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

Physics

Kendall Ortega

5 July, 09:17

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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Jason Velazquez · 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.