Collisions

01

Impulse and a change of momentum

Impulse measures the effect of a force over an interval. Its direction matters: choose the body and positive direction before finding the area or the momentum change.

Momentum is the vector p = mv. In one dimension, use signed velocities along one axis. Newton's second law relates the resultant force component to the rate of change of that momentum component.

For constant mass, Δp = m(v - u). Impulse has units N s = kg m/s. It is force multiplied by time, not force divided by time.

Check what force the graph represents. A contact-force pulse alone gives the net impulse in that direction only when other impulses in that direction are negligible. For a graph that crosses zero, add its signed positive and negative areas; do not add all area magnitudes.

Estimate an impact force

Assume a roughly 0.1 kg ball is stopped from about 10 m/s. Its momentum change has magnitude of order 1 N s. If contact lasts about 0.01 s, the average force has magnitude of order 100 N, opposite the incoming direction.

These rough mass, speed and duration assumptions give a size check, not a universal impact force. A different contact duration changes the average force even if the momentum change is the same.

Average force is not peak force

The average force gives the same impulse when multiplied by the full interval. Its magnitude, 80.0 N, differs from the pulse's 160 N peak.

If the same -4.80 N s momentum change occurred in 0.030 s, the average force would be -160 N. Increasing contact time can reduce the magnitude of average force for a fixed momentum change. It does not, by itself, specify the peak or the exact pulse shape.

If the original ball merely stopped, its momentum change would be 0 - 3.00 = -3.00 N s. Rebounding adds momentum in the opposite direction, increasing the change's magnitude to 4.80 N s.

The ball also exerts an opposite force on the wall. These Newton's-third-law forces act on different bodies. They do not cancel in the ball's individual momentum equation.

Compare a sensor trace with momentum change

To compare a recorded contact pulse with m(v - u), identify the force recipient and use the same positive direction for both measurements. A sensor at a fixed wall may report the ball's force on the wall. With motion towards the wall positive, that force is positive. The force on the ball is opposite, so reverse the reported sign before comparing it with the ball's negative momentum change. Confirm the instrument's calibrated sign convention rather than assuming its displayed sign already refers to the ball.

1. Calibrate and zero the force channel. Check its response to a known force in a known direction, and use a range that captures the pulse without clipping.
2. Record the whole contact interval. Include the start, changing force, peak and return to the baseline. Use actual sample times and a sampling interval short enough to resolve the pulse.
3. Measure the same body's mass and velocities. Use signed velocities just before and just after the contact, so the momentum change and force area refer to the same event.
4. Compare quantities with units and uncertainty. Sum signed force-time areas in N s and compare with m(v - u) in kg m/s. Include or justify neglecting any other impulse in that direction.

Sparse samples can miss a negative peak and underestimate the impulse magnitude. A force-zero shift changes the area throughout the interval. For example, an uncorrected +10 N offset lasting 0.060 s adds +0.60 N s, making this -4.80 N s pulse appear as -4.20 N s. Repeating the same wrongly zeroed measurement does not remove that bias.

A supplied straight-segment model gives an exact geometric area. A sum formed from samples of a real curved pulse is an estimate whose quality depends on timing, response and sampling. The three points below describe the model, not an actual sensor recording.

02

Choose a system for momentum conservation

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:

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.

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 v_A = -2.0 m/s and v_B = +3.0 m/s, with negligible external impulse.

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.

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.

03

Elastic and inelastic outcomes

Momentum conservation is the starting condition, not a classification of the collision. A perfectly elastic collision also conserves total kinetic energy; an inelastic collision does not.

Use the same two-cart model with right positive: A is left of B, with mass 0.60 kg and initial velocity +3.0 m/s; B has mass 0.40 kg and initial velocity -2.0 m/s. Assume a one-dimensional collision with negligible external impulse.

Momentum gives 0.60v_A + 0.40v_B = 1.0. With two unknown final velocities, another condition is needed to predict an outcome. Do not assume kinetic energy is conserved merely to obtain another equation.

Perfectly elastic: use the relative speeds

A perfectly elastic collision conserves total kinetic energy as well as momentum. In a one-dimensional two-body collision, the relative speed of approach equals the relative speed of separation.

Relative speed describes how quickly the distance between the bodies decreases or increases. It is non-negative. The individual velocities in the calculation still retain their signs.

Before contact, A moves right and B moves left, so the gap closes at:

After a separating collision, with B still to A's right, the separation speed is v_B - v_A. For this perfectly elastic outcome:

Substitute v_B = v_A + 5.0 into the momentum equation:

A reverses to the left, while B reverses to the right. Check both conserved quantities:

The final separation speed is 4.0 - (-1.0) = 5.0 m/s, matching the approach speed. Check the bodies' left/right arrangement when forming a relative speed rather than adding or subtracting velocity symbols without reference to the motion.

Inelastic does not require sticking

Consider a supplied alternative outcome: v_A = 0 and v_B = +2.5 m/s. The data provide both final velocities; they are not uniquely predicted by momentum conservation alone.

The total momentum is unchanged, while kinetic energy decreases by 3.5 - 1.25 = 2.25 J. This collision is inelastic. B moves away from stationary A at a relative speed of 2.5 m/s, so the carts separate.

If instead they stick, both velocities are +1.0 m/s, as in the joined-cart calculation. Final kinetic energy is 0.50 J and the kinetic decrease is 3.0 J. Their relative separation speed is zero. Sticking is a perfectly inelastic outcome, not the definition of every inelastic collision.

In these supplied inelastic cases, energy transfers from kinetic energy to other stores or to the surroundings. Total energy is still conserved. The equality of approach and separation speeds belongs to the perfectly elastic case and must not be imposed on either inelastic alternative.