# Force, Impulse and a Changing Mass system

1. Jun 26, 2015

### Beno123

1. The problem statement, all variables and given/known data
A long wagon with mass m0 and inital velocity v0 on a frictionless floor.
On time t=0 we open a faucet which starts dripping water in a given pace Q=Δm\Δt=α.

1. What's the speed of the wagon on time t.
2. Answer 1. if on t=0 the wagon is full of water and it's mass is M0

2. Relevant equations
F=dp\dt

3. The attempt at a solution
It was actually an example shown in a recorded lecture course I'm taking. It's a part of a set of examples about changing mass systems.

F=(dm\dt)v=(-αv) (because α is given and the professor said that the minus sign is because of the diretion of the force acting on the wagon).
F=ma=m(dv\dt) (newton's second law)
⇒ -αv=m(dv\dt)
m = m0 + αt (wagon's initial mass + change)
⇒dv\v=(-αdt)\(m0+αt)
integrate both sides between v0 and v and 0 to t and get:
v=(m0v0)\(m0+αt)

I kind of understand what's going on here, but not really.
Why F=(dm\dt)v instead of deriving both v and m using the chain rule [(dm\dt)v + (dv\dt)m)]? both are changing, no? the wagon gains mass but also loses velocity to the water drops it accelerates.

The wagon's mass is now constant (M0), but there's still a change in the momentum in order to accelerate the new water drops which fall down on it (the excess water fall of the wagon):
F=(dm\dt)v=(-αv)
F=ma=M0(dv\dt)
-αv=M0(dv\dt)
integrate both sides ang get:
V=V0e-αt\M0

I don't understand why the following equation: F=(dm\dt)v=(-αv) is valid in this situtation.
dm\dt = 0 cause the mass of the wagon is constant, no? the only thing that changes is the velocity of the water drops which I'm not sure how to calculate (the wagon accelerates the water drops but how can I know the magnitude? well, it doesn't matter cause it seems it's not the right answer anyway).

The main thing I want to understand is the logic behind all of this. I guess I don't have a good enough grasp of this concept.

Thanks and let me know if what I wrote down here is clear enough!

2. Jun 26, 2015

### Orodruin

Staff Emeritus
I would say this is a confusing explanation at best. Generally, force will be defined as the rate of change in momentum F = dp/dt, which is zero in this case (the added water adds no momentum). Performing the derivative of p=mv gives ma + v(dm/dt) = 0. Taking the second term to the right hand side gives you the appropriate equation for a.

For the second case, you would use that dp/dt is no longer zero, since momentum is being transferred away from the system due to water spilling out and taking momentum with it.

3. Jun 26, 2015

### Orodruin

Staff Emeritus
Just as a side note, people sometimes take the v(dm/dt) term to the right hand side from the start and call it a "force". This gives the minus sign and preserves F=ma.

4. Jun 26, 2015

### SammyS

Staff Emeritus
Hello Beno123. Welcome to PF !

If that is the complete problem statement , then it appears to me to be a poorly constructed problem.

No mention is made of the velocity at which the water expelled from the wagon. -- Direction and magnitude being very important. (Maybe there was an illustration which accompanied the problem.)

As regards Newton's Second Law: It can be more generally written as $\displaystyle \ \vec{F}=\frac{d\vec{p}}{dt}=\frac{d}{dt}(m\vec{v}) \ .\$

In most cases the mass is constant giving $\ \vec{F}=m\vec a \ .\$ In this case the mass changes.

5. Jun 27, 2015

### Beno123

Hi, first of all thank you both for the warm welcoming and helpful replies.

But how can I do that if I don't know how much momentum is being transferred? the professor said that the first case may be possible to derive using conservation of momentum [m0v0=(m0+αt)v] and integration, but not for the second case.

What's the logic behind the way he derived the second answer? should I always take the derivative of the momentum using the chain rule when there's a mass change [(dm\dt)v + (dv\dt)m = F] and work out the math from there? cause there is a mass change, hence the dm\dt and a velocity change because of it, hence dv\dt (or I got it complete wrong?)
I hope to get the concepts and logic behind it right because, the way I see it, understanding what you're doing and each part of the equations is the key to understanding physics. I don't want to just memorize the algorithms for solving questions (which is also the harder and longer way anyway).

I agree, and having the professor skipping stages and not being able to explain himself clearly doesn't help either. I wrote all of the relevant parts from the lecture after watching it 2-3 times.
He drew an illustration but it didn't hold any new information. Just a rectangle as the wagon with an arrow for the velocity and a water tank above it. On the second part of the question he drew the rectangle full of water and water spilling out implying it doesn't change it's mass but no clue about their direction, it seems from the explanation like the water are dropping vertically and thus no velocity is lost in the horizontal direction. There's a momentum change because the wagon accelerates the new water from the tank from 0 velocity the the new velocity V in the horizontal direction.

Am I making sense?
How come momentum is conserved on the first question but not on the second one?
He said you can't solve the second one with conservation of momentum cause some of it is lost due to the water spilling out.

I hope I'm clear enough, cause actually I'm rather confused myself. Feel free to ask for clarifications, I can re-watch specific parts of the lectures if you think I forgot to include something which may be there.
If you know where I can find some questions of this sort to practice on in order to be able to get a good grasp on the concept myself it would be great. I just want that AHA! moment.

Thanks again.

6. Jun 27, 2015

### Orodruin

Staff Emeritus
You cannot. Luckily, you do know (in both cases). In the first case, the water added has no momentum before it hits the wagon (it is not moving!). In the second case, the added water still does not move, but the water spilling out is assumed to have the same velocity as the wagon (without this asumption or a similar one, the problem is not solvable, as noted by SammyS).

I would still write it as dp/dt = (dm/dt) v + ma. In this case there is no change in mass since water is going to be spilling out at the same rate as it is added so dm/dt = 0. The differential equation comes from the change in momentum due to momentum being carried away by the water sipping out.

There is a change in velocity because the new water has to be accelerated to the wagon velocity and momentum is lost by the water already having the velocity of the wagon sipping out of the system, taking momentum with it.

7. Aug 6, 2015

### Beno123

I still don't get how to solve the second question if momentum is obviously not conserved (momentum is carried out by the slipping water).
If I use dp/dt = (dm/dt) v + ma I can't have it equal to zero because there IS a change in momentum. So how do I go on from here? have the change in momentum equal -αv? (the rate of water spilling out times the water's velocity).

8. Aug 6, 2015

### haruspex

The spilt water is part of the total system if you want to be able to use momentum conservation, so yes, write dp/dt = ma + αv = 0.
In general, you might have water dripping in at rate α and horizontal velocity u, and leaking out at rate $\beta$:
$\dot p = \alpha u = (m_0+\alpha t - \beta t) \dot v + (\alpha - \beta )v + \beta v = (m_0+\alpha t - \beta t) \dot v + \alpha v$

Last edited by a moderator: Aug 6, 2015
9. Aug 6, 2015

### Beno123

Hi, thanks for replying and sorry for being rather slow in getting the point.

But why do we care about the total system? I mean, after some water spill out of the cart, they carry out momentum and the velocity of the cart must change, right? The cart is the object we are interested in, so how can I say dp/dt=0 for the cart as well?
The spilling water take away momentum and die on the floor while the "new" water dripping on the cart are replacing them (and thus conserve the mass, but need to be accelerated).

The question didn't say anything about dripping rate so I assume they meant it's the same as the spilling rate α.
You can check out the way the professor solved it at class (a recorded lesson, unfortunately, that's why I have to rely on good guys online to help me out with questions.) He is rather obscure, to say the least.

Thanks again!

10. Aug 6, 2015

### haruspex

You care about the total system if you want to use conservation of momentum. The alternative is to account for momentum added to or removed from the subsystem.
In this problem, yes. I was just showing how it would generalise.

11. Aug 18, 2015

### Beno123

But the water carried out of the wagon loses it's momentum when it hits the floor. The system is no longer closed and isolated.

I still can't get my hand wrapped around this problem.

12. Aug 18, 2015

### haruspex

Does the cart know or care what happens to the water once it has left?

13. Aug 21, 2015

### Beno123

No, so you would use conservation of momentum for both parts of the question?

14. Aug 21, 2015

Yes.