A 9.6-g bullet is fired into a stationary block of wood having mass m = 4.95 kg. The bullet imbeds into the bl?
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A 9.6-g bullet is fired into a stationary block of wood having mass m = 4.95 kg. The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.606 m/s. What was the original speed of the bullet? (Express your answer with four significant figures.) answer in m/s
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Answer:
the velocity of the block + bullet after a completely inelastic impact is v' = (m1v1 + m2v2) / (m1 + m2) where m1, v1 = mass and velocity of the bullet, m2, v2 = mass and velocity of the block 0.606 = (0.0096*v1 + 4.95*0) / (0.0096 + 4.95) --> v1 = 313.075 m/s <--- velocity of bullet OG
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Other answers
the velocity of the block + bullet after a completely inelastic impact is v' = (m1v1 + m2v2) / (m1 + m2) where m1, v1 = mass and velocity of the bullet, m2, v2 = mass and velocity of the block 0.606 = (0.0096*v1 + 4.95*0) / (0.0096 + 4.95) --> v1 = 313.075 m/s <--- velocity of bullet OG
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