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# Conservation of Energy and of Momentum

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A block of mass M= 2.5 kg is suspended in equilibrium by a long cord. A bullet of mass m= 0.055 kg, moving vertically upward with an unknown velocity v, hits the block from below. The block with embedded bullet then rises a total distance H= 1.2 meters above its initial position. Find v, the velocity of the bullet before impact.

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#### Solution Preview

During the collision, momentum must be conserved, so: 'momentum of bullet before collision equals the momentum of block and bullet after collision'.
With parameters:
(1) mv = (m + M) V
in which V is the velocity of the block ...

#### Solution Summary

The conservation of energy and momentum is examined. The initial position and total distance is examined.

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