8867 / 2027
Collisions overview

Lesson 2 of 3

Choose a system for momentum conservation

  • I can state conservation of total momentum for a system with no net external impulse.Syllabus

    Syllabus 8867, 6(b). State conservation of total momentum for a system with no net external impulse.

  • I can apply momentum conservation to inelastic and perfectly elastic interactions between two bodies in one dimension.Syllabus

    Syllabus 8867, 6(c). Apply momentum conservation to inelastic and perfectly elastic interactions between two bodies in one dimension.

  • I can distinguish conserved total momentum in an isolated interaction from the kinetic energy, which usually changes.Syllabus

    Syllabus 8867, 6(e). Distinguish conserved total momentum in an isolated interaction from the kinetic energy, which usually changes.

Make a guess. You are not marked.

Two skaters of different masses stand at rest on smooth ice, facing each other. They push each other apart. What is the total momentum of the two skaters just after the push?

  1. Zero, the same as before they pushed.
  2. Greater than zero, in the direction of the lighter skater.
  3. It depends on who pushed harder.
Show the answer

Zero, the same as before they pushed.

No resultant external force acts on the pair during the push, so the total momentum stays at its starting value, zero. The two momenta are equal in size and opposite in direction.

When a question asks you to state the principle of conservation of momentum, write this sentence.

Principle of conservation of momentum
The total momentum of a system remains constant provided no resultant external force acts on the system.

"No resultant external force" means nothing outside the system gives it a net impulse. So the system's total momentum cannot change. In a short collision, a small external force such as friction gives only a tiny impulse. We can then treat total momentum as conserved. Draw the boundary of your system before writing the equation.

Momentum is p = mv, so add signed momenta along the same chosen direction. In one dimension, two constant-mass bodies give:

mAuA + mBuB
= mAvA + mBvB

This states that total momentum before equals total momentum after, under the negligible-external-impulse condition. More generally, the change in the system's total momentum equals its net external impulse.

Internal interaction forces occur in equal and opposite pairs. Their impulses cancel in the total system account, although each body's momentum changes. A closed or isolated momentum system needs this external-impulse condition; saying only that no matter crosses its boundary is insufficient.

A short collision can make the impulse from a small steady external force negligible compared with the contact impulses. Short duration alone does not guarantee that every external impulse is small.

A ball alone is not an isolated collision system

For the ball rebounding from a wall, the wall's force is external to a ball-only boundary. The ball's momentum changes. Including the wall and Earth changes the boundary: the contact impulses are then internal, and total momentum can be conserved if the remaining external impulse is negligible.

The wall and Earth's large mass can make their velocity change too small to notice. That does not make their momentum transfer zero. Do not infer conservation of an individual body's momentum merely because its interaction partner looks stationary.

Worked two-cart system

Carts that join together

Choose carts A and B together as the system, with right positive. A has mass 0.60 kg and velocity +3.0 m/s; B has mass 0.40 kg and velocity -2.0 m/s. They approach along a straight level track. Assume negligible external horizontal impulse and no relevant mass change.

Choose A + B as the momentum system

Cart A has mass 0.60 kg and cart B has mass 0.40 kg. Take right as positive and assume the net external horizontal impulse is negligible during this one-dimensional collision. Both views use the same system boundary and velocity scale.

Before: the two carts approach

Before: the two carts approachThe dashed boundary selects cart A plus cart B as the momentum system. Cart A is left, mass 0.60 kilograms and velocity positive three metres per second. Cart B is right, mass 0.40 kilograms and velocity negative two metres per second. Right is positive. Their separate blue velocity arrows point towards one another and are 66 and 44 drawing units long. Total momentum is positive one kilogram metre per second and total kinetic energy is 3.50 joules. No force arrows or environmental objects are included inside this system boundary.System: cart A + cart BA0.60 kg+3.0 m/sB0.40 kg-2.0 m/sPositive direction

Total kinetic energy: 3.50 J. The individual momenta are +1.80 and -0.80 kg m/s. Their signed sum is +1.00 kg m/s.

After: the carts move together

After: the carts move togetherThe same dashed A-plus-B system boundary now encloses the two touching joined carts, with A still on the left and B on the right. A single blue arrow, 22 drawing units long, shows their common velocity positive one metre per second. Both masses still belong to the system. Total momentum remains positive one kilogram metre per second, but kinetic energy is now 0.50 joules. The common velocity does not mean either cart retains its original momentum. Cart positions and body sizes are schematic; the velocity-arrow scale matches the before view.System: cart A + cart BA0.60 kgB0.40 kgBoth: +1.0 m/sPositive direction

Total kinetic energy: 0.50 J. A common +1.0 m/s gives the same +1.00 kg m/s total momentum. The 3.0 J decrease in kinetic energy is transferred to other stores and surroundings as appropriate; energy has not disappeared.

All blue cart arrows share the same velocity scale, including the common arrow for joined carts. A zero velocity has no direction arrow. Cart positions and sizes are schematic; these are state comparisons, not force diagrams.

The dashed boundary includes both carts before and after the interaction. The arrows represent signed velocity. After joining, one common arrow describes both carts' final motion.
pinitial,total = 0.60(3.0) + 0.40(-2.0)
= 1.8 - 0.8 = +1.0 kg m/s

Joining supplies an extra condition: both carts have the same final velocity v. Therefore:

(0.60 + 0.40)v = 1.0
v = +1.0 m/s

A's momentum changes from +1.8 to +0.60 kg m/s, while B's changes from -0.80 to +0.40 kg m/s. Their changes are -1.20 and +1.20 N s respectively, which sum to zero. Neither cart keeps its own original momentum.

Use kinetic energy Ek = ½mv2 to check a different quantity:

Ek,initial = ½(0.60)(3.0)2 + ½(0.40)(-2.0)2
= 2.7 + 0.8 = 3.5 J
Ek,final = ½(1.00)(1.0)2 = 0.50 J

Total kinetic energy decreases by 3.0 J while total momentum stays the same. Deformation and internal-energy increases can account for energy within the carts; sound can carry energy to surrounding air. Include the surroundings where needed in a complete energy account. Energy has not disappeared.

Try it step by step

Work it out in steps: Total momentum before in kg m/s, then Combined mass in kg, then Common velocity after in m/s.

  1. Problem 1. A 2.0 kg trolley moving at 3.0 m/s collides with a stationary 1.0 kg trolley and the two move off together. Find their common velocity.
  2. Problem 2. A 1500 kg car moving at 20 m/s runs into a stationary 1000 kg car and the two lock together. Find their common velocity just after the collision.
  3. Problem 3. A 0.80 kg trolley moving at 2.5 m/s catches up with a 1.2 kg trolley moving the same way at 0.50 m/s, and they couple. Find their common velocity.
Show the answer

Problem 1.

Total momentum before in kg m/s: p = 2.0 x 3.0 + 1.0 x 0 = 6.0 kg m/s

Combined mass in kg: m = 2.0 + 1.0 = 3.0 kg

Common velocity after in m/s: v = 6.0 / 3.0 = 2.0 m/s in the original direction

Problem 2.

Total momentum before in kg m/s: p = 1500 x 20 = 30000 kg m/s

Combined mass in kg: m = 1500 + 1000 = 2500 kg

Common velocity after in m/s: v = 30000 / 2500 = 12 m/s (2 sf)

Problem 3.

Total momentum before in kg m/s: p = 0.80 x 2.5 + 1.2 x 0.50 = 2.0 + 0.60 = 2.6 kg m/s

Combined mass in kg: m = 0.80 + 1.2 = 2.0 kg

Common velocity after in m/s: v = 2.6 / 2.0 = 1.3 m/s (2 sf)

Momentum alone is not enough to predict two unknown final velocities. It gives one equation. Joining makes the velocities equal and supplies the extra information here. A perfectly elastic collision supplies a different condition, explained in elastic and inelastic outcomes.

Momentum can stay constant while kinetic energy increases

For a separate interaction, start the same carts at rest with a light compressed spring between them. Suppose they move apart at vA = -2.0 m/s and vB = +3.0 m/s, with negligible external impulse.

pfinal,total = 0.60(-2.0) + 0.40(3.0) = 0
Ek,final = ½(0.60)(2.0)2 + ½(0.40)(3.0)2
= 1.2 + 1.8 = 3.0 J

The total momentum still equals its initial zero value. The kinetic gain comes from stored spring energy; if there are no other changes or losses, that store decreases by 3.0 J. Momentum conservation does not imply constant kinetic energy in every interaction.

Check your understanding

A ball rebounds from a fixed wall. Why can you not conserve the ball's momentum by treating the ball alone as the selected system during contact?

  1. A short collision guarantees conservation, so the ball's momentum should actually be unchanged.
  2. The wall gives the selected ball a non-negligible external impulse.
  3. Momentum is conserved only in perfectly elastic collisions.
Show the answer

The wall gives the selected ball a non-negligible external impulse.

The contact force is external to a ball-only boundary and changes its momentum. A larger ball-wall-Earth system can conserve total momentum if its net external impulse is negligible; the large bodies may have only a tiny velocity change.

Measure the same collision before and after

A cart investigation can compare the momentum and kinetic-energy accounts, but the measurements must describe the same bodies and event:

  1. Define the system and direction. Use both carts for the conservation account. Check that track tilt or external friction is small enough for its impulse during the chosen interval to be negligible.
  2. Measure each moving mass. Include attached sensors, flags, magnets and bumpers in the mass of the cart they move with. Check the balance zero and use a suitable range.
  3. Measure velocities close to contact. Record both carts immediately before and after the same collision. A light gate gives average speed over its blocking interval from flag length / blocking time. Measure the flag's effective length and determine direction separately.
  4. Keep each reading in one state. A gate interval must lie entirely before or entirely after contact, not straddle the collision. Avoid measurements so far from contact that friction appreciably changes the velocity between the gate and collision.
  5. Compare totals and uncertainties. Calculate the signed total momentum and the scalar total kinetic energy on each side. Assess any momentum difference against measurement uncertainty and possible external impulse before drawing a conclusion.

If a force sensor moves with a cart, its mass belongs in that cart's measured total and its channel must be labelled with the force recipient and direction. The force-trace method then compares that cart's impulse with its individual momentum change; the two-cart sum checks the whole system.

If the track is tilted, weight has a component along the track; substantial external friction can also supply impulse along the motion. These effects undermine the isolated-system approximation. Levelling the track and choosing a short, well-defined interval target them. Repeating a trial does not correct an omitted moving mass, a wrong flag length or a systematic force zero.

Practise exam questions

Write your answer first, then tick the mark points it makes.

Exam-style question

Two trolleys collide on a smooth horizontal track. Use Newton's laws to explain why the total momentum of the two trolleys is conserved. [3]

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Exam-style question

State the principle of conservation of momentum. [2]

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