Year 11 PhysicsModule 2⏱ ~40 min5 MC · 3 Short AnswerLesson 13 of 15
Momentum Synthesis
In 1742, Benjamin Robins presented the ballistic pendulum to the Royal Society of London: a bullet of mass 0.0019 kg embeds in a hanging wooden block and swings it upward. From the swing height alone, Robins back-calculated bullet velocity to 968 m/s, using momentum conservation for the impact stage ($m_1 v_1 = (m_1+m_2)v_2$) and energy conservation for the swing stage ($\frac{1}{2}mv^2 = mgh$). This two-stage technique is the foundation of forensic ballistics.
Today's hook: In 1742, Benjamin Robins fired a 0.0019 kg bullet into a hanging block and watched it swing 0.12 m upward. He used two different conservation laws, one for the impact, one for the swing, to calculate bullet speed as 968 m/s. Why can you NOT use the same law for both stages?
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Warm up first
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
A bullet embeds in a pendulum block (Stage 1), and then the combined mass swings up (Stage 2). Before calculating: which conservation law applies in Stage 1, and which in Stage 2? Why can't you use the same law for both stages?
When a bullet embeds in a block (perfectly inelastic), which conservation law is used to find the velocity just after impact?
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Learning Intentions
Learning Intentions
goals
Know
Two-stage synthesis chain for ballistic pendulum
Safety device analysis using $J = \Delta p$ and $F = \Delta p/\Delta t$
How to quantitatively compare elastic vs inelastic collisions
Understand
Why Stage 1 uses momentum conservation (not energy), energy is lost to heat
Why Stage 2 uses energy conservation, no collision, no energy loss
Why safety devices work by increasing $\Delta t$ for the same $\Delta p$
Evaluate the effectiveness of a safety device quantitatively
Compare elastic and inelastic collision outcomes
Cross-lesson links: L13 is the synthesis lesson for Phase 3, combining momentum conservation from L12 with energy conservation from L07. This two-stage approach, momentum for collisions, energy for free motion, mirrors the Phase 2 synthesis chain in L09. Safety device analysis here uses $F = \Delta p/\Delta t$ from L11, connecting back to Newton's Second Law (L04). L14 consolidates all five Phase 3 formulae before the Module 2 capstone in L15.
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The Two-Stage Synthesis Chain
Core Content
01
The Two-Stage Synthesis Chain
+5 XP
A lump of clay is thrown at a hanging pendulum bob and sticks to it. The pendulum swings upward. You want to know: how fast was the clay thrown? You cannot use energy conservation for the whole event, the clay-bob collision was perfectly inelastic and kinetic energy was lost. But once the clay is embedded and the system is swinging freely, energy conservation works perfectly. Two stages, two different laws.
Stage 1: Collision/explosion
Use conservation of momentum. Kinetic energy is NOT conserved, energy is lost to heat, sound, and deformation.
$m_1v_1 + m_2v_2 = (m_1 + m_2)v_f$
Stage 2: After the collision
Use conservation of energy (if no friction). Kinetic energy converts to PE as object rises.
$\frac{1}{2}(m_1+m_2)v_f^2 = (m_1+m_2)gh$
Worked ExampleBallistic Pendulum
A 0.01 kg bullet embeds in a 1 kg stationary block. The combined mass rises 0.2 m. Find the bullet's initial velocity. ($g = 9.8$ m/s²)
Two-stage synthesis: Stage 1 (collision/explosion) uses conservation of momentum ($\Sigma p$); Stage 2 (free motion) uses conservation of energy ($\frac{1}{2}mv^2 = mgh$); often solve Stage 2 first to find post-collision velocity, then substitute that result into the Stage 1 momentum equation.
Pause, copy the highlighted two-stage strategy into your book before moving on.
True or false: In a ballistic pendulum, the kinetic energy immediately after the bullet embeds equals the potential energy at maximum height.
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Safety Device Evaluation
02
Safety Device Evaluation
+5 XP
We just saw that two-stage problems split at the collision boundary, momentum governs the collision, energy governs the motion after. That raises a question: what is the practical purpose of increasing collision time, and how do we quantify it? This card answers it → safety devices increase $\Delta t$ to reduce the average force for the same change in momentum.
Safety devices work by increasing collision time, the same impulse ($\Delta p$) delivered over a longer time means a smaller average force.
To evaluate quantitatively: calculate the force without the safety device (shorter $\Delta t$) and with it (longer $\Delta t$). The reduction in force is the "effectiveness".
Device
Mechanism
Effect on $\Delta t$
Effect on $F$
Airbag
Spreads deceleration over 0.1–0.15 s
Increases $\Delta t$
Reduces $F$
Crumple zone
Car body deforms, extending stopping distance/time
Increases $\Delta t$
Reduces $F$
Seatbelt
Stretchy webbing absorbs energy over 0.1–0.2 s
Increases $\Delta t$
Reduces $F$
Padded helmet
Foam deforms, extending head deceleration time
Increases $\Delta t$
Reduces $F$
All safety devices reduce injury by increasing $\Delta t$: for the same $\Delta p$, $F_{avg} = \Delta p/\Delta t$ decreases as $\Delta t$ increases; effectiveness is quantified as the ratio $F_{without}/F_{with}$; NESA expects you to calculate both forces numerically and state the ratio.
Pause, copy the highlighted safety device rule into your book before moving on.
A helmet increases the contact time of an impact from 5 ms to 20 ms. For the same $\Delta p$, the average force on the head is reduced by a factor of:
Which best explains why an elastic collision conserves kinetic energy while a perfectly inelastic collision does not?
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Safety Device Effectiveness Calculation
Activities
Safety Device Effectiveness Calculation
EvaluateBand 6
A 70 kg cyclist falls and decelerates from 8 m/s to rest. Without helmet: contact time = 3 ms. With helmet: contact time = 18 ms. Calculate average force in each case. Express the effectiveness as a ratio and explain why the helmet is effective using the impulse-momentum theorem.
✓
Show what you have learned
Quick recall, Momentum Synthesis
+5 XP
Short Answer, 10 marks
+5 XP
UnderstandBand 3(3 marks) 1. In the ballistic pendulum, momentum is conserved in Stage 1 (collision) but not kinetic energy. Explain why kinetic energy is NOT conserved when the bullet embeds in the block.
ApplyBand 4(3 marks) 2. A 0.05 kg bullet embeds in a 4.95 kg stationary block. The combined mass then rises 0.5 m. Find the initial speed of the bullet. ($g = 9.8$ m/s²)
EvaluateBand 6(4 marks) 3. A seatbelt test shows a 75 kg dummy at 15 m/s stops in (a) 0.15 s with seatbelt and (b) 0.012 s without seatbelt (hits dashboard). Calculate the average force in each case. What multiple of body weight is each force? Evaluate the seatbelt's effectiveness. ($g = 9.8$ m/s²)
Show all answers
Short Answer, Model Answers
Q1 (3 marks): When the bullet embeds in the block, the collision is perfectly inelastic, the bullet and block deform and permanently stick together. During this process, kinetic energy is converted to heat (from friction between bullet and wood), sound, and the energy used to deform the materials. This energy cannot be recovered as mechanical kinetic energy. Momentum is still conserved because Newton's Third Law requires the impulses on bullet and block to be equal and opposite.
Q3 (4 marks): $\Delta p = 75 \times 15 = 1125\text{ kg m/s}$. (a) With seatbelt: $F = 1125/0.15 = 7500\text{ N}$. (b) Without: $F = 1125/0.012 = 93\,750\text{ N}$. Body weight $= 75 \times 9.8 = 735\text{ N}$. (a) $7500/735 \approx 10.2$ times body weight. (b) $93\,750/735 \approx 127.6$ times body weight. The seatbelt reduces the average force by a factor of 12.5 (from 93,750 N to 7,500 N) by extending the stopping time from 12 ms to 150 ms, a 12.5-fold increase in $\Delta t$ produces a 12.5-fold decrease in $F$.
In Robins's 1742 Royal Society demonstration: Stage 1, the bullet embeds in the block (inelastic collision, KE is lost as heat/deformation, but momentum is conserved: $m_1 v_1 = (m_1+m_2)v_2$). Stage 2, the block swings up (no collision, mechanical energy is conserved: $\frac{1}{2}(m_1+m_2)v_2^2 = (m_1+m_2)gh$). Matching the correct law to each stage is the entire skill, blindly applying the same law throughout gives the wrong answer for one of the stages.