← Unit 3: How do fields explain motion and electricity?

VICPhysicsSyllabus dot point

How do physicists explain motion in two dimensions?

investigate and apply theoretically and practically Newton's three laws of motion in situations where two or more coplanar forces act along a straight line and in two dimensions; apply the concepts of momentum and impulse, including the conservation of momentum in one and two dimensions, and distinguish between elastic and inelastic collisions

A focused answer to the VCE Physics Unit 3 dot point on Newton's laws, momentum and impulse. Covers force, mass and acceleration in two dimensions, impulse as the area under a force-time graph, conservation of momentum in 1D and 2D collisions, and how to tell elastic from inelastic collisions.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answer

Have a quick question? Jump to the Q&A page

What this dot point is asking

VCAA wants you to apply Newton's three laws when more than one force acts, in one or two dimensions, and to use momentum (p=mvp = mv) and impulse (J=FΞ”t=Ξ”pJ = F \Delta t = \Delta p) including the conservation of momentum in collisions. You also need to distinguish elastic collisions (kinetic energy conserved) from inelastic collisions (kinetic energy not conserved).

The answer

Newton's three laws

First law (inertia). An object continues at rest or in uniform straight-line motion unless acted on by a net external force.

Second law. The net force on an object equals its mass times its acceleration: Fnet=maF_{net} = ma. In two dimensions, resolve forces along perpendicular axes and apply Fnet=maF_{net} = ma on each axis independently.

Third law. When object A exerts a force on object B, object B exerts an equal and opposite force on A. The two forces act on different objects, so they never cancel on a single free-body diagram.

Momentum and impulse

Momentum is a vector: p=mvp = mv, units kg m/s.

Impulse is the change in momentum: J=Ξ”p=FnetΞ”tJ = \Delta p = F_{net} \Delta t, units N s (same as kg m/s).

On a force-time graph, impulse is the area under the curve. This is how VCAA tests variable forces (for example a ball striking a bat).

To reduce the average force in a collision, increase the contact time Ξ”t\Delta t for a given Ξ”p\Delta p. This is why airbags, crumple zones and bent knees on landing reduce injury.

Conservation of momentum

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

βˆ‘pbefore=βˆ‘pafter\sum p_{before} = \sum p_{after}

In two dimensions, momentum is conserved independently along each axis. Resolve the velocity vectors into x and y components and apply conservation on each axis.

Elastic vs inelastic collisions

Momentum Kinetic energy
Elastic conserved conserved
Inelastic conserved not conserved
Perfectly inelastic (stick together) conserved not conserved (maximum KE lost)

In VCE, almost all real collisions are inelastic. Steel ball bearings come close to elastic; cars colliding, clay landing on a board, and bullets embedding in blocks are all inelastic.

Worked example with numbers

A 0.500.50 kg ball moving east at 4.04.0 m/s collides with a stationary 0.300.30 kg ball. After the collision the 0.500.50 kg ball moves east at 1.01.0 m/s. Find the velocity of the 0.300.30 kg ball and classify the collision.

Conservation of momentum: (0.50)(4.0)+0=(0.50)(1.0)+(0.30)v(0.50)(4.0) + 0 = (0.50)(1.0) + (0.30) v.

2.0=0.5+0.30v2.0 = 0.5 + 0.30 v, so v=5.0v = 5.0 m/s east.

Kinetic energy before: 12(0.50)(4.0)2=4.0\frac{1}{2}(0.50)(4.0)^2 = 4.0 J. After: 12(0.50)(1.0)2+12(0.30)(5.0)2=0.25+3.75=4.0\frac{1}{2}(0.50)(1.0)^2 + \frac{1}{2}(0.30)(5.0)^2 = 0.25 + 3.75 = 4.0 J.

KE is conserved, so the collision is elastic.

Common traps

Treating Newton's third law as cancelling forces. The action-reaction pair acts on two different objects. In a free-body diagram of one object, only one of the pair appears.

Forgetting that momentum is a vector. When a ball rebounds, Ξ”p\Delta p uses signed velocities. A ball striking a wall at 88 m/s and rebounding at 66 m/s changes momentum by m(8βˆ’(βˆ’6))=14mm(8 - (-6)) = 14m, not 2m2m.

Assuming kinetic energy is conserved. Only momentum is automatically conserved in an isolated collision. KE is conserved only if the collision is explicitly elastic.

Confusing impulse with force. Impulse has units N s, force has units N. Always include Ξ”t\Delta t.

Using speeds instead of velocity components in 2D. Conservation of momentum is applied separately to the x and y axes.

In one sentence

Newton's laws give Fnet=maF_{net} = ma on each axis, impulse is the area under a force-time graph equal to Ξ”p\Delta p, and momentum is conserved (vectorially) in every isolated collision while kinetic energy is conserved only in elastic ones.

Past exam questions, worked

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

2022 VCE4 marksA 1500 kg car travelling east at 12 m/s collides head-on with a 1000 kg car travelling west at 18 m/s. The two cars stick together. Calculate the velocity of the wreckage immediately after the collision and state whether the collision is elastic or inelastic.
Show worked answer β†’

Take east as positive. Apply conservation of momentum: pbefore=pafterp_{before} = p_{after}.

pbefore=(1500)(12)+(1000)(βˆ’18)=18000βˆ’18000=0p_{before} = (1500)(12) + (1000)(-18) = 18000 - 18000 = 0 kg m/s.

Combined mass =2500= 2500 kg, so vafter=0/2500=0v_{after} = 0 / 2500 = 0 m/s. The wreckage is stationary.

The collision is inelastic because the cars stick together (a perfectly inelastic collision). Kinetic energy is not conserved: KEbefore=12(1500)(12)2+12(1000)(18)2=108000+162000=270000KE_{before} = \frac{1}{2}(1500)(12)^2 + \frac{1}{2}(1000)(18)^2 = 108000 + 162000 = 270000 J; KEafter=0KE_{after} = 0 J. The lost kinetic energy is converted to heat, sound and deformation.

Markers reward the signed setup, the conservation statement, and the explicit elastic-vs-inelastic justification using kinetic energy.

2024 VCE3 marksA 0.20 kg ball travelling at 8.0 m/s strikes a wall and rebounds at 6.0 m/s. The ball is in contact with the wall for 0.040 s. Calculate the average force exerted on the ball by the wall.
Show worked answer β†’

Impulse equals change in momentum: J=Ξ”p=FΞ”tJ = \Delta p = F \Delta t.

Take the initial direction of motion as positive, so vi=+8.0v_i = +8.0 m/s and vf=βˆ’6.0v_f = -6.0 m/s.

Ξ”p=m(vfβˆ’vi)=0.20Γ—(βˆ’6.0βˆ’8.0)=0.20Γ—(βˆ’14.0)=βˆ’2.80\Delta p = m(v_f - v_i) = 0.20 \times (-6.0 - 8.0) = 0.20 \times (-14.0) = -2.80 kg m/s.

F=Ξ”pΞ”t=βˆ’2.800.040=βˆ’70F = \frac{\Delta p}{\Delta t} = \frac{-2.80}{0.040} = -70 N.

The magnitude of the average force is 70 N, directed away from the wall (back toward the ball's origin).

Markers reward signs handled correctly, impulse equated to change in momentum (not just mvmv), and a final answer with magnitude and direction.

Related dot points