) = v + \Delta v$, no?

 

[Saitama]

$v(t + \Delta t) = v + \Delta v$, no?

I recently posted a thread (Sand flowing out of the car) where I had to expand v(t+dt) using Taylor series. What you wrote doesn't look like it results from Taylor series. Using Taylor series, $v(t+\Delta t)=v(t)+v'(t)\Delta t$.

 

[voko]

Well, you are just using the more complicated approach. In the limit $\Delta X /\Delta t = X'$ by definition. Anyway, express the momentum of the car at $t + \Delta$ and think what its difference over $\Delta t$ should be equal to.

 

[Saitama]

Well, you are just using the more complicated approach. In the limit $\Delta X /\Delta t = X'$ by definition. Anyway, express the momentum of the car at $t + \Delta$ and think what its difference over $\Delta t$ should be equal to.

I am still confused, I am writing down the equations blindly. :(

$P(t+\Delta t)=(M(t)+\Delta m)(v(t)+\Delta v)=M(t)v(t)+M(t)\Delta v+\Delta m v(t)+\Delta m\Delta v$
I think I can drop the last term. Do I have to calculate $P(t+\Delta t)-P(t)$? But I still don't have an expression for P(t). Sorry if I am missing something obvious.

 

[voko]

You have the expression for $P$. It is $Mv$. What you really want to know is the change of the momentum ($\Delta P$) during $\Delta t$. Then you need to figure out how that change is related to the sand flowing in.

 

[Saitama]

You have the expression for $P$. It is $Mv$. What you really want to know is the change of the momentum ($\Delta P$) during $\Delta t$. Then you need to figure out how that change is related to the sand flowing in.

Why P is Mv at time t? I mean, there is an influx of sand at time t also and you have not taken that into consideration.

I hope you understand what I am trying to say.

 

[voko]

Momentum is $mv$ by definition. The influx of the sand changes the momentum of the car over time, but the instantaneous momentum is given by that simple formula.

 

[Saitama]

Momentum is $mv$ by definition. The influx of the sand changes the momentum of the car over time, but the instantaneous momentum is given by that simple formula.

$$P(t+\Delta t)-P(t)=\Delta P=M(t)\Delta v+\Delta mv$$
Dividing both the sides by $\Delta t$
$$\frac{\Delta P}{\Delta t}=M(t)\frac{\Delta v}{\Delta t}+\frac{\Delta m}{\Delta t}v$$
Is $\Delta P/\Delta t=0$?

 

[voko]

Read #15 and #17 carefully.

 

[haruspex]

Is $\Delta P/\Delta t=0$?

Of course not. The incoming sand is increasing the momentum. What momentum does the incoming sand add in time Δt? Be careful to be consistent by using an inertial frame for this.

 

[TSny]

Why P is Mv at time t? I mean, there is an influx of sand at time t also and you have not taken that into consideration.

It is very important to define your system. You can take the system at time t to include the cart plus the sand that is about to enter during time dt. Or you can choose the system at time t to include only the cart and consider the sand that is about to enter as an external agent that will act on the system during the time dt. Either way is fine, but you just have to be clear on what choice you want to make.

 

[ehild]

In principle, you have inelastic collision between the car and sand, repeated.

First there is a car in rest, having mass M , and you throw sand in it, of mass Δm and velocity u=5 m/s. Apply conservation of momentum. What is the new velocity of the car?

You repeat the same process after Δt second, with the new mass M+Δm, new velocity v1, and the relative velocity of the sand u, and then again and again. The mass of the car increases by Δm in every step.

In this problem, the mass thrown at rate λ=10 kg/s, so the mass of the car is M(t)=M+tλ at time t, when its velocity is v(t). Sand of mass Δm=λΔt is thrown again, with relative velocity u=5m/s. The velocity of the car increases by Δv, so the new velocity is (v+Δv). Write up the momentum equation, and find Δv/Δt.

With the limit Δt-->0 it is dv/dt, and the equation becomes a differential equation for v(t).

ehild

 

[Saitama]

It is very important to define your system. You can take the system at time t to include the cart plus the sand that is about to enter during time dt. Or you can choose the system at time t to include only the cart and consider the sand that is about to enter as an external agent that will act on the system during the time dt. Either way is fine, but you just have to be clear on what choice you want to make.

I am really sorry, I should have read my text carefully before asking the questions here. A similar example in the book shows how to select a system.

I will go with choice A.

At time t, the momentum of system is $M(t)v(t)+\Delta m u$ where M(t) is the mass of car at time t.

At time t+Δt, the momentum of system is $(M(t)+\Delta m)(v(t)+\Delta v)$.

Conserving momentum, $M(t)v(t)+\Delta m u=M(t)v(t)+M(t)\Delta v+\Delta mv(t)+\Delta m\Delta v$
Rearranging (I am writing v instead of v(t)),
$$\Rightarrow \Delta m(u-v)=M(t)\Delta v$$
Dividing both the sides by $\Delta t$ and with the limit $\Delta t \rightarrow 0$, the equation becomes,
$$(u-v)\frac{dm}{dt}=M(t)\frac{dv}{dt}$$
Substituting M(t)=M+(dm/dt)t and solving the D.E, I end up with a strange answer.
$$v=-\frac{(dm/dt)tu}{M}$$

In principle, you have inelastic collision between the car and sand, repeated.

First there is a car in rest, having mass M , and you throw sand in it, of mass Δm and velocity u=5 m/s. Apply conservation of momentum. What is the new velocity of the car?

You repeat the same process after Δt second, with the new mass M+Δm, new velocity v1, and the relative velocity of the sand u, and then again and again. The mass of the car increases by Δm in every step.

In this problem, the mass thrown at rate λ=10 kg/s, so the mass of the car is M(t)=M+tλ at time t, when its velocity is v(t). Sand of mass Δm=λΔt is thrown again, with relative velocity u=5m/s. The velocity of the car increases by Δv, so the new velocity is (v+Δv). Write up the momentum equation, and find Δv/Δt.

With the limit Δt-->0 it is dv/dt, and the equation becomes a differential equation for v(t).

Thanks ehild for the wonderful explanation! I have used your hints in my working above, can you please see if I did it correctly?

 

[voko]

At time t, the momentum of system is $M(t)v(t)+\Delta m u$ where M(t) is the mass of car at time t.

In what frame is the velocity of the car $v$, and in what frame is the velocity of the sand $u$?

 

[Saitama]

In what frame is the velocity of the car $v$, and in what frame is the velocity of the sand $u$?

In the frame of locomotive.

 

[voko]

In the locomotive's frame, $v = 0$.

 

[ehild]

u is the velocity of the sand with respect to the locomotive. The car and the locomotive travel with the same velocity, so u is also the relative velocity of sand with respect to the car. You see the car from the ground, where the velocity of the car is v and the velocity of the sand is v+u.
Use the appropriate velocity for the sand in the equation

At time t, the momentum of system is M(t)v(t)+Δmu

ehild

 

[Saitama]

In the locomotive's frame, $v = 0$.

Oh yes, sorry about that.

Switching to the ground frame and as ehild suggest, the speed of sand is (v+u). Modifying the equation and solving the D.E, I get:
$$v=u\ln\left(\frac{M+(dm/dt) t}{M}\right)$$

But if I work in the locomotive frame and set v=0 in the equation:
$$(u-v)\frac{dm}{dt}=M(t)\frac{dv}{dt}$$
and solve the D.E, I get the same answer. I am confused on this part. I mean I set v=0 and solving gives me some definite value (or expression) for v. It doesn't look right to me.

 

[voko]

But if I work in the locomotive frame and set v=0 in the equation:
$$(u-v)\frac{dm}{dt}=M(t)\frac{dv}{dt}$$

At this stage you should find a way to interpret $\Delta v$ (and $\dot{v}$, too) in this frame.

You could say that the frame in which the equation is formed is instantaneously co-moving, and $v$ is the velocity of the frame. Then you should naturally get the same result.

Note the frame cannot be just always co-moving, because then it is non-inertial and you will need to take that into account in same way.

I would like to attract your attention to the equation you derived in #18. It is independent of this locomotive/car/sand story, and is applicable whenever you deal with a system where mass is not constant. Its differential form can be obtained directly, by differentiating $mv$.

 

[ehild]

Oh yes, sorry about that.

Switching to the ground frame and as ehild suggest, the speed of sand is (v+u). Modifying the equation and solving the D.E, I get:
$$v=u\ln\left(\frac{M+(dm/dt) t}{M}\right)$$

But if I work in the locomotive frame and set v=0 in the equation:
$$(u-v)\frac{dm}{dt}=M(t)\frac{dv}{dt}$$
and solve the D.E, I get the same answer. I am confused on this part. I mean I set v=0 and solving gives me some definite value (or expression) for v. It doesn't look right to me.

If you use the car as system of reference, you can not speak about the change of its velocity. In its own frame of reference,the velocity is identically zero.
It has no sense choosing a system as frame of reference, the motion of which you do not know. You want to find the velocity of the car with respect to the ground as function of time.

As voko said, choosing the car as frame of reference means a non-inertial frame. In that frame, the virtual force -ma acts on every mass m, and the equation you got from conservation of momentum is not valid.



ehild

 

[Saitama]

If you use the car as system of reference, you can not speak about the change of its velocity. In its own frame of reference,the velocity is identically zero.
It has no sense choosing a system as frame of reference, the motion of which you do not know. You want to find the velocity of the car with respect to the ground as function of time.

As voko said, choosing the car as frame of reference means a non-inertial frame. In that frame, the virtual force -ma acts on every mass m, and the equation you got from conservation of momentum is not valid.

Thanks ehild and voko! This was an interesting thread. :)