# Relationship between Newton's 2nd and Momentum?

1. Feb 24, 2010

### Glorzifen

1. The problem statement, all variables and given/known data
Rain is falling vertically into an open railroad cart which moves along a track at a constant speed. The engine must exert extra force on the cart as it collects water (the water is initially stationary be brought up to the speed of the train).

Calculate this force if v = 20m/s
Water collects at rate of 6kg per minute (=0.1kg/s)

2. Relevant equations
This is where I'm a little confused. I missed this lecture and my online notes confuse me. It says that Newton's 2nd Law can be re-written because, at a constant mass, $$\Delta$$(mv) = m *$$\Delta$$v. Based on this: F = $$\Delta$$(mv) if the mass is not constant. Is that right?

3. The attempt at a solution
If it's $$\Delta$$(mv) then won't the answer always be zero since V isn't changing...or am I just making a stupid mistake? Should it be:

(0.2kg/s*20m/s)-(0.1kg/s*20m/s) = 2N?

I appreciate any help I can get on this as I (clearly) don't fully understand it.

Thanks,

Graeme

2. Feb 24, 2010

### Lok

Well this problem has an answer looking like N/s, as the engine has to put more and more force on the cart as it fills. Knowing this is actually the solution. No need for derivates.

3. Feb 24, 2010

### Lok

Bad way of putting it as i look at what I wrote again. Basically you have a stream of 0.1kg of water that hits your cart at 20m/s in the time of one second. So F=m*v/s. Still no need for derivates.

4. Feb 24, 2010

### D H

Staff Emeritus
No, it doesn't. This problem is asking for a force, which has units of Newtons.
No, it doesn't.

5. Feb 26, 2010

### Lok

Yeah I was drunk with another problem where the train had to climb anyway I did correct myself :P

6. Feb 28, 2010

### Glorzifen

Thanks for the replies thus far.

I just want to make sure I understand what's going on here in regards to Newton's second law. The purpose of the question in the lecture slides was to demonstrate how Newton's 2nd can be used as F = dp/dt.

Firstly; my notes say that this is true for NON-CONSTANT mass systems (like this question). Wikipedia says, "Both statements of the second law are valid only for constant-mass systems". Could someone please confirm which is correct?

Secondly, if I am solving this train/rain problem with the F = dp/dt application of Newton's 2nd, then how do I it? Any solution I come up with doesn't seem to make sense since the mass is constantly changing over time...

7. Feb 28, 2010

### Staff: Mentor

Sounds like both statements are a bit ambiguous and could be improved. Better to say that F = dp/dt is true for both constant and non-constant mass systems, while F = ma is true only for constant mass systems.

F = dp/dt = d(MV)/dt = ??

8. Feb 28, 2010

### Glorzifen

Do I need to get the 6kg/min into kg/s? If so, it's 0.1kg/s

That means that the change in momentum (mv) is (0.1*20)/s = 2N of force that needs to be added?

9. Feb 28, 2010

### Staff: Mentor

Right.

Right.

F = d(mv)/dt = v*dm/dt (since v is constant) = 20*.1 = 2 N.

10. Feb 28, 2010

### Glorzifen

Okay. So because the mass changes, you use dm/dt whereas v is just a constant so its just v. I think I understand things a little more now; thanks.