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Skid numerical: A from West, B from South (45, 50)

Short answer

Find post-impact speeds from skid distances using v=sqrt(2 g f d), then use conservation of momentum in x-y components for impact speeds, and add pre-collision skid energy using v0=sqrt(u^2+2 g f d). Original speeds: A = 70.0 km/h, B = 77.9 km/h.

Two vehicles A and B approaching at right angle, A from west and B from south, collide with each other. After the collision, vehicle A skids in a direction 45 deg. north of the west and vehicle B, 50 deg. east of north. The initial skid distance of vehicles A and B are 35 and 25m respectively before collision. The skid distance after collision are 30m and 10m respectively. If the weight of vehicles A and B are 3.5 and 6 tonnes respectively, calculate the original speed of vehicles. The avg. skid resistance of the pavement is found to be 0.55.

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: WA=3.5W_A = 3.5 t, WB=6W_B = 6 t, f=0.55f = 0.55, g=9.81 m/s2g = 9.81\ \text{m/s}^2. Pre-collision skids: dA=35d_A = 35 m, dB=25d_B = 25 m. Post-collision skids: dA′=30d_A' = 30 m, dB′=10d_B' = 10 m.

Vehicle A comes from the West (moving East). Vehicle B comes from the South (moving North). After impact A slides 45 degrees North of West and B slides 50 degrees East of North.

Directions are written as bearings θ\theta measured clockwise from North: θA=90∘\theta_A = 90^\circ, θB=0∘\theta_B = 0^\circ (travel directions) and θA′=315∘\theta_A' = 315^\circ, θB′=50∘\theta_B' = 50^\circ (skid directions after impact).

   before impact          after impact
   A: bearing 90 deg      A': bearing 315 deg
   B: bearing 0 deg       B': bearing 50 deg
   (bearing measured clockwise from North)

Step 1: Speeds just after collision

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

vA′=2(9.81)(0.55)(30)=17.99 m/svB′=2(9.81)(0.55)(10)=10.39 m/s\begin{aligned} v_A' &= \sqrt{2(9.81)(0.55)(30)} = 17.99\ \text{m/s}\\ v_B' &= \sqrt{2(9.81)(0.55)(10)} = 10.39\ \text{m/s} \end{aligned}

Step 2: Momentum balance (vector form)

Take xx = East and yy = North, and let uA,uBu_A, u_B be the speeds just before impact. Weights are proportional to masses, so they can be used directly.

WAuAsin⁡θA+WBuBsin⁡θB=WAvA′sin⁡θA′+WBvB′sin⁡θB′WAuAcos⁡θA+WBuBcos⁡θB=WAvA′cos⁡θA′+WBvB′cos⁡θB′\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:

3.5 uA(1.0000)+6 uB(0.0000)=3.5(17.992)(−0.7071)+6(10.388)(0.7660)=3.2173.5 uA(0.0000)+6 uB(1.0000)=3.5(17.992)(0.7071)+6(10.388)(0.6428)=84.593\begin{aligned} 3.5\,u_A(1.0000) + 6\,u_B(0.0000) &= 3.5(17.992)(-0.7071) + 6(10.388)(0.7660) = 3.217 \\ 3.5\,u_A(0.0000) + 6\,u_B(1.0000) &= 3.5(17.992)(0.7071) + 6(10.388)(0.6428) = 84.593 \end{aligned}

Solving: uA=0.92 m/s=3.3 km/hu_A = 0.92\ \text{m/s} = 3.3\ \text{km/h} and uB=14.10 m/s=50.8 km/hu_B = 14.10\ \text{m/s} = 50.8\ \text{km/h} at the instant of impact.

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

v0=u2+2gfdv_0 = \sqrt{u^2 + 2 g f d}:

vA=(0.92)2+2(9.81)(0.55)(35)=19.46 m/svB=(14.10)2+2(9.81)(0.55)(25)=21.65 m/s\begin{aligned} v_A &= \sqrt{(0.92)^2 + 2(9.81)(0.55)(35)} = 19.46\ \text{m/s}\\ v_B &= \sqrt{(14.10)^2 + 2(9.81)(0.55)(25)} = 21.65\ \text{m/s} \end{aligned}

Answer: original speed of vehicle AA = 70.0 km/h and of vehicle BB = 77.9 km/h (speeds at impact: 3.3 and 50.8 km/h).

✓ Answer checked by Claude Opus 5.5 (high effort)

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