2821
ADVANCED SUBSIDIARY GCE UNIT
PHYSICS A
Forces and Motion
FRIDAY 8 JUNE 2007
Morning
Time: 1 hour
Additional materials:
Electronic calculator
Ruler (cm/mm)
Protractor
INSTRUCTIONS TO CANDIDATES
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•
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Write your name, Centre Number and Candidate number in the boxes above.
Answer all the questions.
Use blue or black ink. Pencil may be used for graphs and diagrams only.
Read each question carefully and make sure you know what you have to do before starting your answer.
Do not write in the bar code.
Do not write outside the box bordering each page.
For Examiner’s Use
WRITE YOUR ANSWER TO EACH QUESTION IN THE SPACE
PROVIDED. ANSWERS WRITTEN ELSEWHERE WILL NOT BE MARKED.
Qu.
Max.
Mark
INFORMATION FOR CANDIDATES
1
11
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2
8
3
9
4
8
5
12
6
12
Total
60
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The number of marks for each question is given in brackets [ ] at the end of
each question or part question.
The total number of marks for this paper is 60.
You will be awarded marks for the quality of written communication where
this is indicated in the question.
You may use an electronic calculator.
You are advised to show all the steps in any calculation.
This document consists of 16 printed pages.
SPA (DR/DR) T15139/7
© OCR 2007 [M/100/3700]
OCR is an exempt Charity
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Data
speed of light in free space,
c = 3.00 × 10 8 m s –1
permeability of free space,
0 = 4 × 10 –7 H m–1
permittivity of free space,
0 = 8.85 × 10 –12 F m–1
elementary charge,
e = 1.60 × 10 –19 C
the Planck constant,
h = 6.63 × 10 –34 J s
unified atomic mass constant,
u = 1.66 × 10 –27 kg
rest mass of electron,
me = 9.11 × 10 –31 kg
rest mass of proton,
mp = 1.67 × 10 –27 kg
molar gas constant,
the Avogadro constant,
R = 8.31 J K –1 mol –1
NA = 6.02 × 10 23 mol –1
gravitational constant,
G = 6.67 × 10 –11 N m 2 kg –2
acceleration of free fall,
g = 9.81 m s –2
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Formulae
uniformly accelerated motion,
s = ut +
1
2
at 2
v 2 = u 2 + 2as
1
sin C
refractive index,
n=
capacitors in series,
1
1
1
=
+
+...
C C1 C2
capacitors in parallel,
C = C1 + C2 + . . .
capacitor discharge,
x = x0e–t/CR
pressure of an ideal gas,
p=
radioactive decay,
x = x0e– λt
1
3
Nm 2
<c >
V
t = 0.693
λ
critical density of matter in the Universe,
relativity factor,
3H02
ρ0 =
8G
= √ (1 –
current,
I = nAve
nuclear radius,
r = r0A1/3
sound intensity level,
© OCR 2007
= 10 lg
v2
)
c2
( )
I
I0
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Answer all the questions.
1
Fig. 1.1 shows a vacuum cleaner of weight W being pushed with a force P. The force P acts at 30°
to the horizontal.
dust
collector
P = 24.0N
30°
floor
W = 65.0N
Fig. 1.1
The weight W is 65.0 N and the magnitude of force P is 24.0 N.
(a) (i)
Calculate
1 the horizontal component of the force P
horizontal component = ................................N
2 the vertical component of the force P.
vertical component = ................................N [3]
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(ii)
Show that the total downward vertical force is 77.0 N.
[1]
(iii)
Hence determine the magnitude of the resultant of the forces W and P.
resultant force = ................................N [2]
(iv)
The vacuum cleaner is not switched on and is pushed in such a way that it travels at a
constant velocity to the left. There are other forces acting on the vacuum cleaner. State
and explain the magnitude of the resultant of these other forces.
...........................................................................................................................................
...........................................................................................................................................
...........................................................................................................................................
.......................................................................................................................................[2]
(b) (i)
The total area of the vacuum cleaner in contact with the floor is 4.2 x 10–3 m2.
Calculate the pressure exerted on the floor by the total downward vertical force.
pressure = ................................Pa [2]
(ii)
State and explain what happens to this pressure if the handle is lifted so that its angle
with the horizontal direction is more than 30°. The force P and the total area in contact
with the floor remain constant.
...........................................................................................................................................
...........................................................................................................................................
.......................................................................................................................................[1]
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[Total: 11]
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2
(a)
Explain the term centre of gravity of an object.
...................................................................................................................................................
...................................................................................................................................................
...............................................................................................................................................[2]
(b) Fig. 2.1 shows a lawn mower which is carried by two people.
A diagram has been removed due to third party copyright restrictions
Details:
A diagram of a lawn mower
Fig. 2.1
(i)
The two people apply forces
lawn mower is 350 N.
A and B at each end of the lawn mower. The weight of the
1 Explain why the weight of the lawn mower does not act in the middle of the lawn
mower, that is 55 cm from each end.
........................................................................................................................................
....................................................................................................................................[1]
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2 Use the principle of moments to show that the force B is 64 N.
[2]
3 Determine the force A.
A = ................................N [1]
(ii)
State and explain what happens to the forces A and B if the person that applies force B
moves his hands along the handle towards the middle of the lawn mower.
...........................................................................................................................................
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...........................................................................................................................................
.......................................................................................................................................[2]
[Total: 8]
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3
Fig. 3.1 shows the path of a tennis ball after passing over the net.
tennis
ball
net
1.20m
line
11.9 m
Fig. 3.1
As the ball passes over the net it is travelling horizontally at a height of 1.20 m. The ball strikes the
ground on a line 11.9 m from the net.
(a) Assume that there is no air resistance acting on the ball.
(i)
Show that the time taken for the ball to reach the line after passing over the net is
0.495s.
[3]
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(ii)
At the instant the ball strikes the line calculate
1 the horizontal component of its velocity
horizontal component = ................................m s–1 [2]
2 the vertical component of its velocity.
vertical component = ................................m s–1 [2]
(b) The mass of the tennis ball is 6.00 x 10–2 kg. Calculate the loss in gravitational potential
energy of the ball from the time it passes over the net until it hits the line.
loss in potential energy = ................................J [2]
[Total: 9]
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(a)
Define the Young modulus .
...................................................................................................................................................
...............................................................................................................................................[1]
(b) Fig. 4.1 shows a violin.
A diagram has been removed due to third party copyright restrictions
Details:
A diagram of a violin
Fig. 4.1
Two of the wires used on the violin, labelled
A and G are made of steel. The two wires are
both 500 mm long between the pegs and support. The 500 mm length of wire labelled
G has
a mass of 2.0 x 10 –3 kg. The density of steel is 7.8 x 10 3 kg m –3 .
(i)
Show that the cross-sectional area of wire
G is 5.1 x 10
–7
m 2.
[2]
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(ii)
The wires are put under tension by turning the wooden pegs shown in Fig. 4.1. The
Young modulus of steel is 2.0 x 1011 Pa.
Calculate the tension required in wire G to produce an extension of 4.0 x 10–4 m.
tension = ................................N [3]
(iii)
Wire A has a diameter that is half that of wire G. Determine the tension required for wire
A to produce an extension of 16 x 10–4 m.
tension = ................................N [1]
(iv)
State the law that has been assumed in the calculations in (ii) and (iii).
.......................................................................................................................................[1]
[Total: 8]
© OCR 2007
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5
Fig. 5.1 shows a vehicle that is used for carrying people ‘off-road’.
An image has been removed due to third party copyright
restrictions
Details:
An image of a 4x4 vehicle
Fig. 5.1
An off-road vehicle is designed so that it can be driven on rough and uneven ground. The mass of
the vehicle and occupants is 3000 kg.
(a)
Explain how the tyres are designed to reduce the pressure of the vehicle on the surface over
which it is travelling.
...................................................................................................................................................
...............................................................................................................................................[1]
(b) The braking distance of the vehicle when travelling on a normal road at 26 ms
–1
is 52 m.
Calculate
(i)
the kinetic energy of the vehicle and occupants before braking occurs
kinetic energy = ................................J [3]
(ii)
the average deceleration of the vehicle when braking
deceleration = .................. unit................. [2]
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(iii)
the average braking force acting during the deceleration.
braking force = ................................N [2]
(c) (i)
State and explain how two different road conditions affect the braking distance of a car.
1. ........................................................................................................................................
...........................................................................................................................................
2. ........................................................................................................................................
.......................................................................................................................................[2]
(ii)
The braking distance for a small car is shorter than for the off-road vehicle described
above when they are tested travelling on the same road surface at the same speed.
Discuss one difference between the small car and the off-road vehicle that could explain
this.
...........................................................................................................................................
...........................................................................................................................................
.......................................................................................................................................[2]
[Total: 12]
© OCR 2007
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6
In this question, two marks are available for the quality of written communication.
A skydiver jumps from an aircraft that is flying horizontally at a height of 5000 m with a constant
speed.
(a) Describe and explain the motion of the skydiver as she descends towards the ground from
the moment she jumps until she opens her parachute. In your description of her motion use
the terms speed, acceleration and force.
...................................................................................................................................................
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...................................................................................................................................................
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...................................................................................................................................................
...............................................................................................................................................[6]
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(b) In some competitions the skydiver attempts to travel at the highest possible speed over a
vertical distance of 1000 m. Discuss and explain the different methods that could be used by
the skydiver to achieve the highest possible speed.
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...............................................................................................................................................[4]
Quality of Written Communication [2]
[Total: 12]
END OF QUESTION PAPER
© OCR 2007
16
PLEASE DO NOT WRITE ON THIS PAGE
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reasonable effort has been made by the publisher (OCR) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the
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© OCR 2007
2821
Mark Scheme
June 2007
Mark Scheme
Unit Code
Session
Year
Version
Page 1 of 6
2821
JUNE
2007
FINAL
Abbreviations,
annotations and
conventions
used in the
Mark Scheme
/
;
NOT
()
Question 1
Expected Answers
1
ecf
AW
ora
=
=
=
=
=
=
=
=
alternative and acceptable answers for the same marking point
separates marking points
answers which are not worthy of credit
words which are not essential to gain credit
(underlining) key words which must be used to gain credit
error carried forward
alternative wording
or reverse argument
Marks
(a) (i) 1
Horizontal component = 24cos30
= 21 (20.8) (N)
C1
A1
2
vertical component = 24sin30
= 12
(12.0) (N)
A1
(ii)
(iii)
(iv)
(b) (i)
vertical force = 65 + 12
= 77
horizontal force = 20.8 (note ecf for 20.8 component)
resultant = [(77)2 + (20.8)2 ]1/2
= 80 (79. 8) (N)
(or by vector triangle need correct labels and arrows
for C1 mark)
M1
A0
80 (79.8)(N) / equal to (iii)
allow ecf
the resultant force needs to be zero or forces need
to balance above value to give no acceleration or
constant velocity
B1
P=F/A
C1
C1
A1
B1
= 77 / 4.2 x 10-3
= 18000 (18333)
(ii)
A1
(Pa)
more / increases
downward / vertical component (of P) will be greater
(for larger angles)
B1
Total: 11
3
2821
Mark Scheme
June 2007
Mark Scheme
Unit Code
Session
Year
Version
Page 2 of 6
2821
JUNE
2007
FINAL
Abbreviations,
annotations and
conventions
used in the
Mark Scheme
/
;
NOT
()
Question
Expected Answers
2
(a)
The(single) point where the weight of an object (may
be taken to) acts
Weak answers score one only e.g. where the weight
acts
(b) (i) 1
The (distribution of the) mass of the lawn mower is
not uniform
2
3
(b)(ii)
ecf
AW
ora
=
=
=
=
=
=
=
=
alternative and acceptable answers for the same marking point
separates marking points
answers which are not worthy of credit
words which are not essential to gain credit
(underlining) key words which must be used to gain credit
error carried forward
alternative wording
or reverse argument
Marks
B2
B1
One correct moment about A stated
B x 110 or 350 x 20
B1
B = (350 x 20) / 110 (moments equated)
B1
B = 63.6 (N)
A0
A = 350 – 63.6 = 286(.4) (N)
A1
A goes down and B goes up
B1
Turning effect of B is less / B needs greater force to
produce the same moment / if distance goes down
force needs to go up (to maintain the same turning
effect)
B1
Total: 8
4
2821
Mark Scheme
June 2007
Mark Scheme
Unit Code
Session
Year
Version
Page 6 of 6
2821
JUNE
2007
FINAL
Abbreviations,
annotations and
conventions
used in the
Mark Scheme
/
;
NOT
()
Question 6
Expected Answers
6
(a)
ecf
AW
ora
=
=
=
=
=
=
=
=
alternative and acceptable answers for the same marking point
separates marking points
answers which are not worthy of credit
words which are not essential to gain credit
(underlining) key words which must be used to gain credit
error carried forward
alternative wording
or reverse argument
Marks
Vertical motion:
• Initial speed (down) is zero
• Initial acceleration (down) is g / 9.8 m s-2
• Initial force (down) is weight / no drag force
• Speed (down) increases
• Acceleration decreases
• Drag force increases (with speed) / force is
weight – air resistance
• Reaches terminal velocity / speed is constant
• Acceleration is zero
• Drag force equals weight
Maximum of 5 marks for vertical motion
Horizontal motion:
• Initial velocity is the same as the aircraft
• Air resistance will reduce this horizontal
velocity
Maximum of six marking points required
(b)
•
•
•
•
B6
Reduce air resistance
Fall head first / fall vertically / reduce area in
direction of fall
Fall with arms to the side / in line with body /
streamline body
Wear tight fitting clothes / smooth surface
Good physics mark
QWC
Maximum of four marking points required
B4
SPAG ( greater than two errors)
B1
TECHNICAL LANGUAGE
B1
Total: 12
5
2821
Mark Scheme
June 2007
Mark Scheme
Unit Code
Session
Year
Version
Page 3 of 6
2821
JUNE
2007
FINAL
Abbreviations,
annotations and
conventions
used in the
Mark Scheme
/
;
NOT
()
Question
Expected Answers
ecf
AW
ora
=
=
=
=
=
=
=
=
alternative and acceptable answers for the same marking point
separates marking points
answers which are not worthy of credit
words which are not essential to gain credit
(underlining) key words which must be used to gain credit
error carried forward
alternative wording
or reverse argument
Marks
s = ut + ½ at2
3 (a) (i)
C1
1.20 = 0 + ½ x 9.81 x t2
(ii) 1
or u = 0 and s = ½ x a x t2
A1
t2 = 2.4 / 9.81 / 0.2446
A1
t = 0.494(6) s
A0
horizontal component = distance / time
C1
= 11.9 / 0.495
= 24.0(6) (m s-1)
2
vertical component = u + at
A1
C1
= 0 + 9.81 x 0.495
= 4.86 (m s-1)
(b)
(loss of / change in) potential energy = mgh
A1
C1
= 6 x 10-2 x 9.81 x 1.2
= 0. 71 (0.706)(J) 2 sf needed
A1
Total: 9
6
2821
Mark Scheme
June 2007
Mark Scheme
Unit Code
Session
Year
Version
Page 4 of 6
2821
JUNE
2007
FINAL
Abbreviations,
annotations and
conventions
used in the
Mark Scheme
/
;
NOT
()
Question
Expected Answers
4
ecf
AW
ora
=
=
=
=
=
=
=
=
alternative and acceptable answers for the same marking point
separates marking points
answers which are not worthy of credit
words which are not essential to gain credit
(underlining) key words which must be used to gain credit
error carried forward
alternative wording
or reverse argument
Marks
(a)
Young modulus = stress / strain
B1
(b) (i)
Density = mass / volume
B1
Area x length = mass / density
Area = (2.0 x 10-3) / (7800 x 0.5) or 2.56 x 10-7 / 0.5
= 5.1(3) x 10-7 m2
(ii)
B1
A0
E = (F x l) / (A x e) / stress = F / A (1.6 x 108 and strain
= e / l (8 x 10-4)
C1
F = (E x A x e) / l
= (2 x 1011 x 5.1 x 10-7 x 4.0 x 10-4) / 0.5
C1
=82 (N) (81.6)
A1
(iii)
Diameter for D is half G hence area is ¼ of G
Extension is 4x greater
Tension required is the same = 82 (N)
A1
(iv)
The extension is proportional to the force / Hooke’s
law (OWTE)
B1
Total: 8
7
2821
Mark Scheme
June 2007
Mark Scheme
Unit Code
Session
Year
Version
Page 5 of 6
2821
JUNE
2007
FINAL
Abbreviations,
annotations and
conventions
used in the
Mark Scheme
/
;
NOT
()
Question
Expected Answers
5
ecf
AW
ora
=
=
=
=
=
=
=
=
alternative and acceptable answers for the same marking point
separates marking points
answers which are not worthy of credit
words which are not essential to gain credit
(underlining) key words which must be used to gain credit
error carried forward
alternative wording
or reverse argument
(a)
Large surface area (of tyres) (in contact with road)
large area in contact with the road
(b)(i)
K = ½ mv2
C1
= 0.5 x 3000 x (26)2
= 1014000 (J)
(ii)
C1
1.01 x 106(J)
A1
v2 = u2 + 2as
a = - 262 / (2 x 52)
A1
deceleration = 6.5
unit:
(iii)
(c)(i)
(ii)
Marks
B1
m s-2
using F = ma
B1
or work done = F x d
= 3000 x 6.5
F = 1.01 x 106 / 52
= 19500 (N)
= 19400 (N)
smooth / rough surface less / more friction longer /
shorter distance
ice surface less friction greater distance
dry / wet more / less friction longer / shorter distance
up slope / down slope refer to weight component
and effect
smaller mass
same force gives larger deceleration
or less braking force (from off-road) B1 gives less
deceleration B1
C1
A1
B2 max
B1
B1
Total: 12
8