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When dealing with momentum, it's important to realize that momentum is not simply a matter of speed. It involves direction. That’s why it's a vector quantity. The given problem is concerned with two cars moving in different directions. Therefore, the total momentum is divided into components. In this case, we identify two components:

• South Component (y-axis): This refers to the momentum directed along the southward direction.
• West Component (x-axis): This is the momentum pointed towards the west.

The total momentum we are given is 7200 kg⋅m/s at an angle of 60° west of south. To handle situations like this, we use trigonometry to split the momentum vector into its respective south and west quantities, which makes understanding the contributions from each car easier.

Momentum is a classic example of a vector quantity, as it involves both magnitude and direction. Vector quantities are represented with arrows where the direction represents the force's direction, while the length represents its magnitude. In our example:

• The convertible’s southward momentum and the SUV's westward movement contribute to the vector sum.
• Both movements are integrated into a single momentum vector that is simplified using trigonometric methods to solve calculations more efficiently.

Because these vectors aren’t aligned along the same axis, resolving these into perpendicular components—such as x and y axes—is necessary for straightforward problem-solving.

Trigonometry is the key tool used to break down vector quantities into components in physics. Here, sine and cosine functions are incredibly useful:

• South Component: The sine of the angle (60°) gives the proportion of the momentum in the south direction. Calculated as 7200 sin(60°), resulting in approximately 6235.38 kgâ‹…m/s.
• West Component: The cosine of the same angle provides the proportion in the west direction. This comes out to be 7200 cos(60°), equaling 3600 kgâ‹…m/s.

Using trigonometry allows us to handle the scenario efficiently by separating a single momentum vector into usable parts aligned with the vehicles’ directions.

Having broken down the momentum vector, we now apply basic momentum equations to find velocities. Momentum (\(p\)) is the product of mass (\(m\)) and velocity (\(v\)): \[ p = mv \]-

For the blue convertible traveling south:

We have the equation \(1500V_b = 6235.38\), so solving this gives the velocity \(V_b = \frac{6235.38}{1500} \approx 4.16\, \text{m/s}\).-

For the red SUV traveling west:

The west direction equation is \(2000V_r = 3600\), leading to \(V_r = \frac{3600}{2000} = 1.8\, \text{m/s}\).These equations ensure the calculated velocities correctly correspond to the given total momentum in each respective direction.