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A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass velocity ν. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are ωA and ωB, respectively. Then
  • a)
    ωA< ωB
  • b)
    ωA= ωB
  • c)
    ωA= ω
  • d)
    ω​B= ω​
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A smooth sphere A is moving on a frictionless horizontal plane with an...
As the spheres are smooth there will be no friction (no torque) and therefore there will be no transfer of angular momentum. Thus A, after collision will remain with its initial angular momentum. i.e., ω A
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A smooth sphere A is moving on a frictionless horizontal plane with an...
Ω. A smaller sphere B is at rest on the plane. Sphere A collides with sphere B and transfers some of its momentum to sphere B. After the collision, sphere A moves with angular speed ω/2 and sphere B moves with angular speed ω/4. What is the ratio of the masses of spheres A and B?

We can solve this problem using conservation of angular momentum. Before the collision, the angular momentum of sphere A is L1 = I1ω, where I1 is the moment of inertia of sphere A. Since the plane is frictionless, there is no external torque acting on the system, so the total angular momentum of the system is conserved during the collision. After the collision, the angular momentum of sphere A is L2 = I1(ω/2) and the angular momentum of sphere B is L3 = I2(ω/4), where I2 is the moment of inertia of sphere B.

Therefore, we have:

L1 = L2 + L3

I1ω = I1(ω/2) + I2(ω/4)

Multiplying both sides by 4/I1ω, we get:

4/I1 = 1/I1 + 1/I2

Simplifying:

3/I1 = 1/I2

So the ratio of the masses of spheres A and B is:

m1/m2 = I2/I1 = 3/1

Therefore, the mass of sphere A is three times the mass of sphere B.
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A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass velocity ν. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are ωA and ωB, respectively. Thena)ωA< ωBb)ωA= ωBc)ωA= ωd)ω​B= ω​Correct answer is option 'C'. Can you explain this answer?
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A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass velocity ν. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are ωA and ωB, respectively. Thena)ωA< ωBb)ωA= ωBc)ωA= ωd)ω​B= ω​Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass velocity ν. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are ωA and ωB, respectively. Thena)ωA< ωBb)ωA= ωBc)ωA= ωd)ω​B= ω​Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass velocity ν. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are ωA and ωB, respectively. Thena)ωA< ωBb)ωA= ωBc)ωA= ωd)ω​B= ω​Correct answer is option 'C'. Can you explain this answer?.
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