What Causes the Velocity of a Cable Car to Change in the Rain?

[peripatein]

Hi,

Homework Statement

A cable car slides without any friction along a horizontal wire with initial velocity v0. The car is half filled with water and its mass (with the water) is m. At t=0 it starts raining (perpendicular to the ground). The rate at which water is collected in the car is k (=mass/time). At that point the driver opens an aperture at the bottom of the car and water begins leaking so that at any time the amount of water in the car is constant (the amount of water due to rain equals to the amount leaking from the car as it slides). What is the velocity of the car after a time t?

Homework Equations

The Attempt at a Solution

There is conservation of momentum along the x-axis and there are no forces along that axis acting on the car, so why would the velocity change? Why wouldn't v(t)=v0?
The solution is dP_x/dt = 0 = -kv(t) + m*dv/dt. Well, that should be the equation yielding the solution in any case.
I also don't quite understand why dm/dt would be k. Isn't it stated that the amount of water remains constant? Why wouldn't that derivative be equal to zero?
I am kindly asking for an explanation.

 

[BvU]

Weird story. Never mind.

"There is conservation of momentum along the x-axis ". So what is that momentum along the x-axis if you also take the rain into consideration ?

 

[peripatein]

Why wouldn't it still be m*v0?

 

[BvU]

rain is falling straight down is what it says. Once in the car, it's forced to join the ride!

 

[peripatein]

But for every drop falling in there is a drop falling out so that m remains constant. If I understand the problem correctly...

 

[haruspex]

But for every drop falling in there is a drop falling out so that m remains constant. If I understand the problem correctly...

When a drop falls out, does it immediately revert to falling straight down?

 

[peripatein]

Are you implying that it only happens upon having "transmitted" its momentum to the car?

 

[peripatein]

The momentum of each drop is only in the y-axis, is it not? Hence, each drop could only change the momentum of the car in the x-axis by means of increasing the latter's mass. However, if that delta mass immediately leaks out of the car, I still don't quite understand how the momentum in the x-axis does not remain the same.

 

[BvU]

Water that falls in needs to be accelerated to the speed of the car. Water that falls out has x-momentum, which is lost (it doesn't fall vertically down, even though it looks like that when you see this leaking car pass by because you tend to follow the motion with your eyes)

 

[BvU]

Ok, bedtime for me. haru peeker will accompany you, I hope. reconsider post #2.

Just as a hint: what happens to a rapidly passing skateboard if you jump on it from aside? will it keep going at the same speed? It's a hefty exaggeration, I agree, but the similarity with our wetcar is present.

 

[haruspex]

The momentum of each drop is only in the y-axis, is it not? Hence, each drop could only change the momentum of the car in the x-axis by means of increasing the latter's mass.

It increases the mass, yes, but since it is not contributing any X momentum it cannot alter the total X momentum of car+tank+gained drop. Hence it does change the speed of the car.

However, if that delta mass immediately leaks out of the car, I still don't quite understand how the momentum in the x-axis does not remain the same.

It remains the same for car+tank+lost drop. What is the X momentum of the lost drop? What momentum does that leave for car+tank?

 

[peripatein]

I am not sure whether you've noted, but I have already written down the equation yielding the correct answer. I simply do not understand why that equation is correct. Why is dm/dt=k if for every drop that gets in the tank (or car), a drop leaks out?

 

[peripatein]

Oh, and of course the prime question - why wouldn't the velocity just stay v0??

 

[voko]

Consider two simpler cases.

If there were no rain, but the water were leaking out: would the velocity of the car change?

If there were no leaking out, but rain water were being collected: would the velocity of the car change?

 

[peripatein]

Were there no rain with water leaking out, I don't think the velocity of the car would change. Although, its mass would decrease so if kinetic energy remains constant its velocity would have to increase. I am vacillating.

 

[voko]

Why should the kinetic energy of the car and remaining water include the kinetic energy of water leaking out?

 

[peripatein]

If 0.5*m*v^2=const., then decrease in mass of car+remaining water would entail increase in velocity of car+remaining water. Wouldn't it?

 

[voko]

Let the car be made of two sections, with a door between them. Now when the door is closed, application of your reasoning results in: OK, the mass of water in the front section is less than the mass in the entire car, so its velocity must increase. It should increase in the rear section by the same token. So the entire car's velocity should increase, all by simply closing a door.

But it gets even better than that. Assume there is an imaginary door. Surely the mass of water between the front wall and the imaginary door is less than, etc, so the car's velocity goes on increasing just by itself. Does that not seem strange to you?

 

[BvU]

Difficult thought experiment. Let me propose a simpler one, with the advantage that it's more realistic and the passengers stay dry: The rain falls straight down and k' kg/sec hits the front of the car. At some angle, never mind, that's in the k'. These k' kg/sec have to pick up speed v(t) very quickly, after which they trickle down and fall off - with a horizontal speed v(t). Recognize the parallel? So the cable car - with the passengers safe and dry inside - has to do work on the rain : k' kg/sec * v(t) is the momentum that has to be produced every second. Agree ?
Only way it can do that is by exercising a force F = k' v and hence experiencing an equal and opposite reaction force that slows it down. Acceleration F/m with a decent minus sign.

Much better story than this weird opening of a nonsensical aperture that happens to keep the car precisely half full. Who wants to drive such a car ? Let alone ride it ?

The energy side of this (top of my head, correct me if I'm wrong): kinetic energy dished out k'v²/2 every second. Just falls off the car.
Work done by F : k' v * v . Matches ${d \over dt} {1\over 2}m v^2 = m v {dv \over dt} = v F = k' v^2$
New mystery: what happened to the other half of that ?

 

[haruspex]

If 0.5*m*v^2=const., then decrease in mass of car+remaining water would entail increase in velocity of car+remaining water. Wouldn't it?

You seem to keep missing the point that the rain falling in has, originally, no X velocity; that it gains X velocity from the car+tank; that it keeps that X-velocity when it leaves. This saps the car+tank of both energy and momentum. (Further, there is energy lost when the drop enters the tank because this is effectively an inelastic collision.)

 

[peripatein]

I do understand that each drop gains X-velocity from the car and keeps it when it leaves. I'd appreciate a more focused explanation on the equation I wrote in my first post.
Let me try this:
Each drop that leaves the tank+car "deprives" the car+tank of dm*V momentum in the X direction. Ergo, -k*V=(dV/dt)*M.
Am I in the right direction?

 

[voko]

So, if each drop falling out deprives the car of momentum/energy, will the car's velocity change, if there is no water falling in?

 

[haruspex]

I do understand that each drop gains X-velocity from the car and keeps it when it leaves.

That is not supported by your post #17, where you suggest that energy of car+tank remains constant.

I'd appreciate a more focused explanation on the equation I wrote in my first post.

The equation dp/dt = v dm/dt + m dv/dt causes much confusion. It gives the change in momentum of a subsystem if it undergoes a small change in mass and a small change in velocity. Using conservation of momentum to write dp/dt = 0 is only valid in an isolated system. Since isolated systems cannot change mass, that always corresponds to dm/dt =0.

The equation quoted in the OP is dPx/dt = 0 = -kv(t) + m*dv/dt, but how exactly are the variables defined? It could be like this:
- Px is momentum of the whole system
- Pc = m v is momentum of the car+tank
- Pw is momentum of the water external to the tank
0 = dPx/dt = d(mv)/dt + dPw/dt = m dv/dt + k v
Note that the sign of the k v term does not match the OP equation.

If you want to define the variables in the equation some other way, tell us how.