Asked by Lesedi
A bullet of mass 40g strikes a stationary wooden block of mass 6kg horizontally at 500m/s the bullet goes through the block,and emerges from from the other side at 200m/s.calculate the velocity of the block once the bullet emerges from it
Answers
Answered by
Damon
original momentum = .04*500
= 20 kg m/s
that will be the final momentum so
6 v + .04*200 = .04*500
6 v = .04*300
v = .04*50
v = 2 m/s
= 20 kg m/s
that will be the final momentum so
6 v + .04*200 = .04*500
6 v = .04*300
v = .04*50
v = 2 m/s
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