Skid numerical: A from North, B from West

A collision on a flat road is solved by (i) finding each vehicle's speed just after impact from its post-collision skid, (ii) applying conservation of linear momentum as a vector equation to get the speeds just before impact, and (iii) adding back the energy lost in the pre-collision skid.

Given: $W_A = 4$ t, $W_B = 6$ t, $f = 0.52$, $g = 9.81\ \text{m/s}^2$. Pre-collision skids: $d_A = 32$ m, $d_B = 18$ m. Post-collision skids: $d_A' = 14$ m, $d_B' = 33$ m.

Vehicle A comes from the North (moving South). Vehicle B comes from the West (moving East). After impact A slides 47 degrees East of North and B slides 56 degrees South of East.

Directions are written as bearings $\theta$ measured clockwise from North: $\theta_A = 180^\circ$, $\theta_B = 90^\circ$ (travel directions) and $\theta_A' = 47^\circ$, $\theta_B' = 146^\circ$ (skid directions after impact).

   before impact          after impact
   A: bearing 180 deg     A': bearing 47 deg
   B: bearing 90 deg      B': bearing 146 deg
   (bearing measured clockwise from North)

Step 1: Speeds just after collision

From work-energy, $v' = \sqrt{2 g f d'}$:

$$\begin{aligned} v_A' &= \sqrt{2(9.81)(0.52)(14)} = 11.95\ \text{m/s}\\ v_B' &= \sqrt{2(9.81)(0.52)(33)} = 18.35\ \text{m/s} \end{aligned}$$

Step 2: Momentum balance (vector form)

Take $x$ = East and $y$ = North, and let $u_A, u_B$ be the speeds just before impact. Weights are proportional to masses, so they can be used directly.

$$\begin{aligned} W_A u_A \sin\theta_A + W_B u_B \sin\theta_B &= W_A v_A' \sin\theta_A' + W_B v_B' \sin\theta_B' \\ W_A u_A \cos\theta_A + W_B u_B \cos\theta_B &= W_A v_A' \cos\theta_A' + W_B v_B' \cos\theta_B' \end{aligned}$$

Substituting the numbers:

$$\begin{aligned} 4\,u_A(0.0000) + 6\,u_B(1.0000) &= 4(11.951)(0.7314) + 6(18.349)(0.5592) = 96.526 \\ 4\,u_A(-1.0000) + 6\,u_B(0.0000) &= 4(11.951)(0.6820) + 6(18.349)(-0.8290) = -58.668 \end{aligned}$$

Solving: $u_A = 14.67\ \text{m/s} = 52.8\ \text{km/h}$ and $u_B = 16.09\ \text{m/s} = 57.9\ \text{km/h}$ at the instant of impact.

Step 3: Original speeds (before the pre-collision skid)

$v_0 = \sqrt{u^2 + 2 g f d}$:

$$\begin{aligned} v_A &= \sqrt{(14.67)^2 + 2(9.81)(0.52)(32)} = 23.27\ \text{m/s}\\ v_B &= \sqrt{(16.09)^2 + 2(9.81)(0.52)(18)} = 21.03\ \text{m/s} \end{aligned}$$

Answer: original speed of vehicle $A$ = 83.8 km/h and of vehicle $B$ = 75.7 km/h (speeds at impact: 52.8 and 57.9 km/h).