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Year 11 Physics Module 2: Dynamics 45 min Synthesis Lesson 13 of 15

Momentum Synthesis

One bullet. One rifle. One block. One rough surface. Every Phase 3 formula in a single chain — and a connection back to Phase 2 at the end. This lesson maps how it all fits together.

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Think First

A 0.008 kg bullet is fired from a 4 kg rifle into a 2 kg stationary wooden block. The bullet embeds in the block. The block then slides 0.8 m along a rough surface before stopping.

Without calculating anything yet — how many different Phase 3 concepts can you spot in this scenario? List every momentum or energy concept that applies, and at which moment.

Type your list below — be as specific as you can about which concept applies at which moment.

Write your list in your book before reading on.

Write your list in your book

Saved

You will map this fully in Card 1.

📐

Key Relationships — This Lesson

$p = mv$

p = momentum (kg m/s) | vector — define positive direction first

$\Sigma p_{before} = \Sigma p_{after}$

Conservation of momentum — applies in each collision or explosion stage

$J = F\Delta t = \Delta p$

Impulse-momentum theorem — links force × time to change in momentum

$W_{net} = \Delta KE$

Work-energy theorem — links the sliding stage to friction force × distance

Multi-stage chain: Stage 1 explosion → Stage 2 collision → Stage 3 sliding stop ($W_f = \Delta KE \;\Rightarrow\; d = KE/\mu mg$)

Misconceptions to Fix

✗

Wrong: Heavier objects fall faster than lighter ones.

✗

Right: In a vacuum, all objects fall at the same rate regardless of mass; air resistance causes differences in real situations.

📚 Core Content

Key Terms

bulletfired from a 4 kg rifle into a 2 kg stationary wooden block

massthe most common error in multi-step momentum problems

block scenariothree completely separate stages

output of each stagethe input for the next

which stage type youin — explosion, collision, or post-collision motion — before writing any equation

Energy and momentumnot separate topics — they are two accounting systems for the same physical events

01Card 1 — Map the Momentum

Card 1 — Map the Momentum

Before calculating anything, label every object and every velocity — before and after each interaction. Missing a velocity or a mass is the most common error in multi-step momentum problems.

The bullet-block scenario is three completely separate stages. Each stage uses a different formula. The output of each stage is the input for the next.

Stage 1

Explosion from rest

0 = m₁v₁' + m₂v₂'

Rifle + bullet at rest. Bullet fires forward. Rifle recoils backward. Total momentum remains zero.

Stage 2

Perfectly inelastic

(m₁+m₂)v_f = m₁v₁

Bullet embeds in block. They move together. Bullet's velocity from Stage 1 is the input here.

Stage 3

Friction stops block

W_net = ΔKE

Block+bullet slide to rest. Phase 2 work-energy theorem. Block's velocity from Stage 2 is the input here.

Key skill Identifying which stage type you are in — explosion, collision, or post-collision motion — before writing any equation. Students who write the wrong equation for the wrong stage lose all marks for that stage even if their arithmetic is correct.

02Card 2 — Full Quantitative Chain

Card 2 — Full Quantitative Chain

The answer from one stage becomes the input for the next. Energy and momentum are not separate topics — they are two accounting systems for the same physical events.

Stage 1

Explosion

Positive direction: direction bullet travels = positive

System at rest before firing → total p = 0

0 = m_bullet × v_bullet + m_rifle × v_rifle 0 = 0.008 × v_bullet + 4 × (−0.6) 0.008 × v_bullet = 2.4 v_bullet = +300 m/s

Impulse check: J = mΔv = 0.008 × 300 = 2.4 N s on bullet

↓ v_bullet = +300 m/s feeds Stage 2

Stage 2

Perfectly inelastic

Bullet embeds in block — they move together

m_bullet × v_bullet + m_block × 0 = (m_bullet + m_block) × v_f 0.008 × 300 = 2.008 × v_f 2.4 = 2.008 × v_f v_f = +1.195 m/s

KE before collision = ½ × 0.008 × 300² = 360 J

KE after collision = ½ × 2.008 × 1.195² = 1.43 J — 358.6 J lost to deformation

↓ v_f = +1.195 m/s feeds Stage 3

Stage 3

W_net = ΔKE

Block+bullet slide to rest against friction

F_N = (m_bullet + m_block) × g = 2.008 × 9.8 = 19.68 N f_k = μ_k × F_N = 0.09 × 19.68 = 1.77 N W_net = ΔKE: −f_k × s = 0 − 1.43 s = 1.43 / 1.77 s = 0.81 m ≈ 0.8 m ✓

The slide distance checks out. Stage 3 confirms Stage 2's velocity output.

The chain connection Notice how 358.6 J of KE is lost in the bullet-block collision (Stage 2), and only 1.43 J remains to be dissipated by friction in Stage 3. The collision itself converted almost all the bullet's kinetic energy to heat and deformation — the slide is just dealing with the tiny remainder. This is why ballistic gelatin (used to test bullets) stops so efficiently.

03Card 3 — Concept Connections

Card 3 — Concept Connections

If you know how every momentum formula connects to every other, you can solve any problem — even ones built from pieces you have never seen combined before.

Phase 3 Formula Connection Map

p = mv

→

× time Δt

→

J = FΔt

(impulse is momentum transferred per unit time)

J = FΔt

→

÷ Δt

→

F = Δp/Δt

(avg. force during collision)

J = Δp

→

applied to system

→

Σp_before = Σp_after

(no net external J → total p constant)

Σp conserved

→

objects stick

→

(m₁+m₂)v_f

(perfectly inelastic special case)

Σp conserved

→

starts from rest

→

m₁v₁' = −m₂v₂'

(explosion special case)

KE = ½mv²

→

compare before/after

→

elastic / inelastic

(classify collision type)

v_f from collision

→

feeds Phase 2

→

W_net = ΔKE

(sliding, friction, stopping distance)

The Cross-Phase Connection

Phase 2 (energy) and Phase 3 (momentum) are not separate — they describe the same events from two different angles. Whenever an object leaves a collision with a known velocity, Phase 2 tools take over to analyse what happens next. The most common bridge is:

Bridge formula After a perfectly inelastic collision → v_f from conservation → KE_f = ½(m₁+m₂)v_f² → this KE feeds W_net = ΔKE for any subsequent sliding, climbing, or braking.

Direction: Momentum (Phase 3) → post-collision velocity → Energy (Phase 2)

Scenario	Phase 3 tool	Output	Phase 2 tool	Final answer
Ball hits block, block slides	Perfectly inelastic	v_f of combined	W_net = ΔKE	Slide distance
Bullet embeds in block	Perfectly inelastic	v_f of block+bullet	W_net = ΔKE	Slide distance or height reached
Car collision, then braking	Perfectly inelastic	v_f of combined cars	W_net = ΔKE	Braking distance
Explosion, then projectile	Explosion from rest	v of each piece	Kinematics	Range or height
Impulse during collision	J = FΔt = Δp	Average force	—	Force on occupants, safety assessment

Interactive: Projectile Launcher

Interactive: Projectile Motion (3D)

Interactive: Projectile Trajectory

✏️ Worked Examples

Worked Example 1 Multi-stage — Collision then Friction

Problem Setup

Problem type: Two-stage. Stage 1 = perfectly inelastic collision. Stage 2 = friction stops combined mass (Phase 2 bridge).

Scenario: A 0.5 kg ball moving at +6 m/s strikes a stationary 1.5 kg block. They stick together and slide on a rough surface (μk = 0.2). Find: (a) velocity just after collision, (b) distance slid before stopping.

m_ball = 0.5 kg, v_ball = +6 m/s | m_block = 1.5 kg, v_block = 0
Perfectly inelastic: v₁' = v₂' = v_f
μk = 0.2 | g = 9.8 m/s²
Positive direction: direction of ball's initial motion = positive

Solution

Stage 1 — conservation:
0.5×6 + 1.5×0 = 2×v_f
3 = 2v_f → v_f = +1.5 m/s

Perfectly inelastic — combined mass = 2 kg. Single velocity after. Momentum conserved: 3 N s before = 3 N s after.

KE just after collision = ½ × 2 × 1.5² = 2.25 J

Bridge to Phase 2. This KE will be dissipated by friction. KE before collision = ½ × 0.5 × 36 = 9 J — 6.75 J lost in the collision itself.

Stage 2 — friction:
f_k = μk × mg = 0.2 × 2 × 9.8 = 3.92 N

Normal force = weight on flat surface. Friction opposes motion (negative direction).

W_net = ΔKE:
−3.92 × s = 0 − 2.25
s = 2.25/3.92 = 0.574 m

Phase 2 bridge. Negative work by friction equals change in KE (from 2.25 J to zero). Slide distance ≈ 0.57 m.

What would change if...

The ball bounced back at −2 m/s instead of sticking. (a) What would the block's velocity be? (b) Would the block slide further or less far than in the original scenario? (c) What does this tell you about which collision type delivers more momentum to the block?

Worked Example 2 Multi-stage — Explosion then Impulse

Problem Setup

Problem type: Two-stage. Stage 1 = explosion from rest (conservation). Stage 2 = impulse analysis (J = FΔt = Δp).

Scenario: A rocket (total mass 500 kg including 100 kg fuel) fires its engine for 8 s. The fuel is ejected backward at 200 m/s relative to the ground. Treat mass as approximately constant at 500 kg. Find: (a) impulse on rocket, (b) change in momentum, (c) average thrust force.

m_fuel = 100 kg, v_fuel = −200 m/s (backward)
m_rocket ≈ 500 kg (simplified)
Δt = 8 s | system starts at rest

Solution

Stage 1 — explosion from rest:
0 = m_fuel × v_fuel + m_rocket × Δv_rocket
0 = 100×(−200) + 500×Δv_rocket

Conservation. Total p = 0 before. Fuel expelled backward at −200 m/s. Positive direction = forward (direction of rocket thrust).

(a) Δp_rocket = −Δp_fuel = −(100×(−200)) = +20 000 N s

Equal and opposite momenta. Fuel goes backward (−20 000 N s), rocket goes forward (+20 000 N s). Total = 0 ✓

(b) J = Δp_rocket = +20 000 N s

Impulse equals change in momentum by the theorem. The engine delivers 20 000 N s of impulse to the rocket over 8 s.

Average thrust force. 2500 N forward — this is the thrust that accelerates the 500 kg rocket. a = F/m = 2500/500 = 5 m/s² (connecting back to Newton 2).

What would change if...

The rocket fired for 16 s with the same total fuel and ejection speed. What would the average thrust force be? What would the final velocity of the rocket be? Does doubling the burn time double the final velocity?

Visual Break

Copy into your books

▼

Multi-step Strategy

Define positive direction FIRST — always
Label every mass and velocity before calculating
Identify stage type before choosing formula
Each stage output is the next stage input

Stage Types

Explosion from rest: 0 = m₁v₁' + m₂v₂'
Perfectly inelastic: (m₁+m₂)v_f = Σp_before
General collision: m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
Post-collision slide: W_net = ΔKE (Phase 2)

Phase 2–3 Bridge

After collision → v_f known → KE_f = ½(m₁+m₂)v_f²
This KE feeds W_net = ΔKE for any sliding
Slide distance: s = KE_f / f_k
Direction: Momentum (p) → velocity → Energy (KE)

Formula Chain

p = mv → ×Δt → J = FΔt
J = Δp → system → Σp conserved
Σp conserved → KE check → elastic/inelastic
v_f from Phase 3 → W_net = ΔKE → Phase 2

🏃 Activities

Activity 01 — Pattern A

Diagram + Full Chain

A 1200 kg car travelling at +15 m/s collides with a stationary 800 kg car. They lock together and skid on a surface with μk = 0.4.

Part A — Draw a three-panel momentum (1) before collision, (2) just after collision, (3) skidding to rest. Label all masses, velocities, and momentum vectors. Arrows should be proportional to momentum magnitude.

Part B — Calculate the post-collision velocity of the combined cars.

Part C — Using the Phase 2 bridge, find the skid distance d.

Type your working for Parts B and C. Draw the diagram in your book.

Draw the diagram and show full working for all three parts in your book.

Draw diagram and show working in your book

Saved

Activity 02 — Pattern A

Classify and Set Up

For each scenario: (a) identify each stage type, (b) name the formula at each stage, (c) write the equation setup (do not need to fully solve).

A 2 kg ball at +5 m/s hits a 3 kg block (at rest). They stick. The combined mass slides 0.4 m to rest. [Two stages]
A rifle (3 kg) fires a bullet (0.01 kg) at rest. The bullet hits a target and stops in 0.02 s. Find the average force on the target. [Two stages: explosion → impulse]
A 500 kg spacecraft at rest ejects 50 kg of gas at −300 m/s. The spacecraft then fires retro rockets, decelerating at 2 m/s² for 10 s. Find the final velocity. [Two stages: explosion → kinematics]
A 60 kg person jumps from a 5 m height onto a mat (stops in 0.3 s) then onto concrete (stops in 0.008 s). Compare forces. [Two sub-problems: conservation into impulse]
Identify all stages in the bullet-block scenario from the Think First — name the formula, list inputs and outputs for each stage.

Type your stage identification and equation setups for all five.

Write stage identification and equation setups in your book.

Write stage identification and equation setups in your book

Saved

✅ Check Your Understanding

Revisit Your Thinking

At the start you were asked to identify every Phase 3 concept in the bullet-block scenario. The full list:

Moment 1 — bullet fired: Explosion from rest (0 = m₁v₁' + m₂v₂'). Impulse on bullet = J = FΔt = Δp. Newton's Third Law — rifle force on bullet = −bullet force on rifle. Rifle recoil velocity from conservation.

Moment 2 — bullet hits block: Perfectly inelastic collision. Conservation of momentum. Maximum KE loss (most energy to heat/deformation). Post-collision velocity from conservation.

Moment 3 — block slides: Phase 2 bridge — W_net = ΔKE. Friction force = μk × Normal force. Slide distance from energy dissipation. Cross-phase connection: momentum gives velocity; energy determines distance.

Compare with your initial list. What did you miss? What surprised you?

Annotate your initial list in your book.

Annotate your initial list in your book

Saved

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?

Multiple Choice

5 random questions from a replayable lesson bank — feedback shown immediately

UnderstandBand 3

7. Draw and label the three-stage momentum flow diagram for the bullet-block scenario from the Think First. For each stage, name the formula that applies and identify one input and one output. 3 MARKS

Draw diagram and label stages in your book

Saved

ApplyBand 5

8. A 3 kg ball travelling at +8 m/s collides with a stationary 5 kg block. They stick together. The combined mass slides on a rough surface with μk = 0.25. (a) Find the velocity just after collision. (b) Find the kinetic energy just after collision. (c) Find the slide distance. 4 MARKS

Answer in your book — show both stages

Saved

EvaluateBand 6

9. A 2 kg ball moving at +8 m/s collides with a stationary 6 kg block. After the collision the ball moves at +2 m/s. (a) Apply the Vector Protocol to find the block's velocity after the collision. (b) Calculate the KE before and after and find the energy lost. (c) Classify the collision type with full justification. (d) The block then slides on a surface with μk = 0.15 — find the slide distance. 4 MARKS

Answer in your book — full Vector Protocol required, all four parts

Saved

Comprehensive Answers

▼

Activity 01 — Car Collision then Skid

Part B: p_before = 1200x15 = 18 000 N s. (2000)v_f = 18 000. v_f = +9 m/s

Part C: KE_f = 1/2 x 2000 x 81 = 81 000 J. f_k = 0.4 x 2000 x 9.8 = 7840 N. -7840 x d = -81 000. d = 81 000/7840 = 10.33 m

The cars skid approximately 10.3 m after the collision. This is why post-crash skid marks are used in accident reconstruction — the skid distance tells investigators the speed at the moment of separation (or impact).

Activity 02 — Stage Identification (key answers)

1. Stage 1: perfectly inelastic → v_f = (2×5)/(2+3) = 2 m/s. Stage 2: Phase 2 bridge → KE_f = ½×5×4 = 10 J. f_k = μk×5×9.8 (need μk). −f_k×0.4 = −10 → f_k = 25 N → μk = 25/49 = 0.51.

2. Stage 1: explosion → 0 = 0.01×v_bullet + 3×v_rifle → if rifle recoil given, find v_bullet. Stage 2: J = m_bullet × v_bullet → F = J/Δt = J/0.02.

3. Stage 1: explosion from rest → 0 = 50×(−300) + 500×v_sc → v_sc = +30 m/s. Stage 2: kinematics → v_f = 30 + (−2)×10 = +10 m/s forward.

4. Impact speed: v = √(2gh) = √(2×9.8×5) = 9.9 m/s. Δp = 60×9.9 = 594 N s. Mat: F = 594/0.3 = 1980 N. Concrete: F = 594/0.008 = 74 250 N — 37.5× larger.

5. Stage 1 (explosion): formula 0 = m_bullet×v_bullet + m_rifle×v_rifle. Input: masses, recoil. Output: bullet velocity. Stage 2 (perfectly inelastic): formula (m_bullet+m_block)v_f = m_bullet×v_bullet. Input: bullet velocity. Output: v_f. Stage 3 (Phase 2): formula W_net = ΔKE. Input: v_f, μk. Output: slide distance.

Multiple Choice

1. B. The sliding stage uses the Phase 2 work-energy theorem. W_net = ΔKE where W_net = −f_k × s. Option A and C are conservation of momentum equations — for the collision stage, not the sliding stage. Option D is explosion from rest.

2. C — ½(m₁+m₂)v_f². After the perfectly inelastic collision, the combined mass (m₁+m₂) moves at v_f. KE = ½×(combined mass)×v_f². Option A uses only m₁ and the original velocity — this is before the collision, not after.

3. A — −2 m/s. Explosion from rest: 0 = 1×10 + 5×v → v = −10/5 = −2 m/s. Negative sign confirms opposite direction to the 1 kg object. Option C (−10) ignores the mass ratio.

4. D. Stage 1: v_f = (2×4)/(2+2) = 2 m/s. KE_f = ½×4×4 = 8 J. f_k = 0.3×4×9.8 = 11.76 N. s = 8/11.76 = 0.68 m. Option A uses the original velocity (4 m/s) not the post-collision velocity (2 m/s). Option C uses v_f = 4 m/s — incorrect.

5. B. Bouncing ball: Δp_ball = m(−v_f − v_i) = larger magnitude than if ball stops. By Newton 3, the impulse on the block equals the impulse on the ball in magnitude — so the bouncing ball delivers MORE impulse to the block. Sticking: the ball's momentum goes from +p to 0 (change = p). Bouncing: ball goes from +p to −p' (change = p + p' > p). The block receives more momentum in the bouncing case.

6. C — 6.25 m/s. Total Δp = 1500 + 1000 = 2500 N s. v = Δp/m = 2500/400 = 6.25 m/s. Momentum changes are additive when the same object receives sequential impulses.

Short Answer — Model Answers

Q8 (4 marks):

(a) Perfectly inelastic: p_before = 3x8 = 24 N s. (3+5)v_f = 24. v_f = +3 m/s

(b) KE just after = 1/2 x 8 x 9 = 36 J

Q9 (4 marks):

(a) Positive = initial direction of ball. 2x8 + 0 = 2x2 + 6xv2'. 16 = 4 + 6v2'. v2' = +2 m/s (forward)

(b) KE_before = 1/2x2x64 = 64 J. KE_after = 1/2x2x4 + 1/2x6x4 = 4+12 = 16 J. KE lost = 48 J

(c) KE not conserved (64J != 16J) -> inelastic. Objects separated -> not perfectly inelastic. Classification: INELASTIC. Momentum check: p_before = 16 N s. p_after = 2x2+6x2 = 4+12 = 16 N s. Conserved.

(d) KE_block = 1/2x6x4 = 12 J. f_k = 0.15x6x9.8 = 8.82 N. s = 12/8.82 = 1.36 m

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