NCEA Level 1 Science 90940 (1.1) — page 1 of 3

SAMPLE ASSESSMENT SCHEDULE

Science 90940 (1.1): Demonstrate understanding of aspects of mechanics

Achievement Criteria

Achievement Merit Excellence

Demonstrate understanding requires the student to provide evidence that will typically show an awareness of how simple facets of phenomena, concepts or principles relate to a described situation. This may include using methods for solving problems involving aspects of mechanics.

Demonstrate in-depth understanding requires the student to provide evidence that will typically show how or why phenomena, concepts or principles relate to given situations.

Demonstrate comprehensive understanding requires the student to provide evidence that will typically show understanding of how or why phenomena, concepts and principles are connected in the context of given situations.

Evidence Statement

One Expected Coverage Achievement Merit Excellence

(a) Calculation of acceleration:

a = v/t

= 12/60

= 0.2 ms-2

TWO of:

in (a) attempts to calculate the acceleration (eg, writes the formula and substitutes values OR finds the acceleration but does not give the unit)

in (b) states at least ONE pair of forces and their relative sizes

in (c) attempts to calculate the distance travelled for ONE section OR completes the graph accurately for at least TWO sections (with the y axis labelled correctly).

TWO of:

in (a) shows understanding of related concepts by calculating the acceleration

in (b) explains why the forces are balanced (eg, push and friction are equal and opposite forces; Fnet=0, therefore forces are balanced. Evidence can come from diagram)

in (c) shows an understanding of how concepts relate to the situation by completing the calculations accurately for each section and attempting to complete the graph.

BOTH of:

in (b) shows an understanding of how the zero net force is connected with the type of motion in Section B of the graph

in (c) draws appropriately shaped graph based on an understanding of the connection between the gradients of speed-time and distance-time graphs (must have correct distance values on axes).

(b) Naming of forces:

weight / gravity downwards

support / reaction upwards

Weight and support are equal and opposite forces:

thrust forward

friction pushes against motion.

Thrust and friction are equal and opposite forces.

Explanation of motion:

Forces are balanced therefore the net force is zero. As the bike is already moving in Section B, it will continue moving at a constant speed as an unbalanced force is required to change its speed.

(c) Calculation of distances:

dA = ½ x 60 x 12 = 360 m

dB = 60 x 12 = 720 m

dC = ½ x 30 x 12 = 180 m

Drawing of graph shapes:

NCEA Level 1 Science 90940 (1.1) — page 2 of 3

Two Evidence Achievement Merit Excellence

(a) Explanation of ‘no work’:

Work is done when a force causes an object to move in the direction of the force. Here the force is not causing the car to move, so no work is being done. It has gained no gravitational potential energy.

TWO of:

in (a), makes an accurate statement concerning the motion of the car (eg, there is no motion, or the force is not causing the movement of the car or similar)

in (b) ONE calculation is undertaken by selecting and substituting into the correct formula and solving it

in (c) makes an accurate statement about mass or weight OR calculates the mass of the car.

TWO of:

in (a) fully explains how the motion of the car is affected by the forces acting on it OR mathematical justification

in (b) shows an understanding of how the TWO formulae that relate to the situation should be applied but fails to give the correct power output or unit

in (c) distinguishes between mass and weight OR calculation plus either mass or weight.

BOTH of:

in (b) shows an understanding of how the TWO formulae relate to the context by accurately calculating the power output

in (c) connects the calculation of the mass of the car with the physics principles underlying the formula used, distinguishing between mass and weight.

(b) Calculation of power output:

W = Fd

= 16 000 x24

= 384 000 J

P = W/t

= 384 000/80

= 4800 W

(c) Calculation of mass:

Fw = mg

so m = Fw/g

= 13 000 / 10

= 1300 kg

Explanation of physics principles:

The car is motionless; therefore the vertical forces are balanced. The weight force on the car must equal the force applied by crane to the car.

Explanation of difference:

Mass is the amount of material (matter) in the object (not the amount atoms / particles).

Weight is the gravitational force on the object (not the amount of gravity).

Three Expected Coverage Achievement Merit Excellence

(a) Calculation of pressure:

P = F/A

Force is given by Rosemary’s weight

Fw = mg = 80 10 = 800 N

Area of skies

A = (1.60 0.10 ) 2 = 0.32 m2

P = 800/0.32 = 2 500 Pa

TWO of:

in (a) calculates the area or weight force correctly

in (b) makes the correct identification and supports it with a simple reason

in (b) calculates the area of boots correctly.

BOTH of:

in (a) shows an understanding of how the TWO formulae that relate to the situation should be applied but fails to give the correct pressure or unit

in (b) shows an understanding of the physics principles involved by justifying the correct identification with reference to the greater pressure exerted even though the

BOTH of:

in (a) shows an understanding of how, force, area, and pressure are connected in the given situation by accurately calculating the pressure

in (b) supports the correct identification with a complete logical argument based on the connections between the physics concepts involved including calculations

(b) Identity: Jacob.

Explanation of physics principles:

(1) The skis have a much bigger area than the boots.

(2) P = F/A, so if A is bigger then the pressure must be smaller or vice versa.

(3) Even though Rosemary is heavier than Jacob, her weight is applied over a larger area. Jacob’s lighter weight is applied over a smaller area. So Jacob puts more pressure on the

NCEA Level 1 Science 90940 (1.1) — page 3 of 3

ground and so sinks more. This can be shown by the calculation

Fw = 77 x 10 = 770N

A = (0.12 x 0.27) x 2

P = 11883 Pa or Nm-2

person is lighter. OR calculation plus simple reason.

Four Expected Coverage Achievement Merit Excellence

(a) Calculation of average speed:

vav = Δd/Δt

= 110 / 36.7

= 3 ms-1

TWO of:

in (a) calculates the average speed

in (b) uses the correct formula in an attempt to calculate the kinetic energy or gives the correct kinetic energy with no unit

in (c) identifies gravitational potential energy as being present at the top of the slope OR recognises that the energy difference is due to energy being converted to heat.

BOTH of:

in (b) shows understanding of related concepts by finding the kinetic energy with unit

in (c) shows understanding of concepts and principles that relate to the given situation by calculating the gravitation potential energy and explaining what happened to the ‘missing energy’.

BOTH of:

in (c) calculates Ep and difference in energy correctly.

in (c) connects the justification for the difference in energy between the top and bottom of the slope with the relevant physics principles showing calculations and discussing frictional forces causing energy loss as heat.

(b) Calculation of kinetic energy:

EK = ½mv2

= ½ 80 82

= 2560 J

(c) Explanation of energy difference:

At the top of slope she has a certain amount of gravitational potential energy and no kinetic energy.

When she reaches the bottom of the slope her gravitational potential energy has changed to kinetic energy.

Ep = mgh = 80 10 4.8 = 3840 J

Difference between Ep & Ek

= 3840 – 2560 = 1280 J (possible follow on from (b))

Some of the kinetic energy is converted into heat due to friction between the skies and the snow and an air resistance between the skier and the air.

Judgement Statement

Achievement Achievement with Merit Achievement with Excellence

Minimum of:

2A

Minimum of:

2M

Minimum of:

2E