NCEA Level 3 Physics (90521) 2010 — page 1 of 3
Assessment Schedule – 2010
Physics: Demonstrate understanding of mechanical systems (90521)
Evidence Statements
Judgements in italics indicate replacement evidence and so are not counted for sufficiency.
Q
ONE
(a)(i)
(ii)
Evidence

Achievement
2
 8.49ζradζs1
0.74
Achievement with
Merit
Achievement with
Excellence
2
Correct answer
2
Correct answer
Clear idea of how
rotational inertia is
affected by mass
distribution.
1
Mass distribution
concept explained
in terms of
EITHER the need
for a radius
estimation
OR the need to
ignore the mass of
both the spokes
and the hub.
1
Mass distribution
concept explained in
terms of both points.
Correct EK(lin)
OR
1
Correct concept of
work =
EK(lin) + EK(rot)
2
Correct answer
consistent with
using the wrong
radius or a
diameter instead
of a radius.
2
Correct answer.
To accelerate the wheel a torque must be
applied. As  = I, and as the mountain bike
has less rotational inertia, less torque needs to
be applied to accelerate it at the same rate.
OR
To accelerate the wheel it must be given
rotational kinetic energy. As EK = ½I2, and
as the rotational inertia of the mountain bike is
smaller, less energy will need to be given to it
to accelerate it to a particular angular speed in
a particular time.
1
Idea that less
torque must be
applied to
accelerate the
mountain bike
wheel.
OR that less energy
has to be supplied
to give the
mountain bike
wheel its rotational
kinetic energy.
1
Concept given for
achievement is
clearly explained.
The only forces acting to bring about the
change in position are internal forces.
1
Good idea of no
external forces
acting.

t

 2 
1
Ek  I  2  0.5  0.64  
2
 0.74 
2
 23.1ζJ
(b)
Use the mass of the wheel as m and use an r
value that is between the rim radius and the
tyre radius.
Assume the mass of the spokes and hub is
negligible.
1
(c)
Work done =  energy = EK(lin) + EK(rot)
= ½mv2 + 2  ½I2,
v
5.5
v  r ΚsoΚ  
 15.19
r
0.724
2
1 2
1
EK  EK  EK  mv  2  m 2
lin
rot
2
2
1


1
2
  68.4  5.5    2  0.64  15.192 
2
2

 

2


 1035  148  1183ΚJ  1180ΚJΚ(3Κs.f.)
(d)
TWO
(a)
NCEA Level 3 Physics (90521) 2010 — page 2 of 3
(b)
11x = 55(13.2  x)
13.2
 11.0Κcm
x=
 11 
 55  1
2
Correct working
(c)
v = (3.5 ms1 + 1.8 m s1)
2
Correct answer.
2
Correct
acceleration /
correct change in
momentum /
correct force using
m = 55 kg.
2
Correct answer.
3.5  1.8
2
=
(d)
2
= 3.9357
1
= 3.94 m s
mv
mv
, OR p = Ft  F =
t
t
v = vi (vf = 0)
F = ma =
 F = 66 
3.9357
= 311 N
0.835
(e)
p = Ft. The vertical speed is half the
horizontal speed, and so change in vertical
momentum is half the change in vertical
momentum. However, the vertical stopping
time is one tenth the horizontal stopping time
and so the upward force needed to stop the
vertical motion will be much greater than the
horizontal friction force needed to stop the
horizontal motion.
1
Shortness of the
vertical stopping
time identified as
the key issue.
OR
2
Correct calculation
of average
horizontal or
vertical force.
1
Short stopping
time linked to the
need for a large
force.
(f)
The crumpling increases the time it takes for
the vertical motion (of the brain (skull / head))
to come to a halt and as the change in
momentum hasn’t changed, there will be a
decrease in the force required.
1
Crumple linked to
an increase in the
stopping time.
1
Crumple
explained in terms
of a valid reason
for a decrease in
the force.
1
Correct
description.
1
Recognition that
the amplitude
decreases
1
Decrease in
amplitude linked
to decrease in
energy.
2
Correct T
2
Correct answer.
THREE As a = 2y, Size of the acceleration must be
(a)
proportional to the size of the displacement.
(b)
(c)
Friction works against the motion, changing
energy to heat. The amplitude depends on the
energy of the motion, so as energy is lost, the
amplitude decreases.
T=
2

=
2
2.40
= 2.61799
1
Answer shows
understanding of how
the difference in
stopping times has
greater significance
than the difference in
momentums.
1
Decrease in amplitude
linked to decrease in
energy which is
linked to the action of
friction changing
energy to heat etc.
83.8
= 32. 01
2.61799
= 32 oscillations.
n=
(d)(i)
(ii)
F = ma = m2y
= 1.65  2.402  0.0270
= 0.2566 = 0.257 N
2
Correct value for a.
OR
1
Correct answer.
2
Correct answer
consistent with
using y in
centimetres.
2
Correct working and
answer.
v = A cos t, y = Asint
y
0.027
 0.18472
  t  sin 1  sin 1
A
0.147
 v = 0.147  2.62  cos[0.183673]
= 0.34680 = 0.35 m s–1
2
Correct t or t.
OR
1
Correct answer.
2
Correct answer
consistent with
using y in
centimetres.
2
Correct working and
answer.
NCEA Level 3 Physics (90521) 2010 — page 3 of 3
Judgement Statement:
Achievement
6 correct answers
including at least 3 from criterion 1
and 2 from criterion 2
Achievement with Merit
8 correct answers
including 5 M or above
and at least
2 from criterion 1
and 1 from criterion 2
Achievement with Excellence
6 correct answers
at merit or above
including at least 2 E,
1 from each criterion