A bullet of mass m moving with a uniform velo | vedclass

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(B) The kinetic energy of the bullet is converted into heat energy upon impact.$1$. The heat required to raise the temperature of the bullet from $25^

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$mS(475 - 25) + mL = \frac{mv^2}{2J}$

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(B) The kinetic energy of the bullet is converted into heat energy upon impact.$1$. The heat required to raise the temperature of the bullet from $25^{\circ}C$ to its melting point $475^{\circ}C$ is $Q_1 = mS(475 - 25)$.$2$. The heat required to melt the bullet at its melting point is $Q_2 = mL$.$3$. The total heat energy required is $Q = Q_1 + Q_2 = mS(475 - 25) + mL$.$4$. According to the principle of conservation of energy,the kinetic energy of the bullet $K = \frac{1}{2}mv^2$ is converted into heat $Q$ such that $K = JQ$,where $J$ is the mechanical equivalent of heat.$5$. Therefore,$\frac{1}{2}mv^2 = J[mS(475 - 25) + mL]$.$6$. Rearranging the terms,we get $mS(475 - 25) + mL = \frac{mv^2}{2J}$.

$A$ bullet of mass $m$ moving with a uniform velocity $v$ stops suddenly after hitting a target. If the whole mass melts,given the specific heat $S$,initial temperature $25^{\circ}C$,melting point $475^{\circ}C$,and latent heat $L$,then $v$ is given by:

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