Particles Moving on a Circle

This problem involves two particles moving on a circular path in opposite directions with given initial speeds. A key event occurs when they cross each other: their speeds are interchanged, but their directions of motion remain the same. We need to find the difference in the total time it takes for each particle to return to the starting point.

Circle Circumference

The particles move along a circle of radius \(R = 10 \text{ m}\). The total distance around the circle is the circumference, \(C\).

The circumference is given by the formula:

\(C = 2\pi R\)

Substituting the given radius:

\(C = 2\pi \times 10 \text{ m} = 20\pi \text{ m}\)

Meeting Point Analysis

The two particles start from the same point and move in opposite directions. Let their initial speeds be \(v_1 = 5 \text{ m/s}\) and \(v_2 = 8 \text{ m/s}\). Since they are moving towards each other along the circle, their relative speed is the sum of their individual speeds.

Relative speed = \(v_1 + v_2 = 5 \text{ m/s} + 8 \text{ m/s} = 13 \text{ m/s}\)

They will meet for the first time when the sum of the distances they have traveled equals the circumference of the circle. Let \(t_{\text{meet}}\) be the time when they first meet.

Distance covered by particle 1 = \(v_1 \times t_{\text{meet}} = 5 t_{\text{meet}}\)

Distance covered by particle 2 = \(v_2 \times t_{\text{meet}} = 8 t_{\text{meet}}\)

At the meeting point, the sum of these distances equals the circumference:

\(5 t_{\text{meet}} + 8 t_{\text{meet}} = 20\pi\)

\(13 t_{\text{meet}} = 20\pi\)

\(t_{\text{meet}} = \frac{20\pi}{13} \text{ s}\)

At the time of meeting, the distance covered by particle 1 from the starting point is:

\(d_1 = v_1 \times t_{\text{meet}} = 5 \times \frac{20\pi}{13} = \frac{100\pi}{13} \text{ m}\)

The distance covered by particle 2 from the starting point is:

\(d_2 = v_2 \times t_{\text{meet}} = 8 \times \frac{20\pi}{13} = \frac{160\pi}{13} \text{ m}\)

Note that \(d_1 + d_2 = \frac{100\pi}{13} + \frac{160\pi}{13} = \frac{260\pi}{13} = 20\pi \text{ m}\), which is the circumference.

Speed Interchange

At the instant they cross each other, their speeds are interchanged, but their directions remain the same. So, after meeting:

• Particle 1's new speed becomes \(v_1' = 8 \text{ m/s}\).
• Particle 2's new speed becomes \(v_2' = 5 \text{ m/s}\).

Their directions are still opposite to each other and the same as their initial directions.

Remaining Journey to Starting Point

Particle 1 traveled a distance of \(\frac{100\pi}{13}\) m from the starting point in its original direction before meeting. To return to the starting point by continuing in the same direction, it needs to cover the remaining part of the circumference from its current position. The remaining distance for particle 1 is:

\(d_1_{\text{rem}} = C - d_1 = 20\pi - \frac{100\pi}{13} = \frac{260\pi - 100\pi}{13} = \frac{160\pi}{13} \text{ m}\)

Particle 2 traveled a distance of \(\frac{160\pi}{13}\) m from the starting point in its original direction before meeting. The remaining distance for particle 2 to return to the starting point by continuing in the same direction is:

\(d_2_{\text{rem}} = C - d_2 = 20\pi - \frac{160\pi}{13} = \frac{260\pi - 160\pi}{13} = \frac{100\pi}{13} \text{ m}\)

Time for Remaining Journey

Now, we calculate the time taken by each particle to cover the remaining distance with their new speeds.

Time taken by particle 1 for the remaining distance (\(t_1_{\text{rem}}\)) with speed \(v_1' = 8 \text{ m/s}\):

\(t_1_{\text{rem}} = \frac{d_1_{\text{rem}}}{v_1'} = \frac{160\pi/13}{8} = \frac{160\pi}{13 \times 8} = \frac{20\pi}{13} \text{ s}\)

Time taken by particle 2 for the remaining distance (\(t_2_{\text{rem}}\)) with speed \(v_2' = 5 \text{ m/s}\):

\(t_2_{\text{rem}} = \frac{d_2_{\text{rem}}}{v_2'} = \frac{100\pi/13}{5} = \frac{100\pi}{13 \times 5} = \frac{20\pi}{13} \text{ s}\)

Total Arrival Times

The total time for each particle to arrive back at the starting point is the sum of the time taken to reach the meeting point and the time taken to travel the remaining distance from the meeting point back to the start.

Total time for particle 1 (\(T_1\)):

\(T_1 = t_{\text{meet}} + t_1_{\text{rem}} = \frac{20\pi}{13} + \frac{20\pi}{13} = \frac{40\pi}{13} \text{ s}\)

Total time for particle 2 (\(T_2\)):

\(T_2 = t_{\text{meet}} + t_2_{\text{rem}} = \frac{20\pi}{13} + \frac{20\pi}{13} = \frac{40\pi}{13} \text{ s}\)

Difference in Arrival Times

The difference between their arrival times at the starting point is:

Difference = \(|T_1 - T_2| = |\frac{40\pi}{13} - \frac{40\pi}{13}| = 0 \text{ s}\)

The difference in their arrival times at the starting point is 0 seconds.

Let's summarize the times:

Both particles take the same total time to return to the starting point. Therefore, the difference in their arrival times is 0 seconds.