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
(a)
(b)
Expected Coverage
Merit
Excellence
Calculation of acceleration:
TWO of:
TWO of:
BOTH of:
a = v/t
= 12/60
-2
= 0.2 ms
 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 (a) shows
understanding of
related concepts
by calculating
the acceleration
 in (b) shows an
understanding of
how the zero net
force is
connected with
the type of
motion in
Section B of the
graph
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)
Achievement
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:
 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).
 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.
 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).
NCEA Level 1 Science 90940 (1.1) — page 2 of 3
Two
(a)
(b)
Evidence
Achievement
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.
Calculation of power output:
W = Fd
= 16 000 x24
= 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
(a)
Calculation of pressure:
P = F/A
Force is given by Rosemary’s weight
A = (1.60  0.10 )  2 = 0.32 m
P = 800/0.32 = 2 500 Pa
(b)
2
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
Excellence
TWO of:
TWO of:
BOTH 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 (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 relate
to the context by
accurately
calculating the
power output
 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) connects
the calculation of
the mass of the
car with the
physics
principles
underlying the
formula used,
distinguishing
between mass
and weight.
 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.
Achievement
Fw = mg = 80  10 = 800 N
Area of skies
Merit
 in (c)
distinguishes
between mass
and weight OR
calculation plus
either mass or
weight.
Merit
Excellence
TWO of:
BOTH of:
BOTH of:
 in (a) calculates
the area or
weight force
correctly
 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 (a) shows an
understanding of
how, force, area,
and pressure are
connected in the
given situation
by accurately
calculating the
pressure
 in (b) makes the
correct
identification and
supports it with a
simple reason
 in (b) calculates
the area of boots
correctly.
 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
 in (b) supports
the correct
identification
with a complete
logical argument
based on the
connections
between the
physics
concepts
involved
including
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
Four
(a)
(b)
Expected Coverage
Achievement
Calculation of average speed:
vav = Δd/Δt
= 110 / 36.7
-1
= 3 ms
Calculation of kinetic energy:
2
EK = ½mv
2
= ½  80  8
= 2560 J
(c)
person is lighter.
OR calculation
plus simple
reason.
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.
Merit
calculations
Excellence
TWO of:
BOTH of:
BOTH of:
 in (a) calculates
the average
speed
 in (b) shows
understanding of
related concepts
by finding the
kinetic energy
with unit
 in (c) calculates
Ep and
difference in
energy correctly.
 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.
 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’.
 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.
Judgement Statement
Achievement
Achievement with Merit
Achievement with Excellence
Minimum of:
2A
Minimum of:
2M
Minimum of:
2E