1) A 12-g bullet with an initial speed of 299 m/s is shot directly at a 1.1-kg w

Question

1) A 12-g bullet with an initial speed of 299 m/s is shot directly at a 1.1-kg wooden block, which rests on a frictionless surface. The bullet passes through the block, after which the block is moving at a speed of 0.172 m/s. What is the speed of the bullet after it has passed through the block?

2) A 12-g bullet with an unknown initial speed is shot directly at a 1-kg wooden block, which rests on a frictionless surface. The bullet passes through the block and emerges with a speed of 300 m/s, after which the block is moving at a speed of 0.157 m/s. What was the speed of the bullet before it hit the block?

Explanation / Answer

1)

by conservation of momentum

momentum before collision = momentum after collision

let speed of the bullet after it has passed through the block = Vf

0.012*299 + 1.1*0 = 1.1*0.172 + 0.012*Vf

Vf = (0.012*299 - 1.1*0.172)/0.012 = 283.23 m/s

2)

momentum before collision = momentum after collision

let the speed of the bullet before it hit the block = Vi

0.012*Vi + 1.0*0 = 1.0*0.157 + 0.012*300

Vf = (1.0*0.157 + 0.012*300)/0.012 = 313.08 m/s

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