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I Net impulse and net work

  1. Dec 28, 2016 #1
    Suppose there is a black box resting on a flat surface, which surface is rigidly attached to the earth. Suppose inside the black box there is a process initiated which causes the black box to lurch to the right 3 centimeters and lets say after one second the box comes to a complete stop. (No observable ejectile exits the black box, nor any energy or radiation. Also, there is no external force acting on the black box that causes it to lurch in one direction and the other direction)

    Now the only external forces acting on the black box after the internal process is triggered are friction forces, acting on the bottom of the box opposite to the direction of the motion of the box. Conservation of momentum would require that after the box comes to a stop that it should move to the left for the same amount of time before coming to a stop, if we assume the magnitude of the friction force remains constant. And this must be true even if the distance the black box travels to the left may be less than 3 centimeters. That is, the work done on the surface of the earth in the right direction may not be the same as the work down on the surface of the earth in the left direction. But the magnitude of the impulse must be the same.

    Put another way, the earth receives an impulse, friction force x time (1 second) in the right direction. In order for the earth to not have a net impulse, there must be an equal and opposite impulse on the earth in the opposite direction, friction force x time (1 second).

    Bottom line. If the box moves to the right for 1 second, it must move to the left for 1 second. Is this conclusion correct? Again, assume the coefficient of static and kinetic friction remain the same such as the case with polytetraflouroethelene(Teflon).
     
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  3. Dec 29, 2016 #2
    Imagine you were put in a box , large enough to stand up in ...

    You could jump in the air at an angle to hit the side wall , the force from this jump would not be enough to overcome static friction , and so would not move the box , but when you hit the wall this sharper impulse would move the box . Repeat this and you can shift the box and yourself across the floor.

    Or you could jump inside the box , while in mid air punch the box sideways , you would come back down close to the opposite side wall , then you walk back to the center of the box and repeat . this would move you and the box as far as you want across the floor.
     
  4. Dec 29, 2016 #3

    A.T.

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    No it doesn't require that.
     
  5. Dec 29, 2016 #4
    ok. Let me be more specific. Lets say in the black box there is a spring attached to one wall. The spring is compressed and a ball lies at rest at the end of the spring. Lets assume there is no friction acting on the box. (The box is in space, away from all gravitational effects.) Thus, the initial momentum of the center of mass of the box-ball system is zero with respect to an inertial reference frame.

    At some point in time the spring is released. The box moves to the left and the ball moves to the right, eventually making an inelastic collision with the opposite wall. Now, since there were no external forces acting on the box-ball system, the box-ball system must come to a complete stop. Or, the momentum of the center of mass of the system must be zero because the initial momentum of the center of mass of the system was zero. Also, with respect to the inertial reference frame, the center of mass of the system did not move.

    Now, lets place the box-ball system on the surface of the earth. We assume the bottom of the box is made of Teflon and the surface likewise. Thus, the coefficient of static and kinetic friction is the same, equal to .04, which would be the case for Teflon. And asssume this coefficient remains constant.
    We repeat the same scenario. Also, assume the rolling friction of the ball remains constant.

    Again, the compressed spring is released, the box moves to the left and the ball moves to the right, making an inelastic collison with the wall opposite where the spring is attached.

    Now in this situation, there is a third body in this system-- the earth. So, we are now dealing with a 3 body system: the box, the ball, and the earth. The initial momentum of the center of mass of this 3 body system is zero, and therefore, the final momentum of the this 3 body system must be zero. That is, if we define the box,ball, and earth as our system, then the friction forces are internal forces, not external forces to the system. Thus, by Euler's first law, these forces cannot change the momentum of the center of mass of the 3 body system. (We can ignore the rolling friction of the ball because this friction is internal to the box-ball system, and therefore, will not affect the momentum of the center of mass of the ball-box system).

    I maintain because of this, any impulse on the earth must be countered by an equal and opposite impulse on the earth. Else, after the box comes to a rest, the earth would have acquired a net impulse in the left direction, and thus, the 3 body system as a whole would have acquired a net increase in momentum in the left direction. But this cannot be because the friction forces are internal forces to the 3 body system.

    Hence, lets focus on the box first. As the box moves to the left, a friction impulse acts on the box to the right, slowing its momentum so that when the ball makes its inelastic collision with the wall, the box will have a mangitude of momentum less than the magnitude of the ball before it makes its collision. (Remember, the rolling friction between the ball and box do not have an impact on the momentum of the center of mass of the ball-box system). Thus, right before the collision of the ball, there will be a net momentum in the right direction of the center of mass of the ball-box system, and thus, the ball-box system must move to the right after the collision, unlike the situation in space where the ball-box system comes to a complete halt after the collision. It comes to a complete halt because in space, where there are no friction forces, there was no change in the momentum of the center of mass of the ball-box system.

    Now, lets focus on the earth, the third member of the 3 body system. As the box moves to the left, there will be an impulse acting on the earth in the left direction. If this impulse is not countered by an opposite impulse, the earth would have acquired a net impulse or change in momentum in the left direction after the box comes to a stop. Again, this is impossible because the friction forces experienced by the box and the earth are internal forces to the 3 body system. Alas, it's because the box moves to the right after the collision of the ball that the earth does experience an impulse in the right direction which must cancel out the impulse it experienced in the left direction. Hence, since the friction force is constant and since the impulses must be equal in magnitude and opposite in direction to cancel to zero, the time the box moves to the right must be equal to the time the box moved to the left.
     
  6. Dec 29, 2016 #5
    Earth receives equal impulse as the ball does and the earth-ball system conserve the momentum
     
  7. Dec 29, 2016 #6

    jbriggs444

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    We are not told that the spring is strong enough and the box-ball are light enough to result in slippage against the coefficient of friction of Teflon on Teflon. Accordingly, it is not certain that the box moves to the left.

    Accordingly, it is not certain that the box moves to the left while it imparts a leftward net impulse to the Earth.

    Accordingly, it is not certain that the impulse imparted while the box is moving left is equal and opposite to the impulse imparted while the box is moving right.

    Any imbalance is accounted for by the fact that a net impulse can be applied while the box is moving neither left nor right.
     
  8. Dec 29, 2016 #7
    Why argue that the scenario isn't necessarily the only possibility when A) it is possible and B) the conclusions are flawed and need correcting? Let's start with the premise as correct and then discuss what is wrong with the conclusions.

    So a box contains a mass on a compressed spring. The box sits on the floor with some coefficient of static and kinetic friction. At time 0 the spring is released. The force of the spring is large enough that the impulse of releasing the spring causes the box to slide on the floor. Nothing wrong with any of that. I'm going to add one more thing to make the end state easier to understand. The spring is long enough that it traps the mass against the far wall of the box so very quickly the inside of the box is static again.

    This is a perfectly reasonable and physically possible scenario. Practically everything the OP said about the result of this scenario is wrong. Conservation of momentum does NOT mean that the objects within a closed system cannot interact and PERMANENTLY move relative to each other. In fact that is how we almost always use conservation of momentum. Cars colliding masses separated by springs etc., in most of these kinds of problems the objects interact and change and move permanently relative to each other. With no external forces we invoke conservation of momentum and that puts constraints on what can happen, but the cars are absolutely not required to uncollide to conserve momentum!

    So let's take your box. First suppose there is NO friction. The box is the system. The spring goes off. The ball is pushed right, the box is pushed left AND THEN THEY BOTH STAY IN THE NEW POSITIONS. How is momentum conserved? It IS conservation of momentum that required that the box move left! The mass moved right. The mass of the box had to move left for the center of mass to stay put. Put another way through the spring the box pushed on the mass. By Newton's law of action and reaction the mass pushed on the box. The actions have to be equal. Why did Newton conclude they had to be equal? No external forces equals no acceleration equals CONSERVATION OF MOMENTUM.

    Ok, let's add the earth and the friction. Now the box doesn't move quite as far left. Why? Because it pushed on the earth and the earth moved a tiny bit left. Now the closed system includes the earth. There are no forces external to THAT system, but there is no reason objects within the system can't move relative to each other. The mass moved right. The box AND the earth moved left. Afterwards they all stayed in their new positions, but the center of mass did not move and momentum is conserved.
     
  9. Dec 29, 2016 #8

    jbriggs444

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    Because A) an alternative is also possible and B) because the conclusion follows in some cases but not in the alternative case.

    The conclusion that I have in mind is, paraphrased:,

    "If the box starts and ends at rest [and has no nasty vertical components to its motion, is finite and settles down to a fixed internal configuration, etc, etc] then the total duration over which the box moves left must be equal to the total duration over which the box moves right.

    That conclusion is almost correct. If the box is moving leftward, it applies a leftward force of fixed magnitude to the ground and by Newton's third law is subject to a rightward force of equal magnitude. Similarly if the box is moving rightward. The net impulse applied to the box is the difference between the leftward and rightward impulses. But both of those are proportional to the durations of the leftward and rightward movement [and with the same constant of proportionality]. The net impulse will be zero if and only if the box spends the same amount of time moving left as it does moving right.

    That argument actually holds... If one ignores the possibility of a non-zero force being delivered by a stationary box. [Or the possibility of a time-varying normal force, or of a coefficient of friction that varies depending on something other than normal force or of an infinite box or of a box that never settles down]

    A way to poke holes in the argument is to use a scenario where a non-zero force is delivered while the box is stationary.
     
  10. Dec 29, 2016 #9

    Dale

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    This is correct. If follows for any interaction, due to Newton's 3rd law, regardless if the interaction is an internal or external force.



    The conclusion is certainly wrong as worded. If the impulses are equal and opposite and the magnitude of the force is constant then the duration of the force must be equal, not the motion. Indeed, the box may only move in one direction.
     
  11. Dec 29, 2016 #10
    No. I am not asserting that the magnitude of the motions or the displacements of the box must be equal, I am only asserting that the time that the box moves to the left and to the right must be equal. As you said, "the duration of the force must be equal." This is what I am saying.

    The box might move to the left, lets say 1 centimeter, before coming to a stop after, lets say .5 seconds. But then the box must move to the right for .5 seconds and it doesn't matter what distance it travels: it could be only .3 centimeters or .1 centimeters. It doesn't matter how far it moves. It only matters that the friction force acts for .5 seconds. This would be possible because the ball-box system after the ball collides with the box will be going slower to the right. Thus, it will still take it .5 seconds to travel the .3 centimeters to the right whereas it took .5 seconds for the box to move 1 centimeter to the left because it was going faster.

    The box cannot move in "one direction" as you said because then the earth would only receive an impulse in one direction. The box must move in the right direction to cancel out the intial impulse on the earth in the left direction. If the box only moved to the left and came to a stop, where would be the equal and opposite impulse on the earth to cancel out this initial impulse to the left on the earth? If there was no cancellation, the center of mass of the earth would acquire a net momentum.
     
  12. Dec 29, 2016 #11

    Dale

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    No, one may be zero and the other non zero.

    And I am pointing out that the duration of the force and the duration of the movement are completely different things.

    This is not correct. Again, force and movement are different things. For a constant mass the force is the derivative of the movement.
     
    Last edited: Dec 29, 2016
  13. Dec 29, 2016 #12
    I think I'm beginning to catch on here. I think you are confusing impulse with movement. impulse is the change in momentum. It is proportional to the derivative of the displacement, not the displacement. If you accelerate to the right and then stop the impulse is toward the right when you accelerate, but toward the left when you stop. The impulse is both directions. The net impulse is zero. But the motion is only toward the right and the displacement is not zero.

    That is what happens with the mass in the box. The mass is accelerated toward the right by the spring and then it hits the wall and is decelerated. (I'm still sticking it to the wall for simplicity). The force is in both directions first right and then left. The impulse is in both directions first right and then left. The net impulse is zero. But the motion, the displacement, is only toward the right. Likewise the box and the Earth are accelerated toward the left. Then when the ball hits the wall and stops they are decelerated and stopped by a force, an impulse to the right. However the movement, the displacement of the box and Earth is only to the left.

    Mass moved right. Box and earth moved left. All impulses were balanced and cancelled. The center of mass never moved.
     
  14. Dec 30, 2016 #13
    Ok. Dale, maybe we are not understanding each other. I would be very happy to have my thinking cleared up if I am in error, and I would be the first to admit it if I am wrong. So be patient with my questions and my assertions.

    Lets just look at the earth for a moment in a free body diagram. When the box slides to the left, there will be an impulse on the earth to the left equal to the normal force of the weight of the ball and box times the coeficient of friction times the duration that box slides to the left on the surface of the earth, relative to an inertial reference frame not attached to the earth. Do you agree with that?

    Now what would the net impulse on the earth be if the box did not slide to the right? Would it be zero?
     
  15. Dec 30, 2016 #14
    I'm sorry to interrupt this question to Dale, and I'm sure he'll answer. I just wanted to point something out. What you say here is true. Have you considered that the impulse you describe on the earth actually causes the earth to start moving? When the mass hits the wall and stops the box the friction with the earth is going to have to stop the earth. There will be an impulse on the earth to the right that would be easy to miss if you forget the earth was accelerated.
     
  16. Dec 30, 2016 #15

    Dale

    Staff: Mentor

    Yes, where the coefficient of friction you refer to is the coefficient of kinetic friction and the time arbitrary.

    For clarity, let's use the term "the system" to refer to the box and the ball together. That way we can distinguish between the movement of the box and the movement of the (center of mass) of the system. The whole purpose of the system arrangement is to decouple the motion of the box and the motion of the system.

    No, it could be anything, with the maximum force being limited by the coefficient of static friction and the time being arbitrary.
     
  17. Dec 30, 2016 #16
    Yes. And there can only be an impulse on the earth to the right if the box momentarily moves to the right after the ball collides. The box cannot come to a dead stop after the ball collides with the inner right wall. If it did, there would be no friction impulse on the earth to the right.
     
  18. Dec 30, 2016 #17

    A.T.

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    Impulse can be transferred without relative movement, through static friction.
     
  19. Dec 30, 2016 #18
    Ok. Now think about this. The box has come to a stop. There is no relative motion between the box and the earth. Now, the earth and box will move at the same velocity to the left relative to an inertial frame because the earth acquired a net impulse greater than zero to the left. This means the center of mass of the earth-box-ball system will be moving to the left, which means there will be momentum to the left of the 3 body system. This cannot happen unless we can account for an "external" force that acted on the 3 body system per Euler's first law.
     
  20. Dec 30, 2016 #19

    jbriggs444

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    The "box" and the "center of mass of the box and ball" are two different things that can be in relative motion.
     
  21. Dec 30, 2016 #20

    Dale

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    No it doesn't. Remember, by design you have decoupled the motion of the box from the motion of the center of mass of the system. They are not the same thing, by design. So the fact that the box is moving left with the earth does not imply that the CoM of the box+ball system is moving left.
     
    Last edited: Dec 30, 2016
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