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Momentum conserved and energy isn't conserved?

  1. Oct 6, 2005 #1
    Suppose a box whose mass m(t) increases at a constant rate moves with constant velocity v.

    The force necessary to keep this object at constant velocity is F = dp/dt = dm/dt*v.

    Now we try to find the work done by the force from time 0 to time t. The force is constant, so W = F*d = F*v*t = dm/dt *v*v*t = (m(t)-m(0))*v^2. (since dm/dt is constant, t*dm/dt = dm).

    Now if we calculate the change in kinetic energy from time 0 to time t, it should be the same as the work done, if energy is conserved.
    So KE = .5*m(t)*v^2 - .5*m(0)*v^2 = .5 (m(t)-m(0))*v^2.

    It's not equal to the work done! Am I making a stupid mistake, or is energy indeed not conserved in this system?
     
  2. jcsd
  3. Oct 6, 2005 #2

    mukundpa

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    I suppose that the mass added is initially at rest. Is it? Is there any other force which is making the mass dm move with velocity V.

    If you put and object on a slowly moving platform how it starts moving.
     
  4. Oct 6, 2005 #3
    The added mass is initially at rest. The force that is making dm move is just the normal between m and dm. In a way I can see how energy isn't conserved, since this could just be a series of inelastic collisions. But is it really the case that energy is not conserved?
     
  5. Oct 6, 2005 #4

    mukundpa

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    If the force between them is only normal then how dm starts moving forward?
     
  6. Oct 7, 2005 #5

    Andrew Mason

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    Your calculations are right. Why they are different is quite a good question. I think you have to look at the mechanism by which the mass increases: inelastic collisions. When two masses collide and stick together, there is always a loss of energy.

    If you consider the addition of mass by a series of inelastic collisions of an infinitessimal billiard ball mass of mass m being dropped from rest and colliding and sticking to another such ball that is moving at speed v (and then both being accelerated together back up to speed v) you will be able to see why the loss occurs:

    The inelastic collision conserves momentum so the speed of the two balls reduces to 1/2 v. This results in a loss of [itex].5mv^2 - .5*2m(.5v)^2 = .25mv^2[/itex] or half the kinetic energy of the first ball. So for every ball that is added to the system (thereby acquiring speed v and kinetic energy [itex]KE=.5mv^2[/itex]) the system loses exactly half that energy. This energy is simply lost and has to be replaced by energy supplied to the system (when the added ball and the ball it collided with are accelerated from 1/2 v to v).

    AM
     
    Last edited: Oct 7, 2005
  7. Oct 7, 2005 #6
    Thanks for the confirmation Andrew.

    Just as an aside, it also seems to me that the work energy theorem is almost never true in the case of systems which involve mass flow (this is because when integrating F dx, we assume constant mass).
     
  8. Oct 7, 2005 #7

    Tide

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    I think you're a little too eager to toss away energy conservation. The work done in accelerating an element of added mass is certainly [itex]F \Delta x[/itex]. However, you must be a bit more careful in writing the force.

    Start with [itex]F =v \frac {\Delta m}{\Delta t}[/itex] so [itex]F \Delta x = v \frac {\Delta m}{\Delta t} \Delta x[/itex]. Now, [itex]\Delta x = \bar v \Delta t[/itex] where we use the average speed of the element of mass being added. That mass is not instantaneously brought up to speed!

    If [itex]\Delta t[/itex] is very small then [itex]\bar v = \frac {1}{2}v[/itex] so that [itex]F \Delta x = \frac {1}{2} \Delta m v^2[/itex]. Energy is conserved! :)
     
  9. Oct 7, 2005 #8

    Andrew Mason

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    There is no question that the element of mass dm acquires [itex]\frac{1}{2}dmv^2[/itex] of kinetic energy. It is just that the force pushing the system inputs twice that energy:

    [tex]dE = Pdt = Fvdt = \frac{dp}{dt}vdt = dp v = dm v^2[/tex]

    AM
     
  10. Oct 7, 2005 #9

    Tide

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    I don't think so. The force provides exactly [itex]dE = F dx = \frac {1}{2} v^2 dm[/itex] of energy since, by Newton, the force on the element of mass is the same as the force on the box.
     
  11. Oct 7, 2005 #10
    Tide,

    I'm not totally understanding your argument. Can you spell it out a bit more?

    I think I'd have to side with what andrew is saying. Just by noting the fact that when two objects collide and stick together, energy is necessarily not conserved, you can see that energy can't be conserved in this case.
     
  12. Oct 7, 2005 #11

    Tide

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    asdf,

    I'm not looking for anyone to take sides. I am just not ready to give up energy conservation quite so quickly. The argument Andrew is making is essentially that the added mass achieves its final speed instantaneously. It doesn't.

    Basically, the energy imparted to the newly added mass is [itex]F \Delta x = \frac {1}{2} \Delta m \times v^2 [/itex]. Andrew argues that the force is also expressibe as the rate of change of momentum [itex]\frac {\Delta p}{\Delta t}[/itex] so [itex]F \Delta x = \frac {\Delta m}{\Delta t} v \Delta x[/itex]. He simply wants to replace [itex]\Delta x[/itex] with [/itex]v \Delta t[/tex] but that assumes the new mass moves at the same speed throughout the entire acceleration process.

    I argue that [itex]\Delta x = \bar v \Delta t[/itex] over the increment of time using the average speed of the new mass which is [itex]v/2[/itex], i.e. half the final speed if the acceleration is uniform over small [itex]\Delta t[/itex]. Therefore, [itex]F \Delta x = \frac {\Delta m}{\Delta t} v \times \frac {1}{2} \times v \Delta t[/itex]. This simplifies to [itex]F \Delta x = \frac {1}{2} (\Delta m) v^2[/itex] as expected.

    With regard to the force applied to the "big mass," we again have to start with [itex]\Delta E = F \Delta x[/itex] for the amount of work done and Andrew wants to express this in terms of power times time. Let's do that: [itex]\Delta E = F v \Delta t = F \frac {\Delta x}{\Delta t} \Delta t[/itex] which is just [itex]\Delta E = \frac {1}{2} \Delta m v^2[/itex].
     
    Last edited: Oct 7, 2005
  13. Oct 8, 2005 #12

    Andrew Mason

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    If energy is conserved, then there is no loss of energy despite the fact that this process seems to involve a series of inelastic collisions.

    An inelastic collision does not necessarily have to dissipate energy - it can store it in a spring, for example. So you could have this system adding mass and kinetic energy at the same rate that power is input, thus conserving energy, but also storing additional energy in the spring. That sounds like an energy creation machine.

    AM
     
  14. Oct 8, 2005 #13

    Tide

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    Andrew & ASDF,

    The trouble is that insufficient information is provided in the original statement of the problem.

    Consider the situation from the perspective of the box. It is stationary and stuff is passing by at speed v. Let the box accumulate mass at a fixed rate and again ask the question how much force is needed to keep the box stationary.

    Clearly, the force has to be [itex]F = \frac {dm}{dt} v[/itex]. Is energy conserved? Work has to be done to decelerate the matter stream and some agent has to act over some distance to achieve that. And we will find that some force times some distance does indeed perform work at a rate of [itex]\frac {1}{2} \frac {dm}{dt} v^2[/itex].

    However, the situation becomes a little cloudy when we try to do a global energy balance in the manner in which it was done in the original problem. We can try to calculate [itex]F dv[/itex] but that is zero in this frame of reference! Our force keeps the box from accelerating but all the work is done on the matter stream.

    Andrew, you were on the right track thinking about elastic vs. inelastic collisions but we need to go a little further. Let's run the box problem in reverse! Say I have a rocket attached to a wall that will prevent it from moving. I ignite the fuel and mass leaves the rocket at some rate (we can take it to be constant for our purposes) with some exhaust velocity v. How much force does the wall exert on the rocket to prevent it from moving? It's the same as before: [itex]F = \frac {dm}{dt} v[/itex]

    Again, the wall does no work since the rocket does not move! There is a hidden agent in this problem and that is the internal chemical energy of the rocket fuel.

    In the end, energy is conserved by taking full account of ALL the energy (internal energy as well as kinetic) but the original problem provides no detail.

    ASDF, I assume that the original problem you had to solve asked only for the force required to keep the box moving at constant speed. We all agree with your solution! You are to be applauded for going beyond that and asking the question of whether energy is conserved. Just remember that energy is always conserved but you have to do a full accounting!
     
  15. Oct 8, 2005 #14

    Doc Al

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    Here's how I understand the problem. I visualize a lump of clay (mass m; speed v) moving along while small bits of clay (mass dm; speed 0) are continually dropped onto the larger mass at the rate dm/dt.

    To see how mechanical energy cannot be conserved, view the situation from a reference frame moving along with the clay mass. In that frame the clay mass is at rest. Bits of clay (mass dm; speed v) are being fired into the larger mass. Since an applied force (v dm/dt) prevents the clay mass from moving, the KE of the clay bits is completely transformed into internal energy.

    Returning to the original frame, we see that the average work done by the applied force in time dt is [itex]F dx = v \ dm/dt \ v dt = v^2 \ dm[/itex], which equals the sum of the increase in internal energy ([itex]1/2 \ dm \ v^2[/itex]) plus the increase in the KE of the incoming clay mass ([itex]1/2 \ dm \ v^2[/itex]). So, as Tide says, energy balance is maintained.

    Note that in applying Newton's second law to the incoming clay bit (a deformable body), the average speed of the center of mass of that clay bit is [itex]v/2[/itex]. Thus the force (F) times the displacement of the center of mass ([itex]v/2 \ dt = dx/2[/itex]) equals the change in the KE of the clay bit [itex]1/2 \ dm \ v^2[/itex]. But this is not equal to the real work done on the clay bit (in the conservation of energy sense); the real work equals the force (F) times the displacement of the point of application of that force (dx = v dt). (The quantity force x distance used in the center of mass equation, [itex]F_{net} \Delta X_{cm} = 1/2 M V_{cm}^2[/itex], is sometimes called pseudowork to distinguish it from the real work.)
     
    Last edited: Oct 8, 2005
  16. Oct 8, 2005 #15

    arildno

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    The original exercise is complete nonsense and cannot be solved.
    This is because information has not been given as to how momentum is added to the system.
    If, for example, the mass increase is done by throwing sticky balls at the box with a velocity component PARALLELL to the box's velocity, so that the collision retards the ball down to the box's velocity, then you would need a restraining force to be applied in order to keep the box moving with constant velocity.

    If, on the other hand, the sticky balls attaches to the box and had initially the SAME HORIZONTAL VELOCITY AS THE BOX, then no horizontal force needs to be applied to keep the box moving at a constant rate.

    To use the equation F=dm/dt*v is, in general false.
    The silly equation isn't even a Galilean invariant.:grumpy:
     
    Last edited: Oct 8, 2005
  17. Oct 8, 2005 #16
    If there is to be an equilibrium situation, ie. constant veloctiy, then you must have no net force. This means there must be a retarding force. This must be provided by either friction or air resistance/ whatever. the energy loss occurs by deforming the air around the box or inducing a current, generating heat, etc.
     
  18. Oct 8, 2005 #17

    Doc Al

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    Assuming that the added mass has a speed of zero (with respect to the ground), of course a force is required to keep the mass moving at constant speed. That was mentioned in the opening post. I wouldn't call it a retarding force; it must be an active force, not merely friction or air resistance. (Unless a strong wind is blowing it along, of course. :smile: )

    I'm a bit puzzled by the title of the thread, since neither momentum nor (mechanical) energy is conserved.
     
  19. Oct 8, 2005 #18

    uart

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    Yes clearly the original problem was too vague about how the mass was added, but from the equations asfd60 used it was clear that he assumed that added mass had no initial momentum, so lets just discuss that particular problem ok.

    In my opinion Andrew correctly explained the situation in his first reply.
     
    Last edited: Oct 8, 2005
  20. Oct 8, 2005 #19

    uart

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    Well that's exactly the point, and exactly why conservation of momentum holds in an inelastic collision while conservation of KE does not. You're the one making the very point that the velocity of the added mass (assumed to be initially zero) does not increase to "v" instantaneously but it happens over some small time period during which time the average speed of the added mass is less than v (for simplicity lets say v/2).

    "the force on the element of mass is the same as the force on the box". Yes and the change in momentum is [tex]\Delta p =F \Delta t[/tex] while the change in Energy is [tex]\Delta W = F \Delta x[/tex]. Now which is the same for both the added mass and the original moving mass during the acceleration of the accrued mass? Is it the time t or is it the distance x? Bingo you got it, the time is the same for both but the distance traveled is different as the velocities are different while the added mass is being brought up to speed!
     
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