Law of conservation of energy problem

[leapoldstotch]

Homework Statement

A .5 kg block is dropped from 15m. A 5 gram bullet hits the block at 233 m/s. At what height must the block be hit to completely stop the blocks freefall momentarily.

Homework Equations

p=mv
1/2mv^2=mgh

The Attempt at a Solution

bullets ke
1/2(.005)(233)^2=.5(9.8)(h)
When the bullets kenetic energy is equal to the blocks potential it will momentarily pause in mid air. Atleast i thought.

bullet's ke = 135.72 joules= blocks pe = 27.698m
The block would have to be 27 meters above the ground to have as much potential energy as the bullet described.

Then i was thinking of solving it using momentum but I am not sure how that would work out.

 

[rl.bhat]

Equating momentums of the block and the bullet, you can find the velocity of the block.
Then by using kinematics you can find h.

 

[tiny-tim]

Hi leapoldstotch!

A .5 kg block is dropped from 15m. A 5 gram bullet hits the block at 233 m/s. At what height must the block be hit to completely stop the blocks freefall momentarily.
…
When the bullets kenetic energy is equal to the blocks potential it will momentarily pause in mid air. Atleast i thought.
…
Then i was thinking of solving it using momentum but I am not sure how that would work out.

Energy is not conserved in a collision unless the question tells you it is …

in this case, the bullet will embed in the block, and have the same speed as it, so this is a perfectly inelastic collision and energy is obviously not conserved.

You can only use conservation of momentum (for the collision itself).