Question

A lead bullet strikes against a steel armor plate with the velocity of 300 meter/second. If the bullet, assuming that the impact is perfectly inelastic find the rise in temperature of the bullet. Assume that the heat produced is shared equally between the bullet and the target, specific heat of lead = $0.03cal/g{/}^{\circ }C$

• A. ${178.6}^{\circ }C$
• B. ${78.6}^{\circ }C$
• C. ${2780.6}^{\circ }C$
• D. ${100.6}^{\circ }C$

Answer 1

The correct option is A ${178.6}^{\circ }C$

Suppose m be the mass of the bullet and T be the rise in temperature.
Heat produced in the bullet
$H=m\times s\times T=m\times 0.03\times Tcal$
Kinetic energy of the bullet
$=\frac{1}{2}m{v}^2=\frac{1}{2}m(300\times {10}^2{)}^{2}ergs$
Half of this energy goes into the bullet on impact, therefore
$W=\frac{1}{2}\times \left[\frac{1}{2}m\right(300\times {10}^2{)}^2]=\frac{1}{4}m\times 9\times {10}^{8}ergsUsingtheformulaW=JH,wehave\frac{1}{4}m\times 9\times {10}^8=(4.2\times {10}^7)(m\times 0.03\times T)T=\frac{9\times {10}^8}{4\times (4.2\times {10}^7)\times 0.03}={178.6}^{\circ }C$