At a crossing a truck travelling towards the north collides with a car travelling towards the east. After the collision the car and the truck stick together and move off at an angle of 30º east of north. If the speed of the car before the collision was 20 ms⁻¹, and the mass of the truck is twice the mass of the car, calculate the speed of the truck before and after the collision.

Introduction

This is a classic two-dimensional collision problem involving conservation of momentum. Since the truck and car stick together after collision, it’s an example of a perfectly inelastic collision. We will use vector components to solve for the unknown initial speed of the truck and the final velocity after the collision.

Given Data

Mass of car = m
Mass of truck = 2m
Initial speed of car (eastward) = 20 m/s
Final direction of combined mass = 30º east of north

Step 1: Use Conservation of Momentum

Let v_t be the speed of the truck before collision (to be calculated).

Let V be the speed of the combined object after collision (to be calculated).

We write x- and y-components of momentum:

Initial x-momentum: m × 20 (car only)

Initial y-momentum: 2m × v_t (truck only)

Final momentum (combined mass 3m): 3m × V, at angle 30° east of north

Step 2: Break Final Momentum into Components

x-direction: 3mV sin(30º) = m × 20 → 3V × 0.5 = 20 → 1.5V = 20 → V = 13.33 m/s

Step 3: Now Solve for Truck’s Speed

y-direction: 3mV cos(30º) = 2m × v_t

cos(30º) ≈ 0.866

3 × 13.33 × 0.866 = 2 × v_t

34.63 = 2 × v_t → v_t = 17.31 m/s

Final Answers

Speed of truck before collision: 17.31 m/s
Speed after collision: 13.33 m/s at 30° east of north

Conclusion

This problem demonstrates how the principle of conservation of linear momentum applies in two-dimensional collisions. By breaking the momentum into components and equating them before and after the collision, we determine the unknown velocities of both the truck and the combined object.