Skid numerical: A from West, B from South (49, 27)

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, $f = 0.55$, $g = 9.81\ \text{m/s}^2$. Pre-collision skids: $d_A = 37$ m, $d_B = 19$ m. Post-collision skids: $d_A' = 15$ m, $d_B' = 36$ m.

Vehicle A (4 t) comes from the West (moving East). Vehicle B (6 t) comes from the South (moving North). After impact A slides 49 degrees North of West and B slides 27 degrees East of North.

Let $\theta$ be the bearing measured clockwise from North: $\theta_A = 90^\circ$, $\theta_B = 0^\circ$ (travel directions) and $\theta_A' = 270^\circ + 49^\circ = 319^\circ$, $\theta_B' = 27^\circ$ (skid directions after impact).

   before impact          after impact
   A: bearing 90 deg       A': bearing 319 deg (49 deg N of W)
   B: bearing 0 deg        B': bearing 27 deg  (27 deg E of N)
   (bearing measured clockwise from North)

Step 1: Speeds just after collision

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

$$\begin{aligned} v_A' &= \sqrt{2(9.81)(0.55)(15)} = 12.72\ \text{m/s}\\ v_B' &= \sqrt{2(9.81)(0.55)(36)} = 19.71\ \text{m/s} \end{aligned}$$

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 ($\sin 319^\circ = -0.6561$, $\cos 319^\circ = 0.7547$):

$$\begin{aligned} 4\,u_A(1.0000) + 6\,u_B(0.0000) &= 4(12.723)(-0.6561) + 6(19.710)(0.4540) = 20.301 \\ 4\,u_A(0.0000) + 6\,u_B(1.0000) &= 4(12.723)(0.7547) + 6(19.710)(0.8910) = 143.777 \end{aligned}$$

Solving: $u_A = 5.08\ \text{m/s} = 18.3\ \text{km/h}$ and $u_B = 23.96\ \text{m/s} = 86.3\ \text{km/h}$ at the instant of impact.

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

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

$$\begin{aligned} v_A &= \sqrt{(5.08)^2 + 2(9.81)(0.55)(37)} = 20.62\ \text{m/s}\\ v_B &= \sqrt{(23.96)^2 + 2(9.81)(0.55)(19)} = 27.91\ \text{m/s} \end{aligned}$$

Answer: original speed of vehicle $A$ = 74.2 km/h and of vehicle $B$ = 100.5 km/h (speeds at impact: 18.3 and 86.3 km/h).