$A$ bullet of mass $0.012\;kg$ and horizontal speed $70\;m\;s^{-1}$ strikes a block of wood of mass $0.4\;kg$ and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also,estimate the amount of heat produced in the block.

(N/A) Mass of the bullet,$m = 0.012\;kg$.
Initial speed of the bullet,$u_b = 70\;m\;s^{-1}$.
Mass of the wooden block,$M = 0.4\;kg$.
Initial speed of the wooden block,$u_B = 0\;m\;s^{-1}$.
Let the final speed of the system (bullet + block) be $v$.
Applying the law of conservation of momentum: $m u_b + M u_B = (m + M) v$.
$0.012 \times 70 + 0.4 \times 0 = (0.012 + 0.4) v$.
$0.84 = 0.412 v \implies v = \frac{0.84}{0.412} \approx 2.039\;m\;s^{-1}$.
For the system of the bullet and the wooden block,applying the law of conservation of energy: $m' g h = \frac{1}{2} m' v^2$,where $m' = m + M = 0.412\;kg$.
$h = \frac{v^2}{2g} = \frac{(2.039)^2}{2 \times 9.8} \approx 0.212\;m$.
The wooden block will rise to a height of $0.212\;m$.
Heat produced = Initial kinetic energy of the bullet - Final kinetic energy of the system.
Heat = $\frac{1}{2} m u_b^2 - \frac{1}{2} (m + M) v^2$.
Heat = $\frac{1}{2} \times 0.012 \times (70)^2 - \frac{1}{2} \times 0.412 \times (2.039)^2$.
Heat = $29.4 - 0.857 = 28.543\;J$.