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Collision-conceptual doubt

  1. Apr 19, 2013 #1
    1. The problem statement, all variables and given/known data

    While solving problems ,during collision ,we neglect impulse of external forces like gravitational force , spring force , acting on the system .

    The cases are

    1) An object falling from a height on a pan attached to a vertically hanging massless spring and colliding inelastically to it.

    2)Two balls,one thrown vertically upwards ,other moving vertically downwards colliding mid way in air .

    Does that mean we can always neglect impulse of all the conservative forces namely spring force,gravitational force,electrostatic force during collision?

    Kindly clear my conceptual doubt .

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 20, 2013 #2

    ehild

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    No, we cannot neglect all forces -conservative or not- during a collision. What we do is to consider the collision instantaneous - happening in a very short time Δt. During the collision, the external forces accelerate the centre of mass, so the overall momentum changes by F(external)Δt. If Δt is very short the change of the overall momentum can be neglected with respect to the change of momentum of the parts and conservation of momentum applies. How fast the collision is it depends on the internal forces between the colliding parts. To decide if it is fast enough you should know the internal forces in detail. Usually you do not know these forces so experience will tell if the model has been appropriate or not.

    When the external force is some kind of constraint that force can change during the collision process and can be very large. In that case you can not apply conservation of momentum.

    ehild
     
  4. Apr 20, 2013 #3
    Hi ehild :) hope you are doing well...

    Yes,we cannot neglect the forces ,but the impulse of the conservative forces during collisions is almost always neglected , and hence momentum is conserved before and after the collision while solving problems The cases I have mentioned before

    1) An object falling from a height on a pan attached to a vertically hanging massless spring and colliding inelastically to it.

    2)Two balls,one thrown vertically upwards ,other moving vertically downwards colliding mid way in air .

    Another case can be

    3)Two blocks colliding on an inclined plane having friction .

    Here we neglect impulse of friction .I cant remember the problem right now.

    Can you explain why we neglect impulse of spring force in 1) , gravitational force in 2) , friction in 3) ?

    How should I determine when to neglect impulse of forces while solving numericals ?
     
  5. Apr 20, 2013 #4

    ehild

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    Have you read my previous post?

    We can neglect the effect of an external force if the collision process is very fast, so the external force does not have enough time to change the momentum of the colliding bodies. You can know from experience when this approximation is valid.

    Try to solve the problem with the two balls colliding in mid-air, if both balls have the same mass of 1 kg, speeds of 2 m/s, and consider the balls elastic so keeping one ball compressed by ΔD force F=kΔD is needed, with k=2500 N/m. Can conservation of momentum assumed during the interaction between the balls? What happens if k=250000 N/m?

    ehild
     
  6. Apr 21, 2013 #5
    Yes I read your post...I understand what you have explained .I know that during collision the contact forces are large enough and the interaction time duration is very small , and hence impulse of external forces at times may be neglected .I had thought that some other concept related to conservative and nonconservative forces was involved , and hence put my doubt on the forum.
     
  7. Apr 21, 2013 #6

    ehild

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    Both the impulse of a force, FΔt and the work of the force FvΔt can be neglected if Δt is very small, no matter if the force is conservative or not.

    It is different if the external force changes during the interaction to a high extent. Such happens when a sand bag is suspended on a string of fix length and a bullet is shot into it. The string resists stretching and exerts reaction force that can be considered infinite. Conservation of momentum can not be applied, you need to use conservation of angular momentum instead.

    ehild
     
  8. May 28, 2013 #7
    Is friction an impulsive force ? Does it depend on the context of the problem i.e it might be impulsive in some cases and non-impulsive in other cases ?

    A sample case is -

    A bullet moving horizontally hits a block placed on a surface with friction . Should we conserve linear momentum in this case ?
     
    Last edited: May 28, 2013
  9. May 28, 2013 #8

    ehild

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    If there is friction you have three interacting bodies instead of two. The bullet interacts with the block, the block interacts with the ground. There are limiting cases when the problem is easy to solve.

    1. The friction is strong, the block is large, so the force of the bullet on the block can not overcome static friction. The block does not move while the block penetrates into it. The bullet moves for some Δt times till it loses all momentum, the impulse from the block is mΔv=FΔt.

    2. The force between bullet and block is greater than static friction. And it is so great that the bullet decelerates inside the block very fast so that the force of kinetic friction does not change the overall momentum appreciably during the time Δt when the bullet is in motion relatively to the block.

    In all other cases you need to know the forces in detail and solve the problem of three interacting bodies.

    ehild
     
  10. May 28, 2013 #9
    Hi ehild

    Thanks for the response :)

    Please have a look at the attachment . The question was asked in a prestigious national examination .The question along with the solution is given.The answer is given to be option c) .

    At other places answer is given to be a) and c) . Different opinions on this question exist .

    I know how to solve the problem .

    Solution 1)If friction is considered to be impulsive ,then we will get option c) .We cant conserve linear momentum .By applying angular impulse angular momentum relation we find , angular velocity to be anticlockwise ,hence friction towards left .

    Solution 2) If friction is considered non-impulsive, then we get both a) and c) . In this case we conserve linear momentum and get the linear speed of the center of the ring to be zero. By applying angular impulse angular momentum relation we find, angular velocity to be anticlockwise, hence friction towards left.

    Now I feel there is a third case which nowhere has been discussed which is as follows –

    There might not be friction present at all i.e the surface might be frictionless. The ring would have been rolling without slipping happily on a frictionless surface .The stationary ring could have been set into motion with an impulse such that it could have started rolling straightaway without slipping.

    In that case the answer would be option a) and d).

    Kindly give your thoughts on this problem especially the third case.
     

    Attached Files:

    Last edited: May 28, 2013
  11. May 28, 2013 #10

    About Case 2- Friction (if present) will be impulsive. This is because friction is always directly proportional to normal reaction and normal reaction is impulsive here (otherwise the ring would go inside the ground!).

    About Case 3- I am not sure but I think that the paper setters didnt think of the possibility you mention.
     
  12. May 28, 2013 #11

    ehild

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    Hi Tania,

    consciousness is right. If we consider the normal force between the ball and ring impulsive so should be the normal force between the ground and the ring. So the force of static friction is as big as needed: it is impulsive. The ball will roll after the collision.

    If the static friction was not great enough, that would have caused slipping during the collision, but the force of friction would be finite and the change of linear momentum would be so small in the very short time of the collision that you could consider the linear momentum to be conserved.

    Your version, (zero friction) can happen in principle. The ground can not exert horizontal force on the ball. Then both the linear momentum and angular momentum are conserved. In the real world, there is friction between the ground and the rolling object, but it can happen that the coefficient of static friction is very small: that again, means the friction is not impulsive.


    ehild
     
  13. May 30, 2013 #12

    utkarshakash

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    I think this was asked in IITJEE. OK as far as zero friction is considered, it is IMPOSSIBLE. You can't make a ring roll on a frictionless surface. It can move but it can't roll. To clarify it let us consider a hypothetical situation in which a ring is rolling without slipping on a frictionless horizontal surface. Let's start by drawing FBD of the ring. Can you tell me which force has a non-zero torque on the ring about CM?
     
    Last edited: May 30, 2013
  14. May 30, 2013 #13
    Absolutely POSSIBLE...The ring can surely roll without slipping on a frictionless surface.

    Place a ring and give an impulse such that the ring starts rolling without slipping immedietely with no friction required.The impulse given will provide the required rotational motion as well as translational motion.Now in order to maintain its motion it doesnt require any external force.
     
  15. May 31, 2013 #14
    When sphere is doing rolling without slipping on a horizontal surface, static friction is not involved, and hence it do no work on the sphere, and sphere rolls maintaining the condition of Vcm = Rω. It does not even mean that the surface was frictionless. In fact there can be pure rolling on a rough surface only(why?). If the surface was all smooth, the sphere would rather slip due to momentary impulse, and static friction could not bring the condition of pure rolling. Static friction is conservative force, remember. Same thing applies for ring.

    utkarshakash is right.

    Tanya, you are correct that when there is pure rolling, no external force is required to maintain it but of course "without friction" it cannot be initiated. Hint: While doing the numericals, why do you take the axis of rotation passing through contact point of sphere with surface ?
     
  16. May 31, 2013 #15

    ehild

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    Imagine that the object (sphere, ring ...) rolls on a rough surface so the condition of Vcm=rω holds, and then it arrives to a frictionless surface. Do either the velocity of the CM or the angular velocity of rotation change?

    ehild
     
  17. May 31, 2013 #16

    haruspex

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    I disagree. Friction could be insufficient to prevent slipping yet still deliver a substantial impulse. If the coefficient of static friction is μ and the normal impulse is J then surely the frictional impulse would be μJ.
    Strike the ring horizontally at the top and impulses will be just right to set the disc rolling without requiring any friction from the ground. Same applies to a disc and a sphere, but with different strike heights.
    In the present question, the impact was below the top of the ring, so a frictional force in the opposite direction is required to achieve rolling. If the impulse had been above the top of the ring (through some lever arm attachment) then the friction would need to act the other way.
     
  18. Jun 1, 2013 #17
    Velocity of CM or the angular velocity of rotation does not change, because as soon as the object starts pure rolling, there is no need of any external force to maintain it. When there was pure rolling, hypothetically there was no friction at that time but the ground was rough.

    Cannot understand.

    WITHOUT FRICTION, pure rolling cannot be initiated, right ? How can it be ? There has to be a sort of grab at the contact point of sphere. Otherwise it would just slip.
     
  19. Jun 1, 2013 #18

    ehild

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    I wrote about the case of non-impulsive friction which is defined that the impulse delivered can be taken zero.

    ehild
     
  20. Jun 1, 2013 #19

    haruspex

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    I should have said kinetic friction, not static.
    Impulses are never truly instantaneous. We just mean there was a large and probably variable force that acted for a very short time, resulting in a momentum change of ∫F.dt. If the normal force is N=N(t) over that interval and the normal impulse is JN = ∫N.dt, and slipping occurred through most of that, then the frictional force was F(t) = μkN(t), so the frictional impulse was ∫μkN(t)dt = μkJN.
    Wrong. Calculate this: a uniform cylinder mass m radius r rests on a smooth horizontal surface. A horizontal impulse J is delivered at height 3r/2. Find the horizontal and rotational speeds thereafter. At what speed, relative to the ground, does the part of the cylinder contacting the ground move?
     
  21. Jun 2, 2013 #20
    Is this site wrong then ? : http://www.astro.ucla.edu/~malkan/astro8/physics1a/rolling.htm

    It says that :

    Other references :

    http://www.phy.davidson.edu/fachome/dmb/PY430/Friction/rolling.html
    http://www.lhup.edu/~dsimanek/scenario/rolling.htm

    Thanks.
     
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