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Concept question, linear momentum and mechanical energy

  1. Apr 17, 2014 #1
    A box containing a pebble is attached to an ideal horizontal spring and is oscillating on a friction-free air table. When the box has reached its maximum distance from the equilibrium point, the pebble is suddenly lifted out vertically without disturbing the box. Will the following characteristics of the motion increase, decrease, or remain the same in the subsequent motion of the box? Justify your answer.
    a. frequency; b. period; c. amplitude; d. maximum kinetic energy of the box; e. maximum speed of the box.

    I correctly reasoned that the frequency will increase, and the period will decrease using the relationships about SHM we've learned, but I'm having issues understanding the other three.

    I know the equation for amplitude has angular frequency in it, and since I know that decreasing the mass will increase the angular frequency, I would think that the amplitude after removing the pebble would be less than the amplitude before, even though intuitively I would think the opposite. The equation I have for A is A = sqrt (x^2 +(v/w)^2) but I don't really understand this equation - what is v? Just linear velocity? What about on a pendulum - how would that work?

    For d., I would of course assume that decreasing the mass in the system would decrease the kinetic energy, but linear conservation of momentum says that it doesn't, so if the mass decreases, the speed must increase to make the total mechanical energy remain the same. But I don't understand how removing mass, and therefore weight, from a system doesn't change its total mechanical energy. Especially if you've already determined that your amplitude is going to increase, and you have E = 1/2*k*A^2 : if your amplitude increased, shouldn't your total mechanical energy increase?

    I'm used to figuring if kinetic energy is conserved for collisions, but this isn't really a collision, so I have no idea what conservation of linear motion applies here.

    Thanks!
     
  2. jcsd
  3. Apr 17, 2014 #2

    Simon Bridge

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    You need to think about how the amplitude got there in the first place.
    Normally the box would just be pulled to some initial position at t=0 and then released - and that is the amplitude.

    So set t=0 at the instant the pebble was removed....

    At the point that the mass is removed, what is the kinetic energy?

    So your answer for one part seems to contradict your answer for another - this suggests that at least one of them must be wrong.
    How would you figure out which one?

    You can usually trust conservation of momentum.
    You could consider what happens for the situation the box does not have a spring attached - it slides at constant velocity and the pebble gets vanished.

    You do have to decide if any momentum carried by the pebble vanishes along with it.
    If you have done problems where a mass is suddenly added to something, then I'd treat this is the reverse of whatever approach was used then.
     
  4. Apr 17, 2014 #3
    When the mass is removed the total kinetic energy would be zero, but that doesn't imply (to me) that changing the mass won't change the overall kinetic energy. But I do see how potential energy is conserved - if it is at its original maximum amplitude when the pebble is removed, then 1/2*k*x^2 will be the same before and after removal. And since potential energy is completely kinetic energy at equilibrium, and hasn't changed value, I guess that would be one way to conclude that the maximum potential energy is the same.

    Which would lead to saying that the amplitude before and after is the same. The book solution online says the amplitude increases afterwards because Af = Ai *sqrt ((m1+m2)/(m1-m2)) where m2 = mass of pebble. I have no idea where they got that relationship, but saying that potential energy is conserved, so must be kinetic, mechanical, and finally amplitude makes sense to me.

    If the box does not have a spring attached, and no friction, you have zero work done if velocity is constant, so final kinetic energy equals initial kinetic energy - so if you removed weight you would have to increase velocity to make that still true. It's difficult not to think of that as adding a negative outside force though. In an inelastic collision kinetic energy is not conserved, so what does that say about the "opposite" of an inelastic collision? Or maybe I should be thinking of mass as a sort of "moment of inertia", so that if you decrease the mass, you increase the ability/willingness of that object to move/accelerate. So in that way I could see how it's conserved. Clearly I need to go back and do some mechanical/kinetic energy concept questions!

    Thanks for your help!
     
  5. Apr 17, 2014 #4

    Simon Bridge

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    ... the KE could have an instant change so at ##t=0^-##, ##E_i=0## and at ##t=0^+##, ##E_f\neq 0##. So ##\Delta E = E_f## ... so what would be the effect of changing the kinetic energy at this point?

    Well done.
    The max PE is unchanged by removing the pebble at that stage - you can see that the max PE term does not depend on the mass.

    Maybe the book is expecting you to make another assumption?
    Is this something you can provide a link for? Which book?

    You assumed that no work was done because the velocity was a constant, but concluded that the velocity has to change ... what does that tell you about your initial assumption?

    I meant that the box is initially travelling at constant velocity... it must end up at a constant velocity too, but it can be different.

    How you do the calculation depends on how the pebble vanishes.

    If you insist on conserved kinetic energy, then ##\frac{1}{2}Mv^2=\frac{1}{2}(M+m)u^2\implies v=\sqrt{(M+m)u^2/M}##
    ... where m is the mass removed and u the initial velocity.

    So there is a change in momentum of: $$\Delta p = \left[\sqrt{M(M+m)}-(M+m)\right]u\neq 0$$... leading to your intuition that there must be some force giving rise to this specific impulse.

    If we prefer to think that the problem calls for no force to act on the box to remove the pebble, then conservation of energy does not describe the situation in hand.

    If you insist on conserved momentum, then ##v=(M+m)u/M##
    ... then there is a change in kinetic energy:

    $$\Delta E = \frac{(M+m)mu^2}{2M}$$

    i.e. The process of removing the mass is inelastic.

    The difference is because we have not said anything about how the mass got removed, it just magically vanished: a non-physical process. Normally you'd use conservation of momentum (usually implied by "for system undisturbed"). Conservation of momentum describes the symmetry of the physical laws in this case - it is not a cause-and-effect type of thing.

    In the situation described in the problem statement, the box is stationary at the instant the pebble is being removed, it is acted on by an unbalanced force (the spring) so it has a non-zero instantaneous acceleration.

    Removal of the pebble changes the instantaneous acceleration doesn't it?
    After the removal, ##-kA=Ma_f##, before the removal ##-kA=(M+m)a_i## after all.

    I don't think this problem is purely a matter of figuring the right equations - it is trying to test your understanding of SHM.
    It also restricts the kind of answers I can give you :(

    Yes.
    Look up "inertia".

    I think you should review how your course has handled examples of conservation of momentum, yes. Particularly where a mass got added or removed without specifying how it happened. There is usually some unspoken assumption involved.
     
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