Collision-conceptual doubt

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In summary, during collision, we often neglect the impulse of external forces like gravitational force, spring force, and friction. This is because the collision process is usually very fast, so the external forces do not have enough time to significantly change the momentum of the colliding bodies. However, in some cases, such as when the external force changes significantly during the interaction, or when there are multiple interacting bodies, conservation of momentum may not apply and we need to consider other factors, such as conservation of angular momentum.
  • #1
Tanya Sharma
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Homework Statement



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 .

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The Attempt at a Solution

 
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  • #2
Tanya Sharma said:

Homework Statement



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?
[/b]

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
 
  • #3
Hi ehild :) hope you are doing well...

ehild said:
No, we cannot neglect all forces -conservative or not- during a collision.
ehild

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 can't 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 ?
 
  • #4
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
 
  • #5
ehild said:
Have you read my previous post?
ehild

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.
 
  • #6
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
 
  • #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 ?
 
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  • #8
Tanya Sharma said:
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 ?

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
 
  • #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 can't 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.
 

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  • #10
Tanya Sharma said:
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 can't 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.


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.
 
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  • #11
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
 
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  • #12
Tanya Sharma said:
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 can't 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.

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?
 
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  • #13
utkarshakash said:
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.

Absolutely POSSIBLE...The ring can surely roll without slipping on a frictionless surface.

utkarshakash said:
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?

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 doesn't require any external force.
 
  • #14
Tanya Sharma said:
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 doesn't require any external force.

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 ?
 
  • #15
sankalpmittal said:
... when there is pure rolling, no external force is required to maintain it but of course "without friction" it cannot be initiated.

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
 
  • #16
ehild said:
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.
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.
sankalpmittal said:
when there is pure rolling, no external force is required to maintain it but of course "without friction" it cannot be initiated
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.
 
  • #17
ehild said:
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

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.

haruspex said:
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.

Cannot understand.

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.

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.
 
  • #18
haruspex said:
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.

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

ehild
 
  • #19
sankalpmittal said:
haruspex said:
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.
Cannot understand.
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.
WITHOUT FRICTION, pure rolling cannot be initiated, right ?
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?
 
  • #20
haruspex said:
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?

Is this site wrong then ? : http://www.astro.ucla.edu/~malkan/astro8/physics1a/rolling.htm

It says that :

An important point to remember is that rolling motion cannot occur on a frictionless surface. The thing will simply slide along. In the case of something rolling down an incline if the angle of the incline is very large or the inclines coefficient of friction is very small the sphere will also slide. To get pure rolling motion the force of friction must be less than the normal force times the coefficient of friction.

Other references :

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

Thanks.
 
  • #22
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.
. I think all of you should consider this statement of sankalpmittal to clarify your doubt.
 
  • #23
I would suggest everyone to take a look at this site
http://www.lhup.edu/~dsimanek/scenario/rolling.htm
 
  • #24
Tanya Sharma said:
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 doesn't require any external force.

You cannot give an impulse such that the ring starts rolling without slipping on a frictionless surface. I would rather like to suggest you refer to HC Verma Part 1. I have sorted out some problems that might make you think on your reasoning once again. Worked Out Examples- Q.No- 25, 29. Exercises- Q.No. - 80, 81, 82.
 
  • #25
utkarshakash said:
I would suggest everyone to take a look at this site
Pls quote the passage you wish to bring to our attention. (Mostly the article seems to be about speaking of friction as opposing motion instead of as opposing relative motion of the surfaces. Indeed, is that error at the root of your objection? You understand that once rolling is established on a horizontal surface there is no frictional force, right?)
You cannot give an impulse such that the ring starts rolling without slipping on a frictionless surface.
As ehild pointed out, you could have rolling occurring on a frictionless surface in consequence of the ring first rolling on a frictional surface then transiting to a smooth one. I suspected that would not satisfy you because friction was still involved in establishing the rolling motion, but it does demonstrate that if the linear velocity just happens to be rω, by whatever means, then you have frictionless rolling. So I set up an example of how exactly the right combination can occur merely by judicious application of an impulse. Did you do the algebra?
 
  • #26
ehild said:
I wrote about the case of non-impulsive friction which is defined that the impulse delivered can be taken zero.
That's not how I read you prior post:
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.
I read that as saying that either the impulsive friction is enough to produce rolling immediately, or it can be neglected. I'm saying that there is a third possibility between the two, and you can prove this by starting with a realistic scenario in which the impact occurs over a short period of time, then letting that period tend to zero.
 
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  • #27
haruspex said:
So I set up an example of how exactly the right combination can occur merely by judicious application of an impulse. Did you do the algebra?
Consider this question
"A thin spherical shell of radius R lying on a rough horizontal surface is hit sharply and horizontally by a cue. Where should it be hit so that the shell does not slip on the surface?"

What do you have to say about this?Why it mentions the word "rough surface"?
 
  • #28
haruspex said:
Pls quote the passage you wish to bring to our attention.

"That last sentence is important. The force due to friction doesn't necessarily act to oppose the motion of the body but acts to oppose slipping or sliding along the contact surface."

When an impulse is given the part of the body which is in contact with the ground has a tendency to slip. This is where frictional force comes into play by opposing the tendecy to slip along the surface. If there were no friction the tendency to slipping couldn't have been countered.
 
  • #29
utkarshakash said:
"That last sentence is important. The force due to friction doesn't necessarily act to oppose the motion of the body but acts to oppose slipping or sliding along the contact surface."
I agree entirely, and it supports my argument. In the special circumstances I've described, there is no tendency to relative motion of the two surfaces, hence no friction.
"A thin spherical shell of radius R lying on a rough horizontal surface is hit sharply and horizontally by a cue. Where should it be hit so that the shell does not slip on the surface?"

What do you have to say about this?Why it mentions the word "rough surface"?
Because the impulse is purely horizontal. This means there is no impulsive downward force, so no impulsive normal reaction. The author of the question appears to be testing the reader's ability to deduce that no matter how rough the surface is it would be unable to prevent slipping if the ball were struck at just any old height. Only if it is struck at exactly the right height will the resulting horizontal impulse and angular impulse be so as to produce rolling motion straight away. In short, the surface might just as well be smooth, but it's up to the reader to figure that out.
 
  • #30
haruspex said:
That's not how I read you prior post:
I read that as saying that either the impulsive friction is enough to produce rolling immediately, or it can be neglected. I'm saying that there is a third possibility between the two, and you can prove this by starting with a realistic scenario in which the impact occurs over a short period of time, then letting that period tend to zero.

You are right, I did not think it over. Friction means finite change of momentum even in case of non-impulsive impact. ehild
 
  • #31
utkarshakash said:
You cannot give an impulse such that the ring starts rolling without slipping on a frictionless surface. I would rather like to suggest you refer to HC Verma Part 1. I have sorted out some problems that might make you think on your reasoning once again. Worked Out Examples- Q.No- 25, 29. Exercises- Q.No. - 80, 81, 82.

In each of these questions we considered rough road and took the axis of rotation to be passing through contact point of sphere, so that we could conserve angular momentum of the sphere system. This truly reveals that friction(Static) is necessary to bring pure rolling.

haruspex said:
Pls quote the passage you wish to bring to our attention. (Mostly the article seems to be about speaking of friction as opposing motion instead of as opposing relative motion of the surfaces. Indeed, is that error at the root of your objection? You understand that once rolling is established on a horizontal surface there is no frictional force, right?)

Friction opposes skidding or slipping. Without it, what thing will prevent the object to slip? Your explanation states that during pure rolling friction does not work and it is not involved. But, that does not even mean that there can be pure rolling initiated without the aid of friction, i.e. on a frictionless road. Does it ?

As ehild pointed out, you could have rolling occurring on a frictionless surface in consequence of the ring first rolling on a frictional surface then transiting to a smooth one. I suspected that would not satisfy you because friction was still involved in establishing the rolling motion, but it does demonstrate that if the linear velocity just happens to be rω, by whatever means, then you have frictionless rolling. So I set up an example of how exactly the right combination can occur merely by judicious application of an impulse. Did you do the algebra?

Yes, Vcm=Rω, the condition of pure rolling once established, does not need friction, but again this condition cannot be without the help of friction. I did the algebra in the questions posted by utkarshakash of H.C. Verma.


haruspex:
Because the impulse is purely horizontal. This means there is no impulsive downward force, so no impulsive normal reaction. The author of the question appears to be testing the reader's ability to deduce that no matter how rough the surface is it would be unable to prevent slipping if the ball were struck at just any old height. Only if it is struck at exactly the right height will the resulting horizontal impulse and angular impulse be so as to produce rolling motion straight away. In short, the surface might just as well be smooth, but it's up to the reader to figure that out.

Friction does not act for a very short time. It keeps the relative motion opposed(in case of kinetic) and it endeavours to bring the condition of relative momentary rest(in case of static). Maybe I am overlooking your post, but I cannot fathom what you said. Can you please explain ?
 
  • #32
sankalpmittal said:
I did the algebra in the questions posted by utkarshakash of H.C. Verma.
I don't have a copy of those, but most likely friction is necessary to establish rolling in those cases. My argument is that there are special cases where it is not. I provided an example.
Friction does not act for a very short time.?
It can be impulsive. If the impact has a vertical component, resulting in an impulsive normal reaction (i.e. a very large force for a very short time) then the kinetic friction will be correspondingly impulsive. However, that's all irrelevant to the question of whether it is possible to initiate rolling motion in the absence of friction.
 
  • #33
utkarshakash said:
You cannot give an impulse such that the ring starts rolling without slipping on a frictionless surface. I would rather like to suggest you refer to HC Verma Part 1. I have sorted out some problems that might make you think on your reasoning once again. Worked Out Examples- Q.No- 25, 29. Exercises- Q.No. - 80, 81, 82.


HCV Rotational Mechanics OBJECTIVE II (multiple correct options)

Q11 A sphere cannot roll on
a) a smooth horizontal surface
b) a smooth inclined surface
c) a rough horizontal surface
d) a rough inclined surface

Answer is (b), not (a)(b) the only required condition for pure rolling is v=rω.


I think that students were supposed to ignore the possibility that Tanya mentioned in her post (case three). As far as I can tell this was an instance of mismanagement and the question was wrong.

See this link for IIT JEE 2011 mistakes-http://www.hindustantimes.com/India-news/NewDelhi/IITs-correct-JEE-error-sparks-more-confusion/Article1-697995.aspx

It is mentioned that one 4 marks physics question was wrong from paper 2.
 
  • #34
consciousness said:
HCV Rotational Mechanics OBJECTIVE II (multiple correct options)

Q11 A sphere cannot roll on
a) a smooth horizontal surface
b) a smooth inclined surface
c) a rough horizontal surface
d) a rough inclined surface

Answer is (b), not (a)(b) the only required condition for pure rolling is v=rω.I think that students were supposed to ignore the possibility that Tanya mentioned in her post (case three). As far as I can tell this was an instance of mismanagement and the question was wrong.

See this link for IIT JEE 2011 mistakes-http://www.hindustantimes.com/India-news/NewDelhi/IITs-correct-JEE-error-sparks-more-confusion/Article1-697995.aspx

It is mentioned that one 4 marks physics question was wrong from paper 2.

Please read the question again. It simply asks whether a sphere can roll or not. Rolling does not always mean "pure rolling". Also it mentions nowhere whether it can be initiated or not.
 
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  • #35
utkarshakash said:
Please read the question again. It simply asks whether a sphere can roll or not. Rolling does not always mean "pure rolling". Also it mentions nowhere whether it can be initiated or not.

Pure rolling is many a times shortened to "rolling" as in this case. As for initiation see haruspex's post #19.
 
<h2>1. What is collision-conceptual doubt?</h2><p>Collision-conceptual doubt is a term used in physics to describe the uncertainty or lack of clarity surrounding the concept of a collision. It refers to the difficulty in accurately defining and understanding the physical interactions between two or more objects during a collision.</p><h2>2. Why is collision-conceptual doubt important?</h2><p>Collision-conceptual doubt is important because it highlights the limitations of our current understanding of collisions and the need for further research and development in this area. It also has practical implications in fields such as engineering and transportation, where collisions are a common occurrence.</p><h2>3. What factors contribute to collision-conceptual doubt?</h2><p>There are several factors that contribute to collision-conceptual doubt, including the complexity of the physical forces involved, the variability of real-world collisions, and the limitations of our current scientific models and theories.</p><h2>4. How can collision-conceptual doubt be addressed?</h2><p>Collision-conceptual doubt can be addressed through continued research and experimentation, as well as the development of more accurate and comprehensive scientific models and theories. Collaborative efforts between scientists and engineers can also help to improve our understanding of collisions and their effects.</p><h2>5. What are some potential consequences of not addressing collision-conceptual doubt?</h2><p>If collision-conceptual doubt is not addressed, it can lead to inaccurate predictions and assumptions in various fields, potentially resulting in safety hazards, design flaws, and other negative consequences. It can also hinder progress and advancements in our understanding of the physical world.</p>

1. What is collision-conceptual doubt?

Collision-conceptual doubt is a term used in physics to describe the uncertainty or lack of clarity surrounding the concept of a collision. It refers to the difficulty in accurately defining and understanding the physical interactions between two or more objects during a collision.

2. Why is collision-conceptual doubt important?

Collision-conceptual doubt is important because it highlights the limitations of our current understanding of collisions and the need for further research and development in this area. It also has practical implications in fields such as engineering and transportation, where collisions are a common occurrence.

3. What factors contribute to collision-conceptual doubt?

There are several factors that contribute to collision-conceptual doubt, including the complexity of the physical forces involved, the variability of real-world collisions, and the limitations of our current scientific models and theories.

4. How can collision-conceptual doubt be addressed?

Collision-conceptual doubt can be addressed through continued research and experimentation, as well as the development of more accurate and comprehensive scientific models and theories. Collaborative efforts between scientists and engineers can also help to improve our understanding of collisions and their effects.

5. What are some potential consequences of not addressing collision-conceptual doubt?

If collision-conceptual doubt is not addressed, it can lead to inaccurate predictions and assumptions in various fields, potentially resulting in safety hazards, design flaws, and other negative consequences. It can also hinder progress and advancements in our understanding of the physical world.

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