Open Car fills with Rain Question

[Shadowness]

An open railroad car is rolling horizontally, without friction, in a heavy rainstorm, where all the raindrops are falling vertically.

 

1. What will happen to the horizontal speed of the car as it fills with rain? What will happen to its momentum?

 

2. When the raindrops hit the car, they lose their vertical momentum. How is momentum conservation working here?

 

** My answers

1. The horizontal cart will only increase in mass and its velocity will remain constant (assuming the rain does not slow down the car). Therefore its momentum increases.

2. My guess is that since the raindrops' momentum is zero after the collision, momentum of the entire system is conserved because the car increases momentum.

 

Are these answers correct or am I missing something?

[NeonBlack]

Are these answers correct or am I missing something?

I will try to help without giving anything away.

 

Remember that momentum is a vector.

 

Also, the famous F=ma is only true when mass is constant. In general, [math]F=ma+v\frac{dm}{dt}[/math]

[Bignose]

I think in both cases you are neglecting forces. You have to remember that the conservation of momentum equation contain common source terms: forces such as gravity, friction, etc. Whereas the conservation of mass equation only has source terms in really exotic situations -- i.e. nuclear events.

 

So, regarding your train and raindrop situation, when the rain falls and hit the car, it's vertical momentum is "used" to impart a force on the bottom of the car when the drop hits. Or, to think about it in an equal but opposite way, the bottom of the train imparts a force in the raindrop when the drop hits, and the net result of the train's upward force is a drop with zero velocity.

 

Momentum cannot just change directions. Each component of the momentum, the vertical and horizontal momentum are basically independent. Without a force acting to turn an object's momentum from vertical to horizontal, it won't just happen.

 

Which, coincidently, is what happens in the first case, the drop coming in has no horizontal momentum. The train will give up a little horizontal momentum by imparting a horizontal force on the drop of water so that the drop and the train will travel at the same speed. I am unsure if the pre-wetted train and the train with rain on it would still have the same momentum. Because the increasing mass and the decreasing velocity are multiplied together, I'm not sure if the two terms come back out. In the real world, there would be friction and inefficiencies of transfer that would definitely have lesser total momentum, but I am not sure about what could happen in a perfect frictionless 100% efficient world.

[Benjamin]

Bignose is on to something when he mentions vectors. Technically, the two vectors (Train car and rain drop(s)) are not in the same direction.

 

Rain pushes south, so to speak, while the train pushes east.

 

This gives you a final vector pointing south east. Now, the train is on a frictionless track, so it cannot go downward. The added down vector by the rain is canceled by the upward force of the car upon it, which has a now greater upward force from the track.

 

This leads me to reason out (although i can hardly believe it) that the train never changes x velocity, because there are no forces present to change x velocity. This would be very cool to test. : )

 

The problem is somewhat similar to a bowling ball rolling out of a plane, with no wind resistance. No matter how fast the ball drops, it still travels with a constant speed horizontally.

 

Also, i am assuming that the train car is not running into rain drops, because that would qualify in my book as a form of air resistance. The car only increases mass.

[Bignose]

This leads me to reason out (although i can hardly believe it) that the train never changes x velocity, because there are no forces present to change x velocity. This would be very cool to test. : )

 

No, the train will lose some x velocity. The rain drop that falls into the train will need to be accelerated up to the speed of the train. The rain drop isn't just going to stand still and leave. (Now a rain drop on the top of your car's roof... it will slide off, but we're talking about a drop of rain that falls into an open container. That drop will have to be accelerated, which is done by the train. Since the train uses some of its momentum to accelerate the drop, the train's speed will drop.

[Benjamin]

Can you explain that mathematically?

 

You are probably right though. Ah! I see my problem. I classified "hitting" drops as wind resistance. Yes, If the rain is falling straight down, the train would slow down.

 

If the rain is falling straight down relative to the train car, (not diaganally) Then the car would not slow down. It depends on who's perspective you are looking from.

[Shadowness]

hey guys, thanks for the helpful answers. I didn't mention this, but I believe this question is supposed to be easy. So I don't think the prof will expect us to concern the fact that the car loses momentum to change the momentum of the raindrops. I'm confused how momentum is to be conserved in this system though since the momenta are in different directions (as some of you guys pointed out).

[Benjamin]

Ask the prof if the rain is falling straight down at the same horizontal speed as the car, or if it is falling straight down realative to the earth.

 

You can never think to much.

[Shadowness]

Ask the prof if the rain is falling straight down at the same horizontal speed as the car, or if it is falling straight down realative to the earth.

 

You can never think to much.

 

From the way its worded in question, I take it as the rain falling straight downward.

[Benjamin]

Then Bignose is right.

 

Although, when i think about it, the Momentum is a trickier subject than speed. Does the increased mass compensate for the lost speed when maintaining inertia?

 

What exactly do you mean by momentum?

[Bignose]

hey guys, thanks for the helpful answers. I didn't mention this, but I believe this question is supposed to be easy. So I don't think the prof will expect us to concern the fact that the car loses momentum to change the momentum of the raindrops. I'm confused how momentum is to be conserved in this system though since the momenta are in different directions (as some of you guys pointed out).

 

Again, you have to remember that forces acting on an object are source/sinks in the "conservation of momentum."

 

Let me give you a very simple example: hold an object, any object completely still some distance above the floor. Momentum ([math]\mathbf{p}[/math]) is [math]\mathbf{p} = m \mathbf{v}[/math] where m = mass of the object, [math]\mathbf{v}[/math] is the velocity. (p.s. Benjamin, this has to be exactly what was meant by momentum, or the teacher is teaching something that is not the otherwise common, accepted definition that is used by all other physicists everywhere.)

 

So, that object is held still, which means that the velocity is zero, hence there is zero momentum. Drop that object. Obviously it gains speed before it hits the ground, so there is a non-zero velocity, which means that there is now some momentum. The object went from zero momentum to some momentum... did we create some? Did we violate the conservation of momentum? No, an outside force, gravity in this case, acted on the object. In this case a force is a source of momentum. Gravity can also be a sink of momentum, if you throw that object straight up, it has some momentum as it leaves your hand, but eventually gravity pulls it back down. There is some instant of time when its velocity is exactly zero... in this case the gravity force acted as a sink taking away momentum; the momentum the object had when it left your hand was reduced to zero when it reached its peak.

 

So, yes, there is a conservation of momentum, but you have to remember that forces can change how much momentum an object has. The total momentum of a system is perfectly conserved if there are no forces. Such as perfectly elastic collisions (billiard balls on a frictionless table are a classic example, noble gas molecule collisions is a better one).

 

Finally, I would personally say that the concept that the train loses momentum so that the raindrop's speed is that same as the train's is exactly the point of acting the problem. I think that it is a very important non-negligible point, but maybe I am getting ahead of myself. Again, I think that the key word here is that the train car is described as "open" which means that it collects the rain, rather than letting the rain roll off the back like what would happen on a closed car.

[swansont]

hey guys, thanks for the helpful answers. I didn't mention this, but I believe this question is supposed to be easy. So I don't think the prof will expect us to concern the fact that the car loses momentum to change the momentum of the raindrops. I'm confused how momentum is to be conserved in this system though since the momenta are in different directions (as some of you guys pointed out).

 

Actually, I think that concerning yourself with the conservation of momentum of the drops + car is exactly what the prof expects. You've gotten several pointers from the different posters (though there have been one or two off-base comments in the mix)

 

If you are interested in the x-motion of the train, you can ignore anything happening in an orthogonal direction, i.e. the vertical force of gravity has no effect. (The presence of gravity means that the y-component of momentum will not be conserved for the raindrops)

 

Whether you consider momentum to be conserved in the x-direction depends on how you define the system. If it's the car + the water in the car only, then there is a force. When a drop of water is added, it must be accelerated to the speed of the car; the car exerts a force on it, so it must exert a force on the car. But if the system is the car + the all of the water, then the total momentum must be conserved in the x-direction; the falling rain has none, so the car + collected water momentum must remain constant.

 

Neonblack gave the equation that applies here.

[igosaur]

I would have thought that the car would slow down as it is having to convert the raindrops into a horizontal momentum.

 

You could imagine that as a raindrop falls down into the car, the car will then have to drag that drop forwards. This will cause friction and therefore it must slow down.

 

Surely, the downward energy of the raindrop would be converted to heat, sound etc. as it would if the car were stationary and so the energy needed to make the raindrop go forwards would have to come from the rail car, thus, slowing it down.

 

An open railroad car is rolling horizontally, without friction, in a heavy rainstorm, where all the raindrops are falling vertically.

 

1. What will happen to the horizontal speed of the car as it fills with rain? What will happen to its momentum?

 

2. When the raindrops hit the car, they lose their vertical momentum. How is momentum conservation working here?

 

** My answers

1. The horizontal cart will only increase in mass and its velocity will remain constant (assuming the rain does not slow down the car). Therefore its momentum increases.

2. My guess is that since the raindrops' momentum is zero after the collision, momentum of the entire system is conserved because the car increases momentum.

 

Are these answers correct or am I missing something?

[D H]

If you are interested in the x-motion of the train, you can ignore anything happening in an orthogonal direction, i.e. the vertical force of gravity has no effect. (The presence of gravity means that the y-component of momentum will not be conserved for the raindrops)

Even the y-component of momentum is conserved if one observes from a proper inertial frame. The Earth-fixed frame is not inertial because it is accelerating toward the raindrops as they fall to the Earth. Consider a single raindrop that is just about to hit the railcar. In the Earth+car+raindrop center of mass frame, the raindrop's downward momentum is balanced by the Earth+railcar's upward momentum. The net vertical component of momentum is zero. After the collision, the y-components of the Earth's and railcar+raindrop's momentum are zero. Momentum is conserved.

[Mr Skeptic]

An open railroad car is rolling horizontally, without friction, in a heavy rainstorm, where all the raindrops are falling vertically.

 

1. What will happen to the horizontal speed of the car as it fills with rain? What will happen to its momentum?

 

2. When the raindrops hit the car, they lose their vertical momentum. How is momentum conservation working here?

 

1. The horizontal momentum of the car+raindrops will remain unchanged, even as the car's mass increases. This means its velocity will decrease.

 

2. The raindrops are pushing the earth downward. This is exactly canceled by the gravitational attraction pulling the earth and raindrops together.

 

You can model this as a series of inelastic collisions; the raindrops and car will lose kinetic energy (which will be converted into heat), and stick together. If you want to be really exact, consider that some drops will also hit the front of the car, slowing it down without filling it up, but you did not give enough information to figure that in.

[pioneer]

Since the rain is falling downward, vertically, and not adding any horizontal force, the addition of the rain, will simply add mass to the train, such that the total mass will increase. If we assume the kinetic energy is conserved, one divides the original knientic energy, by the total mass (rain plus train), and resolves the equation for the new velocity.

 

One way to see what is happening, the rain has no horizontal velocity component when it enters the train. When it enters the train, we need to accelerate this rain up to train speed. Some of the train's kinetic energy will go into the rain, for that acceleration, lowering the original energy of the train. The total kinetic energy will be the same, only now it is distributed in both the train and the rain, causing the combo to go slower.

[swansont]

Since the rain is falling downward, vertically, and not adding any horizontal force, the addition of the rain, will simply add mass to the train, such that the total mass will increase. If we assume the kinetic energy is conserved, one divides the original knientic energy, by the total mass (rain plus train), and resolves the equation for the new velocity.

 

One way to see what is happening, the rain has no horizontal velocity component when it enters the train. When it enters the train, we need to accelerate this rain up to train speed. Some of the train's kinetic energy will go into the rain, for that acceleration, lowering the original energy of the train. The total kinetic energy will be the same, only now it is distributed in both the train and the rain, causing the combo to go slower.

 

On what basis would you assume that kinetic energy remains constant? It's not an inherently conserved quantity, and in fact will not remain the same. (KE = p²/2m, and m is not constant)

[pioneer]

I was using the conservation of energy. One could make this example fancier by subtracting heat due to friction to recalculate the final kinetic energy. The same analysis could be done with momentum, with the total momentum conserved. The final mass is higher so velocity lowers.

 

If our train track was like a roller coaster ride, with hills and valleys, one could also subtract or add, the potential energy of the gravity, to get the more instantaneous velocity, as the rain accumulates in the train.

 

If we start to add turns, since the water is fluid, its momentum will keep going forward while the more solid train corners. If the water didn't spill, we would have to reaccelerate the water into a new direction, causing some of the kinetic energy to be used up, to become the gravity potential within the waves that we would create. This would settle as heat. The conservation of energy makes it easier to scale up to complexity.

[swansont]

I was using the conservation of energy. One could make this example fancier by subtracting heat due to friction to recalculate the final kinetic energy. The same analysis could be done with momentum, with the total momentum conserved. The final mass is higher so velocity lowers.

 

Conservation of energy and conservation of kinetic energy are two very different things. It won't be the same analysis as conservation of momentum; you'll get the wrong answer if you assume KE is conserved.

[pioneer]

You are right. It is strange, that kinetic energy and momentum would result in two different velocity answers, but it does. If we doubled the mass and kept the same momentum, the velocity would fall to half. Or MV=2Mv ; v=V/2.

 

If the did the same for kinetic energy, we would have (1/2)MV2=(1/2) 2Mv2. Or v=V/1.414. So conservation of kinetic energy results in higher final velocity than the conservation of momentum?

 

I think I see the energy problem. When the water enters the train, we need to accelerate it to train speed. This will take energy, so we need to convert some of the velocity based kinetic energy into an acceleration based force-work-energy. Some of the original kinetic energy is converted, so the final kinetic energy is lower. I guess if we subtract the kinetic energy calculated for the two final velocities, i.e., momentum and kinetic, that gives us a handle on the force-work energy.

[Mr Skeptic]

pioneer, keep in mind that thermal energy is basically kinetic energy, so that kinetic energy is also conserved as heat plus kinetic energy. That is, if you modeled each individual particle, you would conserve momentum and kinetic energy, but some of the kinetic energy is in the form of heat (random particle movement as compared to the overall movement of the train+water).