horizontally at 460 m / s is shot through a 1.50 k g wood block suspended on a s
ID: 2308611 • Letter: H
Question
horizontally at 460 m/s is shot through a 1.50 kg wood block suspended on a string 1.00 mlong. Part A If the center of mass of the block rises a distance of 0.500 cm , find the speed of the bullet as it emerges from the block. Express your answer in meters per second to three significant figures. horizontally at 460 m/s is shot through a 1.50 kg wood block suspended on a string 1.00 mlong. Part A If the center of mass of the block rises a distance of 0.500 cm , find the speed of the bullet as it emerges from the block. Express your answer in meters per second to three significant figures. horizontally at 460 m/s is shot through a 1.50 kg wood block suspended on a string 1.00 mlong. horizontally at 460 m/s is shot through a 1.50 kg wood block suspended on a string 1.00 mlong. horizontally at 460 m/s is shot through a 1.50 kg wood block suspended on a string 1.00 mlong. Part A If the center of mass of the block rises a distance of 0.500 cm , find the speed of the bullet as it emerges from the block. Express your answer in meters per second to three significant figures. Part A If the center of mass of the block rises a distance of 0.500 cm , find the speed of the bullet as it emerges from the block. Express your answer in meters per second to three significant figures. Part A If the center of mass of the block rises a distance of 0.500 cm , find the speed of the bullet as it emerges from the block. Express your answer in meters per second to three significant figures.Explanation / Answer
[you didnt give how many grams of bullet, so i am assuming 5.00 g bullet]
Initial K.E= 1/2 m v2 = 0.5 * 0.005 kg * 4602 = 529 J
Energy gained by the block after the impact = m g h = 1.50 * 9.81 * 0.005 m = 0.0736 J
K.E. of bullet after impact = Initial K.E - Energy transferred
= 529 - 0.0736 = 528.93 J
K.E = 1/2 m v2 = 528.93
v2 = 528.93 * 2 / 0.005
v = 460 m/s
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