Mass of the bullet, m = 0.012 kg

The initial speed of the bullet,

$u_b$ = 70 m/s

Mass of the wooden block, M = 0.4 kg

Initial speed of the wooden block, $u_B$ = 0

Final speed of the system of the bullet and the block = ν

Applying the law of conservation of momentum:

${\mathrm{mv}}_{\mathrm{b}} + {\mathrm{Mu}}_{\mathrm{B}} = \left(\mathrm{m} + \mathrm{M}\right)\mathrm{v}$
$0.012 \times 70 + 0.4 \times 0 = \left(0.012 + 0.4\right)\mathrm{v}$
$\Rightarrow \mathrm{v} = \frac{0.84}{0.412} = 2.04 \mathrm{m}/\mathrm{s}$

For the system of the bullet and the wooden block:

Mass of the system, m' = 0.412 kg

Velocity of the system = 2.04 m/s

Height up to which the system rises = h

Applying the law of conservation of energy to this system:

The potential energy at the highest point = Kinetic energy at the lowest point

$\mathrm{m}\text{'}\mathrm{gh} = \frac{1}{2}\mathrm{m}\text{'}{\mathrm{v}}^2$
$\Rightarrow \mathrm{h} = \frac{1}{2}\left(\frac{{\mathrm{v}}^2}{\mathrm{g}}\right)$
$= \frac{1}{2} \times \frac{{\left(2.04\right)}^2}{9.8}$
$= 0.2123 \mathrm{m}$

The wooden block will rise to a height of 0.2123 m.

The heat produced = Kinetic energy of the bullet – Kinetic energy of the system

$= \frac{1}{2}{\mathrm{mu}}^2 - \frac{1}{2}\mathrm{m}\text{'}{\mathrm{v}}^2$
$= \frac{1}{2} \times 0.012 \times {\left(70\right)}^2 - \frac{1}{2} \times 0.412 \times {\left(2.04\right)}^2$
$= 29.4 – 0.857 = 28.54 \mathrm{J}$