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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 0.591 m/s. What was the original speed of the bullet? (Express your answer with four significant figures.)

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Blacklash

Answer:

Original speed of the bullet = 305.21 m/s

Explanation:

Here momentum is conserved.

Mass of bullet = 9.6 g = 0.0096 kg

Mass of wood = 4.95 kg

Let velocity of bullet be v.

Initial momentum = 0.0096 v

Final mass = 0.0096 + 4.95 = 4.9596 kg

Final velocity = 0.591 m/s

Final momentum = 4.9596 x 0.591 = 2.93 kgm/s

Equating both momentum

           0.0096 v = 2.93

                      v = 305.21 m/s

Original speed of the bullet = 305.21 m/s

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