The figure from below shows a projectile whose mass is $m$ goes goes through a block of mass $M=5m$ initially at rest as indicated in the figure from below. The bullet goes out from the block with a speed which is half of the speed when it entered the block. What will be the energy lost expressed in a percentage?

The alternatives given are as follows:

$\begin{array}{ll}

1.&20\,\%\\

2.&30\,\%\\

3.&50\,\%\\

4.&70\,\%\\

5.&90\,\%\\

\end{array}$

I'm confused exactly how should I understand this kind of collision when a bullet goes through a block. Is it an inelastic or an elastic collision?. I'm assuming that it is an inelastic collision as the two objects are joined at least for a brief period.

Then using this logic:

$p_1=p_2$

$m_1v_1=\left(m_1+m_2\right)v_2$

$mv_1= \left(m+5m\right)v_2$

According to this the speed of the block would be

$v_2=\frac{1}{6}v_1$

But I am given the exit speed of the bullet, then:

Initially the energy of the bullet:

$E_k=\frac{1}{2}mv^2$

The energy of the bullet after the collision:

$E_k=\frac{1}{2}m\left(\frac{1}{2}v\right)^2$

Then the subtraction would give the energy lost:

$\Delta E_k = \frac{1}{2}mv^2 - \frac{1}{2}m\left(\frac{1}{2}v\right)^2$

$\Delta E_k = \frac{1}{2}mv^2 - \frac{1}{8}mv^2$

$\Delta E_k = \frac{3}{8}mv^2$

Then the percentage lost would be:

$\frac{\Delta E_k}{E_k} = \frac{\frac{3}{8}mv^2}{\frac{1}{2}mv^2}$

$\frac{\Delta E_k}{E_k} = \frac{3}{4}=0.75$

But this results in $75\,\%$ but this is not among the alternatives. What part I missed the right path?.