Full chapter
Collisions
All 3 topics and the revision summary on one page.
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.
= Δp = pfinal - pinitial
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.
Worked rebound and force pulse
A ball rebounds from a wall
A 0.150 kg ball approaches a fixed wall at +20.0 m/s and rebounds at -12.0 m/s. Take motion towards the wall as positive. Other horizontal impulses during contact are negligible.
pfinal = 0.150 × (-12.0) = -1.80 kg m/s
Δp = -1.80 - 3.00 = -4.80 N s
The impulse is negative because the ball's momentum changes opposite the positive incoming direction. Subtracting speeds, 12 - 20, would miss the reversal.
Keep the same body and positive direction throughout
A 0.150 kg ball rebounds from a wall
Within these two ball views, the blue arrows share one velocity scale. The positions are schematic. The change is final minus initial: -1.80 - (+3.00) = -4.80 kg m/s.
The signed force-time area gives the same impulse
60 ms = 0.060 s. The area is -½ × 160 N × 0.060 s = -4.80 N s. Moving the peak away from the midpoint does not change this triangle's area when its base and height are unchanged.
This is the horizontal force on the ball, with other horizontal impulses negligible. The ball's opposite force on the wall acts on a different body and does not cancel from the ball's own momentum change.
The supplied force model joins (0 ms, 0 N), (20 ms, -160 N) and (60 ms, 0 N) with straight lines. First check its size: a constant -160 N for the whole 0.060 s would give -9.6 N s. The actual triangular pulse has a smaller area magnitude, so |impulse| < 9.6 N s.
Convert milliseconds to seconds before calculating:
Impulse = -½ × 160 × 0.060 = -4.80 N s
This agrees with the ball's momentum change. The peak occurs at 20 ms, not halfway through the 60 ms interval. The triangle's area still uses its full base and perpendicular height.
Average force is not peak force
= -4.80/0.060 = -80.0 N
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.
Optional check A 0.150 kg ball travels towards a wall at +20.0 m/s and rebounds at -12.0 m/s. What is the impulse on the ball in this chosen direction?
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.
- 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.
- 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.
- 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.
- 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.
| Time / ms | Force on ball / N |
|---|---|
| 0 | 0 |
| 20 | -160 |
| 60 | 0 |
02
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.
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.
Initial total kinetic energy = ½(0.60)(3.0)2 + ½(0.40)(2.0)2
= 3.5 J
Momentum gives 0.60vA + 0.40vB = 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.
Three possible outcomes from the same initial account
Initially: A is left, 0.60 kg at +3.0 m/s; B is right, 0.40 kg at -2.0 m/s. Total momentum is +1.00 kg m/s and kinetic energy is 3.50 J. The approach speed is 3.0 - (-2.0) = 5.0 m/s.
Right is positive. Each alternative below assumes negligible net external horizontal impulse for the A + B system, so every outcome retains +1.00 kg m/s total momentum. Their kinetic energies differ.
1. Perfectly elastic
Total kinetic energy: 3.50 J. Kinetic energy is conserved. Separation speed is 4.0 - (-1.0) = 5.0 m/s, equal to the initial approach speed.
2. Inelastic, with the carts separating
Total kinetic energy: 1.25 J. Kinetic energy decreases by 2.25 J even though the carts separate. Their separation speed is 2.5 m/s; the elastic relative-speed equality does not apply.
3. Sticking together
Total kinetic energy: 0.50 J. Kinetic energy decreases by 3.0 J. The carts have one common velocity, so their relative separation speed is zero.
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.
Inelastic does not always mean sticking. Only the perfectly elastic outcome preserves the initial kinetic energy and the equality of approach and separation speeds.
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:
= 3.0 - (-2.0) = 5.0 m/s
After a separating collision, with B still to A's right, the separation speed is vB - vA. For this perfectly elastic outcome:
0.60vA + 0.40vB = 1.0
Substitute vB = vA + 5.0 into the momentum equation:
vA = -1.0 m/s, vB = +4.0 m/s
A reverses to the left, while B reverses to the right. Check both conserved quantities:
Ek,final = ½(0.60)(1.0)2 + ½(0.40)(4.0)2
= 0.30 + 3.20 = 3.50 J
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: vA = 0 and vB = +2.5 m/s. The data provide both final velocities; they are not uniquely predicted by momentum conservation alone.
Ek,final = 0 + ½(0.40)(2.5)2 = 1.25 J
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.
Optional check Cart A (0.60 kg, initially +3.0 m/s) collides with B (0.40 kg, initially -2.0 m/s) with negligible external impulse. Afterwards A is at rest and B moves right at +2.5 m/s. Which description is correct?
Optional check Before contact, A is to the left of B with u_A = +3.0 m/s and u_B = -2.0 m/s. Which relative-speed statement is valid for the perfectly elastic separating outcome?
Revision summary
Name the body or system, the direction and the interval. Distinguish an individual body's momentum change from conservation of the total.
Signed impulse
Δp = m(v - u) for constant mass
Faverage = net impulse / elapsed time
Use signed velocities, particularly for a rebound. Add positive and negative graph areas algebraically. Convert ms to s for an impulse in N s. Peak force is not generally average force.
A named contact force supplies the net impulse only if other impulses in that direction are negligible. Newton's-third-law partners act on different bodies and do not cancel in one body's equation.
In a sensor comparison, check force recipient, sign, calibration, zero, sample times and the whole pulse interval. A force measured on the wall has the opposite sign from the corresponding force on the ball. Missing a peak or shifting the baseline biases the area.
Conserve the selected system's total momentum
= mAvA + mBvB
This requires zero or negligible net external impulse on the system during the interval. Internal impulses cancel in the total, while each body's momentum can change. If external impulse is significant, include it as the change in total momentum.
Use one signed axis. Momentum alone gives one equation for two unknown final velocities. Sticking supplies vA = vB; other outcomes need their own additional information.
For cart measurements, include moving attachments in each mass, obtain signed velocities immediately before/after the same collision, and keep any gate interval entirely on one side of contact. A gate measures an interval-average speed; direction must be known separately. Tilt, friction and misplaced measurement intervals can undermine the ideal account.
Choose the energy and relative-speed conditions
| Outcome | Additional condition |
|---|---|
| Perfectly elastic | Total kinetic energy is conserved. Relative approach and separation speeds are equal. |
| Inelastic, separating | Kinetic energy is not conserved. The bodies need not have the same final velocity; use supplied information about the outcome. |
| Sticking | Both bodies share one final velocity. This is a perfectly inelastic outcome. |
For the stated arrangement with A left of B, approach speed is uA - uB and separation speed is vB - vA. Equality applies to the perfectly elastic separating case. Check the arrangement and signs when using these expressions.
Total momentum can be conserved while kinetic energy decreases or increases in an interaction. Account for the corresponding changes in other stores and transfers. Total energy does not disappear.
| Quantity | Symbol or expression | Units |
|---|---|---|
| Mass | m | kg |
| Initial; final velocity | u; v | m/s |
| Momentum | p = mv | N s = kg m/s |
| Net impulse | Δp | N s |
| Force; time | F; t | N; s |
| Kinetic energy | Ek = ½mv2 | J |
A force-time area is an impulse in N s; a force-displacement area is work in N m = J. Momentum and kinetic energy have different units and different conservation conditions.
Back to impulse