A lead bullet at ${27}^0\mathrm{C}$ 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 (M.P. of lead = ${327}^0\mathrm{C}$, specific heat of lead = 0.03 cal/${\mathrm{gm}}^0\mathrm{C}$, 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

${\mathrm{Q}}_1 = \mathrm{m} \mathrm{c} ∆\mathrm{\theta } + \mathrm{mL} = \mathrm{m}\times 0.03 (327-27) + \mathrm{m} \times 6$

      = 15 m cal = (15 m $\times 4.2 ) \mathrm{J}$

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

At 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 ${\mathrm{Q}}_2 > {\mathrm{Q}}_1$

i.e., $\frac{3}{8}{\mathrm{mv}}^2 \times {10}^{-3} \ge 15 \mathrm{m} \times 4.22 \Rightarrow {\mathrm{v}}_{\min } = 410 \mathrm{m}/\mathrm{s}$