A gas explosion occurred in a house on the first floor. The gathered crowd was mostly amazed that a door had flown a considerable distance. To be calculated: with what force in kg did the exploding gas mixture press the door in the final phase of the explosion? Furthermore, how long did the explosion last, assuming that the pressure acting on the door increased uniformly from 0 to its final high value? The dimensions of the door are: $a=2.20 \mathrm{~m}, b=1.15 \mathrm{~m}$, its mass is $m=30 \mathrm{~kg}$. The door flew $s=80 \mathrm{~m}$ horizontally from its original position, the height of the floor was $h=6 \mathrm{~m}$. The temperature of the exploding gas mixture can be assumed to be $t=1500^{\circ}$, and the expansion coefficient of the gas mixture is $\alpha=\frac{1}{273}$.
For approximate calculations, we can assume that before the explosion, the room was at normal atmospheric pressure (760 mm) and a temperature of $0^{\circ} \mathrm{C}$. The air in the room heated up to $t=1500^{\circ} \mathrm{C}$ as a result of the explosion, while the volume remained constant, so the final internal pressure was

$$
p^{\prime}=760\left(1+\frac{1500}{273}\right)
$$

From this, we need to subtract the pressure acting on the outer surface of the door:

$$
p=760\left(1+\frac{1500}{273}\right)-760=760 \frac{1500}{273} \mathrm{~mm} \text { (mercury). }
$$

Thus, the final value of the force acting on the door is

$$
P=\frac{76 \cdot 1500 \cdot 13.56 \cdot 220 \cdot 115}{273 \cdot 10^{3}} \mathrm{kgs}=143200 \mathrm{kgs} .
$$

To answer the second question, we first need to calculate the initial speed of the door. The door fell for $t$ (sec), so $\frac{g}{2} t^{2}=6$, from which

$$
t=1.1(\mathrm{sec})
$$

Thus, the initial horizontal speed was

$$
v=\frac{80}{1.1}=72.7 \frac{\mathrm{m}}{\mathrm{sec}} \text { . }
$$

According to Newton's third law, the change in momentum is equal to the impulse of the force. Here, the force continuously increased to reach its final large value. For this approximate calculation, we can assume that the increase was uniform, so in the impulse calculation, we can take the average value of the force: $\frac{0+P}{2}$ as a constant acting force, thus

$$
m v=\frac{1}{2} P \cdot t^{\prime}
$$

where $t^{\prime}$ represents the duration of the explosion. Since the force is expressed in practical units, the mass must also be expressed in practical units, so

$$
\begin{aligned}
& m=\frac{30}{g}\left(\frac{\frac{\mathrm{kgs}}{m}}{\sec ^{2}}\right)
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_34222359956db5a92f37g-1.jpg?height=149&width=799&top_left_y=1563&top_left_x=652)

(Oszkár Szilas, Budapest.)