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A 10.1-g bullet is fired into a stationary block of wood having mass m = 5.03 kg. The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.593 m/s. What was the original speed of the bullet? (Express your answer with four significant figures.)

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gilcarusolautaro

Answer:

The original speed of the bullet was of V1= 295.9187 m/s.

Explanation:

m1= 10.1 g = 0.0101 kg

m2= 5.03 kg

V2= 0.593 m/s

V1= ?

m1*V1 = (m2+m1)*V2

V1= [(m2+m1) * V2] / m1

V1= 295.9187 m/s

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