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Homework Help: Re: impulse

  1. Jan 8, 2010 #1
    1. A mass of 5kg moves horizontally across a smooth surface at 1ms-1. Another mass of 5kg is lowered gently onto the first mass.
    a) What is the change in velocity of the mass?
    b) What impulse has been applied to the mass?
    c) Is this an elastic 'collision'?



    2. force x time = change in momentum
    p=mv
    1/2mv2




    3. a) I think there is no change in velocity, as the mass is added downwards gently and it is a smooth surface so no friction force need be considered.
    b) p=mv Doubling the mass, doubles the momentum. Doubling the momentum, doubles the impulse.
    c) Is this a collision, as the mass is lowered gently? If it is, then I think it must be an inelastic collision as Total Kinetic Energy before Collision =! (does not =) Total Kinetic Energy after Collision - 1/2mv2, as the mass has been doubled.


    Any good?
     
  2. jcsd
  3. Jan 8, 2010 #2
    The surface is smooth, but how about the contact between the two masses?

    How could the momentum be doubled unless you applied a force to one of the masses? Reconsider the conclusion you drew.
     
  4. Jan 8, 2010 #3
    impulse = force x time. As there is no force being applied, there is no impulse?
     
  5. Jan 8, 2010 #4
    momentum is a vector quantity - There is no force being applied as the mass is lowered gently, right?
    But surely from the equation p=mv, if you change the mass you change the momentum. If you change the velocity you change the momentum.
     
  6. Jan 8, 2010 #5
    If by lowering the mass onto the other mass, by F=ma, mass x acceleration due to gravity, you have applied a force downward of about 50N. But because the first mass is on a smooth surface it cannot affect the velocity.
    Momentum is a vector quantity - a Force applied in a direction against the momentum direction means that an impulse has been applied downwards.
    But it is an inelastic collision as there has been no velocity change
     
  7. Jan 8, 2010 #6
    Hmm, on second thought, the question is poorly phrased.

    What do they mean by 'lowered gently?'

    Does that mean that it was moving at the same velocity as the original block as it was put on it, or that it was placed on it from rest?

    Each set of assumptions gives us a different answer.

    You've assumed the first. You assumed that they both start at the same velocity, so there's no reason for the velocity to change when they're brought together.
    Therefore there was no impulse applied to the original mass, since its individual velocity remained the same, as did its individual mass!

    Be very careful here. If you assume that both masses start at the same velocity, then your initial momentum is [tex]P=2mv[/tex] ! And each block has an individual momentum of [tex]mv[/tex]

    Since the velocity of each individual block has not changed, nor have their individual masses changed, the total energy remains unchanged just as well!

    Let's explore the second set of assumptions now:

    If, on the other hand, you assume that the second mass starts from rest, then your initial total momentum is [tex]mv[/tex], with the second mass contributing nothing to the total momentum in the system since its velocity is 0.

    You are told that after some process, the two now move as one unit, with mass [tex]2m[/tex]

    For the momentum of the system to change, an external force must be applied. Assuming that by gentle, the meaning is without any resistance from the 'lowering mechanism.' What can you say about the external forces available to provide an impulse to the system?

    Having reached that conclusion, what can you say about the total momentum after the collision, and how does that affect the velocity of the masses and the energy that's in the system in the form of kinetic energy?

    Can you speculate as to what forces are responsible for the result you've found?
     
  8. Jan 8, 2010 #7
    Exploring the second set of assumptions:
    What can you say about the external forces available to provide an impulse to the system?
    The only external force I can think is available is gravitational force, weight.

    What can you say about the total momentum after the collision?
    I'm a bit fuzzy on this one. The force of weight is acting downwards. Momentum is a vector quantity - a Force applied in a direction against the momentum direction means that an impulse has been applied downwards.
    But it is an inelastic collision as there has been no velocity change.

    And how does that affect the velocity of the masses and the energy that's in the system in the form of kinetic energy?
    The force downwards cannot create a frictional force, as it's a smooth surface, so it cannot decrease the velocity of the joined masses.
    In elastic collisions kinetic energy and momentum are conserved. But for this collision we have gained some kinetic energy by gaining mass, so kinetic energy is not conserved, meaning that this is an inelastic collision
     
  9. Jan 8, 2010 #8
    Can you speculate as to what forces are responsible for the result you've found?

    The force of gravity has also been transfered into velocity?
    like gravitational potential energy
     
  10. Jan 8, 2010 #9
    Try and answer my questions in the answer I've posted them, I hope they'll lead you along the path to best understand the problem.

    The impulse due to gravity, in this case, is canceled by the impulse due to the normal reaction force.

    Remember, the force of gravity is a vector quantity as well. Let's call the direction of the velocity the [tex]\hat x[/tex] direction, and the direction of gravity the [tex]\hat y[/tex] direction.

    Can a force applied to an object in the [tex]\hat y[/tex] direction change its [tex]\hat x[/tex] direction velocity? In this case, the answer to that question is no.
     
  11. Jan 10, 2010 #10
    RoyalCat: For the momentum of the system to change, an external force must be applied. Assuming that by gentle, the meaning is without any resistance from the 'lowering mechanism.' What can you say about the external forces available to provide an impulse to the system?

    Lemon: If the impulse due to gravity is canceled by the impulse due to the normal reaction force, then I think the only external force that can provide an impulse to the system is a negative force in the horizontal, acting left, against the momentum of the first brick travelling to the right.

    RoyalCat: Having reached that conclusion, what can you say about the total momentum after the collision, and how does that affect the velocity of the masses and the energy that's in the system in the form of kinetic energy?

    Lemon: Total momentum after collision does not equal total momentum before collision. A force acting in the opposite direction must decrease the momentum.

    RoyalCat: Can you speculate as to what forces are responsible for the result you've found?

    Lemon: The only force I can think of acting here would be the frictional and contact forces between the two masses.
     
  12. Jan 10, 2010 #11
    You are very very close, but you're wrong about the conservation of momentum.

    You are correct to say that a frictional force will act on the surfaces of contact of the two blocks, lowering the velocity of the bottom block.

    But remember Newton's Third Law! Action equals minus reaction!

    Assuming the initial direction of travel was to the right, we'll name the top block #2 and the bottom block #1.
    The surface at the bottom of #2, will apply a frictional force to the top of #1. As a result, block #2 will also experience an equal and opposite force as a result of the friction with block #1.

    The 'block #1+block #2' system experiences no net external force since, as we've just determined, the frictional force is internal to the system and is not external!

    Having come to that conclusion, what can you say about the momentum of the system before and after the collision?

    Hint:
    [tex]\frac{d\vec P}{dt}=\Sigma \vec F_{ext}[/tex]
     
  13. Jan 11, 2010 #12
    There are no external forces acting on the system. All forces are N3 pairs. All acting on different objects with equal magnitudes but in opposite directions. The forces are of the same type(2 normal reaction and 2 gravitational forces)
    The law of Conservation of Momentum tells us that: Whenever objects interact, their total momentum in any direction remains constant, provided that no external force acts on the objects in that direction.
    If all our forces are internal: Total momentum after the interaction = total momentum before the interaction. Momentum is conserved.

    There was no gravitational potential energy. There was a loss in kinetic energy. There was work done against friction. Energy is transfered from kinetic to heat/sound from the friction. Kinetic energy is not conserved.
    In Elastic collisions, no energy is transferred to forms other than KE. As energy has been transfered to other forms here, this is an Inelastic collision.

    So, contact forces are internal forces which act on two objects?
    External forces are those outside the system that create a 'net' force?

    So, there is a negative velocity to the left. I'm not sure how to describe the impulse, but it must be acting horizontally on both objects equally as it is acting on both of the objects and internal. Does that mean it cancels out?
    Impulse = force x time. There is no time here anywhere.
     
  14. Jan 11, 2010 #13
    That's an excellent answer! Very well put. :)

    If we're looking at each block individually, then their individual momentum is not conserved, but looking at the two block system, all forces between the two (The normal reaction force pair, and the frictional force pair), become internal to the system.

    When you have a frictional force between two surfaces, each experiences the same force (But in an opposite direction!), for the duration of the frictional contact. That is to say, both the surfaces in contact receive the same impulse, but with a different sign.

    A fruitful way to look at impulse is not only as force x time, but as the change in momentum!

    If you are unfamiliar with the differential form of force, read on to where it says HERE in bold and underlined.

    Looking at the relationship [tex]F=\frac{dP}{dt}[/tex] we find that we can rearrange it a bit and get:
    [tex]F dt = dP[/tex]

    Integrating over both sides, we see:
    [tex]\int F dt = \Delta P[/tex]

    But we already have a name for [tex]\int F dt[/tex], we call that impulse.
    We've now proven that the impulse provided to a system is the change in its momentum.

    HERE
    For a simpler case, where the force is constant, then the acceleration is also constant.
    For a constant acceleration the following holds true: [tex]a=\frac{\Delta V}{\Delta t}[/tex]

    Looking at the definition of impulse we see the following:

    [tex]J = F \times \Delta t[/tex]

    Using Newton's second law:
    [tex]F=ma=m\frac{\Delta V}{\Delta t}[/tex]

    We substitute and find:

    [tex]J=m\Delta V=\Delta P[/tex] (Provided that the mass remains constant, though this does not have to be the case for this to be valid.)

    (As an aside, we used a limited form of Newton's Second Law in our simple derivation for a constant force. Using the differential form we see:
    [tex]F=\frac{dP}{dt}[/tex]
    [tex]F=\frac{d(mv)}{dt}[/tex]
    [tex]F=m\frac{dv}{dt}+v\frac{dm}{dt}[/tex]

    Since we usually only deal with constant-mass situations, the second term is 0 and we're left with the usual [tex]F=ma[/tex])

    Back to the matter at hand though, looking at the redefinition of impulse as the change in the momentum of the system, you can now examine three objects.

    The first is the system as a whole. What was its momentum before the collision and after, and what that says about the impulse that acted on it.

    The second and third are the individual blocks. They each had their own individual momentum prior to, and after the collision. Their individual momentum changed, so therefore, an impulse acted on them. By finding the change in their momentum, you can find the impulse!

    This is a completely inelastic collision as defined by the question (My understanding is that the two masses continue to move as one unit after the collision), so that means that from conservation of momentum (Which we have proved to be the case here) we find:

    [tex]\vec P_{block #1 initial}+\vec P_{block #2 initial}=\vec P_{block #1 final}+\vec P_{block #2 final}[/tex]

    Since the masses are the same, and one of them starts at rest, the solution for the final velocity of the two-mass system should be especially simple! :)
     
  15. Jan 11, 2010 #14
    Well. When I first looked at Newton's three laws, I thought the third seemed simplest. How mistaken could a person be.
    I owe you a debt of gratitude for your care and guidance.
    Thank you RoyalCat.
     
  16. Jan 11, 2010 #15
    It's been a pleasure. :)

    Just two things before we part ways.

    You considered gravity a force internal to the system, and that's not quite true. The gravitational attraction among the two blocks is internal to the system, but the force of gravity exerted on the blocks by the earth is not!

    Why then did we ignore it in our analysis? Simple, the net external force on our system remained 0! The external force of gravity acting on both blocks, is canceled by the external normal reaction force from the ground, and so, these forces together cannot provide an impulse nor perform work on the system.

    If there's anything more you'd like to talk about, feel free to drop me a PM, I'll be happy to oblige. :)
     
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