--- title: "Skid numerical: A from North, B from West" subject: "Transportation Engineering II" chapter: 1 key: "semester_7/transportation_engineering_ii:skid-collision-north-west-47-56" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/skid-collision-north-west-47-56 site: BCT Engineering --- # Skid numerical: A from North, B from West Transportation Engineering II · Chapter 1: Traffic Engineering - Asked in: 2082 Bhadra, 8 marks - Video: [Find initial Speed | Transportation numerical | Collision of 2 vehicle | TU.PU.KU.POU](https://www.youtube.com/watch?v=ICzAWfNu_Ks) — Sudip Dhakal - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/skid-collision-north-west-47-56 - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1 ## Question Two vehicles A and B approaching at right angles, A from North and B from West collide with each other. After the collision, vehicle A skids in a direction 47° East of North and vehicle B skids 56° South of East. The initial skid distance of the vehicles A and B are 32 and 18 m respectively before collision. The skid distances after collision are 14 and 33 m respectively. If the weights of vehicles A and B are 4 and 6 tonnes respectively, calculate the original speeds of vehicles. The average skid resistance of the pavement is found to be 0.52. ## Short answer Use post-collision skids to get speeds after impact from v = sqrt(2 g f d), apply conservation of momentum, then add pre-collision skid energy. Original speeds: A = 83.8 km/h, B = 75.7 km/h; impact speeds 52.8 and 57.9 km/h. ## Answer 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). ```text 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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In what order is the collision problem solved? A. Pre-collision skid, then momentum, then post-collision skid B. Post-collision skid speeds, then vector momentum balance, then add back the pre-collision skid energy loss C. Momentum balance only, since skid is irrelevant D. Pre-collision skid only Answer: B. Speeds just after impact come from post-skid, momentum gives speeds at impact, and the pre-skid gives original speeds. (Key point: Solution sequence) 2. What formula gives the speed just after impact from the post-collision skid distance d'? A. v' = sqrt(2 g f d') B. v' = g f d' C. v' = sqrt(g f d') D. v' = 2 g f d' Answer: A. Work-energy gives v'^2 = 2 g f d', so v' = sqrt(2 g f d'). (Key point: Speed after collision) 3. Why is momentum conservation applied in two directions (East and North components)? A. The vehicles are of equal mass B. Momentum is a vector, and the vehicles travel and skid in different directions C. Skid resistance changes direction D. Because weights are in tonnes Answer: B. Momentum is a vector, so components along East and North must each balance. (Key point: Momentum as a vector equation) 4. Vehicle A had speed 14.67 m/s at impact after a 32 m pre-collision skid. How is its original speed found? A. v = 14.67 - 2 g f (32) B. v = sqrt(14.67^2 + 2 g f (32)) because the skid removed energy before impact C. v = 14.67 + f(32) D. v = 14.67 / f Answer: B. The skid dissipated energy, so v0^2 = u^2 + 2 g f d, giving 23.27 m/s (83.8 km/h). (Key point: Original speed from impact speed) ## Close variants set in other exams - [Skid numerical: A from West, B from South (51, 61)](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Askid-collision-west-south-51-61) — Transportation Engineering II, chapter 1, asked 2× - [Skid numerical: A from West, B from South (45, 50)](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Askid-collision-west-south-45-50) — Transportation Engineering II, chapter 1, asked 2× ## Related in Transportation Engineering II - [Skid numerical: A from West, B from South (49, 27)](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Askid-collision-west-south-49-27) — Transportation Engineering II, chapter 1 - [Skid numerical: moving vehicle hits parked vehicle](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Askid-moving-vehicle-hits-parked-40m) — Transportation Engineering II, chapter 1 - [Skid numerical: car 430 kg and truck 1299 kg](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Askid-collision-car-430kg-truck-1299kg) — Transportation Engineering II, chapter 1