Level 1 Science (90940) 2011 Assessment Schedule

NCEA Level 1 Science (90940) 2011 — page 1 of 6

Assessment Schedule – 2011 Science: Demonstrate understanding of aspects of mechanics (90940) Evidence Statement

Question Evidence Achievement Achievement with Merit Achievement with Excellence

ONE Dd 2400 • Calculates average speed. v = = = 40 ms-1 (a) Dt 60

(b) Draw and label forces. • Correct statement and / or • Describes ONE force (weight / • Air resistance = 0N at the instant Forces acting are the weight downwards and air diagram. gravity) and explains why an the parachutist leaves the plane. resistance upwards. unbalanced force situation exists. Acceleration at that instant will be 10ms-2 Weight is greater than air resistance when she has just Eg: • Describes TWO forces and jumped. Identifies two forces. explains why they are AND unbalanced. • When falling her air resistance Net force OR < weight / gravity, therefore Net force is in the downwards direction and greater than • Describes TWO forces and Describes net force. unbalanced forces apply (net zero. explains why they are balanced OR (when parachutist is travelling at downward force) which causes Forces are unbalanced. Forces are unbalanced. “terminal velocity”). acceleration (increase in speed). Explanation of motion OR • Describes that unbalanced forces Motion is acceleration towards ground. Students must answer both to gain Acceleration. lead to acceleration An unbalanced force is required to make an object’s point. speed change, therefore, as there is an unbalanced force greater than zero towards the ground, the parachutist’s speed will increase.

(c) The parachute has an increased surface area compared to • Correct statement: • Explains effect of surface area • Explains the link between the parachutist on her own. This increases the size of the air Increased surface area (size and shape of parachute) on increased surface area of resistance / friction / drag force acting in the upwards the forces involved (air resistance parachute and the change to the OR direction making it larger than the force of gravity. This AND weight / gravity) . Needs to forces involved (net force is creates an unbalanced force in the upwards direction (a Unbalanced force acting. mention both forces. upward), and how this reduces negative net force), causing the parachutist to decelerate. OR the speed of fall (deceleration). • That air resistance is slowing parachutist as net force is upwards.

Not achieved Achievement Achievement with Merit Achievement with Excellence NCEA Level 1 Science (90940) 2011 — page 2 of 6

NØ = no evidence N1 = correct N2 = 1 point Q1 or no relevant idea. Eg correct from A3 = 2 points A4 = 3 points M5 = 2 points M6 = 3 points E7 = 1 point E8 = 2 points evidence unit Achievement NCEA Level 1 Science (90940) 2011 — page 3 of 6

TWO F = ma • Calculates F with unit. (a) = 60  0.2 = 12 N

(b) W = Fd • Calculates either • Calculates Power done, but: • Calculates power and gives an = 12  12.5 = 150 J W correct. EITHER appropriate unit. W P = OR Does not give a unit. t P using an incorrect value for W, OR 150 = = 30 W (or J s-1 ) which has not been calculated. Gives an incorrect value of P 5 using a calculated value of W. Possible follow-on error from 2(a).

(c) Dv • Has correct shapes for sections • Either • Graph with correct values on y- a = X or Y on graph, but no values axis and correct shapes for Dt Calculates vf correctly. v - 2 on y-axis. sections X and Y. 0.2 = f OR 5 Shows correct value (3) on y-axis

vf = (0.2 ´ 5) + 2 but graph is incorrect. = 1+ 2 –1 = 3 ms • Calculated vf incorrectly but –1 OR showed 1m s increase using By reasoning (formula not required), ie 5 s at 0.2 m s–2 acceleration. Increase in speed = 1 m s–2 • Showed correct shapes for Started at 2 m s–1 sections X and Y. now at 3 m s–1. NCEA Level 1 Science (90940) 2011 — page 4 of 6

Not achieved Achievement Achievement with Merit Achievement with Excellence

N1 = correct N2 = 1 point NØ = no evidence or formula not Q2 from A3 = 2 points A4 = 3 points M5 = 2 points M6 = 3 points E7 = 1 point E8 = 2 points no relevant evidence correctly used Achievement OR correct unit.

THREE EP = mgh • Correctly calculated height. • Correctly calculated height. (a) 5100 Unit not necessary. Unit necessary Dh = = 8.5m 60 ´10

(b) Name of energy • Identifies type of energy at the • Identifies type of energy at the • Work correctly calculated and Gravitational potential energy. top. top and states that energy is stated to be the same as the Calculation of work • Correct calculation of work. changed to heat. gravitational potential energy changed / gained. Energy W = Fd = 60  10  8 = 4 800 J • Energy difference calculated • Correct calculation of work and difference correctly calculated. OR (allow error in work calculation) correct calculation of energy W = EP = mgh = 60  10  8 = 4 800 J but no reference to work equals difference. • Some of the energy used is Calculation of difference energy. • Correct calculation of energy changed to heat by the muscles of the girl’s body, and so the Work done is energy gained, • Heat is produced. difference (allow consequential so energy difference = 5 100 – 4 800 = 300J error) AND work stated to be gravitational potential energy equal to energy changed. she gains is less than the total Explanation of difference energy that she uses. The difference is due to energy being converted into heat in the muscles of her body.

E7 = both points NØ = no evidence N1 = correct N2 = 1 correct E8 = both attempted, but Q3 or no relevant formula not point from A3 = 2 points A4 = 3 points M5 = 2 points M6 = 3 points points fully some minor evidence correctly used Achievement answered inconsistencies or omissions NCEA Level 1 Science (90940) 2011 — page 5 of 6

FOUR F = mg = 40  10 = 400 N • Calculates • Calculates pressure for one of • Calculates pressures of both, (a) For boot without studs boots. with correct unit. F P = Weight force. A OR 400 = Pressure incorrectly using 0.0165 F = 40 = 24 242 N m-2 (Pa) = 2.43 N cm-2 For boot with studs F P = A 400 = 0.0006 = 666 667 N m-2 (Pa) = 66.7 N cm-2

(b) Comparison • States correct pressure • Explains the relationship between • Links greater pressure created The pressure exerted by the boot with studs is greater comparison. surface area and pressure with by smaller area of studs to more than the boot without studs. Eg, states smaller surface area respect to studded and non- grip / traction when running on Relationship leads to greater pressure or visa studded boots. grass. The smaller the surface area (the force is exerted over), versa. OR • Comparison with the other boot the greater the pressure exerted on the ground. • Shows understanding of physics without stud is provided (ie How it helps on grass • States correct effects comparison principles involved by smaller pressure created by larger area of shoe leads to As the student’s weight is applied over a smaller surface Eg, states studs will sink into identifying greater pressure leads slipping / loss of friction.) area with the boots with studs, it leads to a greater ground more or states studs will to boots sinking into grass. pressure on the ground. This means that the studs will give better grip. sink into the ground more, which will enable the student to get a better grip / traction on grass when running. NCEA Level 1 Science (90940) 2011 — page 6 of 6

N1 = correct unit NØ = no evidence N2 = 1 point OR Q4 or no relevant from A3 = 2 points A4 = 3 points M5 = 1 point M6 = 2 points E7 = 2 point E8 = 3 points correct equation evidence Achievement F = mg, but incorrect calc

Judgement Statement

Achievement Achievement Not Achieved Achievement with Merit with Excellence Score range 0 – 10 11 – 19 20 – 24 25 – 32