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JEE Advanced Physics Center of Mass Momentum and Collision 2023 JEE Advanced 2023 (Paper 1)

JEE Advanced Physics Question (2023) — Solution

Question

A bar of mass M = 1 . 00   kg and length L = 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 . 10   kg is moving on the same horizontal surface with 5 . 00   m   s – 1 speed on a path perpendicular to the bar. It hits the bar at a distance L 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 . Which of the following statement is correct?

Options

  1. A. ω = 6 . 98   rad   s – 1  and  v = 4 . 30   m   s – 1
  2. B. ω = 3 . 75   rad   s – 1  and  v = 4 . 30   m   s – 1
  3. C. ω = 3 . 75   rad   s – 1  and  v = 10 . 0   m   s – 1
  4. D. ω = 6 . 80   rad   s – 1  and  v = 4 . 10   m   s – 1

Answer

A. ω = 6 . 98   rad   s – 1  and  v = 4 . 30   m   s – 1

Step-by-step solution

Before: After: Applying conservation of angular momentum about point O , we get m v 0 L 2 = M L 2 3 ω - m v L 2               . . . i As the collision is elastic,  e = 1 ⇒ velocity of separation(after collision) velocity of approach(before collision) = 1 ⇒ L ω 2 - - v v 0 = 1 ⇒ v = v 0 - L ω 2             . . . ii Using equation i  and ii , we can write m v 0 L 2 = M L 2 3 ω - m v 0 - L ω 2 L 2 Given:  m = 0 . 1   kg ,   M = 1   kg ,   L = 0 . 2   m  and v 0 = 5   m   s - 1 Therefore, 0 . 1 × 5 0 . 2 2 = 1 × 0 . 2 2 3 ω - 0 . 1 5 - 0 . 2 ω 2 0 . 2 2 ⇒ 0 . 05 = 0 . 04 3 ω - 0 . 5 - 0 . 01 ω × 0 . 1 ⇒ 0 . 05 + 0 . 05 = ω 0 . 04 3 + 0 . 001 ⇒ ω = 0 . 1 × 3 0 . 04 + 0 . 003 ≈ 6 . 98   rad   s - 1 Now, ⇒ v = 5 - 0 . 2 × 6 . 98 2 ≈ 4 . 3   m   s - 1

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