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A smooth sphere 'A' is moving on a frictionless horizontal surface with angular speed ω and centre of mass velocity v. 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)ωB = 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 surface with ...
To answer the question, we need to use the concept of rotational kinetic energy.
The rotational kinetic energy of a rotating object can be calculated using the formula:
KE_rot = (1/2) * I * ω^2
where KE_rot is the rotational kinetic energy, I is the moment of inertia of the object, and ω is the angular speed of the object.
For a solid sphere, the moment of inertia (I) is given by:
I = (2/5) * m * R^2
where m is the mass of the sphere and R is the radius of the sphere.
Since the sphere is smooth and moving on a frictionless horizontal surface, there is no external torque acting on it. Therefore, its angular speed remains constant.
Let's assume the initial angular speed of the sphere is ω_0.
The initial rotational kinetic energy of the sphere is given by:
KE_rot_0 = (1/2) * I * ω_0^2
Now, let's calculate the final rotational kinetic energy when the angular speed becomes 3ω_0:
KE_rot_final = (1/2) * I * (3ω_0)^2
Simplifying:
KE_rot_final = (1/2) * I * 9ω_0^2
Since the angular speed is 3 times the initial angular speed, the final rotational kinetic energy is 9 times the initial rotational kinetic energy.
Therefore, the final rotational kinetic energy is 9 times greater than the initial rotational kinetic energy.
The rotational kinetic energy of a rotating object can be calculated using the formula:
KE_rot = (1/2) * I * ω^2
where KE_rot is the rotational kinetic energy, I is the moment of inertia of the object, and ω is the angular speed of the object.
For a solid sphere, the moment of inertia (I) is given by:
I = (2/5) * m * R^2
where m is the mass of the sphere and R is the radius of the sphere.
Since the sphere is smooth and moving on a frictionless horizontal surface, there is no external torque acting on it. Therefore, its angular speed remains constant.
Let's assume the initial angular speed of the sphere is ω_0.
The initial rotational kinetic energy of the sphere is given by:
KE_rot_0 = (1/2) * I * ω_0^2
Now, let's calculate the final rotational kinetic energy when the angular speed becomes 3ω_0:
KE_rot_final = (1/2) * I * (3ω_0)^2
Simplifying:
KE_rot_final = (1/2) * I * 9ω_0^2
Since the angular speed is 3 times the initial angular speed, the final rotational kinetic energy is 9 times the initial rotational kinetic energy.
Therefore, the final rotational kinetic energy is 9 times greater than the initial rotational kinetic energy.
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A smooth sphere A is moving on a frictionless horizontal surface with angular speed ωand centre of mass velocity v. 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 < ωBb)ωA = ωBc)ωA= ωd)ωB = ωCorrect answer is option 'C'. Can you explain this answer?
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A smooth sphere A is moving on a frictionless horizontal surface with angular speed ωand centre of mass velocity v. 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 < ωBb)ωA = ωBc)ωA= ωd)ωB = ωCorrect 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 surface with angular speed ωand centre of mass velocity v. 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 < ωBb)ωA = ωBc)ωA= ωd)ωB = ωCorrect 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 surface with angular speed ωand centre of mass velocity v. 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 < ωBb)ωA = ωBc)ωA= ωd)ωB = ωCorrect answer is option 'C'. Can you explain this answer?.
A smooth sphere A is moving on a frictionless horizontal surface with angular speed ωand centre of mass velocity v. 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 < ωBb)ωA = ωBc)ωA= ωd)ωB = ωCorrect 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 surface with angular speed ωand centre of mass velocity v. 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 < ωBb)ωA = ωBc)ωA= ωd)ωB = ωCorrect 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 surface with angular speed ωand centre of mass velocity v. 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 < ωBb)ωA = ωBc)ωA= ωd)ωB = ωCorrect answer is option 'C'. Can you explain this answer?.
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ample number of questions to practice A smooth sphere A is moving on a frictionless horizontal surface with angular speed ωand centre of mass velocity v. 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 < ωBb)ωA = ωBc)ωA= ωd)ωB = ωCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
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