Asked by Zoey
A 8.05- g bullet from a 9-mm pistol has a velocity of 346.0 m/s. It strikes the 0.785- kg block of a ballistic pendulum and passes completely through the block. If the block rises through a distance h = 16.04 cm, what was the velocity of the bullet as it emerged from the block?
Answers
Answered by
Damon
initial momentum = .00805 * 346
final momentum just after collision
= M V + m v
so
.00805 = MV + m v
we can do MV from how far up it goes
(1/2) M V^2 = M g h =M (9.81)(.1504)
cancel M and find V
now we have it
initial momentum = final momentum
.00805 * 346 = .785 V + .00805 v
we know V
solve for v
final momentum just after collision
= M V + m v
so
.00805 = MV + m v
we can do MV from how far up it goes
(1/2) M V^2 = M g h =M (9.81)(.1504)
cancel M and find V
now we have it
initial momentum = final momentum
.00805 * 346 = .785 V + .00805 v
we know V
solve for v
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