Getting different velocity from momentum conservation and work energy theorem

Question

Consider a system of bullet (of mass 5 kg) and block (of mass 20 kg). Initially the block is at rest and the bullet moves towards it at 30 m/s. After penetrating the block, the bullet comes out with speed 10 m/s. What is the velocity of block?

My attempt
As there is no external force in the system, momentum of the system of particles remain conserved.
Initial momentum= Final momentum
$5\times 30 = 20\times v+ 5\times 10$
$\implies v = 5 m/s$

Another approach
When block penetrates the block it experiences a retarding force $f$ from the block and the block experiences the same force which accelerates it.
For bullet, $10=30-\frac{F}{5}t$.
$\implies Ft=100$

For block, $v=\frac{F}{20}t$
$\implies v=5 m/s$

Using work energy theorem
Work done against on the bullet by the force f = Initial KE of bullet - Final KE of bullet
= $\frac{5}{2}(30^2-10^2)=2000J$
This energy basically goes as KE of the block.
2000=Final KE of block - Initial KE of block
$\implies \frac{20}{2}v^2=2000$.
$\implies v^2=200$
$\implies v=10\sqrt 2$

The correct answer is 5 m/s.
Why am I getting incorrect answer with the Work energy theorem?

Answer 1

Why am I getting incorrect answer with the Work energy theorem?

The collision is not elastic, therefore KE is not conserved, it is lower after the collision. The missing KE is converted to heat and sound and deformation of the block. It is precisely this missing KE that makes bullets deadly.

Answer 2

Kinetic energy of the bullet-block system isn't a conserved quantity. i.e. not all of the kinetic energy lost by the bullet is transferred into kinetic energy of the block.