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 s - 2 )

Question

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 s - 2 ), 841 . 4 m s - 1, 811 . 4 m s - 1, 831 . 4 m s - 1, 821 . 4 m s - 1

MARKS App · Accepted Answer

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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