A 1570 kg car skidding due north on a level frictionless icy road at 156 km/h collides with a 2245.1 kg car skidding due east at 120 km/h in such a way that the two cars stick together. At what angle (−180◦ ≤ θ ≤ +180◦ ) East of North do the two coupled cars skid off at?

θ = 47.75º East of North.

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.

Tons?