5. (15 points) The figure shows the dependence of gas pressure on its density in two processes. It is known that in the process $1-2$, work $A_{1-2}$ was done on the gas. Determine the work done on the gas in the process $3-4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ddf1918985530a395e14g-04.jpg?height=583&width=988&top_left_y=414&top_left_x=611)
Answer: $A_{3-4}=\frac{1}{2} A_{1-2}$

Solution. $1-2$ and $3-4$ are isothermal processes.

The work in an isothermal process can be found as the area under the graph plotted in $p-V$ coordinates.

![](https://cdn.mathpix.com/cropped/2024_05_06_ddf1918985530a395e14g-04.jpg?height=688&width=954&top_left_y=1478&top_left_x=628)

(2 points)

It is noteworthy that for any value of pressure, the volume in process $1-2$ is exactly twice the volume in process $3-4$.

That is, if we consider a small change in volume in process $1-2$, it turns out to be exactly twice the small change in volume in process $3-4$, taken at the same pressure: $\Delta V_{3-4}=\frac{1}{2} \Delta V_{1-2}$.

We obtain that the area under the graph of $3-4$ is exactly half the area under the graph of $1-2$. That is, $A_{3-4}=\frac{1}{2} A_{1-2}$.

Note. A solution to the problem where the student gets the correct answer by writing the equation $A=p \Delta V$ without analyzing the areas under the graphs in $p-V$ coordinates is considered incorrect!

6. (10 points) On a horizontal surface, there are two identical small stationary blocks, each with mass $M$. The distance between them is $S$. A bullet of mass $m$ hits and gets stuck in the left block, moving horizontally. What should be the speed of the bullet so that the final distance between the blocks is also equal to $S$? The collision between the blocks is perfectly elastic. The mass of the bullet is much smaller than the mass of the block $m<<M$. The coefficient of friction between the blocks and the horizontal surface is $\mu$, and the acceleration due to gravity is $g$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ddf1918985530a395e14g-05.jpg?height=166&width=1176&top_left_y=1162&top_left_x=517)

Answer: $v=\frac{2 M}{m} \sqrt{\mu g S}$

Solution. The law of conservation of momentum for the collision of the bullet with the left block: $m v=(M+m) u=M u$.

In the case of a perfectly elastic collision between bodies of the same mass, there is an "exchange of velocities".

The law of conservation of energy for the subsequent motion of the blocks:

$\frac{M u^{2}}{2}=A_{m p}=2 \mu M g S$

As a result, the speed of the bullet: $v=\frac{2 M}{m} \sqrt{\mu g S}$.