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,Q1=mcΔθ+mL=m×0.03(327−27)+m×6     =15m cal =(15m×4.2)JNow when bullet is stopped by the obstacle, the loss in its mechanical energy =12m×10−3v2J                                                                                                               As m is in gm=m×10−3kgAs 25% of this energy is absorbed by the obstacle, the energy absorbed by the bullet,Q2=75100×12mv2×10−3=38mv2×10−3JNow the bullet will melt if Q2≥Q1i.e.,  38mv2×10−3≥15m×4.2⇒vmin=410m/s