A 10.0-g bullet moving at 300 m/s is fired into a 1.00-kg block at rest. the bullet emerges (the bullet does not get embedded in the block with half of its original speed. what is the velocity of the block right after the collision?

Question

Keith Blackwell

Physics

14 January, 06:58

A 10.0-g bullet moving at 300 m/s is fired into a 1.00-kg block at rest. the bullet emerges (the bullet does not get embedded in the block with half of its original speed. what is the velocity of the block right after the collision?

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Answers (1)

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Leandro Chase · Answer 1

Momentum before the hit:

p = mv = 0.01 * 300 + 1 * 0

Momentum after the hit:

p = 0.01 * 150 + 1 * v

Momentum is conserved:

0.01 * 300 = 0.01 * 150 + v

3 = 1.5 + v

v = 1.5

The velocity of the block after the collision is 1.5 m/s.