---
title: "Skid numerical: A from West, B from South (51, 61)"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:skid-collision-west-south-51-61"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/skid-collision-west-south-51-61
site: BCT Engineering
---

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

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2078 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-west-south-51-61
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

Two vehicles approaching at right angles, A from West and B from South, collide with each other. After the collision, vehicle A skids in a direction 51° North of West and vehicle B, 61° East of North. The initial skid distances of the vehicles A and B are 39m and 21m respectively before collision. The skid distances after collision are 15m and 36m respectively. If the weights of vehicles B and A are 6 and 4.4 tonnes respectively, calculate the original speeds of the vehicles. Take average skid resistance as 0.55.

## Short answer

Find post-collision speeds from skid distances, apply conservation of momentum to get impact speeds, then add pre-collision skid energy. Original speeds are 92.6 km/h for vehicle A and 81.2 km/h for vehicle B.

## 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:** $W_A = 4.4$ 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).

```text
   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).

## Check yourself

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

1. In the vector solution, what bearings (clockwise from North) are used for A before and after impact?
   A. 90 degrees before, 321 degrees after
   B. 0 degrees before, 51 degrees after
   C. 270 degrees before, 51 degrees after
   D. 90 degrees before, 51 degrees after
   Answer: A. A travels East (90 degrees) and slides 51 degrees North of West, i.e. bearing 321 degrees. (Key point: Bearings of motion)

2. The post-collision speed of B (36 m skid, f = 0.55) is:
   A. 12.7 m/s
   B. 19.7 m/s
   C. 25.7 m/s
   D. 16.8 m/s
   Answer: B. v_B' = sqrt(2 x 9.81 x 0.55 x 36) = 19.71 m/s. (Key point: Post-impact speed)

3. Why can the weights 4.4 t and 6 t be used directly in the momentum equation?
   A. Weights are proportional to masses, so g cancels
   B. Weights equal speeds
   C. Friction cancels
   D. Because both vehicles are equal
   Answer: A. W = m g, and g appears on both sides, so weights can replace masses. (Key point: Use of weights)

4. The solved speeds just before impact are:
   A. u_A = 55.8 km/h, u_B = 60.5 km/h
   B. u_A = 92.6 km/h, u_B = 81.2 km/h
   C. u_A = 60.5 km/h, u_B = 55.8 km/h
   D. u_A = 15.5 km/h, u_B = 16.8 km/h
   Answer: A. u_A = 15.50 m/s and u_B = 16.81 m/s from the x and y balances. (Key point: Speeds at impact)

5. Which equation gives the original speeds, and what are they?
   A. v0 = sqrt(u^2 + 2 g f d): A = 92.6 km/h, B = 81.2 km/h
   B. v0 = u - 2 g f d: A = 20 km/h
   C. v0 = u: A = 55.8 km/h
   D. v0 = sqrt(2 g f d): A = 60 km/h
   Answer: A. The pre-collision skid energy is added back to the impact speed: 25.71 m/s and 22.56 m/s. (Key point: Original speeds)

## 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×

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