How Fast Was the Bullet After Passing Through the Ballistic Pendulum?

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SUMMARY

The discussion focuses on calculating the velocity of a bullet after it passes through a ballistic pendulum. A 9.05 g bullet from a 9-mm pistol, traveling at 331.0 m/s, strikes a 0.705 kg block, which rises 19.47 cm after the impact. The solution involves applying the principles of conservation of momentum and energy. The initial kinetic energy of the bullet and the potential energy of the block are equated to find the bullet's final velocity after emerging from the block.

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1. A 9.05- g bullet from a 9-mm pistol has a velocity of 331.0 m/s. It strikes the 0.705- kg block of a ballistic pendulum and passes completely through the block. If the block rises through a distance h = 19.47 cm, what was the velocity of the bullet as it emerged from the block? i know that this is supposed to be done using conservation of energy and momentum. but i can't get it
 
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It's still a conservation of energy.

Initial kinetic energy of bullet + initial potential energy of block (which is zero) -> final ke of bullet + final potential energy of block.
 

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