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VCE Physics (Year 12) · Motion and Forces · 20m read

Momentum, Impulse and Collisions

Ever wonder why airbags save lives while rigid dashboards don't? It's all about momentum — the "weight" of motion — and how quickly you can change it. You'll learn that momentum is always conserved during collisions and explosions, and that impulse (the force applied over time) is the key to slowing things down safely. Master these ideas and you'll unlock real-world scenarios from car crashes to rocket launches, plus ace VCAA's favourite applied questions.

Learning objective: Apply conservation of momentum and the impulse–momentum theorem to collision, explosion and vehicle-safety scenarios.

By the end of this lesson you can:

  • I calculate momentum as a vector $\vec{p} = m\vec{v}$ with correct sign.
  • I apply the impulse-momentum theorem $J = \Delta p = F_{avg} \Delta t$.
  • I apply $\sum p_{before} = \sum p_{after}$ to elastic, inelastic and perfectly inelastic collisions.
  • I explain how vehicle safety features (airbags, crumple zones) reduce force by extending collision time.
  • I verify collision type by calculating kinetic energy before and after.

Collisions are pool-table physics. Total momentum before equals total momentum after — even when balls crash and bounce.

Momentum — the "quantity of motion"

Momentum $\vec{p}$ is the product of mass and velocity:

$$\vec{p} = m\vec{v} \ \quad [\text{kg m/s} = \text{N s}]$$

Momentum is a vector — direction matters. A 1500 kg car at 20 m/s east has a different momentum from one moving west at the same speed.

### Why momentum matters

Newton's 2nd law in its most general form:

$$\vec{F}_{net} = \frac{d\vec{p}}{dt}$$

This is more fundamental than $F = ma$ — it works even when mass changes (e.g. rockets). For constant mass, $F = ma$ is recovered.

### Conservation of momentum

In an isolated system (no external net force), the total momentum is constant:

$$\sum \vec{p}_{before} = \sum \vec{p}_{after}$$

This is a consequence of Newton's 3rd law: in any interaction, the forces are equal and opposite, so the net impulse on the system is zero. This makes momentum conservation one of the most powerful tools in physics.

> Isolated system means no net external forces. In collisions, external forces (friction, gravity) often act, but if the collision time $\Delta t$ is very short, the impulse from external forces is negligible compared to internal collision forces — momentum is approximately conserved.

Impulse — force applied over time

Impulse $\vec{J}$ is the change in momentum:

$$\vec{J} = \Delta \vec{p} = m\vec{v}_f - m\vec{v}_i$$

Since $\vec{F} = \frac{\Delta \vec{p}}{\Delta t}$, we can also write:

$$\vec{J} = \vec{F}_{avg} \cdot \Delta t \ \quad [\text{N s} = \text{kg m/s}]$$

This is the impulse–momentum theorem: the impulse equals the change in momentum.

### Why vehicle safety features work

A car hitting a wall at $20 \ \text{m/s}$ and stopping has a fixed $\Delta p = mv_f - mv_i$ regardless of how it stops. The impulse cannot be changed. But the *force* can:

$$F_{avg} = \frac{\Delta p}{\Delta t}$$

Increasing $\Delta t$ while keeping $\Delta p$ constant → reduces $F_{avg}$.

| Feature | Mechanism | Effect |
|---------|-----------|--------|
| Airbag | Cushion inflates, slows occupant over longer time | Reduces peak force on head/chest |
| Crumple zone | Front of car deforms progressively | Extends collision time |
| Seatbelt | Stretches slightly under impact | Distributes force, extends stopping time |
| Helmet foam | Compresses on impact | Extends head deceleration time |

### Impulse on a v–t graph

For a variable force:
$$J = \int F \ dt = \text{area under } F\text{–}t \text{ graph}$$

For a constant average force, it's simply the area of a rectangle.

Types of collisions

All collisions conserve momentum. They differ in what happens to kinetic energy:

### Elastic collisions
Both momentum and kinetic energy are conserved.

$$\sum KE_{before} = \sum KE_{after}$$

Real-world approximation: billiard balls, superball bouncing, gas molecules. True elastic collisions are rare at human scale.

In elastic collisions between equal masses where one is initially at rest: the moving ball stops and the stationary one moves off at the original speed. (Demonstrated in Newton's cradle.)

### Inelastic collisions
Momentum is conserved; kinetic energy is not fully conserved (some converted to heat, sound, deformation).

$$\sum p_{before} = \sum p_{after} \ \text{ but } \ \sum KE_{before} > \sum KE_{after}$$

Most real-world collisions are inelastic (car crashes, ball catching).

### Perfectly inelastic collisions (objects stick together)
A special case of inelastic: objects join and move with a common final velocity $v_f$.

$$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v_f$$

Maximum kinetic energy is lost (but momentum is still conserved).

### Explosions (reverse of perfectly inelastic)
Objects start together, fly apart. Total initial momentum = 0 (if at rest). Final momenta are equal and opposite:

$$m_1 v_1 + m_2 v_2 = 0 \implies m_1 v_1 = -m_2 v_2$$

Momentum and Impulse Formulas

  • Momentum: \vec{p} = m\vec{v} (Vector — always include direction)
  • Impulse: \vec{J} = \Delta \vec{p} = m\vec{v}_f - m\vec{v}_i ($\Delta p = J$ always)
  • Average force: F_{avg} = \dfrac{\Delta p}{\Delta t} (Longer $\Delta t$ → smaller $F$)
  • Conservation: \sum p_{before} = \sum p_{after} (Isolated system only)
  • Perfectly inelastic: m_1u_1 + m_2u_2 = (m_1+m_2)v_f (Objects stick together)
  • KE check: KE = \tfrac{1}{2}mv^2 (Compare before/after to find loss)

Collision problem protocol

  • 1. Choose +ve direction: Write it down. Assign $+$ or $-$ to every velocity.
  • 2. Calculate p_before: Sum $m_1u_1 + m_2u_2 + \ldots$
  • 3. Write p_after: Use the collision type (separate velocities or combined mass).
  • 4. Set equal and solve: $p_{before} = p_{after}$. This gives you $v_f$ or the unknown.
  • 5. Find impulse if needed: $J = \Delta p = p_f - p_i$. Then $F = J / \Delta t$.
  • 6. Check collision type: Calculate $KE_{before}$ and $KE_{after}$. If equal: elastic. If less after: inelastic.

Worked example (Easy)

A $1200 \ \text{kg}$ car moves east at $20 \ \text{m/s}$. Find its momentum. If it stops in $4.0 \ \text{s}$, what is the average braking force?

  1. Calculate momentum. Taking east as positive: p = mv = 1200 \times 20 = +24{,}000 \ \text{kg m/s (east)}
  2. Calculate impulse (change in momentum). Final velocity = 0 (stops): \Delta p = mv_f - mv_i = 0 - 24{,}000 = -24{,}000 \ \text{N s}
  3. Average braking force. F = \frac{\Delta p}{\Delta t} = \frac{-24{,}000}{4.0} = -6000 \ \text{N}
  4. Interpret sign. Negative sign means force is westward (opposing eastward motion). Magnitude = 6000 N.

Answer: Momentum $p = 24{,}000 \ \text{kg m/s east}$; average braking force $= 6000 \ \text{N}$ (opposing motion).

Worked example (Medium)

A $1200 \ \text{kg}$ car moving east at $20 \ \text{m/s}$ collides head-on with an $800 \ \text{kg}$ car moving west at $15 \ \text{m/s}$. They stick together (perfectly inelastic). Find their combined velocity after the collision.

  1. Sign convention. East = positive, west = negative. u_1 = +20 \ \text{m/s}, \ u_2 = -15 \ \text{m/s}
  2. Total momentum before. p_{before} = m_1 u_1 + m_2 u_2 = 1200(+20) + 800(-15) = 24{,}000 - 12{,}000 = +12{,}000 \ \text{kg m/s}
  3. Apply conservation (perfectly inelastic). They stick together, total mass = $1200 + 800 = 2000 \ \text{kg}$: (m_1 + m_2)v_f = p_{before}
  4. Solve for $v_f$. v_f = \frac{12{,}000}{2000} = +6.0 \ \text{m/s (east)}

Answer: $v_f = 6.0 \ \text{m/s east}$. The combined wreck moves east because the larger car had more momentum.

Worked example (Hard)

A $75 \ \text{kg}$ driver's head (mass $4.0 \ \text{kg}$, model it as this) decelerates from $15 \ \text{m/s}$ to rest. (a) Without airbag, the head stops in $0.010 \ \text{s}$. With airbag, it stops in $0.080 \ \text{s}$. Calculate the average force on the head in each case. (b) What is the percentage reduction in force?

  1. Find change in momentum ($\Delta p$). Same in both cases — head goes from $15 \ \text{m/s}$ to $0$: \Delta p = m \Delta v = 4.0 \times (0 - 15) = -60 \ \text{N s}
  2. (a) Force without airbag. F_{no\ airbag} = \frac{\Delta p}{\Delta t} = \frac{60}{0.010} = 6000 \ \text{N}
  3. (a) Force with airbag. F_{airbag} = \frac{\Delta p}{\Delta t} = \frac{60}{0.080} = 750 \ \text{N}
  4. (b) Percentage reduction. \% \text{reduction} = \frac{6000 - 750}{6000} \times 100 = \frac{5250}{6000} \times 100 = 87.5\%

Answer: Without airbag: $F = 6000 \ \text{N}$. With airbag: $F = 750 \ \text{N}$. The airbag reduces impact force by $87.5\%$ — a factor of 8.

Common mistakes

  • Using speeds (magnitudes) instead of velocities (vectors) in momentum conservation Momentum is a vector. A ball moving left has negative momentum. Adding $m_1 u_1 + m_2 u_2$ with both as positive when they move in opposite directions gives the wrong $p_{before}$. Fix: Always establish a positive direction. Assign negative values to objects moving in the negative direction before summing.
  • Claiming momentum is not conserved because kinetic energy decreases Conservation of momentum and conservation of kinetic energy are independent laws. Inelastic collisions conserve momentum but not kinetic energy — the lost KE goes to heat, sound and deformation. Fix: Momentum is always conserved in isolated systems regardless of the type of collision. Only elastic collisions also conserve KE.
  • Using $F = \Delta p / \Delta t$ with $\Delta p$ as just $mv$ instead of $m(v_f - v_i)$ $\Delta p$ must be the change — final minus initial. If an object bounces back, $\Delta p$ is NOT just the magnitude of the final momentum, it includes the reversal. Fix: Write $\Delta p = p_f - p_i = mv_f - mv_i$. For a ball bouncing back, $v_f$ has the opposite sign to $v_i$, so $|\Delta p| > |p_i|$.

Noether's theorem — why momentum conservation is not just a rule but a law of nature

Conservation of momentum is not an empirical accident — it is a deep consequence of the translational symmetry of space. In 1918, mathematician Emmy Noether proved a remarkable theorem: every continuous symmetry of a physical system corresponds to a conserved quantity.

  • Space is the same everywhere (translational symmetry) → momentum is conserved
  • Space looks the same in every direction (rotational symmetry) → angular momentum is conserved
  • The laws of physics don't change over time (time symmetry) → energy is conserved

This means that if you performed a collision experiment on Earth, on Mars, or in intergalactic space, momentum would be conserved in exactly the same way — because the laws of physics don't depend on where you are.

Elastic vs inelastic at a deeper level: In an elastic collision, both momentum and kinetic energy are conserved. In an inelastic collision, momentum is conserved but kinetic energy is converted to thermal energy (molecular vibrations), sound, and deformation. A perfectly inelastic collision (objects stick together) loses the maximum kinetic energy while still conserving momentum.

Particle physics application: At CERN's Large Hadron Collider, two protons collide at nearly the speed of light. Conservation of momentum and energy (now relativistic: $E^2 = (pc)^2 + (mc^2)^2$) determines what particles can be produced. The Higgs boson was discovered by searching for specific momentum signatures in the debris of proton–proton collisions — momentum conservation is the detective's primary tool.

Practice questions

  1. A vehicle airbag reduces the force on a driver during a collision by:

    1. Increasing the collision time, keeping $\Delta p$ the same
    2. Reducing the change in momentum
    3. Increasing the driver's inertia
    4. Absorbing momentum from the driver

    (1 marks)

  2. A $0.20 \ \text{kg}$ tennis ball moving at $+30 \ \text{m/s}$ is hit back at $-25 \ \text{m/s}$. The impulse on the ball is:

    1. $-11 \ \text{N s}$
    2. $+1.0 \ \text{N s}$
    3. $-1.0 \ \text{N s}$
    4. $+11 \ \text{N s}$

    (2 marks)

  3. In a perfectly inelastic collision between a $3.0 \ \text{kg}$ ball moving at $+8.0 \ \text{m/s}$ and a $5.0 \ \text{kg}$ stationary ball, what is the combined velocity after collision?

    1. $+3.0 \ \text{m/s}$
    2. $+4.0 \ \text{m/s}$
    3. $+5.0 \ \text{m/s}$
    4. $+8.0 \ \text{m/s}$

    (3 marks)

  4. A $1000 \ \text{kg}$ car travelling at $18 \ \text{m/s}$ strikes a barrier and stops in $0.12 \ \text{s}$ with a crumple zone, or $0.020 \ \text{s}$ without. Calculate the force in each case and explain why crumple zones save lives.

    (3 marks)

  5. Two ice skaters initially at rest push apart. Skater A (mass $60 \ \text{kg}$) moves at $+3.0 \ \text{m/s}$. Find skater B's velocity if her mass is $50 \ \text{kg}$. Is kinetic energy conserved?

    (4 marks)

Summary

  • Momentum $\vec{p} = m\vec{v}$ is a vector. Always define positive direction.
  • Impulse $J = \Delta p = F_{avg} \Delta t$. Same $\Delta p$, longer $\Delta t$ → smaller force.
  • Conservation: $\sum p_{before} = \sum p_{after}$ in any isolated system.
  • Elastic: KE and $p$ conserved. Inelastic: only $p$ conserved. Perfectly inelastic: objects stick, maximum KE lost.
  • Explosions: start at rest (total $p = 0$), pieces fly apart with equal and opposite momenta.
  • Vehicle safety features (airbags, crumple zones) extend $\Delta t$, reducing peak force on occupants.

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