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

We start with the bullet mass $$m = 5\;\text{gram}$$ and its speed $$v = 210\;\text{m s}^{-1}$$. For uniformity we change the speed to centimetre-second (cgs) units because the energy conversion factor given at the end is in ergs:

$$210\;\text{m s}^{-1} = 210 \times 100\;\text{cm s}^{-1} = 2.10 \times 10^{4}\;\text{cm s}^{-1}.$$

The kinetic-energy formula is first stated:

$$\text{K.E.} = \dfrac{1}{2} m v^{2}.$$

Substituting the values (keeping the mass in grams and the speed in centimetres per second so that the answer comes out in ergs):

$$ \text{K.E.} \;=\; \dfrac{1}{2}\,(5\;\text{g})\,(2.10 \times 10^{4}\;\text{cm s}^{-1})^{2} \;=\; \dfrac{1}{2}\times 5 \times (2.10)^{2} \times 10^{8} \;=\; 2.5 \times 4.41 \times 10^{8} \;=\; 1.1025 \times 10^{9}\;\text{ergs}. $$

The question tells us that one half of this kinetic energy goes into heating the wooden target. Therefore the remaining half heats the bullet itself. So the heat energy actually absorbed by the bullet is

$$ Q = \dfrac{1}{2}\,\text{K.E.} = \dfrac{1}{2}\times 1.1025 \times 10^{9}\;\text{ergs} = 0.55125 \times 10^{9}\;\text{ergs}. $$

We convert this energy into calories because the specific heat is given in calories per gram per degree Celsius. The conversion factor stated in the question is $$1\;\text{calorie} = 4.2 \times 10^{7}\;\text{ergs}$$, therefore

$$ Q = \dfrac{0.55125 \times 10^{9}\;\text{ergs}}{4.2 \times 10^{7}\;\text{ergs calorie}^{-1}} = \dfrac{0.55125}{4.2}\times 10^{2}\;\text{calories} = 0.13125 \times 10^{2}\;\text{calories} = 13.125\;\text{calories}. $$

The specific heat of the bullet’s material is given as $$c = 0.030\;\text{cal}\,(\text{gram}\,^{\circ}\text{C})^{-1}$$. The temperature rise $$\Delta T$$ is obtained from the heat equation

$$ Q = m c \Delta T \quad\Longrightarrow\quad \Delta T = \dfrac{Q}{m c}. $$

Substituting the numerical values:

$$ \Delta T = \dfrac{13.125\;\text{cal}}{(5\;\text{g})(0.030\;\text{cal g}^{-1}{}^{\circ}\text{C}^{-1})} = \dfrac{13.125}{0.15} = 87.5^{\circ}\text{C}. $$

Hence, the correct answer is Option A.