Finding Change in Vertical Height of a Suspended Block After Bullet Impact

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A 0.05 kg bullet traveling at 200 m/s embeds into a 1.0 kg stationary wood block, prompting a discussion on calculating the change in vertical height of the block after impact. The initial approach involved using work-energy principles, but confusion arose regarding the work of non-conservative forces and the initial and final velocities. The correct method involves calculating the bullet's kinetic energy, applying conservation of momentum to determine the block's speed post-collision, and then using conservation of mechanical energy to find the height. Ultimately, the solution reveals that the change in vertical height of the block is 4.63 meters. This demonstrates the importance of understanding energy conservation in inelastic collisions.
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Homework Statement


A 0.05 kg bullet traveling at 200 m/s embeds into a 1.0 kg stationary wood block suspended by two light strings. Find the change in vertical height of the block.

Homework Equations


Wnc = 0 ?
mgho + .5mvo^2 = mghf + .5mfv^2

Wnc != 0 ?
.5mvf^2 + mghf - .5mvo^2 - mgho

The Attempt at a Solution


I really have no idea where to start. I'm not entirely sure if the work of the non conservative forces is equal to zero or not. I'm not even sure what I would put as the initial and final velocities ( 200 and 0 respectively??).

If it's Wnc = 0, then I did this:

.5 ( 1.05 ) ( 200^2) = (1.05) (9.8) hf
2100 = 10.29 hf
2100/10.29 = 204.1m = hf

But that seems like a rather high number
 
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This problem can be solved in 3 steps...

1. Find the kinetic energy of the bullet before it hits the block
2. Use conservation of momentum to find the speed of the block after the collision.(Remember that mechanical energy is not conserved in an inelastic collision.)
3. Use conservation of mechanical energy and solve for the height of the block.
 
Thank you very much! I finally got the answer, 4.63m!
 
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