6. (10 points) On a horizontal surface, there are two identical 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. 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_7a4bc87492c06fc34cd6g-11.jpg?height=97&width=694&top_left_y=748&top_left_x=767)
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 equal mass, there is an "exchange of velocities".

( $\mathbf{\text { point) }}$

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

$\frac{M u^{2}}{2}=A_{\text {fr }}=2 \mu M g S$.

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