Skid numerical: A from West, B from South (51, 61)

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: $W_A = 4.4$ t, $W_B = 6$ t, $f = 0.55$, $g = 9.81\ \text{m/s}^2$. Pre-collision skids: $d_A = 39$ m, $d_B = 21$ m. Post-collision skids: $d_A' = 15$ m, $d_B' = 36$ m.

Vehicle A comes from the West (moving East); its weight is 4.4 t. Vehicle B comes from the South (moving North); its weight is 6 t. After impact A slides 51 degrees North of West and B slides 61 degrees East of North.

Directions are written as bearings $\theta$ measured clockwise from North: $\theta_A = 90^\circ$, $\theta_B = 0^\circ$ (travel directions) and $\theta_A' = 321^\circ$, $\theta_B' = 61^\circ$ (skid directions after impact).

   before impact          after impact
   A: bearing 90 deg      A': bearing 321 deg
   B: bearing 0 deg       B': bearing 61 deg
   (bearing measured clockwise from North)

Step 1: Speeds just after collision

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

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

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:

$$\begin{aligned} 4.4\,u_A(1.0000) + 6\,u_B(0.0000) &= 4.4(12.723)(-0.6293) + 6(19.710)(0.8746) = 68.202 \\ 4.4\,u_A(0.0000) + 6\,u_B(1.0000) &= 4.4(12.723)(0.7771) + 6(19.710)(0.4848) = 100.837 \end{aligned}$$

Solving: $u_A = 15.50\ \text{m/s} = 55.8\ \text{km/h}$ and $u_B = 16.81\ \text{m/s} = 60.5\ \text{km/h}$ at the instant of impact.

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

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

$$\begin{aligned} v_A &= \sqrt{(15.50)^2 + 2(9.81)(0.55)(39)} = 25.71\ \text{m/s}\\ v_B &= \sqrt{(16.81)^2 + 2(9.81)(0.55)(21)} = 22.56\ \text{m/s} \end{aligned}$$

Answer: original speed of vehicle $A$ = 92.6 km/h and of vehicle $B$ = 81.2 km/h (speeds at impact: 55.8 and 60.5 km/h).