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    A bullet of mass 10 g moving horizontally with a velocity of 400 ms–1 strikes of wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be

  • 160 ms–1
  • 100 ms–1
  • 80 ms–1
  • 120 ms–1
  • Explanation:

    During the collision, apply moementum conservation (0.01)(400) + 0 = (2)V + (0.01)V'

    where V = √2gh

    V = √2 ×10× 0.1

    V = √2 solving

    V' = 120 m/sec. Ans.