Question

A lead bullet strikes a target with a velocity of $480m{s}^{-1}$. If the bullet falls dead, the rise in temperature of bullet (c = 0.03 cal/gram), assuming that heat produced is equally shared between the bullet and target is

Answer 1

The kinetic energy possessed by the bullet, K.E. = $\frac{1}{2}m{v}^2$ will be converted to heat, since the heat is being equally shared, heat developed in bullet will be $\frac{1}{2}\times \frac{1}{2}m{v}^2$.
Now, $mc\Delta \theta =\frac{1}{2}\times \frac{1}{2}m{v}^2\to \Delta \theta =\frac{\frac{1}{2}\times \frac{1}{2}m{v}^2}{mc}=\frac{v^2}{4c}$
$1calkg=4.2\times {10}^{3}kg=\frac{480\times 480}{4\times 0.03\times 1000\times 4.2}$ = 457.14