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6. A 10.0 g bullet is fired into a station kg). The the block of wood (m a 5.00 relative motion of bullet stops inside the block. The speed of the Dullet-plus-block combination immediately after the collision is 0.600 m/s. a) (5 pts.) What was the speed of the bullet before the collision? b) (5 pts. How much energy was lost in the collision?
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Answer #1

a) the velocity of the block + bullet after a completely inelastic impact is
v' = (m1v1 + m2v2) / (m1 + m2) where m1, v1 = mass and velocity of the bullet, m2, v2 = mass and velocity of the block
0.60 = (0.010*v1 + 5*0) / (0.010 + 5)
--> v1 = 300.6 m/s

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