Question

A bullet of mass $5\text{g}$, travelling with a speed of $210\text{m/s}$, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet, if the specific heat of its material is $0.030\text{cal/(g°C)}$ $(1\text{cal}=4.2\times {10}^7\text{ergs}$) close to

• A. $87.5°\text{C}$
• B. $83.3°\text{C}$
• C. $119.2°\text{C}$
• D. $38.4°\text{C}$

Answer 1

The correct option is A $87.5°\text{C}$

According to question, one half of its kinetic energy is converted into heat in the wood.
$\frac{1}{2}m{v}^2\times \frac{1}{2}=ms\Delta T$
$\Rightarrow \Delta T=\frac{v^2}{4\times s}=\frac{210\times 210}{4\times 4.2\times 0.03\times 1000}=87.5°\text{C}$