A lead bullet strikes against a steel armor plate with the velocity of 300 meter

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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 =

0.03 c a l / g /^{\circ} C

Options:

A . 178.6∘C

B . 78.6∘C

C . 2780.6∘C

D . 100.6∘C

Answer: Option A
:
A
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 T c a l

Kinetic energy of the bullet

= \frac{1}{2} m v^{2} = \frac{1}{2} m (300 \times 10^{2})^{2} e r g s

Half of this energy goes into the bullet on impact, therefore

W = \frac{1}{2} \times [\frac{1}{2} m (300 \times 10^{2})^{2}] = \frac{1}{4} m \times 9 \times 10^{8} e r g s U s i n g t h e f o r m u l a W = J H, w e h a v e \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

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