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 (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 downQ1 = m c + m L = m x 0.03 (327 – 27) + m x 6    = 15 m cal     Now when bullet is stopped by the obstacle, the loss in its mechanical energy  (As  )As 25% of this energy is absorbed by the obstacle,The energy absorbed by the bullet Now the bullet will melt if  i.e.