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Think First

Before you begin this lesson, take a moment to think about what you already know about this topic. Jot down your ideas — you will revisit them at the end.

Understand the core concepts covered in this lesson.

2

Apply your knowledge to solve problems and explain phenomena.

3

Evaluate and analyse scientific information and data.

Year 11 Physics Module 2: Dynamics 50 min ★ Consolidation Lesson 14 of 15

Phase 3 Consolidation

No new content. Five formulae. Six common errors. Ten mixed practice questions. Three timed extended responses. This is your Phase 3 exam preparation.

📐

Key Relationships — Phase 3

p = mv
p = momentum (kg m/s) | vector — always define positive direction
J = FΔt = Δp
J = impulse (N s) | F = average force (N) | Δt = contact time (s)
Σp_before = Σp_after
Conservation of momentum — applies when no net external force acts on the system
KE = ½mv²
Compare KE before and after to classify collision type (elastic / inelastic / perfectly inelastic)
W_net = ΔKE
Sliding stage after collision: friction work = KE lost → d = KE/(μmg)

Misconceptions to Fix

✗

Wrong: Action and reaction forces cancel each other out on the same object.

✗

Right: Action-reaction pairs act on DIFFERENT objects, so they never cancel; this is why motion occurs.

📐 Phase 3 Formula Sprint

How to use this: For each card, try to recall the use condition, the trap, and the connection — before clicking to reveal. If you cannot recall all three without looking, that formula needs more work before the quiz.
p = mv
When do you use it? What is the trap? What does it connect to?
Use whenCalculating the quantity of motion of any object. Momentum is a vector — always in the direction of velocity.
TrapDefine positive direction before substituting. Objects moving opposite to positive have negative momentum. Missing this sign convention is the single most common Phase 3 error.
Connects toMultiply by Δt → J = FΔt. Use in Σp_before = Σp_after for all collision types.
J = FΔt = Δp
When do you use it? What is the trap? What does it connect to?
Use whenA force acts over a time interval and you need the total momentum transferred — or when force and time are given and you need the impulse.
TrapF must be the net or average force during contact — not just the applied force. For a bouncing object, Δp = m(v_f − v_i) using signed velocities — direction reversal dramatically increases |Δp|.
Connects toDivide by Δt → F = Δp/Δt. Area under F-t graph = impulse. Safety applications: extend Δt to reduce F for same Δp.
F = Δp / Δt
When do you use it? What is the trap? What does it connect to?
Use whenFinding the average force during a collision or impact, given the momentum change and contact time.
TrapConvert contact time to seconds before substituting. Collision times are often given in milliseconds (1 ms = 0.001 s). Dividing by ms instead of s gives an answer 1000× too small.
Connects toSafety devices reduce F by extending Δt. Newton's Third Law: force on ball = −force on bat (same Δt, opposite direction).
Σp_before = Σp_after
When do you use it? What is the trap? What does it connect to?
Use whenAny collision or explosion in a closed system (no net external force). Covers perfectly inelastic, elastic, inelastic, and explosion from rest.
TrapObjects moving in opposite directions get opposite signs. Friction or gravity acting during the collision can violate conservation — for short collisions, these external impulses are negligible and conservation is a valid approximation.
Connects toPerfectly inelastic: (m₁+m₂)v_f = Σp. Explosion from rest: m₁v₁' = −m₂v₂'. v_f feeds Phase 2 bridge → W_net = ΔKE for sliding.
KE = ½mv² → Phase 2 bridge
When do you use it? What is the trap? What does it connect to?
Use whenClassifying collision type (compare KE before and after) or bridging to Phase 2 after finding v_f (post-collision sliding, climbing, or stopping distance).
TrapKE ∝ v² — doubling speed quadruples KE. After a perfectly inelastic collision, calculate KE_after using v_f (post-collision speed), not the original speed of the moving object. KE can never increase in a collision.
Connects toElastic: KE_before = KE_after. Inelastic: KE_after < KE_before. Post-collision: KE_f = ½(m₁+m₂)v_f² → feeds W_net = ΔKE for sliding stage.

🔬 Common Error Clinic

Instructions: Read the student's working. Find the error — write it in your own words before clicking "Show fix". Then compare with the correct explanation.
1
No positive direction defined — bouncing ball
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Student's working A 0.16 kg ball travels at 40 m/s and bounces off a wall at 35 m/s in the opposite direction.
"Δp = m(v_f − v_i) = 0.16 × (35 − 40) = 0.16 × (−5) = −0.8 N s."
Write your answer in your book
Correct explanation The student subtracted speeds (35 − 40 = −5 m/s) instead of using signed velocities. Without defining a positive direction, the direction reversal cannot be captured. If we define toward the wall as positive: v_i = +40 m/s, v_f = −35 m/s (opposite direction). Δv = −35 − 40 = −75 m/s. Δp = 0.16 × (−75) = −12 N s. The correct answer is 15 times larger than the student's answer — a major error that would give an incorrect force in any subsequent calculation.
2
Contact time not converted from milliseconds
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Student's working A ball undergoes a momentum change of 12 N s during a collision lasting 2 ms.
"F = Δp/Δt = 12/2 = 6 N."
Write your answer in your book
Correct explanation The contact time 2 ms = 2 × 10⁻³ s = 0.002 s. The student divided by 2 (milliseconds) instead of 0.002 (seconds), giving an answer 1000× too small. Always convert contact time to SI units (seconds) before substituting. Correct: F = 12/0.002 = 6000 N. The correct force is typical of a high-speed collision — the student's answer of 6 N would be far less than a person's weight and physically implausible for a ball collision.
3
Speed difference used instead of signed Δv
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Student's working A 0.3 kg ball hits a wall at 40 m/s and bounces back at 35 m/s.
"Δv = v_f − v_i = 35 − 40 = −5 m/s. Δp = 0.3 × (−5) = −1.5 N s."
Write your answer in your book
Correct explanation The student used the speeds 35 and 40 in the formula without assigning signs based on direction. When a ball bounces back, the final velocity is in the opposite direction to the initial velocity — it must have the opposite sign. Defining toward wall = positive: v_i = +40 m/s, v_f = −35 m/s. Δv = −35 − (+40) = −75 m/s. Δp = 0.3 × (−75) = −22.5 N s. The correct answer is 15× larger. This is errors 1 and 3 appearing in the same calculation — both stem from failing to define and apply a positive direction.
4
Applying conservation when external forces act
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Student's working A football is kicked and rolls across a grass field. The student uses Σp_before = Σp_after to find the ball's velocity after 5 seconds, with friction = 20 N.
"Momentum is conserved, so the ball still has the same momentum after 5 seconds."
Write your answer in your book
Correct explanation Conservation of momentum applies only to closed systems with no net external force. Friction is an external force — it transfers momentum from the ball to the ground. Over 5 seconds, friction does significant work on the ball, removing momentum continuously. The correct approach for this problem is either Newton's Second Law (F_net = ma → a = −F_friction/m) or the work-energy theorem (W_net = ΔKE). Conservation of momentum cannot be applied here. The approximation that external forces are negligible is only valid for very short collision durations — not for 5 seconds of rolling.
5
Using initial velocity for post-collision KE
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Student's working A 3 kg ball at 8 m/s collides with a stationary 5 kg block. They stick together.
"KE_after = ½ × (3 + 5) × 8² = ½ × 8 × 64 = 256 J."
Write your answer in your book
Correct explanation After the perfectly inelastic collision, the combined mass does not move at 8 m/s — it moves at v_f, which must be found using conservation. v_f = (3 × 8)/(3 + 5) = 24/8 = 3 m/s. KE_after = ½ × 8 × 3² = ½ × 8 × 9 = 36 J. The student's answer (256 J) is greater than the initial KE (½ × 3 × 64 = 96 J) — this is physically impossible. KE can never increase in a collision. This is a key self-check: if your calculated KE_after exceeds KE_before, you have made an error.
6
Explosion: equal speeds instead of equal momenta
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Student's working A 4 kg rifle fires a 0.008 kg bullet. Both start at rest.
"After firing, momentum is equal and opposite, so the rifle and bullet move at the same speed in opposite directions."
Write your answer in your book
Correct explanation Equal and opposite momenta means m₁v₁' = −m₂v₂' — the momenta are equal in magnitude, not the speeds. Since the rifle is 500× heavier than the bullet (4 kg vs 0.008 kg), it moves 500× slower. If the bullet leaves at 300 m/s, the rifle recoils at 0.008 × 300 / 4 = 0.6 m/s — not 300 m/s. Equal momenta with very different masses always produces very different speeds. The lighter object gets the higher speed; the heavier object barely moves.

🔢 Mixed Practice

Instructions: No formula sheet. Define positive direction before any calculation. Show all working. Band 6 questions require both calculation and explanation.
ApplyMomentumBand 31 MARK

1. Calculate the momentum of a 1500 kg car travelling at 28 m/s north. State the direction.

Answer in your book
Saved
ApplyImpulseBand 33 MARKS

2. A 0.4 kg ball hits a wall at 12 m/s and bounces straight back at 10 m/s. Contact time = 0.008 s. Apply the Vector Protocol. Find: (a) the change in momentum, (b) the impulse, (c) the average force on the ball.

Answer in your book — apply Vector Protocol
Saved
UnderstandConservationBand 32 MARKS

3. State the law of conservation of momentum. Identify two conditions required for it to apply and give one example where it does not apply.

Answer in your book
Saved
ApplyPerfectly InelasticBand 4/53 MARKS

4. A 2 kg ball moving at +6 m/s collides with a stationary 4 kg block. They stick together. (a) Find v_f. (b) Calculate the KE before and after. (c) Calculate the KE lost.

Answer in your book
Saved
ApplyExplosionBand 4/53 MARKS

5. A 5 kg rifle fires a 0.01 kg bullet. Both start at rest. The bullet leaves at +450 m/s. (a) Find the rifle's recoil velocity. (b) Calculate the KE of each object after firing. (c) Where did this kinetic energy come from?

Answer in your book
Saved
AnalyseClassifyBand 4/54 MARKS

6. A 0.25 kg ball at +8 m/s collides with a 0.75 kg ball at −4 m/s. After the collision, the 0.25 kg ball moves at −6 m/s. (a) Apply the Vector Protocol to find the velocity of the 0.75 kg ball. (b) Calculate KE before and after. (c) Classify the collision with justification.

Answer in your book — full Vector Protocol
Saved
ApplyImpulse — SafetyBand 4/53 MARKS

7. A 60 kg person falls 4 m and lands on a crash mat, stopping in 0.4 s. (a) Find their speed just before impact using energy conservation. (b) Find the average force on the mat. (c) If they landed on concrete instead (stops in 0.005 s), find the force and explain the safety implication using J = FΔt = Δp.

Answer in your book
Saved
EvaluateFull SynthesisBand 64 MARKS

8. A 3 kg ball at +10 m/s collides with a stationary 5 kg ball. After the collision the 3 kg ball is at rest. (a) Find the velocity of the 5 kg ball. (b) Calculate KE before and after. (c) Classify the collision. (d) Is this outcome physically possible? Justify using both conservation laws.

Answer in your book — all four parts
Saved
EvaluateTwo-StageBand 64 MARKS

9. A 1000 kg car at +20 m/s rear-ends a stationary 1500 kg van. They lock together and skid 8 m to rest. Using the Phase 3 → Phase 2 chain: (a) find v_f just after the collision, (b) find μk from the skid distance using W_net = ΔKE. Show all stages clearly.

Answer in your book — label both stages
Saved
EvaluateDerivationBand 64 MARKS

10. A mass m₁ moves at velocity v and collides perfectly inelastically with a stationary mass m₂. Derive an algebraic expression for the fraction of kinetic energy lost. Show that it simplifies to m₂/(m₁ + m₂) and explain what this tells you about the mass ratio and energy loss.

Answer in your book — show full algebraic derivation
Saved

⏱️ Timed Exam Block

Instructions: Work under exam conditions — no notes. Set the timer for each question. Write full working in your book. Check answers only after time is called.
Exam Timer
08:00
ApplyBand 5

Q11 4 MARKS

Suggested time: 8 minutes | Concepts: J = FΔt = Δp, F = Δp/Δt, Newton's Third Law

A 0.16 kg cricket ball bowled at 38 m/s is struck by a bat and returns at 42 m/s in the opposite direction. Contact time = 0.0018 s.

  1. Define a positive direction and calculate the change in momentum of the ball.
  2. Calculate the impulse delivered by the bat to the ball.
  3. Calculate the average force exerted by the bat on the ball during contact.
  4. Using Newton's Third Law, describe the force that acts on the bat during the same contact time. Calculate its magnitude and state its direction.

Attempt under timed conditions first — then type your working.

Answer in your book under timed conditions
Saved
EvaluateBand 6

Q12 4 MARKS

Suggested time: 8 minutes | Concepts: perfectly inelastic → skid → impulse on passenger

A 600 kg car travelling at +18 m/s collides with a stationary 900 kg truck. They stick together. μk = 0.35. A 70 kg passenger is wearing a seatbelt.

  1. Find the velocity of the combined vehicles just after the collision.
  2. Find the kinetic energy lost in the collision.
  3. Using W_net = ΔKE, find the skid distance.
  4. The passenger goes from 7.2 m/s to rest. Calculate the force on them with seatbelt (Δt = 0.25 s) vs without (Δt = 0.05 s). Explain the safety significance.
Answer in your book under timed conditions
Saved
EvaluateBand 6

Q13 4 MARKS

Suggested time: 8 minutes | Concepts: general collision → classify → elastic case comparison

A 2 kg ball moving at +5 m/s collides with a stationary 3 kg ball. After the collision, the 2 kg ball moves at +1 m/s.

  1. Apply the Vector Protocol and find the velocity of the 3 kg ball after the collision.
  2. Calculate the total KE before and after the collision.
  3. Is mechanical energy conserved? Classify the collision type with full justification.
  4. If this collision were perfectly elastic instead, use conservation of both momentum and KE to find what the velocities would be. (You may use the standard result for elastic collisions between unequal masses, or solve the two simultaneous equations.)
Answer in your book under timed conditions
Saved

Comprehensive Answers

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Mixed Practice Q1–Q7

Q1: p = 1500 × 28 = 42 000 kg m/s north.

Q2: Positive = toward wall. (a) Δp = 0.4 × (−10 − 12) = 0.4 × (−22) = −8.8 N s. (b) J = −8.8 N s. (c) F = −8.8/0.008 = −1100 N (away from wall). The negative sign confirms the wall pushes the ball back.

Q3: Law — in a closed system with no net external force, the total momentum before any interaction equals the total momentum after. Conditions: (1) no net external force on the system; (2) the system is closed — no mass enters or leaves. Counter-example: a ball rolling to a stop due to friction — friction is an external force, so momentum decreases.

Q4: (a) (2+4)v_f = 12 → v_f = +2 m/s. (b) KE_i = ½×2×36 = 36 J. KE_f = ½×6×4 = 12 J. (c) KE lost = 24 J (67% of initial).

Q5: (a) 0 = 0.01×450 + 5×v_rifle → v_rifle = −4.5/5 = −0.9 m/s. (b) KE_bullet = ½×0.01×202500 = 1012.5 J. KE_rifle = ½×5×0.81 = 2.025 J. Total KE = 1014.5 J. (c) The kinetic energy came from the chemical energy stored in the gunpowder (explosive). Before firing, KE = 0; after, KE ≈ 1015 J — all converted from chemical potential energy.

Q6: Positive = initial direction of 0.25 kg ball. (a) 0.25×8 + 0.75×(−4) = 0.25×(−6) + 0.75×v₂'. 2 − 3 = −1.5 + 0.75v₂'. −1 + 1.5 = 0.75v₂'. v₂' = 0.5/0.75 = +0.667 m/s. (b) KE_before = ½×0.25×64 + ½×0.75×16 = 8 + 6 = 14 J. KE_after = ½×0.25×36 + ½×0.75×0.444 = 4.5 + 0.167 = 4.67 J. (c) KE_after < KE_before → inelastic. Objects separate → not perfectly inelastic. Momentum check: p_before = 2 − 3 = −1 N s. p_after = 0.25×(−6) + 0.75×0.667 = −1.5 + 0.5 = −1 N s. ✓

Q7: (a) v = √(2×9.8×4) = √78.4 = 8.85 m/s. (b) Δp = 60×8.85 = 531 N s. F_mat = 531/0.4 = 1327.5 N. (c) F_concrete = 531/0.005 = 106 200 N. The concrete force is 80× larger — 106 kN vs 1.3 kN. This is the difference between a survivable fall and a fatal one. The crash mat extends Δt by a factor of 80, reducing the peak force by the same factor for the same change in momentum.

Mixed Practice Q8–Q10

Q8: (a) 3×10 = 5×v₂' → v₂' = +6 m/s. (b) KE_before = ½×3×100 = 150 J. KE_after = ½×5×36 = 90 J. (c) KE_after < KE_before → inelastic. Objects separate → not perfectly inelastic. (d) Physically possible? p_before = 30 N s. p_after = 3×0 + 5×6 = 30 N s. Momentum conserved ✓. KE_after < KE_before ✓. Yes — this outcome is physically possible and is consistent with an inelastic collision.

Q9: Stage 1: v_f = 1000×20/2500 = +8 m/s. Stage 2: KE_f = ½×2500×64 = 80 000 J. f_k × 8 = 80 000 → f_k = 10 000 N. μk = 10 000/(2500×9.8) = 10 000/24 500 = 0.408.

Q10: v_f = m₁v/(m₁+m₂). KE_before = ½m₁v². KE_after = ½(m₁+m₂)×[m₁v/(m₁+m₂)]² = m₁²v²/[2(m₁+m₂)]. KE_lost = ½m₁v² − m₁²v²/[2(m₁+m₂)] = (m₁v²/2)[1 − m₁/(m₁+m₂)] = (m₁v²/2)[m₂/(m₁+m₂)]. Fraction = KE_lost/KE_before = m₂/(m₁+m₂). Interpretation: when m₂ ≫ m₁ (hitting a very large stationary object), fraction → 1 (almost all KE lost). When m₂ ≪ m₁ (hitting a very small object), fraction → 0 (little KE lost). This is why a car hitting a concrete wall loses far more KE than a car hitting a small cardboard box at the same speed.

Timed Block Q11–Q13

Q11:

(a) Positive = toward bat. v_i = +38, v_f = -42. delta p = 0.16 x (-42 - 38) = 0.16 x (-80) = -12.8 N s
(b) J = delta p = -12.8 N s (impulse directed away from bat)
(c) F = J/dt = -12.8 / 0.0018 = -7111 N (approx -7100 N)

(d) Newton's Third Law: while the bat exerts −7100 N on the ball (away from bat), the ball exerts +7100 N on the bat (toward bat, opposing bat's swing). Same magnitude, same contact time, opposite direction. This is why cricket bats have thick grips and players wear gloves — the reaction force on the bat is identical in magnitude to the force on the ball.

Q12:

(a) p_before = 600x18 = 10 800 N s. 1500 x v_f = 10 800. v_f = +7.2 m/s
(b) KE_i = 1/2x600x324 = 97 200 J. KE_f = 1/2x1500x51.84 = 38 880 J. KE_lost = 58 320 J
(c) f_k = 0.35x1500x9.8 = 5145 N. -5145xd = -38 880. d = 7.55 m

(d) Δp_passenger = 70×(0−7.2) = −504 N s. With seatbelt: F = 504/0.25 = 2016 N ≈ 2.0 kN. Without seatbelt: F = 504/0.05 = 10 080 N ≈ 10 kN. The seatbelt reduces force by 5× by extending the stopping time from 0.05 s to 0.25 s. Without the seatbelt, the passenger continues at 7.2 m/s and hits the dashboard — which stops them in ~50 ms. The 10 kN force is equivalent to a tonne of weight pressing on the occupant, causing severe injury. The seatbelt distributes this over 5× longer, reducing peak force to survivable levels.

Q13:

(a) Positive = initial direction of 2 kg ball. 2x5 + 0 = 2x1 + 3xv2'. 10 = 2 + 3v2'. v2' = +8/3 = +2.67 m/s
(b) KE_before = 1/2x2x25 = 25 J. KE_after = 1/2x2x1 + 1/2x3x7.11 = 1 + 10.67 = 11.67 J
(c) KE not conserved (25J vs 11.67J, loss = 13.33J). Objects separated -> not perfectly inelastic. Classification: INELASTIC. Momentum check: p_before = 10, p_after = 2+8 = 10 N s. Conserved.
(d) Elastic case: v1' = (m1-m2)v/(m1+m2) = (2-3)x5/5 = -1 m/s (2 kg ball BOUNCES BACK). v2' = 2m1v/(m1+m2) = 2x2x5/5 = +4 m/s. KE check: 1/2x2x1 + 1/2x3x16 = 1+24 = 25 J = KE_before. Conserved.

The elastic case is dramatically different — the lighter ball bounces backward at 1 m/s instead of continuing forward at 1 m/s. In the real (inelastic) collision, 53% of the kinetic energy is lost. In the elastic case, 0% is lost. Real collisions between soft objects are always inelastic; elastic collisions are approximations valid for very hard objects (billiard balls, atomic nuclei).

Interactive: Forces-Energy-Momentum Integrator
Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

Key Terms
ForceA push or pull acting on an object that can cause it to accelerate.
NewtonThe SI unit of force; 1 N = 1 kg·m/s².
WeightThe force due to gravity acting on a mass; W = mg.
Normal ForceThe perpendicular contact force exerted by a surface on an object.
FrictionA force that opposes relative motion between two surfaces in contact.
Net ForceThe vector sum of all forces acting on an object.
MC

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Consolidation Game

Phase 3 Consolidation

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