Skid numerical: car 430 kg and truck 1299 kg

After the collision the two vehicles move together, so the common post-impact speed comes from the skid distance, the impact speeds from momentum conservation in two directions, and the original speeds from the pre-collision skids.

Given: car $m_1 = 430$ kg moving East, truck $m_2 = 1299$ kg moving South, $f = 0.5$, pre-collision skids 15 m (car) and 10 m (truck), post-collision skid $d' = 8$ m at $65^\circ$ South of East.

        N
        ^
        |      car (East) ->
  W ----+----> E
        |\ 65 deg
        | \  locked pair
        v  v (S of E)
        S   truck comes moving South

Step 1: Speed of the locked vehicles after impact

$$v' = \sqrt{2gfd'} = \sqrt{2(9.81)(0.5)(8)} = 8.859\ \text{m/s}$$

Step 2: Momentum conservation

Total mass $M = 430 + 1299 = 1729$ kg. East component (only the car had eastward momentum) and South component (only the truck had southward momentum):

$$\begin{aligned} m_1 u_1 &= M v'\cos 65^\circ \Rightarrow u_1 = \frac{1729(8.859)(0.4226)}{430} = 15.054\ \text{m/s}\\ m_2 u_2 &= M v'\sin 65^\circ \Rightarrow u_2 = \frac{1729(8.859)(0.9063)}{1299} = 10.687\ \text{m/s} \end{aligned}$$

Step 3: Original speeds

$$\begin{aligned} v_1 &= \sqrt{u_1^2 + 2gf(15)} = \sqrt{(15.054)^2 + 2(9.81)(0.5)(15)} = 19.33\ \text{m/s}\\ v_2 &= \sqrt{u_2^2 + 2gf(10)} = \sqrt{(10.687)^2 + 2(9.81)(0.5)(10)} = 14.57\ \text{m/s} \end{aligned}$$

Answer: initial speed of the car = 69.6 km/h and of the truck = 52.5 km/h (impact speeds 54.2 and 38.5 km/h).