A water tanker with a hole
A water tanker is rolling on a horizontal, frictionless surface and continuously leaks water from a hole in the bottom.
Over time, the mass of the tanker decreases and its speed __________ .
Assume
that the water drips such that it doesn't affect the tires.
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12 solutions
On this problem, I thought I could use the conservation of momentum, so, since the net external force is zero, I could think of it as m1 v1 = m2 v2 (momentum before equals momentum after). Since water leaks from the tanker, the mass decreases over time, which means m2<m1 and, therefore, v2>v1. What is wrong with this line of though?
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force depend on rate of change of momentum and this rate is the sum of rate of change of mass into velocity plus rate of change of velocity into mass.Here mass is changing so velocity must change at equal rate to make net force zero. Then how can we say velocity is not changing
As Mike Perry pointed out, while the mass of the water in the tanker decreases, the mass of the overall system , which is the tanker and all of the water, does not change. Once the water reaches the floor, no net force acts upon it, so it retains its previous horizontal velocity it had while in the tanker. Therefore, by conservation of momentum, the tanker with its remaining water should also retain its velocity.
Yes, the momentum is conserved but the water that comes out the hole carries its momentum with it that is why the tank keeps moving at a constant speed even when its mass is decreasing.
To elaborate further, the velocity of the leaking water wrt to tank is vertical and hence does not exert any force in the horizontal direction.
@Thales Conta- The water that drips out also keeps moving, so the mass of the total system actually stays the same.
If a change in Mass has no effect on velocity, if the leak was fixed and a similar tanker was running above it at the same speed, leaking into the tanker we are discussing, it would follow that we could move an infinitely large Mass with the same energy.
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Well, an object can't h e kinetic energy without some work being done on it. The implicit assumption here (which isn't necessary to solve the problem) is that some external force had been acting on the tanker, but it stopped at some point in time.
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I agree. So the Mass of the tanker is in movement because of the original external force. That force being sufficient to create the current momentum. Why would increasing the Mass have a different effect to decreasing it, If decreasing the Mass has no effect?
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@Ian Dunbar – The same reasoning works for increasing mass as well. If the extra mass being that is being added has the same horizontal velocity as the tanker, than speed of the tanker would indeed remain constant. If the horizontal velocity of the extra mass is not the same as that of the tanker, then the velocity of tanker would change.
@Thales Conta, at the moment the water drips out, the droplet had the same velocity v. You can imagine the droplet of water continue to move at v till it hits the ground. So momentum is conserved this way
The problem doesn't state that the velocity is constant. We know that the force acting on the tanker is NOT gravity because it is on a horizontal surface. Some force must be pulling the tanker in order for it to move. If the force pulling on the tanker is constant, the tanker speed will increase since the same force required to begin motion at mass A, velocity will increase at that same force at some mass A minus mass water. The only way for the tanker to maintain constant velocity is for the force that is causing motion to decrease along with the mass, i.e., the force must be variable. The problem statement doesn't state that either. I propose this solution is incorrect. Why even ask this question if you have to assume the velocity is constant? The whole question becomes pointless! You could ask this question about anything, as long as you assume the velocity is constant, the answer will ALWAYS be Velocity Remains The Same (i.e., velocity is constant). If you assume the velocity is constant, the question is redundant and doesn't really makes sense. All you are asking at that point is if the speed changes when the velocity is constant. My 6 year old knows that. Not a Brilliant question.
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We can have something moving at a constant velocity without any net forces acting on it, if it was accelerated to that velocity in some way. For example, somebody could've pushed the tanker across the floor, then stopped after some time. Net force becomes zero, but the tanker doesn't suddenly stop. A force is only required to accelerate an object.
The track is frictionless, not the tanker. Wouldn't the tanker resistance slow itself down
Conservation of Linear Momentum: m * v = P a constant. If m is reduced then v, the velocity, has to increase in order to keep P unchanged.
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As @Steven Yuan has pointed out, if you take the system to be the water in the tanker + the water that has just leaked out, then the net force on the system is zero, so momentum is conserved. Even after the water is leaked out of the tanker, it will carry with it the velocity of the tanker. Note that in this case since the mass and momentum of the system both remain constant, the velocity will also remain constant.
If you are looking at only the water in the tanker, then the water that is exiting from the hole is exerting some horizontal force on the water in the tanker. The net force on the system is not zero, so we cannot apply conservation of momentum directly.
If the water had been exiting through a hole parallel to the horizontal plane through which the tanker moves can we say that the velocity will change? As the initial momentum is worth pv in that direction and must be maintained. If some momentum is being propagated negatively through this plane then the original momentum must compensate for the lost mass by an increased velocity, no?
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If you only looking at water in the tanker, then it is a variable mass system. Using Newton's second law accounting for the varying mass, we get F e x t = m d t d v + v r e l d t d m . If you consider the horizontal components only, then F e x t and v r e l = 0 are both zero. This implies that d t d v is also zero, which means that v is constant.
You say the surface is frictionless, so the water keeps moving and therefore momentum is conserved. But what about the adhesive force of the water to the ground- that would slow down the water and eventually make it stop.
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Adhesion acts in the vertical direction, not the horizontal direction. Therefore, it does not affect the horizontal velocity of the water. If you were talking about friction, it's stated in the problem there is none.
Just a stupid question if the power implemented on the tank doesn't change then it should go faster right? because there we no longer need the same power for the lower amount of mass we have so to say more power implementation pro unit of mass because is escaping example what happens if u use a 100 kg of fuel to accelerate to objects one of a mass 20 kg and second of 10 kg one should accelerate twice as fast or maintain twice as long the constant speed? ( let's say that the fuel is not counted as an extra mass like by rockets it's only the object self that on which we implementing the power/force ) or I am just stupid?
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The tank is moving freely without burning any fuel. There is no external force on the tank in the horizontal direction.
yes ,man ur right
Since there is no motive force, positive or negative, (other than inertia and air resistance) acting upon the tank, the tank will slowdown, due to the air resistance. (The framers of the question forgot that minor detail). I answered "Remain the same", because that is the correct answer if you don't have air resistance. Unfortunately for this particular problem, there is air, and therefore air resistance, as evidenced by the fact that the water does not boil immediately upon leaving the tank. And yes, I know "air" doesn't have to mean the breathable stuff. Any gaseous fluid will do.
True. However, considering that the exercise already demands the virtually impossible scenario of frictionless contact, it's easy to infer that air resistance is also magicked away to make this a perfect study case of Newton's first law and inertia.
force depend on rate of change of momentum and this rate is the sum of rate of change of mass into velocity plus rate of change of velocity into mass.Here mass is changing so velocity must change at equal rate to make net force zero. Then how can we say velocity is not changing
Isn't air resistance a form of friction? (Air friction?|)
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Nope, that argument falls through its own vanishing point. it's not a part of the specific "frictionless surface" upon which said tank is rolling. There was no mention that there would be "Frictionless air" also. As I stated earlier, YOU MUST ASSUME THE EXISTENCE OF A NON-VACUUM ENVIRONMENT, otherwise the entire premise for this question crumbles into space dust.
The defined situation is of a frictionless surface only and does not suggest a vacuum. As David has pointed out for the water to exist there must be an atmosphere, which must produce some resistance. In consequence, the correct answer should be that the tank slows down.
When water drops go out of the tanker, it remains the same speed as the tanker, that means water drops doesn't affect the tanker to remains the same speed.
Imagine a rock breaking in half. Neither half of the rock would accelerate or decelerate as a result of that. The same is true of the tanker and the water. They are simply separating from each other, each taking their contribution to momentum with them and keeping the velocity constant.
I agree with Eric. Would be nice to get an explanation on why that answer would be wrong.
In a closed system, energy is constent. Here this system has an kynetic energy(KE) KE= m ((mass) x (velocity)^2) / 2 So, if mass decrease, the second factor (velocity) will increase. Therefore, the answer must be "increased" about velocity.
could it be stated that kinetic energy " 1/2 . m . v^2 " varies even though there isn't force producing a work on the mass?
force depend on rate of change of momentum and this rate is the sum of rate of change of mass into velocity plus rate of change of velocity into mass.Here mass is changing so velocity must change at equal rate to make net force zero. Then how can we say velocity is not changing
I also believe it is incorrect. I reported the problem and suggest others do the same. Eric's math is absolutely correct here.
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The answer needs an explanation from the questionnaire
I thought that as the weight decreases the tank would go faster.
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well, it lost energy at the same time cuz water drops still move front.
Isn't it commonly known that an object sliding down a hill will travel at a constant speed unless weigh is added or taken away? In this case, the water decreases thus making the weight lighter and increasing the speed
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No. It is not true that an object sliding down a hill will travel at a constant speed. It will accelerate and a constant rate, provided the slope is constant and downhill, due to the accelerating force of gravity. But in this case nothing is moving down hill. This is across a flat surface, I believe.
Also, just for the sake of variety, as the water leaves some of the tank, wouldn't some of the rest slosh against the walls of the tanker thus pushing the tanker forward and increasing the speed?
The frictionless surface made the resisting force zero. Since the body is acted by zero force ( i.e. F=0) so, according to Newton's 1st law of motion, the state of motion remains the same and continues to move with constant velocity.
The Net Force on the body remains 0, but its mass is changing. d t d m v + d t d v m = 0
d t − d m v = d t d v m
In this case , d t d m < 0 , m > 0 , and v > 0
Thus, d t d v > 0
I think the answer to this problem is incorrect. The velocity increases.
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Yes, the momentum is conserved but the water that comes out the hole carries its momentum with it that is why the tank keeps moving at a constant speed even when its mass is decreasing.
To elaborate further, the velocity of the leaking water wrt to tank is vertical and hence does not exert any force in the horizontal direction.
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I see now... I was hung up on the changing mass, but missed the fact that the ejecta is normal to the direction of motion. Thanks Rohit.
But the net force on the body is not zero! The water that leaves the tanker exerts force on the water inside the tanker such that even though the mass in the tanker decreases, the speed remains constant.
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I have changed my position on the problem, but I don't agree with your statement: "But the net force on the body is not zero!" The Net Force on the body must surely be zero.
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@Eric Roberts – I agree, I was mistaken. The net force on the body is indeed zero.
This is actually quite a complex question despite its seeming simplicity. I have taken the time to spell out my thoughts on the question. I hope this helps.
Intuitively:
We are told that tanker holds the body of water and that entire system is moving at a constant velocity. Therefore in the rest frame of the tanker and the water, we simply have a tanker containing water at rest. Basically imagine that this tanker is a fish bowl on a train. If it is allowed to drain, does the bowl begin moving to right or to the left? It does not. In this picture we clearly see that the water remains under the bowl so that it moves at the same velocity as the bowl (once it has stopped spreading out).
More technically, there are two possible answers here as far as I can see: the typical one which requires a considerable amount of detail to justify and is still rather unsatisfactory, and another which is more elegant. I will provide both.
Elegant solution:
Consider the following thought experiment. Let another tanker be rolling under the original such that both have the same velocity, and let them be connected by means of some tube. Since there are no dissipative forces, the total horizontal momentum of the system is conserved so that the motion of the centre of mass is unchanged. Here we are faced with two possible outcomes- either the tankers change velocity (one goes faster and the other slower) such that the total horizontal momentum of the combined system is conserved or the tankers do not change velocity so that the total horizontal momentum is once again conserved.
How do we decide which situation occurs? We call to our aid the conservation of angular momentum about the centre of mass. Initially the centre of mass of each tanker-water system, say m1 and m2 and the combined centre of mass M, are at rest in the rest frame of the centre of mass such that the initial total angular momentum is 0. Since M lies vertically between m1 and m2, if m1 moves to the right and m2 moves to the left then we have a non-zero contribution from both m1 and m2 in the same direction so they cannot be assumed to cancel out (they both move either anticlockwise or clockwise about M). Therefore in this situation angular momentum is not conserved so may discard it leaving behind the situation where both tankers move at the same velocity as before (m1 and m2 are at rest relative to M in this situation so angular momentum remains 0) .
Typical solution:
No external horizontal force* acts on the tanker (or the water for that matter) and therefore we can call Newton's First Law to our aid and deduce that horizontal velocity of the tanker and the water remains constant.
*Answering various objections to this claim:
It helps if instead of water we imagine a collection of smooth spheres being released (since we are ignoring dissipative effects). How is a sphere "held in place" in the cart? Is there a force holding it in place? No!
Since it is moving at the same velocity as the tanker, no force is exerted by the tanker to keep a sphere moving at a constant velocity i.e. in the horizontal direction. The only force exerted by the tanker on a sphere is the vertical normal reaction force counteracting gravity. If this force suddenly disappears then the horizontal motion of the tanker and the ball should not be affected. What about the effect of the balls knocking around inside the tanker?
Well here the law of conservation of momentum comes to our aid so that we can be certain that no matter what complicated dynamics ensue, the horizontal momentum of the combined system is always conserved but we have to be a little careful here. Imagine if one of the falling balls hits the side of the tanker and rebounds directly out of the hole. Then the speed of the tanker would change despite the fact that the combined momentum is still conserved.
Here we have to return to the model of the water so that we can assume that there is no imbalance in the forces exerted on the sides of the tanker at any point so that at all points the net horizontal force exerted by the water is zero. In other words, when we return to the model of the water, the situation I described should never occur and there should never be an imbalance of the forces acting on the tanker. (Technically there is no imbalance only when we're dealing with hydrostatic fluids but this is reading way too far into it and for this reason I find this line of reasoning rather unsatisfactory)
Please post your comments- I deleted my other account as I wanted to link it to FB instead.
M
Great analysis of the setup! I love your elegant solution, it's a nice and creative approach to this question.
Verbose over complication of a simple problem.
According to Newton’s first law – every body preserves its state of rest unless some external force compels it to change its state of rest.
Example--[ When a bus suddenly starts, the passengers sitting or standing in the bus tend to fall backward. This is due to inertia of rest and can be explained as follows: when the bus suddenly starts, the lower part of the body of the passenger which is in contact with the bus moves along with the bus while the upper part of the body tends to retain its state of rest due to inertia. As a result, the passenger falls backward.]
so, it will remain same.
Assuming no external force is being applied to the tanker it should continue at the same velocity forever. The mass of the tanker has no bearing on anything unless there is friction. In fact if it's a frictionless surface it doesn't even matter if the axles create friction as the wheels would not turn.
All the water that has leaked is all around the tank as it slides along (although slowly spreading out after their direction changed from downwards to sideways).
Regardless I want to see this irl.
Most arguments seem purely academic, not taking into account the reality of air resistance. As the tank lightens, air resistance has an increasing impact and slows the tanker.
All laws of physics work the same in an inertial frame of reference. If the tanker was still, there would be no change in velocity. Hence, the velocity should not change here. I'm surprised no one else wrote this
The speed of Tanker would not be affected as velocity or speed is independent of mass as Velocity = Distance/Time
Since there is no presence of any contact for es or field forces we can use Newton's first law .ie.law of inertia to justify our answers for the following problem.
The surface is frictionless not depend on weight of tanker
change in mass might imply a change in speed. But by newtons laws of motion the speed will remain the same. the only thing that changes here physically is the tankers momentum.
But doesn't conversation of momentum indicate that the product of mass and velocity remain the same? So if the mass of the tanker goes down, Shouldn't the velocity increase?
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no, conservation of momentum is satisfied. the water when drained moves with a velocity v equal to the velocity of the tanker......the water when drained is not simply at rest when it touches the surface. as the momentum is distributed between the water and tank the total momentum is conserved.
let mass of the tanker with water be m t w P t w = m t w v ⇒ P t w = ( m t + m w ) v ⇒ P t w = m t v + m w v ⇒ P t w = P t + P w
Relevant wiki: Newton's First Law
Newton's First Law states that an object that is moving at a constant velocity will stay at that velocity unless acted upon by an external force. When the water leaks from the tanker, it does not provide a net force on the tanker itself. The only change in forces on the tanker is the force of gravity, but that is negated by the normal force provided by the floor. Therefore, the speed of the tanker will remain the same .
A note about the momentum argument: (copied from one of my comments) As Mike Perry pointed out in the comments, while the mass of the water in the tanker decreases, the mass of the overall system , which is the tanker and all of the water, does not change. Once the water reaches the floor, no net force acts upon it, so it retains its previous horizontal velocity it had while in the tanker. Therefore, by conservation of momentum, the tanker with its remaining water should also retain its velocity.