Why is so much kinetic energy lost when a bullet hits a block?

In a perfectly inelastic collision the fraction of kinetic energy retained is m divided by m plus M where m is the bullet mass and M is the block mass. For a 10g bullet and 2kg block only 0.5 percent of the energy survives. The rest becomes heat sound and deformation. Momentum is always conserved but kinetic energy is not in inelastic collisions.

How do you find the maximum spring compression after a bullet embeds in a block?

Use momentum conservation to find the post-collision velocity v1 equals m times v0 divided by m plus M. Then use energy conservation for the spring compression: one half times m plus M times v1 squared equals one half times k times x max squared plus mu times m plus M times g times x max. Without friction x max equals v1 times the square root of m plus M divided by k.

Bullet–Block–Spring — Where Does 99% of the Energy Go?

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Bullet–Block–Spring