---
title: "Skid numerical: A from West, B from South (45, 50)"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:skid-collision-west-south-45-50"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/skid-collision-west-south-45-50
site: BCT Engineering
---

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

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2080 Bhadra, 6 marks
- 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-west-south-45-50
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

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.

## 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.

## 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:** $W_A = 3.5$ t, $W_B = 6$ t, $f = 0.55$, $g = 9.81\ \text{m/s}^2$. Pre-collision skids: $d_A = 35$ m, $d_B = 25$ m. Post-collision skids: $d_A' = 30$ m, $d_B' = 10$ 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 $\theta$ measured clockwise from North: $\theta_A = 90^\circ$, $\theta_B = 0^\circ$ (travel directions) and $\theta_A' = 315^\circ$, $\theta_B' = 50^\circ$ (skid directions after impact).

```text
   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, $v' = \sqrt{2 g f d'}$:

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

#### Step 2: Momentum balance (vector form)
Take $x$ = 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:

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

Solving: $u_A = 0.92\ \text{m/s} = 3.3\ \text{km/h}$ and $u_B = 14.10\ \text{m/s} = 50.8\ \text{km/h}$ at the instant of impact.

#### Step 3: Original speeds (before the pre-collision skid)
$v_0 = \sqrt{u^2 + 2 g f d}$:

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

**Answer:** original speed of vehicle $A$ = **70.0 km/h** and of vehicle $B$ = **77.9 km/h** (speeds at impact: 3.3 and 50.8 km/h).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the speed of each vehicle just after impact found from its post-collision skid?
   A. v' = sqrt(2 g f d')
   B. v' = g f d'
   C. v' = d'/t only
   D. v' = sqrt(g d'/f)
   Answer: A. Work-energy: the kinetic energy is dissipated by friction over the skid, giving v' = sqrt(2 g f d'). (Key point: Post-collision speed)

2. Why must conservation of momentum be applied in x and y components here?
   A. Because the vehicles approach at right angles and leave in different directions
   B. Because speed is a vector quantity only after impact
   C. Because friction changes momentum direction
   D. Because weights differ
   Answer: A. Momentum is a vector; the vehicles come from perpendicular directions and slide off at different angles, so each direction is balanced separately. (Key point: Momentum as a vector)

3. For post-collision skid of 30 m with f = 0.55, the speed of A just after impact is about:
   A. 10.4 m/s
   B. 18.0 m/s
   C. 25.4 m/s
   D. 32.4 m/s
   Answer: B. v_A' = sqrt(2*9.81*0.55*30) = 17.99 m/s. (Key point: Speed of A after impact)

4. Solving the momentum equations gives the speeds just before impact as approximately:
   A. u_A = 50.8 km/h and u_B = 3.3 km/h
   B. u_A = 3.3 km/h and u_B = 50.8 km/h
   C. u_A = 70 km/h and u_B = 78 km/h
   D. u_A = 17.99 m/s and u_B = 10.39 m/s
   Answer: B. The x-balance gives u_A = 0.92 m/s (3.3 km/h) and the y-balance gives u_B = 14.10 m/s (50.8 km/h). (Key point: Speeds at impact)

5. How are the original speeds obtained from the speeds at impact?
   A. Subtract 2 g f d from u^2
   B. v0 = sqrt(u^2 + 2 g f d), adding the energy lost in the pre-collision skid
   C. Add the skid distance to u
   D. Multiply u by the friction coefficient
   Answer: B. The vehicle lost energy skidding before impact, so v0^2 = u^2 + 2 g f d, giving about 70.0 km/h for A and 77.9 km/h for B. (Key point: Adding back pre-collision skid)

## Close variants set in other exams

- [Skid numerical: A from North, B from West](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Askid-collision-north-west-47-56) — Transportation Engineering II, chapter 1, asked 3×

## Related in Transportation Engineering II

- [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 (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