Answer:
θ = 47.75º East of North.
Step-by-step explanation:
- Assuming no external forces acting during the collision, total momentum must be conserved.
- Since momentum is a vector, if we project it along E-W and N-S axes, the momentum components along these axes must be conserved too.
- So, for the N-S axis, we can write the following equation:

- Since the car moving due east has no speed component along the N-S axis, the initial momentum along this axis is simply:

where m₁ = 1570 kg, v₁₀ = 156 km/h
- In order to work with the same units, we need to convert the speed in km/h to m/s, as follows:

- Replacing by the values in the left side of (1), we get:

- Since the collision is inelastic, both cars stick together, so we can write the right side of (1) as follows:

- From (4) and (5) , we can solve for VfNorth:

- We can repeat exactly the same process for the E-W axis:

- Since the car moving due north has no speed component along the E-W axis, the initial momentum along this axis is simply:

- As we did with v₁₀, we need to convert v₂₀ from km/h to m/s, as follows:

- Replacing by the values in the left side of (7), we get:

- Since the collision is inelastic, both cars stick together, so we can write the right side of (7) as follows:

- From (10) and (11) , we can solve for VfEast:

- In order to find the angle East of North of the velocity vector, as we know the values of the horizontal and vertical components, we need just to apply a little bit of trigonometry, as follows:

- The angle θ, East of North, is simply tg⁻¹ (1.1):
θ = tg⁻¹ (1.1) = 47.75º E of N.