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)

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Infinity Learn · Accepted Answer

If mass of the bullet is m gm, then total heat required for bullet to just melt $$downQ1=mc$$Δθ$$+mL=m$$×$$0.03(327$$−$$27)+m$$×$$6$$     $$=15mcal=(15m$$×$$4.2)JNow$$ 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.  $$38mv2$$×$$10$$−$$3$$≥$$15m$$×$$4.2$$⇒$$vmin=410m/s$$

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