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JEE Advanced 2023 · Paper 1

JEE Advanced 2023 Physics question on Rotational Motion

A bar of mass M=1.00M = 1.00 kg and length L=0.20L = 0.20 m is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass m=0.10m = 0.10 kg is moving on the same horizontal surface with 5.005.00 m s−1^{-1} speed on a path perpendicular to the bar. It hits the bar at a distance L/2L/2 from the pivoted end and returns back on the same path with speed v. After this elastic collision, the bar rotates with an angular velocity ω\omega. Which of the following statement is correct?
  1. A.
    ω=6.98\omega = 6.98 rad s−1^{-1} and v =4.30= 4.30 m s−1^{-1}
  2. B.
    ω=3.75\omega = 3.75 rad s−1^{-1} and v =4.30= 4.30 m s−1^{-1}
  3. C.
    ω=3.75\omega = 3.75 rad s−1^{-1} and v =10.0= 10.0 m s−1^{-1}
  4. D.
    ω=6.80\omega = 6.80 rad s−1^{-1} and v =4.10= 4.10 m s−1^{-1}
Show answer

Correct option: A

[JEE Advanced 2023 · Paper 1]

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