Asked by Anne
A 15g bullet is fired horizontally into a block of wood with a mass of 2.5 k and embedded in the block. Initially the block of wood hangs vertically and the impact causes the block o swing so that its center of mass rises 15cm. find the velocity of the bullet just before the impact.
Answers
Answered by
Damon
increase of potential energy of block = m g h
= (2.515)(9.81)(.15) = 3.70 Joules
so
(1/2) m v^2 = 3.7 Joules at bottom
(1/2)(2.515)v^2 = 3.7
v = 1.72 m/s at bottom after collision
momentum after collision
= 2.515 * 1.72 = 4.31 g m/s
momentum before collision = .015 v
so
v = 4.31 / .015 = 288 m/s
= (2.515)(9.81)(.15) = 3.70 Joules
so
(1/2) m v^2 = 3.7 Joules at bottom
(1/2)(2.515)v^2 = 3.7
v = 1.72 m/s at bottom after collision
momentum after collision
= 2.515 * 1.72 = 4.31 g m/s
momentum before collision = .015 v
so
v = 4.31 / .015 = 288 m/s
Answered by
Anonymous
He knows the mass of the bullet is 85 g. He fires the gun into a 1.3 kg block and finds the speed of the block/bullet combo to be 67 m/s. How fast was the bullet traveling before it hit the block?
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