Topic 2 of 3
Choose a system for momentum conservation
Total momentum is conserved when the selected system has zero or negligible net external impulse over the interval. Draw the boundary 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:
= 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
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
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.
= 1.8 - 0.8 = +1.0 kg m/s
Joining supplies an extra condition: both carts have the same final velocity v. Therefore:
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:
= 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.
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.
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.
Optional check 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?
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:
- 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.
- 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.
- 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.
- 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.
- 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.