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

Norah

29 July, 14:43

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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Lilia · 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.