A lead bullet at 27,^{circ}C just melts when | vedclass

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(A) The kinetic energy of the bullet is $K.E. = \frac{1}{2}mv^2$.Since $25\%$ of the heat is absorbed by the obstacle,$75\%$ of the kinetic energy is

Vedclass · Accepted Answer

$410$

Vedclass · Answer

(A) The kinetic energy of the bullet is $K.E. = \frac{1}{2}mv^2$.Since $25\%$ of the heat is absorbed by the obstacle,$75\%$ of the kinetic energy is used to heat and melt the bullet.Heat required to raise the temperature of the bullet to its melting point: $Q_1 = mc\Delta T = m 	imes 0.03 	imes (327 - 27) = m 	imes 0.03 	imes 300 = 9m\,cal$.Heat required to melt the bullet: $Q_2 = mL = m 	imes 6 = 6m\,cal$.Total heat required: $Q = Q_1 + Q_2 = 15m\,cal$.Converting to Joules: $Q = 15m 	imes 4.2 = 63m\,J$.Equating $75\%$ of $K.E.$ to $Q$: $0.75 	imes \frac{1}{2}mv^2 = 63m$.$0.375v^2 = 63$.$v^2 = \frac{63}{0.375} = 168$.Wait,recalculating with units: $Q = 15m\,cal = 15m 	imes 4.2 = 63m\,J$. $0.75 	imes 0.5 	imes v^2 = 63 \implies 0.375v^2 = 63 \implies v^2 = 168$. This suggests a unit mismatch in the provided solution. Re-evaluating: $15m\,cal = 15 	imes 4.2 = 63\,J$. If $m$ is in grams,$K.E. = \frac{1}{2} 	imes (m 	imes 10^{-3}) 	imes v^2$. Thus $0.75 	imes 0.5 	imes m 	imes 10^{-3} 	imes v^2 = 15m 	imes 4.2$.$0.375 	imes 10^{-3} 	imes v^2 = 63$.$v^2 = \frac{63}{0.375} 	imes 1000 = 168000$.$v = \sqrt{168000} \approx 409.87 \approx 410\,m/s$.

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