Experimental determinations have shown that the projectile undergoes almost uniform acceleration in the gun barrel. For the new French warship's guns, the barrel length from the projectile's center of gravity to the end is $12.2 \mathrm{~m}$, the propellant charge is $90 \mathrm{~kg}$, the projectile's mass is $292 \mathrm{~kg}$, and the projectile leaves the barrel at a speed of $800 \frac{\mathrm{m}}{\mathrm{sec}}$. What is the projectile's acceleration in the barrel? How long does it take to travel through the barrel? What is the projectile's energy? What is the energy yield per kg of propellant? If this projectile strikes an armor plate at this given speed, most of its energy is converted into heat energy. Assuming that $70 \%$ of this energy remains in the projectile itself, to what temperature in Celsius will it heat up, given that the specific heat of its material is $0.12 ?$
According to the theorems of uniformly accelerated motion, the acceleration of the projectile in the barrel is:

$$
a=\frac{800^{2}}{2 \times 12.2}=26229.5 \mathrm{~m} \mathrm{sec}^{-2}
$$

With this acceleration, the projectile travels the length of the barrel in $t=\frac{800}{26229.5} \mathrm{sec}=0.0305 \mathrm{sec}$.

The kinetic energy of the projectile is: $E=\frac{1}{2} \cdot 292 \cdot 800^{2}=93.44 \cdot 10^{6} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{sec}^{-2}=93.44 \cdot 10^{6}$ joules $=934.4 \cdot 10^{12} \mathrm{erg}$.

Since 1 joule $=\frac{1}{9.81} \mathrm{~m} \mathrm{~kg}=0.102 \mathrm{~m} \mathrm{~kg}$, this energy is also equivalent to $9.5309 \times 10^{6} \mathrm{~m} \mathrm{~kg}$ of work. And considering that $428 \mathrm{~m} \mathrm{~kg}$ is equivalent to $1 \mathrm{~kg}$ calorie of heat, the above energy is also $=22.268$ calories. 70% of this is 15.5876 calories, and this is the amount of energy that remains in the projectile after the collision, so each $\mathrm{kg}$ of the projectile receives 53.4 calories. Since it takes only 0.12 calories to raise the temperature of 1 $\mathrm{kg}$ of iron by $1^{\circ}$ Celsius, the projectile will heat up by $\frac{53.4}{0.12}=445^{\circ}$ Celsius.

Power is the work done per unit of time (1 second). If we neglect the friction between the barrel wall and the projectile, the only result of the gunpowder's work is the kinetic energy that the projectile acquires. Thus, the power of the gunpowder is obtained by calculating the kinetic energy that it could have created in 1 second, which is

$$
\frac{93.44 \times 10^{6}}{0.0305} \cdot \frac{\text { joule }}{\text { sec }}=3.064 \cdot 10^{6} \text { watts }
$$

per $\mathrm{kg}$ of gunpowder, it is $34 \cdot 10^{6}$ watts.

(Póka Gyula, Losoncz.)

The problem was also solved by: Aczél F., Bartók I., Bayer B., Hirschfeld Gy., Kertész F., Jánossy J., Lázár L., Pölczer P., Sasvári J., Szmodics H., Tóbiás J. L.