# A 12.0 g rubber bullet travels at a velocity of 150 m/s, hits a stationary 8.5 kg concrete block resting on a frictionless surface, and ricochets in the opposite direction with a velocity of -100. m/s. How fast will the concrete block be moving?

Use the to answer this question.
Calculate the total momentum before the collision.
momentum of bullet = mass x velocity = 0.012 x 150 = 1.8 N.s
(note the mass of the bullet is in grams)
momentum of block = mass x velocity = 8.5 x 0 = 0 N.s
total momentum = 1.8 + 0 = 1,8 N.s
Then find the total momentum after the collision
momentum of bullet = 0.012 x (-100) = -1.2 N.s
(note sign tells us it is in the opposite direction)
momentum of block = 8.5 x v
Total momentum = -1.2 + 8.5v
But the total momentum is always conserved
so
1.8 = -1.2 + 8.5v
3 = 8.5v
v = 3/8.5 = 0.35 m/s

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A 12.0 g rubber bullet travels at a velocity of 150 m/s, hits a stationary 8.5 kg concrete block resting on a frictionless surface, and ricochets in the opposite direction with a velocity of -100. m/s. How fast will the concrete block be moving?
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