# Problem № 6 (10 points)

Five identical balls are rolling towards each other on a smooth horizontal surface. The speeds of the first and second are \( v_{1} = v_{2} = 0.5 \) m/s, while the others are \( v_{3} = v_{4} = v_{5} = 0.1 \) m/s. The initial distances between the balls are the same, \( l = 2 \) m. All collisions are perfectly elastic. How much time will pass between the first and last collisions in this system?

![](https://cdn.mathpix.com/cropped/2024_05_06_3676ffec89688f4f2c70g-6.jpg?height=183&width=277&top_left_y=702&top_left_x=221)

1

![](https://cdn.mathpix.com/cropped/2024_05_06_3676ffec89688f4f2c70g-6.jpg?height=162&width=272&top_left_y=707&top_left_x=595)

2

![](https://cdn.mathpix.com/cropped/2024_05_06_3676ffec89688f4f2c70g-6.jpg?height=166&width=254&top_left_y=705&top_left_x=935)

3

![](https://cdn.mathpix.com/cropped/2024_05_06_3676ffec89688f4f2c70g-6.jpg?height=171&width=297&top_left_y=700&top_left_x=1279)

4

![](https://cdn.mathpix.com/cropped/2024_05_06_3676ffec89688f4f2c70g-6.jpg?height=186&width=365&top_left_y=701&top_left_x=1665)

5

## Time: 10 min

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# Solution and Evaluation Criteria:

In the case of a perfectly elastic collision, identical balls "exchange" velocities.

Therefore, the situation can be considered as if the balls pass through each other with unchanged speeds. The first collision occurs between the second and third balls. The last collision will occur at the moment when the first ball "passes" by the fifth.

(4 points)

The corresponding time: \( t = \frac{3 l}{v_{1} + v_{5}} = \frac{3 \cdot 2}{0.6} = 10 \text{s} \)

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