A model train set with a total mass of 390 kg is heading east on the same tracks that a second train set is using. The second train, with a total mass of 350 kg, is heading west at a rate of 8.3 m/s. Before the collision, the first train has a rate of 7.86 m/s. What is the final velocity (F) of the second train after the collision if the final velocity of the first train was determined to be was 1.82 m/s?

To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of an isolated system remains constant if no external forces are acting upon it.

Mathematically, it's expressed as:

[tex] \Large\boxed{\boxed{\sf m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' }}[/tex]

Where:

[tex]\sf m_1 [/tex] and [tex]\sf m_2 [/tex] are the masses of the first and second train sets, respectively.
[tex]\sf v_1 [/tex] and [tex]\sf v_2 [/tex] are the initial velocities of the first and second train sets, respectively.
[tex]\sf v_1' [/tex] and [tex]\sf v_2' [/tex] are the final velocities of the first and second train sets, respectively.

Given:

[tex]\sf m_1 = 390 [/tex] kg
[tex]\sf m_2 = 350 [/tex] kg
[tex]\sf v_1 = 7.86 [/tex] m/s (initial velocity of the first train)
[tex]\sf v_1' = 1.82 [/tex] m/s (final velocity of the first train)
[tex]\sf v_2 = -8.3 [/tex] m/s (since it's heading west, velocity is negative)

We need to find [tex]\sf v_2' [/tex], the final velocity of the second train.

First, we need to correct the signs. The second train is heading west, so its velocity should be negative. Thus, [tex]\sf v_2 = -8.3 [/tex] m/s.

Now, we plug in the values into the conservation of the momentum equation:

[tex]\sf (390 \times 7.86) + (350 \times -8.3) = (390 \times 1.82) + (350 \times v_2') [/tex]

[tex]\sf 3065.4 - 2905 = 709.8 + 350v_2' [/tex]

[tex]\sf 160.4 = 709.8 + 350v_2' [/tex]

[tex]\sf 350v_2' = -709.8 - 160.4 [/tex]

[tex]\sf 350v_2' = -870.2 [/tex]

[tex]\sf v_2' = \dfrac{-870.2}{350} [/tex]

[tex]\sf v_2' = −2.486285714285 \text{ m/s} [/tex]

[tex]\sf v_2' = −2.49 \textsf{ (in 2 d.p.)} \text{ m/s} [/tex]

The negative sign in [tex]\sf v_2' [/tex] indicates that the second train continues to move westward after the collision.

So, the final velocity of the second train after the collision is [tex]\sf 2.49 [/tex] m/s.