Endorsed by
University of Cambridge International Examinations
Physics Revision Guide Sarah Lloyd
OXFORD U N I V E R S I T Y PRESS
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Tips for effective revision
4
1 General physics 1.1 L e n g t h , v o l u m e a n d t i m e
7
1.2 Speed, velocity a n d acceleration
8
1.3 Mass a n d w e i g h t
15
1.4 Density
16
1.5 Forces
18
1.6 Energy, w o r k a n d p o w e r
33
1.7 Pressure
41
2 Thermal physics 2.1
3
Kinetic t h e o r y
50
2.2 T h e r m a l properties
56
2.3 Transfer of t h e r m a l e n e r g y
67
Properties of waves 3.1 General w a v e properties
80
3.2
85
Light
3.3 Sound
4
96
Electricity and magnetism 4.1
Magnetism
105
4.2
Electrical quantities
110
4.3
Electric circuits
118
4 . 4 Dangers of electricity
129
4.5
131
Electromagnetic effects
4 . 6 C a t h o d e ray oscilloscope
5
139
Atomic physics 5.1 Radioactivity
150
5.2 The nuclear a t o m
158
How to use this revision guide This b o o k is designed t o be used w i t h Complete
Physics for IGCSE. It offers brief notes a n d
simplified explanations, along w i t h practice questions, t o help y o u understand t h e physics principles required f o r t h e C a m b r i d g e IGCSE syllabus. The notes, examples, s u m m a r y questions and e x a m i n a t i o n questions are divided into sections t h a t relate t o t h e syllabus areas. The e x a m i n a t i o n questions t h a t a c c o m p a n y each subsection (e.g. transfer of t h e r m a l energy, w h i c h is p a r t o f t h e t h e r m a l physics section) w i l l a l l o w y o u t o test yourself a t regular intervals. There are also questions a t t h e e n d of each section so t h a t y o u can test y o u r k n o w l e d g e a n d u n d e r s t a n d i n g o f t h e w h o l e t o p i c . In this w a y y o u can revise t o p i c by t o p i c until y o u have covered t h e entire syllabus.
KEY
IDEAS
J
Begin y o u r revision w e l l in advance o f t h e e x a m i n a t i o n
Find a q u i e t place t o w o r k , f r e e f r o m distractions
/
Stud y in e x a m - l e n g t h sessions b r o k e n u p by 1 0 m i n u t e breaks, w h e n y o u leave y o u r desk
/
Process t h e i n f o r m a t i o n t h a t y o u are t r y i n g t o learn: m a k e y o u r o w n notes , d r a w a t a b l e a n d categorise t h e i n f o r m a t i o n , d r a w m i n d m a p s (spider diagrams) t o s u m m a r i s e t h e key p o i n t s
•/
Test yourself o f t e n using e x a m i n a t i o n questions
Active Revision for Physics Making
revision
notes
D o n ' t j u s t w r i t e o u t y o u r notes. Try t o m a k e t h e m as brief as possible, j u s t p i c k i n g o u t t h e essential p o i n t s . This is q u i t e c h a l l e n g i n g ! You c o u l d tr y w r i t i n g y o u r revision notes o n t o cue cards. They are t h e n m o r e p o r t a b l e t o read o n t h e b u s / t r a m t o school or o n t h e w a y t o see y o u r friends . Drawing
revision
mind maps (spider
diagrams)
This is a g o o d w a y t o visually s u m m a r i s e i n f o r m a t i o n . You can link ideas, w h i c h w i l l help y o u r u n d e r s t a n d i n g of t h e t o p i c . Simplifying d i f f i c u l t concepts i n t o d i a g r a m s w i l l help y o u t o reduce t h e i n f o r m a t i o n t h a t y o u have t o learn f o r t h e e x a m . D r a w i n g o u t a m i n d m a p o n a large sheet of paper will a l l o w y o u t o p u t in m o r e diagrams a n d h i g h l i g h t t h e i m p o r t a n t points in colour. Belo w is an e x a m p l e o f a simple m i n d m a p o n radioactivity.
Radioactive particles ionise atoms in cells, which leads to cell damage or cell death. Can cause cancer in the future.
Dangerous!
Alpha, tx (helium nucleus).
Beta, 0 (electron).
Gamma, -y (electromagnetic wave).
J Unstable nucleus, emits radioactive particles.
-#-
Background radiation is all around us.
RADIOACTIVITY
1
People who work with radiation are monitored to check that they have not been exposed to unsafe levels. They wear a film badge.
Useful!
Gamma emitters are used as tracers in medicine. Beta emitters are used to measure the thickness of paper in factories. Alpha emitters are used in smoke alarms.
Sources of background radiation
cosmic rays 10.0% ,
medical 12.0%
fallout 0.3% industrial 0.7%
the body 12.0%
gammas from rocks etc, 14.0%
radon and thoron 51.0%
Writing your own
test
W r i t e d o w n questions as y o u g o t h r o u g h this Revision Guide. W h e n y o u reach t h e e n d o f t h e t o p i c y o u can use t h e questions t o test y o u r k n o w l e d g e a n d u n d e r s t a n d i n g a n d check y o u r progress. Getting
someone
else to test you
Y o u c o u l d give o n e of y o u r classmates y o u r revision notes a n d ask t h e m t o test y o u . Or y o u c o u l d get t h e m t o ask y o u a b o u t parts of t h e syllabus t h e y d o n ' t u n d e r s t a n d . This is a g o o d test o f y o u r u n d e r s t a n d i n g o f t h a t t o p i c ! Using your
syllabus
Y o u can d o w n l o a d a copy o f t h e syllabus f r o m t h e CIE w e b s i t e . M a k e check lists f o r revision using t h e syllabus t o indicate w h e n y o u have m a d e revision notes, practised e x a m questions etc. D o n ' t f o r g e t t o include a c o l u m n t o tick w h e n y o u feel y o u have u n d e r s t o o d a t o p i c . Doing past exam
questions
There are lots o f e x a m i n a t i o n questions t h r o u g h o u t this b o o k f o r y o u t o try. Try revising a small part o f a t o p i c , say f o r a b o u t 3 0 m i n u t e s a n d t h e n test yourself o n t h e e x a m i n a t i o n style questions provided here. W h e n y o u feel ready, full past papers are available o n t h e HE website Making
revision
posters
If y o u m a k e y o u r revision notes into posters a n d p u t t h e m in places w h e r e y o u w i l l see t h e m o f t e n , y o u w i l l read t h e m w i t h o u t even realising! This w i l l help t o keep topics fresh in y o u r m i n d . Your f a m i l y a n d friends w i l l see w h a t y o u have been revising a n d m i g h t talk t o y o u a b o u t it, w h i c h w i l l help t h e i n f o r m a t i o n in stick y o u r m i n d .
Essential quantities and units
Time
second (s)
t
Force
F
newton (N)
Weight
W
)< newton (N)
Velocity
V
metres per second (m/s)
Speed
s
; metres per second (m/s)
Distance
dor s
- mètre (m)
Acceleration
a
metres per second squared (m/s )
Mass
m
kilogram (kg)
Moment
M
2
'•} newton metre (Nm)
Energy
E
joule (J)
Work done
W
joule (J)
Power
P
1 watt(W)
Current
1
ampere (A)
Potential difference (voltage)
V
volt (V)
Resistance
R
ohm(0)
Charge
Q
coulomb (C)
Frequency
f
hertz (Hz)
Pressure
P
newton per metre squared (N/m )
Temperature
T
degrees Celsius (°C)
Density
p or
2
d
g/cm or kg/m 3
3
Wavelength
A
metre (m)
Specific heat capacity
c
joules per kilogram degree Celsius
Specific latent heat
/
joules per kilogram (J/kg)
Formulae and magic triangles These magic triangles are a useful w a y o f r e m e m b e r i n g h o w t o use an e q u a t i o n , b u t are n o s u b s t i t u t e f o r r e m e m b e r i n g t h e e q u a t i o n itself. To use a m a g i c t r i a n g l e , cover t h e q u a n t i t y y o u are t r y i n g t o f i n d . The relationship left b e h i n d s h o w s y o u h o w t o calculate it. For e x a m p l e , cover up s p e e d , u in t h e first t r i a n g l e a n d y o u are left w i t h j .
/ d \
/ ù \
/ f \
X
x
ax
t
t
F=
777 x a
PE=
W X h
X M=
FX
d
WD=Fxd
Q=
p =
mXv
2
F=pXA
I x t
m = px
KE = \ x
V=
IXR
V
v=
f x \
pgh
E=
mc&T
E=pXt
P =
E=
IXV
Ami
1 General physics
. / I U > f K j i ; f l , yrj|ijrr>f-' .;]fK.I i i r r K !
/
L e n g t h is measure d w i t h a ruler, t a p e measure , vernier callipers o r m i c r o m e t e r screw g a u g e
S
V o l u m e m a y be measure d w i t h a m e a s u r i n g cylinder
/
T i m e is measured w i t h a s t o p clock o r s t o p w a t c h
L e n g t h m e a s u r e m e n t s can be m a d e m o r e p r e c i s e by using an i n s t r u m e n t w i t h a vernier scale such as vernier callipers or a m i c r o m e t e r screw g a u g e . A ruler measures t o t h e nearest m i l l i m e t r e , vernier callipers t o 1/10 m m a n d a m i c r o m e t e r 1/100 m m . Lengt h m e a s u r e m e n ts can be m a d e m o r e a c c u r a t e by m e a s u r i n g multiples, such as t h e thickness o f 5 0 0 sheets o f paper, t h e n divide by 5 0 0 t o g e t t h e thickness o f o n e sheet. Check t h e r e l i a b i l i t y o f y o u r m e a s u r e m e n t s by repeating t h e m . If t h e repeate d results are similar, t h e y are reliable. W h e n m e a s u r i n g t h e l e n g t h , / o f a p e n d u l u m as in t h e f i g u r e (right), y o u need t o measure t o t h e centre o f gravity. O n e w a y o f d o i n g this is t o t a k e t w o m e a s u r e m e n t s a n d average t h e m : f r o m t h e fixed e n d o f t h e string t o t h e b e g i n n i n g o f t h e b o b , /, a n d f r o m t h e fixed e n d o f t h e string t o t h e far e n d of t h e b o b /
gap ben ig measured
fixed scale
fixed
mm 0
10 _u_L
J_L
p-trr
Read the highest scale See where divisions coincide. division before f : Read this on sliding scale, puting a decm i al point in front: I 0.4 I
Add: A
Read the highest scale division that can be seen:
Read the scale on the barrel, puting a decm i al point in front: I 0.32
5.5
I
7.4 mm
Reading a vernier
scale on revolving barrel
Add: A
L_
I
5.82 mm
Reading a micrometer
The v o l u m e o f a regular solid can be f o u n d by measuring its dimensions. For e x a m p l e, recording t h e l e n g t h , w i d t h a n d h e i g ht o f a c u b o i d (box) a n d m u l t i p l y i n g these m e a s u r e m e n t s together. The v o l u m e o f an irregularly shaped objec t can also be measured using a measuring cylinder w i t h a eureka c a n . Place t h e objec t in t h e eureka can a n d use t h e m e a s u r i n g cylinder t o measure h o w m u c h w a t e r is displaced. The v o l u m e of w a t e r displaced is equal t o t h e v o l u m e o f t h e object. A d i g i t al s t o p w a t c h measures t i m e t o a p r e c i s i o n of 0.01 s. This is far m o r e precise t h a n h u m a n error w i l l allow, w h i c h is a b o u t 0.2 s. If possible, t i m e over as l o n g a p e r i o d as possible. W h e n t i m i n g t h e t i m e p e r i o d f o r a simple p e n d u l u m , f o r e x a m p l e , it is m o r e a c c u r a t e t o t i m e 1 0 0 s w i n g s w i t h a s t o p w a t c h a n d t h e n divide by t h e n u m b e r of s w i n g s. This reduces t h e error d u e t o h u m a n reaction t i m e . Repeat t i m e m e a s u r e m e n ts t o check f o r reliability of data.
A
Eureka can and measuring cylinder
KEY y
IDEAS
Speed = d | t a n ç e time M
*
/ /
, i i c h a n g e in velocity Acceleration = ^ Velocity a n d acceleration can have b o t h positive a n d negative values
. . . . Speed (m/s) K
distance (m) , \ t i m e (s)
d t
Worked examples 1. A car travels at a speed of 2 0 m/s f o r 3 0 s. H o w far does it travel in this t i m e ?
2.
A cyclist travels 1 0 0 0 m in 3 m i n u t e s . W h a t is his speed?
3.
A girl w a l k s 3 k m at 1.5 m/s. H o w l o n g does her j o u r n e y take?
Answers 1. d = u x
t
= 20 X 30 = 600 m
2. u = d/f = 1000 T ( 3 X
60)
= 1 0 0 0 + 180 = 5.56 m/s
3.
t =
dlu
= (3 X 1 0 0 0 ) -
1.5
= 3 0 0 0 •¥ 1.5 = 2 0 0 0 s (33 m i n 2 0 s) N o t e : in e x a m p l e 2 , t i m e in m i n u t e s m u s t be c o n v e r t e d t o s e c o n d s . In e x a m p l e 3, distance in kilometres m u s t be c o n v e r t ed t o m e t r e s .
Distance-time graphs A j o u r n e y can be represented o n a g r a p h by p l o t t i n g t h e distance travelled o n t h e y - a x i s a n d t h e t i m e t a k e n o n t h e x-axis. The shape of t h e g r a p h describes t h e j o u r n e y .
y step The gradient of the graph
distance/m
_ y step ~ x step = speed
xstep time/s
A
Distance-time graph
Examples of distance-time graphs 1.
In a crash test, a car travels at steady speed a n d t h e n stops s u d d e n l y as it hits a w a l l
The gradient suddenly changes to zero (horizontal line), which means the speed is zero and the object has stopped.
There is a constant gradient because the object travels at steady speed. distance/m
time/s
2.
A r u n n e r sets o f f in a race, increasing her speed until she reaches her m a x i m u m s p e e d.
The gradient gradually increases because the speed is increasing.
distance/m Initially, the gradient is low, which means the speed is low.
time/s
Revision guide: Physics
Velocity and acceleration V e l o c i t y is a v e c t o r q u a n t i t y ; it is e q u a l t o s p e e d in a p a r t i c u l a r d i r e c t i o n . Speed is a m e a s u r e m e n t o f h o w fast an o b j e c t is m o v i n g . A c c e l e r a t i o n is also a vector quantity. It is equal t o t h e c h a n g e in v e l o c i t y p e r s e c o n d . W h e n an o b j e c t is s l o w i n g d o w n , if its velocity is d e c r e a s i n g , t h e acceleration is n e g a t i v e . W e say t h a t it is d e c e l e r a t i n g . See section 1.5 (scalars a n d vectors). A c c e l e r a t i o n ( m / s ) = c h a n g e in velocity (m/s) 2
t i m e (s)
__ v - u a - — r
initial velocity = u, f i n a l velocity = v, t i m e t a k e n f o r t h e c h a n g e in velocity = c h a n g e in velocity = v — u =
t,
Av
Worked examples 1. A car increases its velocity f r o m 10 m/s t o 2 0 m/s in 5 s. W h a t is its acceleration?
2.
A r u n n e r has an acceleration o f 10 m/s . H o w l o n g does it t a k e h i m t o reach a speed o f 2
5 m/s f r o m rest? ( N o t e 'rest' means zero velocity.)
3.
A t r a i n accelerates at 9 m/s f o r 5 s. If its initial velocity is 5 m/s, w h a t is its final 2
velocity?
Answers 1. a =
Av+t
= (20 = 2 m/s
2.
t=
10) ^ 5 2
Av + a
= 5 + 10 = 0.5 s
3.
Av=
a X f =
9 X 5
= 4 5 m/s v = u + Av = 5 + 45 = 50 m/s
Speed-time graphs
The gradient of the speed-time „ . y step 9 P 3FItep r a
speed m/s
=
= acceleration
y step
The area under the graph = distance travelled.
h
xstep time/s
A
Speed-time graph.
The area u n d e r t h e g r a p h is a calculation involving t h e units o n t h e t w o axes. It is n o t a physical area.
Examples of speed-time graphs 1. A car accelerating until it reaches its m a x i m u m speed.
At the maximum speed the graph is horizontal (gradient is zero) indicating the car has reached a steady speed.
The car starts with its greatest acceleration (the gradient is highest). The acceleration decreases with time as the car approaches its maximum speed (the gradient decreases).
time/s
2.
A r u n n e r w h o accelerates w i t h c o n s t a n t acceleration t o his m a x i m u m speed a n d t h e n decelerates steadily t o a s t o p a t t h e e n d o f t h e race.
As the runner reaches a steady speed, the gradient becomes zero (horizontal line on graph).
At the start of the race the acceleration of the runner is greatest (gradient is maximum). This acceleration is constant and so the gradient of the graph is constant (a straight line).
At the end of the race, the runner decelerates (negative gradient) to rest (zero velocity). The straight line shows that the deceleration is constant.
time/s
3.
A skydiver f r o m t h e t i m e she j u m p s f r o m a helicopte r until t h e m o m e n t she reaches t h e g r o u n d .
Initially the skydiver accelerates but the acceleration decreases with time as air resistance increases (gradient is positive but decreasing).
When she opens her parachute, the acceleration changes from positive to negative as she decelerates (negative gradient). speed Finally, the deceleration decreases and she reaches a steady velocity (zero gradient). This is the t e r m i n a l velocity. time/s
ni.- .'.
'Revision guide: Physics
Examination style questions 1.
r
The g r a p h abov e is f o r a 6 0 m i n u t e car j o u r n e y . a.
B e t w e e n w h i c h t i m e s is t h e car speed a t its highest?
b.
Calculate t h e t o t a l t i m e f o r w h i c h t h e car is s t o p p e d .
c.
State w i t h o u t calculation h o w t h e g r a p h c o u l d be used i)
t o f i n d t h e distance travelled in t h e first 12V2 m i n u t e s .
ii)
t o f i n d t h e average speed f o r t h e j o u r n e y . G r a p h a d a p t e d f r o m C I E 0 6 2 5 J u n e '05 P a p e r 2 Q 2
A s t o n e falls f r o m t h e t o p o f a b u i l d i n g a n d hits t h e g r o u n d a t a speed of 3 2 m/s. The air resistance f o r c e o n t h e s t o n e is very small a n d m a y be n e g l e c t e d . a.
Calculate t h e t i m e o f fall.
b.
C o p y a n d d r a w t h e s p e e d - t i m e g r a p h f o r t h e f a l l i n g stone . 40
30
speed 20 m/s
10
1
2
3
4
time/s
C I E 0 6 2 5 N o v e m b e r "06 P a p e r 3 Q 1 a
General physics
3.
The f i g u r e b e l o w s h o w s t h e s p e e d - t i m e g r a p h f o r a j o u r n e y travelled by a tractor.
0
50
100
170
200
time/s a.
J
Use t h e g r a p h t o describe t h e m o t i o n o f t h e t r a c t o r d u r i n g t h e sections OP, PQ, QR a n d RS.
b.
W h i c h t w o points o n t h e g r a p h s h o w w h e n t h e t r a c t o r is stationary?
c.
State t h e greatest speed reached by t h e tractor.
d.
For h o w l o n g w a s t h e t r a c t o r travelling a t c o n s t a n t speed?
e.
State h o w t h e g r a p h may be used t o f i n d t h e t o t a l distance travelled d u r i n g t h e 2 0 0 s j o u r n e y . Do n o t a t t e m p t a calculation . C I E 0 6 2 5 N o v e m b e r '06 P a p e r 2 Q 3
4.
Palm trees are g r o w i n g every 2 5 m alongside t h e h i g h w a y in a holiday resort.
T h e IGCSE school bus drives a l o n g t h e h i g h w a y . a.
It takes 2 s f o r t h e bus t o travel b e t w e e n p a l m t r ee 1 a n d p a l m tree 2 . Calculate t h e average speed of t h e bus b e t w e e n tree 1 a n d tree 2 .
b.
It takes m o r e t h a n 2s f o r t h e bus t o travel f r o m t r e e 2 t o tree 3. State w h a t this i n f o r m a t i o n indicates a b o u t t h e speed of t h e bus.
c.
The speed o f t h e bus c o n t i n u e s t o d o w h a t y o u have said in (b). State h o w t h e t i m e t a k e n t o g o f r o m tree 3 t o t r e e 4 compares w i t h t h e t i m e in (b). C I E 0 6 2 5 N o v e m b e r '05 P a p e r 2 Q 2
•jy
I Revision guide: Physics
5.
In a t r a i n i n g session, a racing cyclist's j o u r n e y is in t h r e e stages. Stage 1
He accelerates u n i f o r m l y f r o m rest t o 12 m/s in 2 0 s.
Stage 2
He cycles a t 12 m/s f o r a distance of 4 8 0 0 m .
Stage 3
He decelerates u n i f o r m l y t o rest.
The w h o l e j o u r n e y takes 5 0 0 s. a.
Calculate t h e t i m e t a k e n f o r stage 2.
b.
C o p y t h e g r i d b e l o w a n d d r a w a s p e e d - t i m e g r a p h of t h e cyclist's ride.
0
100
200
300
400
500
time/s
)
c.
S h o w t h a t t h e t o t a l distance travelled by t h e cyclist is 5 4 0 0 m .
d.
Calculate t h e average speed o f t h e cyclist. C I E 0 6 2 5 J u n e '07 P a p e r 2 Q 2
'1,3 M'ji-r; wk.I
y./r:Hg:l'ti:
# f E V / D £ > Ï S /
Mass is a q u a n t i t y related t o t h e inertia o f a n o b j e c t , m e a s u r e d in k g
/
W e i g h t is t h e f o r c e , in N, o n a mass d u e t o a g r a v i t a t i o n a l f i e l d
/
O n Earth, t h e g r a v i t a t i o n a l f i e l d s t r e n g t h is 10 N/kg
M a s s is t h e a m o u n t o f m a t t e r t h a t makes up an object. It is measure d in kilogram s (kg). All masses have a quality called " i n e r t i a " , t h e t e n d e n c y t o keep m o v i n g if already m o v i n g a n d stay still if already still. For e x a m p l e, a car in a crash test: w h e n t h e car hits t h e w a l l , it decelerates t o rest in a s h o r t t i m e . The crash test d u m m y has inertia d u e t o its mass a n d so it keeps m o v i n g f o r w a r d s a t t h e same speed as b e f o r e t h e car hit t h e w a l l . A l t h o u g h it looks as t h o u g h t h e d u m m y has been t h r o w n f o r w a r d , t h e r e is no net f o r w a r d f o r c e o n it.
" *
%
fc*
, i t T T mmrnrnt
W e i g h t is t h e f o r c e o n a mass d u e t o gravity. It is measure d in n e w t o n s (N). w e i g h t (N) = mass (kg) X g r a v i t a t i o n a l f i e l d s t r e n g t h (N/kg ) W
=
m
X
g
O n Earth, t h e g r a v i t a t i o n al field s t r e n g t h , g = 10 N / k g . This is also called t h e acceleration d u e t o gravity or t h e acceleration o f freefall a n d has an alternative u n i t o f m/s . 2
Examination style question Some IGCSE s t u d e n t s w e r e asked t o w r i t e s t a t e m e n t s a b o u t mass a n d w e i g h t . Their s t a t e m e n t s are p r i n t e d b e l o w . C h o o se t h e t w o correct s t a t e m e n t s . Mass and w e i g h t are the s a m e thing. M a s s is m e a s u r e d in k i l o g r a m s . W e i g h t is a t y p e o f f o r c e . W e i g h t is t h e a c c e l e r a t i o n c a u s e d b y g r a v i t y . CIE 0 6 2 5 N o v e m b e r "06 P a p e r 2 Q 2
KEY /
IDEAS
Density = 3
"? , m e a s u r e d in g / c m or k g / m volume a s s
3
3
D e n s i t y is a q u a n t i t y related t o h o w closely p a c k e d t h e particles in a material are as w e l l as h o w m u c h t h e particles w e i g h . i , . density ( g / c m ) =
mass (g)
3
P
v
-
o
|
u
m
e
<
g
r
f
)
m t/
> '
A simple m e t h o d o f m e a s u r i n g t h e density o f an o b j e c t (if its density is greater t h a n t h a t of w a t e r ) :
measuring cylinder
<— new volume reading
—
—
reading
—
water
object
r
Find t h e mass o f t h e o b j e c t using a balance. A p p r o x i m a t e l y half fill a m e a s u r i n g cylinder w i t h water. Read t h e v o l u m e of w a t e r f r o m t h e m e a s u r i n g cylinder scale. Place t h e o b j e c t in t h e m e a s u r i n g cylinder a n d t a k e t h e n e w v o l u m e r e a d i n g . Calculate ( n e w v o l u m e reading - initial v o l u m e reading) t o f i n d t h e v o l u m e of t h e object. Calculate t h e density by d i v i d i n g t h e mass by t h e v o l u m e .
Worked examples 1. The mass of a s t o n e is f o u n d o n a t o p p a n balance. It has a mass o f 1 2 0 . 0 2 g . A m e a s u r i n g cylinder is filled t o a v o l u m e of 6 0 c m w i t h w a t e r . W h e n t h e ston e is placed 3
in t h e m e a s u r i n g cylinder, t h e n e w w a t e r level is 9 5 c m . Find t h e density of t h e s t o n e . 3
2.
A n o b j e c t o f k n o w n density of 2.7 g / c m is placed i n t o a m e a s u r i n g cylinder o f water. 3
The level of t h e w a t e r rises f r o m 4 5 c m t o 7 2 c m . W h a t is t h e mass of t h e object? 3
3.
3
A metal block of density 3.2 g / c m a n d mass 9 0 g is placed in a m e a s u r i n g cylinder 3
c o n t a i n i n g 65 c m o f w a t e r . W h a t is t h e n e w w a t e r level? 3
Answers 1. V o l u m e = 9 5 -
60 = 35 c m
= [R =
= 3.4 g /
p
c m
3
3
35 2.
Volume = 72 - 45 = 27 c m
3
m = V X p = 2 7 X 2.7 = 73 g
3.
\/=m = | |
= 28cm
3
N e w w a t e r level = 6 5 + 2 8 = 93 c m
I
3
Examination style question 1. A s t u d e n t is given a spring balance w i t h a N e w t o n scale. She is t o l d t h a t t h e acceleration d u e t o gravity is 10 m/s . 2
Describe h o w she c o u l d f i n d t h e mass of a t o y car. Describe h o w she c o u l d g o o n t o f i n d t h e average density o f t h e t o y car. A d a p t e d f r o m C I E 0 6 2 5 N o v e m b e r "05 P a p e r 3 Q 1 b
2.
A s t u d e n t used a suitable m e a s u r i ng cylinder a n d a spring balance t o f i n d t h e density o f a sample o f a s t o n e . i)
Describe h o w t h e m e a s u r i ng cylinder is used, a n d state t h e readings t h a t are t a k e n .
ii)
Describe h o w t h e sprin g balance is used, a n d state t h e reading t h a t is t a k e n .
iii)
W r i t e d o w n an e q u a t i o n f r o m w h i c h t h e density o f t h e s t o n e is c a l c u l a t e d .
iv)
The s t u d e n t t h e n wishes t o f i n d t h e density of cork. Suggest h o w t h e a p p a r a t us a n d t h e m e t h o d w o u l d need t o be c h a n g e d . CIE 0 6 2 5 N o v e m b e r '06 P a p e r 3 Q 1 b
3.
Fig. (a) s h o w s a m e a s u r i n g cylinder, c o n t a i n i n g s o m e w a t e r, o n a balance. Fig. (b) s h o w s t h e s a m e a r r a n g e m e n t w i t h a s t o n e a d d e d t o t h e w a t e r .
a.
W h i c h t w o readings s h o u l d be s u b t r a c t e d t o give t h e v o l u m e o f t h e stone?
b.
W h i c h t w o readings s h o u l d be s u b t r a c t e d t o give t h e mass o f t h e stone?
c.
In a certain e x p e r i m e n t , mass of s t o n e = 57.5 g , v o l u m e of s t o n e = 2 5 c m . 3
i)
W r i t e d o w n t h e e q u a t i o n linking density, mass a n d v o l u m e .
ii)
Calculate t h e density of t h e stone . CIE 0 6 2 5 J u n e '06 P a p e r 2 Q 3
Practical question A n IGCSE s t u d e n t is d e t e r m i n i n g t h e density o f a meta l alloy. The s t u d e n t is p r o v i d e d w i t h several metal rods, as s h o w n o n t h e right.
1.
Measure w i t h a ruler t h e l e n g t h , /, o f o n e of t h e rods.
2.
The s t u d e n t measured t h e d i a m e t e r o f o n e o f t h e rods w i t h a ruler a n d f o u n d it t o be 0.6 c m . Calculate t h e cross-sectional area, A, of t h e r o d .
3.
Use this value t o calculate t h e v o l u m e , V, o f o n e r o d a n d hence t h e w h o l e b u n d l e . The s t u d e n t used a balance t o f i n d t h e mass of t h e b u n d l e a n d f o u n d it t o be 59.1 g . Calculate t h e density o f t h e metal alloy.
KEY IDEAS /
Forces are v e c t o r q u a n t i t i e s t h a t can c h a n g e t h e shape o f an o b j e c t , accelerate it o r c h a n g e its d i r e c t i o n
J
M o m e n t o f a f o r c e a b o u t a p i v o t = f o r c e x p e r p e n d i c u l ar distance f r o m t h e p i v o t
S
A n o b j e c t is in e q u i l i b r i u m if t h e r e is zero n e t f o r c e o n it a n d zero n e t m o m e n t
S
A n o b j e c t w i l l n o t f a l l o v e r if t h e line o f a c t i o n o f its w e i g h t acts t h r o u g h its base; t h i s d e p e n d s o n t h e p o s i t i o n o f t h e c e n t r e o f mass
/"
T h e resultant o f t w o forces can be f o u n d by d r a w i n g a scale d i a g r a m
Forces can p r o d u c e a c h a n g e in s i z e o r s h a p e o f a n o b j e c t . For e x a m p l e , l o a d i n g a spring w i l l increase its l e n g t h .
An experiment to find how the extension of a spring varies with the force applied •
M e a s u r e t h e original positio n of t h e sprin g using t h e p o i n t e r a n d ruler, t o t h e nearest m m .
•
A d d a 100 g mass h a n g e r a n d measure t h e n e w p o s i t i o n .
•
Repeat 6 t i m e s , a d d i n g a 1 0 0 g mass each t i m e .
•
Calculate t h e extension by s u b t r a c t i n g t h e original positio n f r o m each s u b s e q u e n t position reading.
•
Plot a g r a p h of extension against f o r c e , w h e r e force = mass x 10 N/kg.
Graph of results Conclusion From 0 t o P, extension is directly proportional t o the force applied. B e y o n d P, t h e extensions are larger extension/m
f o r t h e same increase in f o r c e . A t E, t h e elastic limit is reached. Beyond this p o i n t , t h e sprin g w i l l force/N
n o t retur n t o its original l e n g t h w h e n t h e f o r c e is r e m o v e d .
Hooke's Law If a material obeys H o o k e ' s Law, t h e extension is directly p r o p o r t i o n a l t o t h e a p p l i e d f o r c e , provided t h a t t h e elastic l i m i t is n o t e x c e e d e d , F=kx W h e r e F = a p p l i e d f o r c e (N), k = f o r c e c o n s t a n t f o r o b j e c t u n d e r test ( N / m ) , x = extension (m)
Force, mass and acceleration A f o r c e can a c c e l e r a t e an o b j e c t . The larger t h e f o r c e o n t h e o b j e c t , t h e greater t h e acceleration if t h e mass stays c o n s t a n t . The larger t h e mass o f t h e o b j e c t , t h e smaller t h e acceleration if t h e f o r c e stays c o n s t a n t. Force (N) = mass (kg) X acceleration ( m / s ) 2
F
=
m
x
a
Worked examples 1. A force o f 10 N acts o n an o b j e c t o f mass 5 k g . W h a t is t h e acceleration o f t h e object?
2.
A force o f 15 N causes an o b j e c t t o accelerate at 2 m/s . W h a t is its mass?
3.
A mass o f 3 kg has a deceleration o f 5 m/s . W h a t f o r c e acts o n it?
2
2
Answers 1.
a = F + m = 10 5 = 2 m/s
2.
m = F =
-r-
2
a
I l 2
= 7.5 k g
3.
F = m X a = 3 X
-5
= -15 N N o t e : in e x a m p l e 3, t h e o b j e c t is decelerating so acceleration is negative.
'Revision guide: Physics
Resultant force The r e s u l t a n t f o r c e a c t i n g o n an o b j e c t is t h e n e t or o v e r a l l f o r c e w h e n t h e s i z e a n d d i r e c t i o n o f all t h e forces a c t i n g are t a k e n i n t o a c c o u n t . F o r c e is a v e c t o r q u a n t i t y.
Examples 1 — • 2N
Resultant f o r c e = 5 N + 2 N =
7N
The forces are a d d e d t o g e t h e r because t h e y act in t h e s a m e d i r e c t i o n .
Resultant f o r c e = 15 N — 10 N = 5 N (to t h e left) The forces are s u b t r a c t e d because t h e y act in o p p o s i t e directions. A f o r c e can cause an o b j e c t t o c h a n g e d i r e c t i o n . The o b j e c t w i l l m o v e in a circle if t h e f o r c e acts p e r p e n d i c u l ar (at a r i g h t angle) t o t h e direction t h a t t h e o b j e c t is travelling .
The f o r c e acts at r i g h t a n g l e s t o t h e d i r e c t i on t h a t t h e o b j e c t is m o v i n g . The f o r c e does n o t d o any w o r k o n t h e o b j e c t because t h e o b j e c t does n o t m o v e in t h e d i r e c t i o n o f t h e f o r c e . The f o r c e c o n s t a n t l y changes t h e d i r e c t i on o f t h e objec t w h i c h means its v e l o c i t y c h a n g e s b u t its s p e e d s t a y s c o n s t a n t . The o b j e c t accelerates t o w a r d s t h e centre of t h e circle. The f o r c e w h i c h causes an o b j e c t t o m o v e in a circle is called t h e c e n t r i p e t a l f o r c e . The c e n t r i p e t al force increases if: •
t h e mass o f t h e o b j e c t increases
•
t h e speed o f t h e o b j e c t increases
•
t h e radius o f t h e circle decreases
If t h e f o r c e w h i c h is p r o v i d i n g t h e centripetal acceleration is suddenly r e m o v e d , t h e o b j e c t w i l l m o v e o n a t a n g e n t t o t h e original circle.
Examination style questions 1. £.3
90 ZU
length/cm
15 m
• n
()
05
10
15
20
25
30
Ic ad/N T h e g r a p h a b o v e w a s o b t a i n e d by a s t u d e n t w h o loaded a spring f r o m 0.5 N t o 3.0 N a n d measured its l e n g t h . a.
Find t h e l e n g t h o f t h e spring w h e n a load o f 2.0 N is applied
b.
Find t h e load required t o stretch t h e spring t o a l e n g t h o f 18 c m
c.
Find t h e extension o f t h e spring w h e n it is stretched by a load o f 1.5 N A d a p t e d f r o m C I E 0 6 2 5 N o v e m b e r "05 P a p e r 2 Q1
2.
A mass o f 5.0 k g accelerates at 3 m/s in a s t r a i g h t line. 2
a.
State w h y acceleration is described as a vector quantity.
b.
Calculate t h e f o r c e required t o accelerate t h e mass. A d a p t e d f r o m C I E 0 6 2 5 J u n e "05 P a p e r 3 Q 3
3.
The l e n g t h o f a spring is measured w h e n various loads f r o m 1 .ON t o 6.ON are h a n g i n g f r o m it. The g r a p h b e l o w gives t h e results.
1
1
j
1
! ! 1
20
i
)
I
_
j
\ l 1 1
i
1
j
!
4„
1
I
_t_.,
i ; | i
1
1
1 1
i i
15 length/cm
i
i
1
i
I
i
i [
i
1 1
_}„.
S
I
i l
I
i
1 _
T"
i I
1
[ 1
!
i
1 *~
I 1 [ 1
1 1
1
1 1
i
i i
1 I
t
3
4
load/N
Use t h e g r a p h t o f i n d a.
t h e l e n g t h o f t h e sprin g w i t h no load a t t a c h e d ,
b.
t h e l e n g t h o f t h e spring w i t h 4 . 5 N a t t a c h e d ,
c.
t h e extension caused by a 4 . 5 N l o a d. CIE 0 6 2 5 N o v e m b e r ' 0 5 P a p e r 2 Q1
(Revision guide: Physics
4.
In an e x p e r i m e n t , forces are a p p l i e d t o a spring as s h o w n in (a). The results o f this e x p e r i m e n t are s h o w n in (b).
a. b.
W h a t is t h e n a m e given t o t h e p o i n t m a r k e d Q in (b)? For t h e p a r t OP of t h e g r a p h , t h e sprin g obeys Hooke's Law. State w h a t this means.
c.
The spring is stretche d until t h e f o r c e a n d extension are s h o w n by t h e p o i n t R o n t h e g r a p h . C o m p a r e h o w t h e spring stretches, as s h o w n by t h e p a r t o f t h e g r a p h O Q , w i t h t h a t s h o w n by QR.
d.
The p a r t OP o f t h e g r a p h s h o w s t h e spring stretchin g a c c o r d i n g t o t h e expression F=
kx.
Use values f r o m t h e g r a p h t o calculate t h e value of k. CIE 0 6 2 5 N o v e m b e r '06 P a p e r 3 Q 2
5. The points p l o t t e d o n t h e g r i d b e l o w w e r e o b t a i n e d f r o m a s p r i n g - s t r e t c h i n g experiment.
100
80
60
length/mm
40 é*
20
load/N
a.
Using a s t r a i g h t e d g e , d r a w a s t r a i g ht line t h r o u g h t h e first 5 p o i n t s . Extend y o u r
b.
Suggest a reason w h y t h e sixth p o i n t does n o t lie o n t h e line y o u have d r a w n .
line t o t h e e d g e o f t h e g r i d .
c.
Calculate t h e extension caused by t h e 3 N load .
d.
A small o b j e c t is h u n g o n t h e u n l o a d e d s p r i n g , a n d t h e l e n g t h o f t h e spring becomes 6 2 m m . Use t h e g r a p h t o f i n d t h e w e i g h t o f t h e o b j e c t . CIE 0 6 2 5 N o v e m b e r "06 P a p e r 2 Q 9
Practical question A n IGCSE class is investigating t h e e f f e c t of a load o n a m e t r e rule a t t a c h e d t o a s p r i n g . The apparatus is s h o w n in t h e d i a g r a m b e l o w .
zero end of rule taped to bench
f Revision guide: Physics
The zero e n d of t h e m e t r e rule is t a p e d t o t h e bench t o s t o p it s l i p p i n g . T h e spring is a t t a c h e d t o t h e rule at t h e 4 0 . 0 c m m a r k a n d t h e masses are a t t a c h e d at t h e 9 0 . 0 c m m a r k . The masses are a d d e d 10 g a t a t i m e a n d t h e angle, 6, b e t w e e n t h e b e n c h a n d t h e rule measured w i t h a protractor. O n e student's results are s h o w n b e l o w
0
29
10
28
20
26
30
25
40
22
50
19
1.
C o m p l e t e t h e c o l u m n headings.
2.
O n e s t u d e n t suggests t h a t m a n d 6 s h o u l d be directly p r o p o r t i o n a l t o each other. Plot a g r a p h o f 6 (y-axis) against m (x-axis). Using y o u r g r a p h s h o w w h e t h e r this p r e d i c t i o n is correct. State y o u r reason.
Turning effect and equilibrium The t u r n i n g e f f e c t or m o m e n t o f a f o r c e a b o u t a pivot is equal t o t h e f o r c e m u l t i p l i e d by its perp e n d i c u l ar distance f r o m t h e p i v o t .
M o m e n t ( N m ) = Force (N) x distance (m) M
=
F
X
d
If an object is in e q u i l i b r i u m t h e r e is n o resultant t u r n i n g e f f e c t a n d n o resultant f o r c e .
Worked example 1. A f o r c e of 2.0 N acts a t distance o f 3.0 m f r o m a pivot. Find t h e m o m e n t of t h e f o r c e .
2.
A f o r c e o f 5.0 N provides a m o m e n t o f 15 N m a b o u t a pivot. W h a t is t h e distance of t h e f o r c e f r o m t h e pivot?
3.
A f o r c e provides a m o m e n t o f 2 0 N m a b o u t a pivot at a distance of 2.0 m. W h a t is t h e size of t h e force?
General physics I
Answers
1.
M = F X d = 2.0 X 3.0 = 6.0 N m
2.
d =
f
= 15. 5.0 = 3.0 m
3.
F =
M d
=
20 2.0
= 10 N
Examples of objects in equilibrium 20 N 30 N 5N
2N r+1 m * ~ « -
1N -3m
-4m-4m-
-6 m-
—Xr-
k-1 ITU 7 \
6N 25 N
3N Clockwise moment = 2 X 3 = 6 Nm Anticlockwise moment = 1 X 6 = 6 Nm Net moment = 6 — 6 = 0 Nm
24 N
Clockwise moment = 20 X 1 = 20 Nm Anticlockwise moment = 5 X 4 = 20 Nm Net moment = 20 - 20 = 0 Nm
Taking m o m e n t s a b o u t t h e pivot in each case.
An experiment to show that there is no net moment on an object in equilibrium
I
spring balance
horizontally balanced metre rule
7
2 10
C)
O
O
20
30
40
9N weight
«1 50 \ ^
o _ 60
O
O
O
70
80
90
horizontal pivot
A n t i c l o c k w i s e m o m e n t d u e t o t h e 9 N w e i g h t = 9 x 0.4 = 3.6 N m Reading o n t h e spring balance = 12 N M o m e n t due t o t h e f o r c e o f t h e spring balance = 12 X 0.3 = 3.6 N m C o n c l u s i o n : in e q u i l i b r i u m , clockwis e m o m e n t = anticlockwis e m o m e n t
100
Clockwise moment = 24 X 1 = 24 Nm Anticlockwise moment = 6 X 4 = 24 Nm Net moment = 24 - 24 = 0 Nm
Revision guide: Physics
Examination style questions 1. a. b.
State t h e t w o factor s o n w h i c h t h e t u r n i n g e f f e c t of a f o r c e d e p e n d s . Forces F, a n d F are a p p l i e d vertically d o w n w a r d s at t h e ends of a b e a m resting o n 2
a pivot P. Th e b e a m has w e i g h t W. Th e b e a m is s h o w n in t h e d i a g r a m b e l o w .
A .
p
V
w
C o m p l e t e t h e s t a t e m e n t s a b o u t t h e t w o requirement s f o r t h e b e a m t o be in e q u i l i b r i u m .
ii)
1.
There m u s t be n o resultant
2.
There m u s t be n o resultant
The b e a m is in e q u i l i b r i u m . F is t h e f o r c e exerted o n t h e b e a m by t h e p i v o t P. C o m p l e t e t h e f o l l o w i n g e q u a t i o n a b o u t t h e forces o n t h e b e a m . F=
iii) W h i c h o n e o f t h e f o u r forces o n t h e b e a m does n o t exert a m o m e n t a b o u t P? CIE 0 6 2 5 N o v e m b e r ' 0 6 P a p e r 2 Q 5
2.
The d i a g r a m b e l o w s h o w s a p p a r a t u s f o r investigating m o m e n t s o f forces.
F 1 spring r j i j l balance
horizontally balanced . metre rule
V 0 10
0
O
O
20
30
40
6.0 N weight
50 \
O
O
O
O
60
70
80
90
100
horizontal pivot
The u n i f o r m m e t r e rule s h o w n is in e q u i l i b r i u m . a.
W r i t e d o w n t w o c o n d i t i o n s f o r t h e m e t r e rule t o be in e q u i l i b r i u m .
b.
S h o w t h a t t h e value o f t h e reading o n t h e spring balance is 8.0 N.
c.
The w e i g h t o f t h e u n i f o r m m e t r e rule is 1.5 N. Calculate t h e f o r c e exerted by t h e p i v o t o n t h e m e t r e rule a n d state its d i r e c t i o n . CIE 0 6 2 5 N o v e m b e r ' 0 5 P a p e r 3 Q 2
Practical question T h e IGCSE class is d e t e r m i n i n g t h e w e i g h t o f a m e t r e rule. B e l o w is a d i a g r a m o f t h e apparatus
• newton meter
j ^ l I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I | I I hf• metre
rule
bench
\ m e t r e rule is s u p p o r t e d at o n e e n d by a p i v o t t h r o u g h t h e 1.0 c m m a r k . The o t h e r e n d is s u p p o r t e d at t h e 9 1 . 0 c m m a r k by a n e w t o n m e t e r h a n g i n g f r o m a c l a m p .
1-
Describe h o w y o u w o u l d check t h a t t h e metre rule is h o r i z o n t a l . You may d r a w a d i a g r a m if y o u w i s h .
2.
The s t u d e n t s record t h e f o r c e F s h o w n o n t h e n e w t o n m e t e r a n d t h e distance d f r o m t h e pivo t t o t h e 91 c m m a r k . They t h e n repeat t h e e x p e r i m e n t several t i m e s using a range o f values o f t h e distance d. The readings are s h o w n in t h e t a b l e .
0.74
0.900
0.78
0.850
0.81
0.800
0.86
0.750
0.92
0.700
C o p y t h e t a b l e . Calculate a n d record o n y o u r t a b l e t h e values of 1/d
3.
a.
On g r a p h paper, p l o t a g r a p h o f F/N (y-axis) against l / i (x-axis). Start t h e y-axis at 0.7 a n d t h e x-axis at 1.0.
4.
b.
D r a w t h e line o f best f i t o n y o u r g r a p h .
c.
D e t e r m i n e t h e g r a d i e n t G o f t h e line.
Calculate t h e w e i g h t of t h e m e t r e rule using t h e e q u a t i o n 1 / 1 / w h e r e k = 0.490 m.
Revision guide: Physics
Centre of mass T h e c e n t r e o f mass o f a n o b j e c t is t h e p o i n t o n t h e o b j e c t w h e r e t h e mass c a n be considered t o be c o n c e n t r a t e d a n d hence w h e r e t h e w e i g h t o f t h e o b j e c t can be c o n s i d e r e d t o act. T h e centre o f mass o f a very t h i n o b j e c t (a lamina) can be f o u n d by e x p e r i m e n t ;
Push a pin t h r o u g h a p o i n t o n t h e e d g e o f t h e lamina a n d a l l o w it t o s w i n g freely. Use a p l u m b line (a small mass o n a piece of string) t o m a r k a vertical line f r o m t h e pin p o i n t across t h e l a m i n a . Repeat f o r a s e c o n d p o i n t o n t h e e d g e of t h e lamina. W h e r e t h e t w o lines cross is t h e positio n o f t h e centre o f mass. The position of t h e centre o f mass affects t h e s t a b i l i t y o f an o b j e c t
For example: If o b j e c t 1 is t i l t e d t h r o u g h a small a n g l e , t h e w e i g h t w i l l a c t o u t s i d e t h e b a s e . There w i l l be a net m o m e n t o n o b j e c t 1 t h a t w i l l cause it t o fall over. If o b j e c t 2 is t i l t e d t h r o u g h a small angle t h e w e i g h t w i l l still a c t i n s i d e t h e b a s e . There w i l l be a net m o m e n t o n o b j e c t 2 t h a t w i l l cause it t o g o back t o its original positio n a n d it w i l l n o t fall over.
Examination style questions 1. A piece o f stiff c a r d b o a r d is stuck t o a p l a n k o f w o o d by means o f t w o sticky-tape " h i n g e s " . This is s h o w n b e l o w .
a.
Initially, t h e c a r d b o a rd is f l a t o n t h e plank of w o o d . A box o f matches is placed o n it. The c a r d b o a r d is t h e n s l o w l y raised at t h e left h a n d e d g e , as s h o w n b e l o w . State t h e c o n d i t i o n f o r t h e box of matches t o fall over.
i)
C o m p l e t e t h e sentence b e l o w , using either t h e w o r d s " g r e a t e r t h a n " or " t h e same a s " or "less t h a n " .
In (b), t h e a n g l e t h r o u g h w h i c h t h e c a r d b o a r d can be lifted befor e t h e b o x of matches falls is
t h e a n g le
b e f o r e t h e box o f matches falls in (a).
ii)
Give a reason f o r y o u r a n s w e r t o (i). A d a p t e d f r o m C I E 0 6 2 5 J u n e "07 P a p e r 2 Q 3
Revision guide: Physics
2.
a.
A ligh t vertical t r i a n g u l a r piece of rigid plastic PQR is p i v o t e d a t c o r n e r P. A h o r i z o n t al 5N f o r c e acts a t Q, as s h o w n in b e l o w .
Describe w h a t , if a n y t h i n g , w i l l h a p p e n t o t h e piece of plastic, b.
O n a n o t h e r occasion, t w o h o r i z o n t a l 5 N forces act o n t h e piece of plastic, as s h o w n in b e l o w .
i)
Describe w h a t , if a n y t h i n g , w i l l h a p p e n t o t h e piece of plastic.
ii)
C o p y t h e d i a g r a m a b o v e a n d m a r k t h e f o r c e t h a t t h e pivot exerts o n t h e piece o f plastic. S h o w t h e direction of t h e f o r c e by means of an a r r o w a n d w r i t e t h e m a g n i t u d e of t h e f o r c e next t o t h e a r r o w . CIE 0 6 2 5 J u n e ' 0 5 P a p e r 2 Q 3
General physics
Scalars and vectors A s c a l a r q u a n t i t y has s i z e only. A v e c t o r q u a n t i t y has s i z e a n d d i r e c t i o n .
tSBSÊMSiSÊSMSÊÊISR Mass
Velocity
Energy
Acceleration
Time
Force
Resultants To calculate t h e resultant (overall) f o r c e o n a p o i n t acte d o n by t w o forces, f, a n d F y o u can 2
d r a w a scale d i a g r a m . Choose a scale e.g. a line of 1 cm represents a force of 1 N
^ ^ Y P l ^ T ^ ^ ^ ^ ^ p i . n n p 0 mm
10
20
30
40
50
60
70
80
|
;„,|„„,
p
90
100
110
120
130
140
•„„
90
100
110
120
130
140
1 N = 1 cm Draw the forces F, and F acting on the point 2
Put the forces "nose to tail" i.e. so the arrows follow on.
Draw the resultant force line from the tail of one arrow to the head of the second arrow.
Measure the length of the resultant force line in cm and then convert this value back to newtons.
0 mm
10
20
30
40
50
60
70
1cm = 1N
80
i
Revision guide: Physics
Examination style question 1. A s t u d e n t sets up t h e apparatus s h o w n b e l o w in order t o f i n d t h e resultant of t h e t w o tensions 7", a n d T a c t i n g a t P. W h e n t h e tensions T 2
v
T a n d T are b a l a n c e d , t h e angles 2
3
b e t w e e n 7" a n d t h e vertical a n d T a n d t h e vertical are as m a r k e d o n t h e d i a g r a m . 1
2
D r a w a scale d i a g r a m o f t h e forces 7", a n d T . Use t h e d i a g r a m t o f i n d t h e resultant o f t h e 2
t w o forces. State: a.
The value o f t h e resultant
b.
The d i r e c t i on o f t h e resultant
c.
The scale used in t h e d r a w i n g CIE 0 6 2 5 J u n e ' 0 6 P a p e r 3 Q 2
KEY S
IDEAS
Energy ca n be transformed from one type to another but it cannot be created or destroyed
S
Different types of stored energy can be transformed into electrical energy in power stations
/
W o r k done = force x distance, powe r = energy/time kinetic energy = j
J
mass x velocity
2
chang e in gravitational potential energy = mass X gravitational field X change in height
Energy A n object may have energy because it is moving or because of its position. Energy can be t r a n s f e r r e d from one place to another, t r a n s f o r m e d from one type to another or s t o r e d . The unit of energy is the j o u l e (J).
Types of energy The energy gained as an object is moved away from the Earth e.g. a book being lifted onto a shelf The energy an object has due to its movement e.g. a person running Stored energy that can be released in a chemical reaction e.g. a battery, fuel such as coal The energy stored when an object changes shape e.g. a stretched rubber band •
• • •
The energy carried by an electric current The energy canied by a sound wave :
-I'M
; The total kinetic and potential energies of all of the particles in an object The energy released when the temperature of a hot object decreases due to a decrease in its internal energy Stored energy that can be released in a nuclear reaction e.g. energy stored in the sun. ; Energy given off, for example, by very hot objects
Heat energy can be t r a n s f e r r e d from a hot object to a cooler one. Kinetic energy can be t r a n s f e r r e d from one car to another in a collision.
Energy transformations In an energy transformation, energy is converted from one type to another. For example: Light bulb
Waterfall
gravitational potential
Bow and arrow
Cell
chemical
• electrical
> kinetic
Revision guide: Physics
Principle of conservation of energy Energy c a n n o t be created or destroyed. It is t r a n s f o r m e d f r o m o n e f o r m t o another.
Example Heat energy out = 98 J Electrical energy in = 100 J Light energy out = 2 J 100 J in = (98 J + 2 J) o u t
Examination style questions 1. A piece of f r u i t is f a l l i n g f r o m a t r e e.
T h e list b e l o w contains t h e names o f s o m e d i f f e r e n t f o r m s o f energy, chemical electrical g r a v i t a t i o n a l (PE) internal ( t h e r m a l ) k i n e t i c (KE) light sound strain a.
W h i c h f o u r f o r m s o f energy are possessed by t h e falling f r u i t ?
b.
W h i c h f o r m o f energy increases as t h e f r u i t falls?
c.
W h i c h f o r m o f e n e r g y decreases as t h e f r u i t falls?
d.
W h i c h f o r m of energy is stored in t h e b o d y o f a person as a result of eating t h e fruit? CIE 0 6 2 5 J u n e '06 P a p e r 2 Q 4
m
A child is sittin g o n an oscillating s w i n g , as s h o w n b e l o w . A t t h e t o p o f t h e oscillation, t h e child a n d s w i n g are m o m e n t a r i l y at rest.
i)
Use t h e names of a p p r o p r i a t e types o f e n e r g y t o c o m p l e t e t h e f o l l o w i n g w o r d equation.
gravitational p o t e n t i a l energy
ii)
a t t h e t o p of t h e
energy at the
oscillation
bottom of the
b o t t o m of the
oscillation
oscillation
energy at t h e
e n e r g y losses
The child c o n t i n u e s t o sit still o n t h e s w i n g . The a m p l i t u d e o f t h e oscillations s l o w l y decreases. Explain w h y this h a p p e n s . C I E 0 6 2 5 N o v e m b e r "06 P a p e r 2 Q 4 b a n d c
'Revision guide: Physics
Energy resources Energy resources are used t o p r o d u c e electrical energy f r o m o t h e r f o r m s o f energy. Coal, oil and gas fired power stations
11
The chemical energy in the fuel is released by burning. The chemical energy is transformed to heat energy which is used to heat water and increase its internal energy, turning it into steam. The steam turns turbines, transferring its k i n e t i c energy to them. The kinetic energy is transformed to electrical energy in the generator.
Geothermal power stations
Water is pumped underground and gains heat energy from the hot rocks deep underground. The heat energy is then converted to kinetic energy in the turbines, which turn the generator to produce electrical energy.
Hydroelectric power stations
The gravitational potential energy of the falling water is transformed to kinetic energy as the water passes through the turbines. The turbines turn the generator to produce electrical energy.
The light energy from the Sun is transformed to electrical energy in the solar cell.
Solar cells
Solar power station
The heat energy from the Sun is concentrated by a series of curved mirrors, which focus the energy into one place. This heat energy converts water to steam, which turns the turbines, giving them kinetic energy. The kinetic energy is transformed to electrical energy in the generator.
Wave power
As the turbines bob on the surface of the sea, gravitational potential energy is transformed to kinetic energy. The kinetic energy is then transformed into electrical energy in the generators.
Tidal power
As the tide comes in, the water builds up behind the dam and gains gravitational potential energy. When the water is released the gravitational potential energy is transformed to kinetic energy and then to electrical energy in the generator.
Nuclear power
i , ; j •
The nuclear energy stored in uranium-235 is released when the uranium nuclei split in a process called nuclear fission. The nuclear energy is transformed to heat energy which is used to turn water to steam. The steam turns the turbines and then this kinetic energy is transformed
j i «
to electrical energy in the generator.
Nuclear fusion The process of nuclear f u s i o n is carried o u t in t h e Sun. H y d r o g e n nuclei collide a t great speed in t h e Sun a n d fuse t o g e t h e r t o f o r m h e l i u m nuclei. This releases energy in t h e f o r m of heat a n d light.
Examination style question 1. The d i a g r a m b e l o w represents a hydroelectric system f o r g e n e r a t i n g electricity.
A n s w e r t h e f o l l o w i n g questions, using w o r d s f r o m this list. chemical electrical gravitational internal (heat) :
kinetic light nuclear sound strain
a.
W h a t sort o f energy, possessed by t h e w a t e r in t h e reservoir, is t h e main source of energy f o r thi s system?
b.
W h e n t h e w a t e r f l o w s d o w n t h e pipe, it is m o v i n g . W h a t sort o f energy does it
c.
The w a t e r makes t h e t u r b i n e s in t h e p o w e r station r o t a t e . W h a t sort of e n e r g y d o
possess because o f this m o v e m e n t ?
t h e t u r b i n e s possess because of t h e i r r o t a t i o n ? d. e.
W h a t sort o f energy does t h e p o w e r s t a t i o n generate? N o n e o f t h e e n e r g y transfer processes is perfect. In w h a t f o r m is m o s t o f t h e w a s t e d e n e r g y released? CIE 0625 J u n e '05 P a p e r 2 Q 4
Revision guide: Physics
Work and power The w o r k d o n e in j o u l e s by a f o r c e a c t i n g o n an o b j e c t = f o r c e X distance m o v e d by t h e o b j e c t in t h e d i r e c t i o n o f t h e f o r c e . w o r k d o n e (J) = f o r c e (N) X distance (m) W
F X d
=
The p o w e r in w a t t s is t h e w o r k d o n e per second or t h e e n e r g y t r a n s f o r m e d per s e c o n d .
power
' - ! Worked examples 1. A car e n g i n e produces a f o r c e of 2 0 0 0 N w h i l e accelerating t h e car t h r o u g h a distance o f 2 0 0 m in a t i m e o f 10 s. a.
W h a t is t h e w o r k d o n e o n t h e car by t h e e n g i n e force?
b. W h a t is t h e p o w e r d e v e l o p e d by t h e engine?
Answers a.
W=
Fx
d
= 2000 X 200 = 400 000 J = 4 0 0 kJ b.
P = =
| 400 000 10
= 40 000 W = 40 kW
Kinetic energy K i n e t i c e n e r g y can b e calculated f r o m t h e f o r m u l a :
w h e r e m = mass in k g ; v = velocity in m/s W h e n an o b j e c t is lifted higher above t h e Earth's surface w o r k m u s t be d o n e . Since w o r k = f o r c e X distance a n d f o r c e = w e i g h t of t h e object w o r k d o n e = w e i g h t x h e i g h t lifted w h e r e w e i g h t = mass x gravitational fiel d =
mg.
Gravitational potential energy It f o l l o w s t h a t t h e c h a n g e in g r a v i t a t i o n a l p o t e n t i a l e n e r g y ( t h e w o r k d o n e in l i f t i n g t h e o b j e c t ) is given by t h e f o r m u l a : PE =
mgh
w h e r e m = mass in k g ; g = acceleration d u e t o gravity 1 0 m / s ; 2
h -
I
c h a n g e in h e i g h t in m
Worked examples 1. A car o f mass 1 0 0 0 kg is travelling at a velocity o f 2 0 m/s. Calculate its kinetic energy.
2.
Calculate t h e c h a n g e in p o t e n t i a l e n e r g y o f a 7 0 kg parachutist as she falls t h r o u g h a h e i g h t of 100 m .
3.
A ball o f mass 0.5 kg is d r o p p e d f r o m rest at a h e i g h t o f 5 m above t h e g r o u n d . Find its velocity w h e n it hits t h e g r o u n d .
Answers 1. KE =
±mv
2
= 1 X
1000 X 2 0
2
= 200 000 J = 2 0 0 kJ
2.
PE
=
mgh
= 7 0 X 10 x
100
= 70 000 J = 7 0 kJ
3.
PE =
mgh
= 0.5 X 10 X 5 = 25 J loss o f PE = gain o f KE v = V(2K£/m) = V(2 X 2 5 / 0 . 5 ) = 10 m/s
Examination style questions 1. A n electric p u m p is used t o raise w a t e r f r o m a w e l l , as s h o w n b e l o w .
pump ground
a.
The p u m p does w o r k in raising t h e water . State an e q u a t i o n t h a t c o u l d be used t o calculate t h e w o r k d o n e in raising t h e w a t e r.
Revision guide: Physics
b.
The w a t e r is raised t h r o u g h a vertical distance o f 8.0 m . The w e i g h t o f w a t e r raised in 5.0 s is 100 N. i)
Calculate t h e w o r k d o n e in raising t h e w a t e r in this t i m e .
ii)
Calculate t h e p o w e r t h e p u m p uses t o raise t h e water.
iii) T h e e n e r g y transferred by t h e p u m p t o t h e w a t e r is greater t h a n y o u r a n s w e r t o (i). Suggest w h a t t h e a d d i t i o n a l e n e r g y is used for. C I E 0 6 2 5 J u n e '06 P a p e r 3 Q 3
2.
A s t u d e n t wishes t o w o r k o u t h o w m u c h p o w e r she uses t o lift her b o d y w h e n c l i m b i n g a f l i g h t o f stairs. Her b o d y mass is 50 k g a n d t h e vertical h e i g h t of t h e stairs is 4 . 0 m . She takes 2 0 s t o w a l k up t h e stairs. a.
b.
Calculate i)
t h e w o r k d o n e in raising her b o d y mass as she climbs t h e stairs,
ii)
t h e o u t p u t p o w e r she develops w h e n raising her b o d y mass.
A t t h e t o p of t h e stairs she has gravitational p o t e n t i a l energy. Describe t h e energy t r a n s f o r m a t i o n s t a k i n g place as she w a l k s back d o w n t h e stairs a n d stops a t the bottom. CIE 0 6 2 5 J u n e '07 P a p e r 3 Q 3
3.
The d i a g r a m b e l o w s h o w s w a t e r f a l l i n g over a d a m .
a.
The vertical h e i g h t t h a t t h e w a t e r falls is 7.0 m . Calculate t h e p o t e n t i a l e n e r g y lost by 1.0 k g of w a t e r d u r i n g t h e fall.
b.
A s s u m i n g all this p o t e n t i a l e n e r g y loss is c h a n g e d t o kinetic e n e r g y o f t h e w a t e r,
c.
The vertical speed of t h e w a t e r is less t h a n t h a t calculated in b. Suggest o n e reason
calculate t h e speed o f t h e w a t e r , in t h e vertical d i r e c t i o n , at t h e e n d o f t h e fall.
f o r this. CIE 0 6 2 5 N o v e m b e r ' 0 6 P a p e r 3 Q 3
1,/ FVf^ijrî* KEY
IDEAS
^
Pressure =
te
J
Pressure is m e a s u r e d in Pascals (Pa) w i t h 1 Pa =
/
A t m o s p h e r i c pressure c a n b e m e a s u r e d using a b a r o m e t e r
/
T h e pressure a t a d e p t h h u n d e r t h e surface o f a liquid o f density p =
/
Th e pressure o f a gas c a n b e m e a s u r e d using a m a n o m e t e r
pgh
T h e p r e s s u r e o n a surface d u e t o a f o r c e is t h e f o r c e o n 1 m of t h e surface. 2
,KU « f o r c e (N) pressure (Nftrf) -
he u n i t of pressure, N / m is also k n o w n as t h e p a s c a l (Pa) 2
Worked examples 1. A force of 10 kN acts o n t h e surface o f a l i q u i d , o f area 0.08 m . W h a t is t h e pressure 2
o n t h e surface of t h e liquid?
2.
A person o f w e i g h t 6 0 0 N exerts a pressure of 2 0 0 kPa o n t h e g r o u n d . W h a t is t h e area o f t h e i r feet?
3.
T h e area o f a d o g ' s p a w is 10 c m . The pressure u n d e r t h e p a w is 50 kPa w h e n it 2
exerts half o f its b o d y w e i g h t o n t h e p a w . W h a t is its w e i g h t ?
Answers 1.
P = £ 10 0 0 0 0.08 125 0 0 0 Pa 125 kPa
2. ^ = £ 600 200 000 0.003 m
3. F = p
2
XA
= 50 0 0 0 X (10 + 10 0 0 0 ) = 50 N Total w e i g h t = 2 x 50 = 1 0 0 N N o t e : In q u e s t i on 1, 1 kN = 1 0 0 0 N In q u e s t i o n 3, area in c m is c o n v e r t e d t o area in m by d i v i d i n g by ( 1 0 0 x 100) ie 2
by 10 0 0 0
2
'Revision guide: Physics
A t m o s p h e r i c (air) pressure can be measured w i t h a b a r o m e t e r . r The glass tube is evacuated and so the mercury can move up inside. The greater the air pressure, the higher the mercury rises up the tube. The height of the mercury column is proportional to the outside air pressure.
> <
air pressure
• vacuum
There is pressure on the surface of the mercury due to the force of the air molecules hitting the surface.
air pressure
_Jv' The air pressure on the surface forces the mercury up the glass tube.
mercury
T h e pressure a t t h e base o f t h e m e r c u r y c o l u m n = density o f m e r c u r y x g x =
height
pgh
w h e r e p = density o f m e r c u r y in k g / m
3
g = acceleration d u e t o gravity in m / s
2
h = h e i g h t o f c o l u m n in m The pressure b e l o w t h e surface o f all liquids increases p r o p o r t i o n a l l y t o d e p t h . For e x a m p l e , t h e f o r m u l a above can be used t o calculate t h e pressure u n d e r t h e surface o f t h e sea. A m a n o m e t e r can be used t o measure t h e pressure o f a gas:
The gas whose pressure is to be measured enters the manometer here
This end of the tube is open to the air.
The air exerts a pressure on the surface of the liquid here.
The gas exerts a pressure on the surface of the liquid.
T h e pressure d u e t o t h e gas = a t m o s p h e r i c pressure +
pgh
W h e r e p = density o f liquid in k g / m ; h = d i f f e r e n c e in h e i g h t o f t w o l i q u id surfaces in m 3
Examination style questions 1.
a.
The d i a g r a m b e l o w s h o w s t w o examples o f f o o t w e a r b e i n g w o r n by people of equal w e i g h t a t a W i n t e r Olympics c o m p e t i t i o n . W h i c h f o o t w e a r creates t h e greatest pressure b e l o w it, a n d w h y ?
b.
Drivers o f high-sided vehicles, like t h e o n e b e l o w , are s o m e t i m e s w a r n e d n o t t o drive w h e n it is very w i n d y .
Suggest w h y t h e y receive this w a r n i n g . CIE 0 6 2 5 N o v e m b e r "05 P a p e r 2 Q 3
2.
The d i a g r a m b e l o w s h o w s a p o n d t h a t is k e p t a t a c o n s t a n t d e p t h by a pressureo p e r a t e d valve in t h e base.
water
pressure-operated valve spring
The p o n d is k e p t a t a d e p t h o f 2.0 m . The density o f w a t e r is 1 0 0 0 k g / m . 3
Calculate t h e w a t e r pressure o n t h e valve. The f o r c e required t o o p e n t h e valve is 50 N. T h e valve w i l l o p e n w h e n t h e w a t e r d e p t h reaches 2.0 m. Calculate t h e area o f t h e valve. The w a t e r supply is t u r n e d o f f a n d t h e valve is held o p e n so t h a t w a t e r drains o u t t h r o u g h t h e valve. State t h e energy changes o f t h e w a t e r t h a t occur as t h e d e p t h of t h e w a t e r drops f r o m 2.0 m t o zero. C I E 0 6 2 5 N o v e m b e r "05 P a p e r 3 Q 3
Revision guide: Physics
Summary questions on unit 1 1. Fill in t h e blanks The speed of an o b j e c t can be calculated by w o r k i n g o u t divided by The u n i t o f speed is
of a
. Speed is equal t o t h e
and
d i s t a n c e - t i m e g r a p h . Velocity is equal t o speed in a certain
.. T he acceleration o f an o b j e c t is equal t o
it has t h e same
o f velocity a n d has units o f
the
o f a v e l o c i t y - t i m e g r a p h . The
A c c e l e r a t i o n can be calculated f r o m t h e area u n d e r a v e l o c i t y - t i m e g r a p h is equal t o t h e
.
2.
Calculate t h e speed o f an o b j e c t t h a t travels 10 m in 5 s.
3.
The initial velocity o f a car is 10 m/s. It reaches a velocity o f 25 m/s in 5 s. W h a t is its acceleration?
4.
W h a t is t h e g r a d i e n t o f a d i s t a n c e - t i m e equal t o ?
5.
W h a t is t h e area u n d e r a v e l o c i t y - t i m e g r a p h equa l t o ?
6. W h a t is t h e d i f f e r e n c e b e t w e e n mass a n d w e i g h t ?
7.
A p e b b l e o f mass 100 g is placed in a m e a s u r i ng cylinder c o n t a i n i n g 50 m l o f water. The w a t e r level rises t o 7 5 m l . W h a t is t h e density o f t h e pebble?
8.
Fill in t h e blanks A f o r c e can stretch an object, or c h a n g e its
. If small forces are a p p l i e d t o
a s p r i n g , t h e extension p r o d u c e d by a load is a p p l i e d . This is t r u e u p t o t h e limit o f an o b j e c t by causing it t o
t o the force . A f o r c e can c h a n g e t h e velocity o f
. The acceleration is
if t h e f o r c e
o n t h e o b j e c t is d o u b l e d a n d t h e mass stays c o n s t a n t . The n e t or overall f o r c e o n an o b j e c t is called t h e t h e y are
f o r c e . W h e n forces act in o p p o s i t e directions t o f i n d t h e resultant f o r c e . W h e n forces act in t h e same
d i r e c t i o n , t h e y are the
t o f i n d t h e resultant f o r c e . A f o r c e can c h a n g e
in w h i c h an o b j e c t is travelling w i t h o u t c h a n g i n g its
This happens w h e n t h e f o r c e is travelling a n d it causes t h e o b j e c t t o m o v e in a
9.
t o t h e direction in w h i c h t h e o b j e c t is .
A f o r c e o f 10 N produces an extension o f 2 0 c m . W h a t extension w o u l d be p r o d u c e d by a f o r c e o f 2.5 N?
.
10. Fill in t h e blanks _. The w e i g h t o f an o b j e c t is
The mass of an o b j e c t is measure d in a type of
. The w e i g h t can be
a n d is measured in
calculated by m u l t i p l y i n g t h e
by t h e .
The density o f an o b j e c t is equal t o its mass d i v i d ed by its _ . The u n i t of density is
or
.
11. M a t c h t h e devices t o t h e e n e r g y t r a n s f o r m a t i o n s
kettle
chemical to electrical
generator
kinetic to electrical
battery
light to electrical
solar cell
electrical to thermal
12. Calculate t h e kinetic e n e r g y o f a ball o f mass 0.2 kg t h r o w n a t a speed o f 7 m/s.
13. Calculate t h e increase in gravitational p o t e n t i a l energy of a b o y o f mass 50 k g c l i m b i n g a f l i g h t o f stairs o f h e i g h t 5 m.
14. Calculate t h e distance travelled by an o b j e c t if t h e w o r k d o n e o n it by a force of 2 0 N is 100 J.
15. W h a t is t h e p o w e r o f a kettl e t h a t t r a n s f o r m s 100 J o f electrical e n e r g y t o internal energy in 0.5 s?
16. W h a t is t h e pressure 3 m u n d e r t h e surface o f t h e sea? Sea w a t e r has a density o f 1200 k g / m . 3
17. Calculate t h e pressure u n d e r a block of mass 0.3 kg w h e n resting o n an area o f 5 c m
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Revision guide: Physics
Examination style questions on unit 1 1. A solid plastic sphere falls t o w a r d s t h e Earth. B e l o w is t h e s p e e d - t i m e g r a p h o f t h e fall up t o t h e p o i n t w h e r e t h e sphere hits t h e Earth's surface.
140 i
i
120 -
100
80 speed m/s 60
/ Q
40
20 7i J\ p 10
20
30
40
50
60
70
80
90
100
time/s
a.
Describe in detail t h e m o t i o n o f t h e sphere s h o w n by t h e g r a p h .
b.
C o p y t h e d i a g r a m b e l o w a n d , d r a w a r r o w s t o s h o w t h e directions o f t h e forces a c t i n g o n t h e sphere w h e n it is a t t h e position s h o w n by p o i n t S o n t h e g r a p h . Label y o u r a r r o w s w i t h t h e names of t h e forces.
c. d.
Explain w h y t h e sphere is m o v i n g w i t h c o n s t a n t speed at S. Use t h e g r a p h t o calculate t h e a p p r o x i m a t e distance t h a t t h e sphere falls i)
b e t w e e n R a n d T,
ii)
b e t w e e n P a n d Q. CIE 0 6 2 5 J u n e ' 0 5 P a p e r 3 Q1
110
The d i a g r a m b e l o w s h o w s a simple p e n d u l u m t h a t s w i n g s b a c k w a r d s a n d f o r w a r d s b e t w e e n P a n d Q.
The t i m e t a k e n f o r t h e p e n d u l u m t o s w i n g f r o m P t o Q is a p p r o x i m a t e l y 0.5 s. Describe h o w y o u w o u l d d e t e r m i n e this t i m e as accurately as possible. i)
State t h e t w o vertical forces a c t i n g o n t h e p e n d u l u m b o b w h e n it is at
ii)
Th e p e n d u l u m b o b moves a l o n g t h e arc of a circle. State t h e d i r e c t i on o f t h e
p o s i t i o n R.
resultant of t h e t w o forces in (i). The mass o f t h e b o b is 0.2 k g . D u r i n g t h e s w i n g it moves so t h a t P is 0.05 m highe r t h a n R. Calculate t h e increase in p o t e n t i a l e n e r g y of t h e p e n d u l u m b o b b e t w e e n R a n d P. CIE 0 6 2 5 J u n e '05 P a p e r 3 Q 2
A bus travels f r o m o n e bus s t o p t o t h e next. The j o u r n e y has t h r e e distinct parts. Stated in order t h e y are u n i f o r m acceleration f r o m rest f o r 8.0 s, u n i f o r m speed f o r 12 s, n o n - u n i f o r m deceleratio n f o r 5.0 s. T h e g r a p h b e l o w s h o w s only t h e deceleratio n o f t h e bus.
10 H speed m/s
10
20 time/s
a.
C o p y t h e g r a p h a n d c o m p l e t e it t o s h o w t h e first t w o parts o f t h e journey.
b.
Calculate t h e acceleration of t h e bus 4 . 0 s a f t e r leaving t h e first bus s t o p .
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Revision guide: Physics
c.
Use t h e g r a p h t o estimate t h e distance t h e bus travels b e t w e e n 2 0 s a n d 2 5 s.
d.
O n leaving t h e s e c o n d bus s t o p , t h e u n i f o r m acceleration o f t h e bus is 1,2 m / s . 2
The mass o f t h e bus a n d passengers is 4 0 0 0 k g . Calculate t h e accelerating f o r c e t h a t acts o n t h e bus. e.
The acceleration of t h e bus f r o m t h e s e c o n d bus s t o p is less t h a n t h a t f r o m t h e first bus s t o p . Suggest t w o reasons f o r t h i s. C I E 0 6 2 5 J u n e '06 P a p e r 3 Q1
4.
The d i a g r a m b e l o w s h o w s a m o d e l car m o v i n g clockwise a r o u n d a h o r i z o n t a l circular track.
a.
A f o r c e acts o n t h e car t o keep it m o v i n g in a circle. i)
C o p y t h e d i a g r a m a n d d r a w an a r r o w t o s h o w t h e direction o f this f o r c e .
ii)
T h e speed o f t h e car increases. State w h a t happens t o t h e m a g n i t u d e o f this
i)
The car travels t o o quickly a n d leaves t h e t r a c k a t P. O n y o u r d i a g r a m , d r a w an
force. b.
a r r o w t o s h o w t h e d i r e c t i o n o f travel a f t e r it has left t h e track, ii) c.
In t e r m s of t h e forces a c t i n g o n t h e car, suggest w h y it left t h e t r a c k a t P.
T h e car, starting f r o m rest, c o m p l e t e s o n e lap of t h e t r a c k in 10 s. Its m o t i o n is s h o w n graphically in b e l o w .
'•I
i) ii)
Describe t h e m o t i o n b e t w e e n 3.0 s a n d 10.0 s a f t e r t h e car has s t a r t e d . Use t h e d i a g r a m a t t h e b e g i n n i n g o f t h e q u e s t i o n t o calculate t h e c i r c u m f e r e n c e o f t h e track.
iii)
Calculate t h e increase in speed per s e c o n d d u r i n g t h e t i m e 0 t o 3.0 s. C I E 0 6 2 5 J u n e '07 P a p e r 3 Q 1
T h e d i a g r a m b e l o w s h o w s a steam safety valve. W h e n t h e pressure gets t o o h i g h , t h e steam lifts t h e w e i g h t W a n d allows steam t o escape.
u.i m
,
^ r
I
• pivot
w
1
force of steam
a.
Explain, in t e r m s o f m o m e n t s of forces, h o w t h e valve w o r k s .
b.
The m o m e n t o f w e i g h t W a b o u t t h e pivot is 12 N m . Th e p e r p e n d i c u l ar distance of t h e line o f a c t i o n o f t h e f o r c e o f t h e s t e a m o n t h e valve f r o m t h e pivot is 0.2 m . The area o f t h e piston is 0 . 0 0 0 3 m . 2
Calculate i)
t h e m i n i m u m steam f o r c e n e e d e d f o r t h e s t e a m t o escape,
ii)
t h e m i n i m u m steam pressure f o r t h e steam t o escape. C I E 0 6 2 5 J u n e '07 P a p e r 3 Q 2
