A block of mass 1 . 9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0 . 1 kg collides with the block and sticks to it. If the velocity of the bullet is 20 m s - 1 in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take g = 10 m s - 2 . Assume there is no rotational motion and loss of energy after the collision is negligible.]

Question

A block of mass 1 . 9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0 . 1 kg collides with the block and sticks to it. If the velocity of the bullet is 20 m s - 1 in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take g = 10 m s - 2 . Assume there is no rotational motion and loss of energy after the collision is negligible.], 21 J, 20 J, 19 J, 23 J

MARKS App · Accepted Answer

Conservation of linear momentum 0 . 1 × 20 = 0 . 1 + 1 . 9 × v v = 1 m s - 1 Using work energy theorem W g = ∆ k 2 × g × 1 = k - 1 2 × 2 × 1 2 ∴ k = 21 J

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