A lead bullet at 27oC 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 = 327oC, specific heat of lead :0.03 cal/gmoC, latent heat of fusion of lead - 6 cal/gm and J = 4.2J/cal)

If mass of the bullet is m gm, then total heat required for bullet to just melt downQ1=mcΔθ+mL=m×0.03(327−27)+m×6     =15mcal=(15m×4.2)JNow when bullet is stopped by the obstacle, the loss in its mechanical energy =12m×10−3v2J(As m gm=m×10−3kg)As 25% of this energy is absorbed by the obstacle,The energy absorbed by the bulletQ2=75100×12mv2×10−3=38mv2×10−3JNow the bullet will melt if Q2≥Q1 i.e.  38mv2×10−3≥15m×4.2⇒vmin=410m/s