A 1.35 kg block of wood sits at the edge of a table, 0.782 m above the floor. A

Question

A 1.35 kg block of wood sits at the edge of a table, 0.782 m above the floor. A 0.0105 kg bullet moving horizontally with a speed of 715 m/s embeds itself within the block. What horizontal distance does the block cover before hitting the ground?

Explanation / Answer

The mass of block M = 1.35 kg The mass of bullet m = 0.0105 kg height h = 0.782 m speed of bullet v = 715 m according to conservation of energy       m v + M v' = ( M + m ) V The initial speed of the block v ' = 0 m/s So that we have           m v =( M + m ) V ==> therefore final speed of the velocity - blocksystem is             V = m v / ( M + m)                 = 0.0105 * 715 / ( 1.35 + 0.0105)                 = 5.518 m/s The time taken to hit the ground is              t = 2h / g                = 2 * 0.782 / 9.8                = 0.399 s Therefore the horizontal distance traveled by the block = V * t = 5.518 * 0.399 = 2.204 m

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