Physics • Year 11 • Module 2: Dynamics • Lesson 14
Phase 3 Consolidation
Lock in the five key formulas, the vector sign convention, and the six common errors before attempting harder problems.
1. Term–definition match
The definitions below are shuffled. In the right-hand column write the matching term from this list: momentum, impulse, conservation of momentum, elastic collision, perfectly inelastic collision, inelastic collision, impulse-momentum theorem, contact time, closed system, kinetic energy. 10 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | The product of an object's mass and velocity; a vector quantity measured in kg m/s. | |
| 1.2 | The product of a net force and the time interval over which it acts; equals the change in momentum. | |
| 1.3 | The principle that the total momentum of a system remains constant when no net external force acts on it. | |
| 1.4 | A collision in which both momentum and kinetic energy are conserved; objects do not stick together. | |
| 1.5 | A collision in which the colliding objects stick together and move as one combined mass afterwards. | |
| 1.6 | A collision in which momentum is conserved but kinetic energy is not; objects separate after the collision. | |
| 1.7 | The relationship J = F Δt = Δp, which links force, time, and change in momentum. | |
| 1.8 | The duration of a collision or impact; extending this reduces the average force for the same momentum change. | |
| 1.9 | A system in which no mass enters or leaves and no net external force acts; the condition for conservation of momentum. | |
| 1.10 | The energy associated with motion, calculated as ½mv²; used to classify collision types. |
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)
2.1 Momentum is a scalar quantity because it depends only on speed, not direction. T / F
2.2 When a ball bounces off a wall and reverses direction, the magnitude of its momentum change is larger than if the ball had simply stopped. T / F
2.3 In a perfectly inelastic collision, both momentum and kinetic energy are conserved. T / F
2.4 Extending the contact time of a collision reduces the average force on an object for the same change in momentum. T / F
2.5 Conservation of momentum can always be applied to any situation, regardless of whether external forces are present. T / F
2.6 In an explosion from rest, the two fragments move in opposite directions with equal and opposite momenta. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 8 marks (1 per blank)
Word bank:
closed · conserved · contact time · direction · force · kinetic energy · momentum · velocity
Momentum is defined as the product of mass and ___________, making it a vector quantity that depends on both speed and ___________. The impulse-momentum theorem states that a net force acting over a time interval changes the ___________ of an object by the same amount. To reduce the peak ___________ during a collision, safety devices such as seatbelts and crash mats extend the ___________. Momentum is ___________ in any ___________ system where no net external force acts. To classify a collision as elastic or inelastic, compare the total ___________ before and after the collision.
4. Formula function recall
Answer each question in 1–2 sentences using precise terms from the lesson. 8 marks (2 each)
4.1 What does the formula p = mv tell you, and why must you always define a positive direction before using it?
4.2 Explain what the “Phase 2 bridge” means: how does the result from a perfectly inelastic collision feed into a work–energy calculation?
4.3 Why does a bouncing object experience a larger impulse than one that simply comes to rest, even if both have the same initial speed?
4.4 State the condition under which Σpbefore = Σpafter is valid. Give one example where it cannot be applied.
5. Build a concept map
Draw labelled arrows between the six terms below to show how they connect. Each arrow must carry a linking phrase (e.g. “equals”, “reduces”, “is used to classify”). Aim for at least 6 labelled arrows. 6 marks (1 per valid labelled arrow)
Supplied terms: momentum · impulse · contact time · force · kinetic energy · collision type.
Q1 — Term–definition match
1.1 momentum • 1.2 impulse • 1.3 conservation of momentum • 1.4 elastic collision • 1.5 perfectly inelastic collision • 1.6 inelastic collision • 1.7 impulse-momentum theorem • 1.8 contact time • 1.9 closed system • 1.10 kinetic energy.
Q2 — True / false with correction
2.1 False. Momentum is a vector quantity — it depends on both speed (magnitude) and direction. Scalar quantities have magnitude only (e.g. mass, speed).
2.2 True. When the ball bounces back, the velocity changes sign (direction reverses), giving Δv = vf − vi where vf and vi have opposite signs. The magnitude of Δp is therefore greater than if the ball simply stopped (where vf = 0).
2.3 False. In a perfectly inelastic collision, momentum is conserved but kinetic energy is not. Kinetic energy is lost (converted to thermal energy, sound, deformation). Only in an elastic collision are both conserved.
2.4 True. J = FΔt = Δp. For the same Δp, increasing Δt means F must decrease. This is the principle behind safety devices such as seatbelts, airbags, and crash mats.
2.5 False. Conservation of momentum applies only when there is no net external force on the system (i.e. in a closed system). A ball slowing due to friction has an external force acting on it — momentum is not conserved.
2.6 True. From rest, total momentum = 0. After the explosion, m1v1 + m2v2 = 0, so m1v1 = −m2v2. The momenta are equal in magnitude and opposite in direction.
Q3 — Cloze paragraph
In order: velocity / direction / momentum / force / contact time / conserved / closed / kinetic energy.
Q4.1 — p = mv and positive direction
p = mv gives the momentum of an object: the product of its mass (kg) and velocity (m/s). A positive direction must be defined because momentum is a vector — objects moving in opposite directions must be given opposite signs. Without a sign convention, the vector nature of momentum is lost and calculations for collisions or bouncing objects will be wrong.
Q4.2 — Phase 2 bridge
After a perfectly inelastic collision, the post-collision velocity vf is found using conservation of momentum. This vf is then used to calculate the kinetic energy just after the collision: KE = ½(m1+m2)vf². The work-energy theorem (Wnet = ΔKE) then gives the skid distance or height climbed in the post-collision sliding stage — bridging Phase 3 (momentum) into Phase 2 (work-energy).
Q4.3 — Bouncing vs stopping
For an object that stops: Δp = m(0 − vi) = −mvi. For one that bounces back at the same speed: Δp = m(−vi − vi) = −2mvi. The bouncing object undergoes a direction reversal, doubling the magnitude of the change in momentum. Because J = Δp, the impulse — and therefore the force (for the same contact time) — is also doubled.
Q4.4 — Condition for conservation of momentum
Conservation of momentum requires a closed system: no net external force acts on the system and no mass enters or leaves. Example where it cannot be applied: a ball decelerating due to friction while rolling across a floor — friction is an external force that continuously removes momentum from the ball, so Σpbefore ≠ Σpafter.
Q5 — Sample concept map
Correct maps should include arrows such as:
- impulse — equals change in → momentum
- impulse — equals → force × contact time
- contact time — increasing it reduces → force
- kinetic energy — is compared before and after to determine → collision type
- momentum — is conserved in a → (closed system) collision, affecting collision type classification
- force — acting over time produces → impulse
Award 1 mark per valid labelled arrow (minimum 6, maximum 6 marked).