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

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(A) Let the mass of the bullet be $m\ g$.The total heat required for the bullet to just melt is given by $Q_1 = mc\Delta T + mL$.Substituting the valu

Vedclass · Accepted Answer

$410$

Vedclass · Answer

(A) Let the mass of the bullet be $m\ g$.The total heat required for the bullet to just melt is given by $Q_1 = mc\Delta T + mL$.Substituting the values: $Q_1 = m 	imes 0.03 	imes (327 - 27) + m 	imes 6 = m 	imes 0.03 	imes 300 + 6m = 9m + 6m = 15m\ cal$.Converting this to Joules: $Q_1 = 15m 	imes 4.2 = 63m\ J$.When the bullet is stopped,the loss in its kinetic energy is $\Delta K = \frac{1}{2} m_{kg} v^2 = \frac{1}{2} (m 	imes 10^{-3}) v^2\ J$.Since $25\%$ of the heat is absorbed by the obstacle,$75\%$ of the kinetic energy is absorbed by the bullet.Energy absorbed by the bullet $Q_2 = 0.75 	imes \frac{1}{2} 	imes m 	imes 10^{-3} 	imes v^2 = \frac{3}{8} 	imes 10^{-3} 	imes m v^2\ J$.For the bullet to just melt,$Q_2 = Q_1$.$\frac{3}{8} 	imes 10^{-3} 	imes m v^2 = 63m$.$v^2 = \frac{63 	imes 8 	imes 1000}{3} = 21 	imes 8000 = 168000$.$v = \sqrt{168000} \approx 409.87\ m/s \approx 410\ m/s$.

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