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How do Newton's laws explain motion under forces, and how is momentum conserved?

Newton's three laws of motion, force as a vector ($F = ma$), free-body diagrams, momentum $p = mv$ and impulse $\Delta p = F \Delta t$, and conservation of momentum in collisions

A focused answer to the VCE Physics Unit 2 key knowledge point on Newton's laws and momentum. The three laws (inertia, $F = ma$, action-reaction), free-body diagrams, momentum $p = mv$, impulse $J = F \\Delta t = \\Delta p$, and conservation of momentum in elastic and inelastic collisions.

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What this dot point is asking

VCAA wants you to apply Newton's three laws and the concept of momentum to motion problems, draw free-body diagrams, and use conservation of momentum in collision problems.

Newton's three laws

First law (inertia). An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted on by a net external force.

Second law. Fnet=maF_{\text{net}} = m a. The net force on an object equals its mass times its acceleration. Force and acceleration are vectors in the same direction.

Third law. For every action, there is an equal and opposite reaction. If A exerts a force on B, then B exerts an equal-magnitude, opposite-direction force on A.

The forces in the third law act on different bodies; they do not cancel.

Forces commonly encountered

  • Gravity. W=mgW = mg where g=9.8g = 9.8 m sβˆ’2^{-2}. Always downward.
  • Normal force. Perpendicular to a surface, from the surface on the object.
  • Friction. Parallel to a surface, opposing relative motion. f≀μNf \leq \mu N for static; f=ΞΌkNf = \mu_k N for kinetic.
  • Tension. Along a rope or string. Always pulling.
  • Applied force. Pushed or pulled by something external.

Free-body diagrams

Draw the object as a dot or simple shape. Show every force acting on the object as a vector arrow originating from the dot. Label each force.

Apply Newton's second law: sum forces vectorially (component by component), set equal to mama.

For an object on a surface with friction:

  • Normal force balances gravity perpendicular to surface.
  • Net force parallel to surface = applied minus friction = mama.

Momentum and impulse

Momentum p⃗=mv⃗\vec{p} = m \vec{v}. Vector quantity, in the direction of velocity.

Impulse J⃗=F⃗Δt=Δp⃗\vec{J} = \vec{F} \Delta t = \Delta \vec{p}. The impulse on an object equals the change in its momentum.

The impulse-momentum theorem follows from Newton's second law: F=ma=mΞ”v/Ξ”t=Ξ”p/Ξ”tF = ma = m \Delta v / \Delta t = \Delta p / \Delta t, so FΞ”t=Ξ”pF \Delta t = \Delta p.

Force-time graph

The area under a force-time graph equals the impulse (= change in momentum).

For non-constant forces, integrate (or estimate area under the curve graphically).

Conservation of momentum

In an isolated system (no external forces), total momentum is conserved.

For a collision between objects 1 and 2:

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

where uu = before-collision velocity, vv = after-collision velocity.

Elastic vs inelastic collisions

Elastic collision. Both momentum and kinetic energy are conserved. Examples: collisions between hard balls (close to elastic in practice).

Inelastic collision. Momentum conserved; kinetic energy not (some lost to heat, sound, deformation).

Perfectly inelastic. The two objects stick together after collision. Maximum kinetic energy loss for given initial conditions.

Worked example: elastic collision

A 2 kg ball at 5 m/s collides elastically with a stationary 3 kg ball. Find final velocities.

Conservation of momentum: 2(5)+3(0)=2v1+3v22(5) + 3(0) = 2 v_1 + 3 v_2, so 10=2v1+3v210 = 2 v_1 + 3 v_2.

Conservation of KE: 12(2)(5)2=12(2)v12+12(3)v22\frac{1}{2}(2)(5)^2 = \frac{1}{2}(2) v_1^2 + \frac{1}{2}(3) v_2^2, so 25=v12+1.5v2225 = v_1^2 + 1.5 v_2^2.

From momentum: v2=(10βˆ’2v1)/3v_2 = (10 - 2 v_1)/3. Substitute and solve.

Result: v1=βˆ’1v_1 = -1 m/s (ball reverses direction); v2=4v_2 = 4 m/s.

Check KE: 12(2)(1)2+12(3)(16)=1+24=25\frac{1}{2}(2)(1)^2 + \frac{1}{2}(3)(16) = 1 + 24 = 25 J. Conserved.

Applications

Car safety. Crumple zones extend the time of a collision, reducing the force (impulse-momentum theorem: F=Ξ”p/Ξ”tF = \Delta p / \Delta t). Larger Ξ”t\Delta t means smaller FF.

Rocket propulsion. Reaction mass expelled backwards gives forward momentum to the rocket (Newton's third law / conservation of momentum).

Sports. Follow-through in tennis or golf extends contact time to deliver more impulse.

Common errors

Confusing FF and mama. F=maF = ma relates net force to acceleration. The total force on the object equals mama; the acceleration is in the direction of net force.

Including internal forces. In conservation-of-momentum problems, only external forces affect total momentum. Internal forces (between parts of the system) cancel by Newton's third law.

Momentum is a vector. Direction matters. In 1D, sign indicates direction.

Confusing elastic and inelastic. Always check kinetic energy separately to determine elasticity.

Action-reaction pair on same object. The two third-law forces act on different bodies. They do not cancel; they obey Newton's third law together.

In one sentence

Newton's three laws of motion (inertia, F=maF = ma, action-reaction) describe how forces produce changes in motion; the impulse-momentum theorem J=FΞ”t=Ξ”pJ = F \Delta t = \Delta p relates force application over time to momentum change; momentum is conserved in any isolated system, allowing collision problems to be solved by m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2, with elastic collisions also conserving kinetic energy and inelastic ones losing some to heat, sound or deformation.

Past exam questions, worked

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

Year 11 SAC4 marksA $1500$ kg car travelling at $20$ m s$^{-1}$ collides with a stationary $1000$ kg car. After the collision they move together. (a) Find the velocity after collision. (b) Is kinetic energy conserved?
Show worked answer β†’

(a) Conservation of momentum.

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

1500Γ—20+1000Γ—0=2500Γ—v1500 \times 20 + 1000 \times 0 = 2500 \times v

30000=2500v30000 = 2500 v

v=12v = 12 m sβˆ’1^{-1}.

(b) Kinetic energy.

KE before: 12(1500)(20)2=300,000\frac{1}{2}(1500)(20)^2 = 300,000 J.

KE after: 12(2500)(12)2=180,000\frac{1}{2}(2500)(12)^2 = 180,000 J.

Lost: 120,000120,000 J. Kinetic energy is NOT conserved; this is an inelastic collision (energy dissipated as heat, sound, deformation).

Momentum is conserved in all isolated-system collisions, but kinetic energy is conserved only in elastic collisions.

Markers reward the momentum equation, the explicit kinetic-energy check, and the distinction between elastic and inelastic.

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