← Unit 2: Linear motion and waves

QLDPhysicsSyllabus dot point

Topic 1: Linear motion and force

Define linear momentum and impulse, and apply the principle of conservation of momentum to one-dimensional collisions and explosions, distinguishing between elastic and inelastic collisions

A focused answer to the QCE Physics Unit 2 dot point on momentum and impulse. Defines $p = mv$ and $J = F \Delta t = \Delta p$, walks through conservation of momentum in one-dimensional collisions and explosions, and distinguishes elastic from inelastic by whether kinetic energy is conserved. Works the QCAA two-cart collision standard problem.

Generated by Claude OpusReviewed by Better Tuition Academy7 min answer

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What this dot point is asking

QCAA wants you to define momentum and impulse, link them through Newton's second law, and apply conservation of momentum to one-dimensional interactions. The QCAA EA tests this both as a calculation (two-body collision) and as a qualitative discussion (elastic versus inelastic, identifying isolated systems).

Momentum

Linear momentum is the product of mass and velocity:

p⃗=mv⃗\vec{p} = m \vec{v}

SI unit: kg m sβˆ’1^{-1} (equivalently N s). Momentum is a vector. Sign matters.

Impulse

Impulse is the product of net force and the time it acts, and it equals the change in momentum:

Jβƒ—=Fβƒ—net Δt=Ξ”pβƒ—=mvβƒ—fβˆ’mvβƒ—i\vec{J} = \vec{F}_{\text{net}} \, \Delta t = \Delta \vec{p} = m \vec{v}_f - m \vec{v}_i

This follows directly from F=ma=mΞ”v/Ξ”tF = ma = m \Delta v / \Delta t. For a variable force, impulse equals the area under the force-time graph.

Impulse is why crumple zones, airbags, padded gloves and bent knees on landing reduce injury: stretching out Ξ”t\Delta t reduces the peak force needed to deliver the same Ξ”p\Delta p.

Conservation of momentum

For an isolated system (no external net force), total momentum is conserved:

βˆ‘mivβƒ—i,before=βˆ‘mivβƒ—i,after\sum m_i \vec{v}_{i,\text{before}} = \sum m_i \vec{v}_{i,\text{after}}

For a one-dimensional collision between two bodies, this becomes:

m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

Conservation of momentum holds in every type of collision, including the most dramatic inelastic ones.

Elastic and inelastic collisions

Type Momentum Kinetic energy Example
Elastic Conserved Conserved Ideal billiard balls, hard atomic collisions
Inelastic Conserved Not conserved (some becomes heat, sound, deformation) Most real collisions
Perfectly inelastic Conserved Maximum KEKE loss Bodies stick together

For perfectly inelastic collisions where the bodies stick:

m1u1+m2u2=(m1+m2)vm_1 u_1 + m_2 u_2 = (m_1 + m_2) v

Explosions

An explosion is the time-reverse of a perfectly inelastic collision. A single body at rest separates into two bodies with equal and opposite momenta:

0=m1v1+m2v20 = m_1 v_1 + m_2 v_2

The lighter fragment moves faster, in the opposite direction to the heavier fragment.

Worked example

A 15001500 kg car travelling east at 2020 m sβˆ’1^{-1} collides head-on with a stationary 10001000 kg car. They lock together after impact. Find the velocity immediately after, and the kinetic energy lost.

Take east as positive. Conservation of momentum:

(1500)(20)+(1000)(0)=(2500)v(1500)(20) + (1000)(0) = (2500) v

v=30000/2500=12v = 30000 / 2500 = 12 m sβˆ’1^{-1} east.

KEbefore=12(1500)(20)2=300 000KE_{\text{before}} = \frac{1}{2}(1500)(20)^2 = 300\,000 J.

KEafter=12(2500)(12)2=180 000KE_{\text{after}} = \frac{1}{2}(2500)(12)^2 = 180\,000 J.

KEKE lost =120 000= 120\,000 J (turned into heat, sound, deformation). Momentum is conserved (both before and after equal 30 00030\,000 kg m sβˆ’1^{-1}); kinetic energy is not. The collision is inelastic.

Common traps

Treating kinetic energy as conserved in every collision. KEKE is only conserved in elastic collisions. Always check.

Dropping the sign on velocity. In a head-on collision the bodies have opposite signs of velocity. Mixing them up gives a momentum total that is too large by a factor of two or three.

Applying momentum conservation to a system with an external force. If a car is being pushed by a person during the collision, the system is not isolated. For typical QCAA problems, the collision time is short enough that external forces (friction, weight) deliver negligible impulse.

Confusing impulse with force. Impulse has units of N s (or kg m sβˆ’1^{-1}). Force has units of N. They are equal only when multiplied or divided by Ξ”t\Delta t.

In one sentence

Momentum p⃗=mv⃗\vec{p} = m \vec{v} is conserved in any isolated one-dimensional collision or explosion, kinetic energy is only conserved in elastic collisions, and impulse J⃗=F⃗Δt=Δp⃗\vec{J} = \vec{F} \Delta t = \Delta \vec{p} explains why stretching the collision time (airbags, crumple zones) reduces the peak force on a body.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

Year 11 SAC5 marksA $0.20$ kg ball travelling at $8.0$ m s$^{-1}$ to the right collides head-on with a stationary $0.30$ kg ball. After the collision the $0.20$ kg ball moves at $1.6$ m s$^{-1}$ to the left. (a) Find the velocity of the $0.30$ kg ball after the collision. (b) Is the collision elastic? Justify.
Show worked answer β†’

Take right as positive.

(a) Conservation of momentum. pbefore=pafterp_{\text{before}} = p_{\text{after}}.

(0.20)(8.0)+(0.30)(0)=(0.20)(βˆ’1.6)+(0.30)v(0.20)(8.0) + (0.30)(0) = (0.20)(-1.6) + (0.30) v

1.6=βˆ’0.32+0.30v1.6 = -0.32 + 0.30 v

v=1.92/0.30=6.4v = 1.92 / 0.30 = 6.4 m sβˆ’1^{-1} to the right.

(b) Kinetic energy check.

KEbefore=12(0.20)(8.0)2=6.4KE_{\text{before}} = \frac{1}{2}(0.20)(8.0)^2 = 6.4 J.

KEafter=12(0.20)(1.6)2+12(0.30)(6.4)2=0.256+6.144=6.4KE_{\text{after}} = \frac{1}{2}(0.20)(1.6)^2 + \frac{1}{2}(0.30)(6.4)^2 = 0.256 + 6.144 = 6.4 J.

KEKE is conserved, so the collision is elastic.

Markers reward the explicit sign convention, both kinetic-energy substitutions, and the conclusion that follows from the comparison.

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