A railway wagon of mass 4000 kg is moving at 3 m/s along a level track when it collides with a stationary wagon of mass 2000 kg. After the collision the two wagons join together and move as one. Explain what happens to the total momentum of the two wagons during this collision, and describe how the velocity of the joined wagons after the collision compares with the velocity of the first wagon before the collision.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Momentum is defined as the product of an object's mass and its velocity. The principle of conservation of momentum states that, provided no external forces act, the total momentum of a system before a collision equals the total momentum after the collision.
Model answer (5 marks)
The total momentum of the system is conserved during the collision – it stays the same because no external force acts on the wagons (the track is level and smooth).
After the collision the two wagons are joined, so the total mass becomes 4000 kg + 2000 kg = 6000 kg. Because momentum is conserved and the mass has increased, the common velocity of the joined wagons must be less than the initial velocity of the first wagon.
Applying conservation of momentum:
4000 kg × 3 m s⁻¹ = 6000 kg × v
v = (4000 × 3)/6000 = 2 m s⁻¹.
Thus the joined wagons move together at 2 m s⁻¹, which is lower than the 3 m s⁻¹ of the first wagon before the collision.
After the collision the two wagons are joined, so the total mass becomes 4000 kg + 2000 kg = 6000 kg. Because momentum is conserved and the mass has increased, the common velocity of the joined wagons must be less than the initial velocity of the first wagon.
Applying conservation of momentum:
4000 kg × 3 m s⁻¹ = 6000 kg × v
v = (4000 × 3)/6000 = 2 m s⁻¹.
Thus the joined wagons move together at 2 m s⁻¹, which is lower than the 3 m s⁻¹ of the first wagon before the collision.
Examiner tips
- Show the conservation of momentum equation explicitly; include the mass values. State that the track is level and smooth to justify no external force. Calculate the final velocity and give the numerical value. Use correct units (kg, m s⁻¹).
Common mistakes
- Forgetting to add the masses of the two wagons. Using the wrong sign for velocity or neglecting the fact that the second wagon was initially at rest. Leaving out the units or mis‑calculating the final velocity.
Mark scheme (5 marks)
- The total momentum of the system is conserved (stays the same / does not change) during the collision
- This is because no external (resultant) force acts on the system / the track is level and smooth
- The total mass after the collision is greater than the mass of the first wagon alone (combined mass is 6000 kg)
- Because momentum is conserved and mass has increased, the velocity of the joined wagons must be less than 3 m/s
- The velocity of the joined wagons is 2 m/s (correctly applying conservation of momentum: 4000 × 3 = 6000 × v, so v = 2 m/s)
Key terms in this question
momentum · velocity · mass · collision
Related
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