A lead bullet at 27 ^oC just melts when stoop | vedclass

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Question

If mass of the bullet is $\mathrm{m}\, \mathrm{gm}$.

then total heat required for bullet to just melt down

$\mathrm{Q}_{1}=\mathrm{m} \mathrm{c}

Vedclass · Accepted Answer

$410$

Vedclass · Answer

If mass of the bullet is $\mathrm{m}\, \mathrm{gm}$.

then total heat required for bullet to just melt down

$\mathrm{Q}_{1}=\mathrm{m} \mathrm{c} \Delta \mathrm{q}+\mathrm{mL}=\mathrm{m} 	imes 0.03(327-27)+\mathrm{m} 	imes 6$

$=15 \mathrm{m} \mathrm{cal}=(15 \mathrm{m} 	imes 4.2) \mathrm{J}$

Now when bullet is stopped by the obstacle, the

loss in its mechanical energy $=\frac{1}{2}\left(\mathrm{m} 	imes 10^{-3}ight) \mathrm{v}^{2} \mathrm{J}$

$\left(\mathrm{As}\, \mathrm{mg}=\mathrm{m} 	imes 10^{-3} \mathrm{kg}ight)$

As $25 \%$ of this energy is absorbed by the obstacle, The energy absorbed by the bullet

$\mathrm{Q}_{2}=\frac{75}{100} 	imes \frac{1}{2} \mathrm{mv}^{2} 	imes 10^{-3}=\frac{3}{8} \mathrm{mv}^{2} 	imes 10^{-3} \mathrm{J}$

Now the bullet will melt if $\mathrm{Q}_{2} \geq \mathrm{Q}_{1}$

i.e., $\frac{3}{8} \mathrm{mv}^{2} 	imes 10^{-3} \geq 15 \mathrm{m} 	imes 4.2 \Rightarrow \mathrm{v}_{\min }=410 \mathrm{m} / \mathrm{s}$

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