science-2018:P.F.7
science-2018:P.F.7 Momentum, impulse and conservation of momentum
Learning goal: Define momentum as a vector, p = mv; define impulse as the change in momentum caused by an average net external force acting over a time interval, Δp = F_avg Δt; use conservation of momentum for a closed system to predict the outcome of collisions and separations in one dimension; distinguish elastic, inelastic and totally inelastic collisions by what happens to kinetic energy; and apply impulse to explain why extending a stopping time reduces the force, as in vehicle safety design.
Momentum. The momentum of an object is its mass times its velocity, p = mv, measured in kg·m/s. Momentum is a vector in the same direction as the velocity, so in one dimension it carries a sign: a 1200 kg car moving at +20 m/s has p = +24,000 kg·m/s, and the same car moving at −20 m/s has p = −24,000 kg·m/s. A slow truck and a fast bicycle can have the same momentum. Momentum is what a force changes: Newton's second law can be written as F_net = Δp/Δt, the rate of change of momentum, which for constant mass is the familiar ma.
Impulse. Rearranging gives the impulse-momentum relation: the change in momentum equals the average net external force times the time over which it acts, Δp = p_f − p_i = F_avg Δt. Impulse is the total momentum transferred into or out of a system, and it can only come from a force exerted by something outside the system. For a given change in momentum, a large force for a short time and a small force for a long time deliver the same impulse. A 2000 kg truck moving at 15 m/s has 30,000 kg·m/s of momentum to lose. Braking to a stop over 6.0 s needs an average force of 30,000 ÷ 6.0 = 5000 N. Stopping against a wall in 0.10 s needs 300,000 N. Same impulse, sixty times the force. Cushioning and restraint features such as crumple zones, airbags, seat belts that stretch and padded barriers work largely by extending Δt so that F_avg is smaller (other safety features work differently, for example by preventing the crash or keeping occupants inside the car).
When can a force be ignored? The impulse relation also says when a force does not matter. During a 1.0 ms collision between a bat and a ball, friction and gravity act too, but their impulse over one millisecond is tiny compared with the bat's, so the momentum of the ball just before and just after the hit is governed by the bat alone. A baseball of mass 0.145 kg arriving at 40 m/s and leaving at 50 m/s in the opposite direction has Δp = 0.145 × (−50 − 40) = −13.05 kg·m/s, so the average force over 1.0 ms is 13,000 N; the ball's weight, 1.4 N, changed its momentum by 0.0014 kg·m/s in that time, one part in ten thousand.
Conservation of momentum. Consider a system of objects that interact only with each other, with no net external force; such a system is called closed here. The forces between the objects are third-law pairs, equal and opposite, acting for the same time, so the impulses they give each other cancel and the total momentum of the system cannot change. The total (net) momentum before an interaction equals the total momentum after. This holds for every collision, whatever happens to the energy, as long as external forces are negligible during the interaction: elastic, inelastic and totally inelastic collisions all conserve momentum. It also holds for separations, such as two carts pushed apart by a spring or a person stepping off a boat. Calculations in this course are in one dimension; the same principle applies in two dimensions, treated qualitatively only.
A fully worked model: a totally inelastic collision. A 1200 kg car moving at 20 m/s runs into a stationary 1800 kg truck and the two lock together. Momentum before: 1200 × 20 + 1800 × 0 = 24,000 kg·m/s. After, the combined 3000 kg mass moves at one velocity v: 3000v = 24,000, so v = 8.0 m/s in the car's original direction. Check the momenta after: car 1200 × 8.0 = 9600, truck 1800 × 8.0 = 14,400, total 24,000. Kinetic energy before: ½ × 1200 × 20² = 240,000 J; after: ½ × 3000 × 8.0² = 96,000 J. Momentum was conserved exactly; 144,000 J of kinetic energy became deformation, sound and heat. Total energy is conserved in every collision, but kinetic energy is conserved only in an elastic one.
Kinds of collision. In an elastic collision the objects separate and the total kinetic energy after equals the total before; collisions between hard steel balls or between air-track carts with magnetic bumpers come close. In an inelastic collision some kinetic energy is transformed into other forms; when the objects still separate afterwards it is sometimes called partially inelastic. In a totally inelastic collision, the extreme case of an inelastic collision, the objects stick and move off together, which is the case that loses the most kinetic energy consistent with momentum conservation. Two special results are useful for prediction: in an elastic head-on collision between equal masses, the objects exchange velocities, so a moving ball stopping dead as it hits an identical resting ball, which then moves off at the original speed, is the signature of an elastic collision; and in a totally inelastic collision the final velocity is the total momentum divided by the total mass. Given the initial motions, qualitative predictions follow: a heavy object hitting a light one at rest keeps moving forward in any head-on collision; a light one hitting a heavy one at rest bounces back if the collision is elastic or close to it (if they stick, both move forward slowly); equal masses sticking together, one of them initially at rest, move at half the first one's speed.
Separations. Two carts, 2.0 kg and 3.0 kg, rest with a compressed spring between them. Released, the 2.0 kg cart moves off at 1.5 m/s. Total momentum before was zero, so after it is still zero: 2.0 × 1.5 + 3.0 × v = 0, and v = −1.0 m/s; the heavier cart moves the other way at 1.0 m/s. Momentum in one direction is balanced by momentum in the other; kinetic energy was zero before and positive after, because the spring supplied it.
Design: stop a vehicle and keep the egg whole. A 0.60 kg cart carrying a raw egg on its front travels at 1.5 m/s toward a wall. Its momentum, 0.90 kg·m/s, must go to zero. Stopping in 0.020 s against the bare wall means an average force of 45 N on the cart, and a large share of that lands on the egg. Stopping in 0.20 s, ten times longer, means 4.5 N. A crumple zone of folded card or a spring bumper, a braking system that drags the cart to a stop before the wall, or a padded barrier all lengthen Δt. The design portfolio for such a project records each trial's stopping distance, an estimate of the stopping time from video, the computed average force, what was changed and why. The same physics underlies the redesign of a dangerous stretch of road: a wider curve reduces the centripetal force demanded of the tires, a longer run-off extends stopping time, and a barrier that deforms lowers the peak force on a car that hits it. Road conditions matter through friction: suppose (as example values) tires give μ = 0.70 on dry road, 0.40 wet and 0.10 on ice; from 20 m/s the minimum stopping distances v²/(2μg) are 29 m, 51 m and 204 m. The driver's momentum is the same in all three cases; what changes is the force the road can supply to remove it, and therefore the time and distance needed.
Investigation: momentum before and after. Materials: two low-friction carts on a track (one with a hook-and-loop bumper for sticking, one pair with magnetic bumpers for near-elastic collisions), a balance, two photogates or a phone video with a meter stick beside the track, and slotted masses. Steps: measure each cart's mass; launch cart A toward cart B at rest; record the speed of A just before impact and the speeds of both after, using the photogate timing over a measured flag length or video frames; repeat for the sticking bumpers and the magnetic bumpers, and with 0.50 kg added to B. Record: masses, velocities before and after with signs, momenta before and after, kinetic energies before and after. A result that supports the claim: total momentum after equal to total momentum before, within the uncertainty of the speed and mass measurements (work it out for your equipment), in every trial, with kinetic energy nearly conserved for magnetic bumpers and largely lost for sticking ones. Safety: keep fingers away from the track ends and catch carts before they fall. What the model cannot show: track friction is a small external force, so momentum measured a moment after the collision is slightly less than before, and the gap grows the longer you wait; the closed-system assumption is an approximation whose quality your data reveal.
Worked example 1 — test a collision for kinetic energy. Cart A (2.0 kg) moving at 3.0 m/s hits cart B (1.0 kg) at rest. After the collision A continues at 1.0 m/s in the same direction. Find B's velocity. Momentum before: 2.0 × 3.0 + 0 = 6.0 kg·m/s. After: 2.0 × 1.0 + 1.0 × v = 6.0, so v = 4.0 m/s. Kinetic energy before: ½ × 2.0 × 9.0 = 9.0 J; after: ½ × 2.0 × 1.0 + ½ × 1.0 × 16 = 1.0 + 8.0 = 9.0 J. This particular collision happens to be elastic; had A continued at 1.5 m/s, B would move at 3.0 m/s and the kinetic energy after would be 2.25 + 4.5 = 6.75 J, an inelastic case.
Worked example 2 — elastic collision between equal masses. A 0.20 kg steel ball moving at 2.0 m/s strikes an identical ball at rest head-on. Momentum before: 0.40 kg·m/s; kinetic energy before: ½ × 0.20 × 4.0 = 0.40 J. If the first ball stops and the second moves off at 2.0 m/s, momentum after is 0.40 kg·m/s and kinetic energy after is 0.40 J: both conserved. This exchange of velocities is the only outcome that conserves both for equal masses in a head-on collision, and it is what a desk-toy of swinging balls shows.
Worked example 3 — impulse on a bouncing ball. A 0.40 kg ball hits a floor at 6.0 m/s downward and rebounds at 4.0 m/s upward, in contact for 0.050 s. Take up as positive: Δp = 0.40 × (4.0 − (−6.0)) = 4.0 kg·m/s upward. Average net force: 4.0 ÷ 0.050 = 80 N upward. The floor must supply that plus the ball's 3.9 N weight, about 84 N upward; because 3.9 N is small beside 80 N, gravity is often ignored during a bounce. A ball that stopped dead instead of rebounding would have Δp = 2.4 kg·m/s, a net force of 48 N and a floor force of about 52 N; bouncing back requires more impulse than stopping.
Worked example 4 — a person and a boat. A 60 kg person standing in a 90 kg boat at rest steps off toward the dock at 2.0 m/s relative to the water. Total momentum before is zero, so 60 × 2.0 + 90 × v = 0 and v = −1.3 m/s: the boat moves away from the dock at 1.3 m/s. The person's push on the boat and the boat's push on the person are a third-law pair, and the impulses cancel for the system.
Representation task: a momentum ledger. For the car-and-truck collision in the lesson (1200 kg at 20 m/s into 1800 kg at rest, locking together), make a table with rows for the car, the truck and the total, and columns for mass, velocity before, momentum before, velocity after, momentum after, kinetic energy before and kinetic energy after. Fill it in (the momenta before are 24,000 and 0; after, 9600 and 14,400; kinetic energies 240,000 J and 96,000 J), then draw a bar chart of the four momenta and write two sentences: one stating what stayed the same and why, and one stating what changed and where it went.
Independent performance: design, test and document an egg-safe stop. Build a device that lets a rolling cart carrying a raw egg on its front stop before a wall without breaking the egg. Keep a design portfolio. Success criteria: the cart's momentum before stopping is measured (mass and speed) for every trial; the stopping time is estimated from video and the average force computed with Δp = F_avg Δt; each revision names the change made, the physics reason (longer Δt, lower peak force, or removing momentum earlier through braking), and the result; the final report discusses one advantage and one disadvantage of the braking approach versus the crumple-zone approach; and the portfolio includes a short note on how the same principle applies to a real road feature, such as a deforming barrier or a run-off area.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
A 0.80 kg skateboard rolls at 12 m/s. Find its momentum, and the momentum of a 60 kg rider on it at the same speed. What is the momentum of the pair?
Skateboard: p = 0.80 × 12 = 9.6 kg·m/s. Rider: 60 × 12 = 720 kg·m/s. Pair: 730 kg·m/s (729.6), in the direction of motion. Momentum is a vector; if the rider and board were moving in opposite directions the momenta would subtract.
A 1500 kg car moving at 20 m/s collides with a 1000 kg car at rest and the two stick together. Find their common velocity and the fraction of kinetic energy lost.
Momentum before: 1500 × 20 = 30,000 kg·m/s. After: 2500v = 30,000, so v = 12 m/s. Kinetic energy before: ½ × 1500 × 400 = 300,000 J; after: ½ × 2500 × 144 = 180,000 J. Lost: 120,000 J, which is 40 percent, transformed into deformation, sound and heat. Momentum is fully conserved; kinetic energy is not, because the collision is totally inelastic.
A 0.15 kg ball moving at 25 m/s is caught by a glove and stopped in 0.020 s. Find the impulse and the average force. How does a padded glove change these?
Impulse: Δp = 0.15 × (0 − 25) = −3.75 kg·m/s, opposite to the ball's motion. Average force: 3.75 ÷ 0.020 = 190 N. A padded glove makes the ball stop over a longer time, say 0.060 s; the impulse is unchanged at 3.75 kg·m/s because the ball still goes from 25 m/s to rest, but the average force falls to 63 N.
Cart A (1.5 kg) at 2.0 m/s hits cart B (0.50 kg) at rest. After the collision A moves at 1.0 m/s in its original direction. Find B's velocity and decide whether the collision was elastic.
Momentum before: 3.0 kg·m/s. After: 1.5 × 1.0 + 0.50 × v = 3.0, so v = 3.0 m/s. Kinetic energy before: ½ × 1.5 × 4.0 = 3.0 J. After: ½ × 1.5 × 1.0 + ½ × 0.50 × 9.0 = 0.75 + 2.25 = 3.0 J. Kinetic energy is conserved, so the collision was elastic. Momentum was conserved regardless; only the energy check distinguishes the kind.
Explain, using Δp = F_avg Δt, why a crumple zone reduces injury in a crash even though the car still comes to a complete stop.
The change in momentum of the car and its occupants is fixed by the mass and speed: from mv to zero. A crumple zone deforms during the crash, so the car stops over a longer distance and time. With Δp fixed and Δt larger, the average force F_avg = Δp/Δt on the car, and on the occupants held by their belts, is smaller. The same impulse is delivered by a smaller force over more time.
Two skaters at rest push apart. The 40 kg skater moves off at 1.5 m/s. Find the velocity of the 60 kg skater and explain why the total momentum is still zero.
Before the push the total momentum is zero. After, 40 × 1.5 + 60 × v = 0, so v = −1.0 m/s: the heavier skater moves at 1.0 m/s in the opposite direction. The two skaters exert equal and opposite forces on each other for the same time (a third-law pair), so the impulses are equal and opposite and the system's total momentum cannot change. Kinetic energy rose from zero, supplied by the skaters' muscles.
In the cart investigation a group finds total momentum of 1.20 kg·m/s before a sticking collision and 1.12 kg·m/s after. Is the principle of momentum conservation violated? Explain, and say what further measurements would help.
Not on this evidence. Conservation applies to a closed system with no net external force, and the gap is 0.08 kg·m/s, about 7 percent of the starting value. Several explanations are possible and the data given cannot choose between them: track friction acting on the carts between the impact and the point where the after-collision speed was taken; uncertainty in the timing or in the flag length used to get the speeds; and uncertainty in the masses. Further measurements would separate them: state the uncertainty of each momentum and see whether 0.08 kg·m/s lies inside it; measure the speed closer to the moment of impact, or on a lower-friction track, and see whether the gap shrinks (as it should if friction is the cause); and re-weigh the carts. The result tests the closed-system assumption and the measurements, not the principle itself.