A lead bullet at 27°C just melts when stopped by an obstacle. Assuming that 25% of heat is absorbed by the obstacle, then the velocity of the bullet at the time of striking is (M.P. of lead = 327°C, specific heat of lead = 0.03 cal/gm°C, latent heat of fusion of lead = 6 cal/gm and J = 4.2 joule/cal)

If mass of the bullet is m gm, then total heat required for bullet to just melt down,

$\begin{array}{l}{\mathrm{Q}}_1=\mathrm{mc\Delta \theta }+\mathrm{mL}=\mathrm{m}\times 0.03(327-27)+\mathrm{m}\times 6\\ =15\mathrm{m}\text{ cal }=(15\mathrm{m}\times 4.2)\mathrm{J}\end{array}$

Now when bullet is stopped by the obstacle, the loss in its mechanical energy $=\frac{1}{2}\left(\mathrm{m}\times {10}^{-3}\right){\mathrm{v}}^2\mathrm{J}$

                                                                                                              $\left(\text{ As }\mathrm{m} \mathrm{is} \mathrm{in} \mathrm{gm}=\mathrm{m}\times {10}^{-3}\mathrm{kg}\right)$

As 25% of this energy is absorbed by the obstacle, the energy absorbed by the bullet,

${\mathrm{Q}}_2=\frac{75}{100}\times \frac{1}{2}{\mathrm{mv}}^2\times {10}^{-3}=\frac{3}{8}{\mathrm{mv}}^2\times {10}^{-3}\mathrm{J}$

Now the bullet will melt if ${\mathrm{Q}}_2\ge {\mathrm{Q}}_1$

i.e., $\frac{3}{8}{\mathrm{mv}}^2\times {10}^{-3}\ge 15\mathrm{m}\times 4.2\Rightarrow {\mathrm{v}}_{min}=410\mathrm{m}/\mathrm{s}$