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JEE Main202116 Mar 2021Evening ShiftPhysicsCenter of Mass, Momentum and CollisionActual

A large block of wood of mass M = 5 . 99 kg is hanging from two long massless cords. A bullet of mass m = 10 g is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their center of mass rising a vertical distance h = 9 . 8 cm before the (block + bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before the collision is: (Take g = 9 . 8 m

Options

  1. A841 . 4   m   s - 1
  2. B811 . 4   m   s - 1
  3. C831 . 4   m   s - 1
  4. D821 . 4   m   s - 1

Correct answer

C. 831 . 4   m   s - 1

Step-by-step solution

From energy conservation, [after bullet gets embedded till the system comes momentarily at rest] M + m g h = 1 2 M + m v 1 2 [ v 1 is velocity after collision] ∴   v 1 = 2 σ h Applying momentum conservation, (just before and just after collision) m v = M + m v 1 v = M + m m v 1 = 6 10 × 10 - 3 × 2 × 9 . 8 × 9 . 8 × 10 - 2 ≈ 831 . 55   m   s - 1

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