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ExamsJEE AdvancedPhysics

A bead of mass m is constrained to slide along a rough rail. A spring (natural length zero) connects the bead to a point that slides freely along a second frictionless rail, with the spring always remaining perpendicular to the smooth rail. The two rails meet at a vertex with angle theta between them, the setup is in gravity-free space, and the bead starts at the vertex with initial speed v0. The kinetic friction coefficient is muₖ. Which of the following statements is/are correct?

  1. The friction force on the bead when it has moved a distance x along the rough rail is muₖ * k * x * sin(theta) * cos(theta).
  2. The bead comes to rest after traveling a distance sqrt(m * v0² / (k * (sin²(theta) + muₖ * sin(theta) * cos(theta)))) along the rough rail.
  3. The bead will return to the vertex if tan(theta) > muₖ.
  4. The friction force on the bead when it has moved a distance x along the rough rail is muₖ * k * x * sin(2 * theta).

Correct answer: The friction force on the bead when it has moved a distance x along the rough rail is muₖ * k * x * sin(theta) * cos(theta).

Solution

When the bead is at distance x along the rough rail, the spring length is x sin(theta) and the spring force component perpendicular to the rough rail provides the normal force. Options A and B and C are correct; option D has a factor of 2 error since sin(2theta) = 2 sin(theta)cos(theta).

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