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The Revision Cuide Exam Bo ard: Edexcel
Hitors: Amy Boutal,Sarah Hilton, Alan Rix,Julie WakelingrSaruh Williams
@ntributors larrc Cartwng;ht, Peter Cecil, Mark A. Edwards, Barbara Mascetfi, John Myers, Zoe Nye, Moira Steven, Andy Williams Piogfreaders: 1I,1:
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fublished by Coordination Group Publications Ltd.. ::.r:,:.'. ":, :"
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Many thanks to Professor Peter Watkins at the University of Birmirrgham for his kind perinission fo r.eproduce the photqg ruphs used on p4gg 34.
ISBN:
g 9781 84762 270 ,,..-,' r'
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Lrrooly website:'w^w w.cgpbooks.co.uk Jolly'bits of clipart from CorelDMW@ printed by fhnders Hindson Ltd,Newcastle upon Tyne. Based on the classic CGP s+vle created hv Richard Parson.s.
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Contenfs How Science Works
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Heat and Temperature .o...............................,..,. 36 lnternal Energy .. .........
o
...
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..... ....... .....
o... . ...
....
.... 3B
.....
...
....
.... 42
4 6 B
Exponential Law of Decay
..
.....
t0 l2 t4 t5 IB
20
22
Measuring Astronomical Distances.................... 52
24
Luminosity and the Hertzspruns-Rus sell Diagram ,.. .. . .. ..... .. .. ... .. . .. .. o. . ... .... o. o. 54
26 Anti particles Quarks
o. .
.. .. .. .,. ... o .. . o.... o.. o......
..... ..... ..... ... .. 2B
30 33
The Life-Cycle
of
Stars ................. o..................... 5 6
Hubble and the Big Nucle ar Fission and
\ang.......... :...,
Fusion.:...
......... 5B
............ 60
Answering Exam Qugstions ....................o..o...:... . 62
How SctElvcr WoRKs
How
Scrrruc
r WoBrcs
How Sclrlvcr
Woinxs
UNn 4: SrcrloN
Uxr 4: SrcnoN 1
Funrurn MrcHANtcs
1
Funrurn MrcHANtcs
ln reality, collisions usually happen in more than one dimension. Momentum is still conserved the only difference is that you have to resolve the velocity vectors of the colliding objects to find the components that affect the collision.
-
1)
Resolve the velocity vector into the components parallel and perpendicular to tha line of the collision.
3) The n you can ignore,vrrnd ,!" v, to work out the new velocity of the helium nucleus just like you normally would.
,}
,$'
(^
Momentum before = Momentum after
,f ..."'
Line of collision
(tx3.5) +(4x0) =(1 x0) +(4xv)
v=5ms-l Use
,
2)
On
t,
Iti,e,pal
dllel components
,:: ,,,,The perpendicular
3.5 = 4v ll = 0.9 mstl
trig,to resolve:
= v. x cos (45') v2= v x sin(45")
V,
interact duringthe
colli-sion.
componenti don't change..
4) -
And,finally, drl^i'A ai;ag remembering'to'inclu de
and
the
new
,m
v,
velocity vector.
,
AII the momentum calculations you'll encounter in your exam will be in one or two dimensions .Remember dont have to - you you;ti these four tips for solving problerns'*na Oa,tin", i) Sketch a diagram things are much easier to understand when you can see exactly whatrs going on. z) The component of-the momentum in the x-direction (i.e. across the page) and the compon*, i*6"
worry about things happening in three dimensions.
-,,lir,y-direction(i.e.upanddownthepage)will i
3),,
,,'
bothbeconservedbefoieinarft"r,n"-."iiiri;.''"'':",r',,
Only the component of the velocity vector that is parallet to the tine of the collision will have any effect.
+)
The component of the velocity vector that is perpendicutar to the line of the collision
Ql Q2 Q3 Q4
What is the equation for momentum? what unit is momentum measured in? Explain how rocket propulsion is an example of the conservation of momentum.
will
,
i
!i
have no effect.
cive two other examples of conservation of momentum in practice. What do you have to do differently when solving momentum problems in two dimensions?
Exam Questions
Ql
toy i:rain of *]::-0;Jt-gl trave[ing at,0.3 ms-r, collides with a stationary toy carriage of mass 0.4 kg. The two toys couple together. Whaiis their new velocity?
A.
Momentum will never be an endan ered s
o' q
-
it's alwars conserved...
lo, Sress what - momentum is conserved in collisions. tf you forget that you'll really scupper your chances of getting lots of marks from solving momentum problems. So remember, momentum is mass times velocitf, and momentum is conserved. See, it's easy momentum is your friend.
-
UNtr
4: srcnoN I
Funrutn MtcHANtcs
You did most o,f ,this at CCSEI forget this stuff in the e:xam
but that doesn't mean you can just skip over it now. Yotu'll be kicking yourself if you easy
marks...
:
If mass is constant, this can be written as the well-known equation:
it crops up all over the place in A2 Physics. Learn this And learn what it means too:
-
1) 2)
lt says that the more force you have acting on a certain mass, the more acceleration you get. lt says that for a given force the more mass you have, the less acceleration you get.
1)
The
resultant force is the vector sum of ali'thiiot.t!
5) The aqceleiation
is alwayi in the safie direition as."..-" the resultant force, .
.j.
..
.r r..
:
. :t'
,
, ,:',
-'
,,,'
',1',
I
The principle of conservation of ene r1y says that:
it depends on the mass As you'll probably remember, kinetic energy is the energy of anything moving you need to learn that for kinetic energy two equations Thereire is moving. whatever and velocity of momentum: of in terms form written and a love, know and form you'll that the standard
-
where v is the velocity the object is travelling at and m is its mass. You need to be able to derive the second form of the equation from
-
where p is the momentum of the object an d m is its mass. it's not as scary as it sounds.
Simon liked to derive kinetic energy attop speed.
Uxr 4: StcrloN I
FunrulR MECHANI:S
1)
As long as there's no friction, you know that momentum is always conserved in a collision (you have the same total
momentumafteracollisionasyouhadbefore)(seepages+andS). \ \ ) \ \ t t t / / ./ '
After a collision, objects sometimes stick together, and sometimes bounce apart. - , ln the real world, some energy's / I -' - duEp lost in a collision. Sometimes, --- *,Either way the momentum is still conserved.if the energyloss is small' ifs okay J 3) But the kinetic energy is not always conserved. Usually, some of it gets converted into sound or ^T"litl"collision "ftTl ..A collision where kinetic energy is the same after a collision is called an elastic collision. [otal A collision where the tdt{ kinetic energy is less after a cottision is called an inelastic coilision.
2)
heafenergy.
'i
i
=
f-*fl$*i*** **
$ ..'
ffi -: a) 'i-., ,,,.',
usrng conservatron conservation of 'Using ot 'momentum momentum (pages 4_5): 4-5): total momentum before = total momentum after
:;,.:
gF.-11 tir..y.--Fr:; 1.1,:, ;"" rl l,',rr'-+1,{,1 lt=f (0.05x20)+(0.06x0)=(0.05x8)+(0.06xv-) j=ft
,
.
[+i,r*ffffi*$s* ,€#*tp"+g#ffi,.r|Illl
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l' {=uTlrru'r=' .r\ \,
Sreffifit'i,",'.1"),')}
'ii:r.i, j
l=o.o+o.06vrlvr=0.6+0.06=rbir-, Before: KE =1/z x 0.05 x 202 = l0 I
c/ ,ii,.:, -.1.,''" o
rhe I ne collision colllslon must De be inelastic, tnelasttc, because the total ,ooi kinetic i'n"r,. energy is reduced in the ,n.7it,J^i-,.' collision.
-
converled into
',r-r"ruced
"nu.
other'forms-of
' --
*t"F)l ##"dr.g+ ,f}iif;l.. 5 j..j, rFB+-sggg.-'*='
State the principle of conservation of energy.
What are the two equations for calculating kinetic ener gy? What's always conserved in a collision (if there's no frictionX What's the difference between elastic and inelastic collisions? Exam Questions
A skateboarder is on a half-pipe. He rolls freely down one side of the ramp and up the other.
Q1
(a)
The height of the ramp is 2 m. Take g as 9.81 Nkg-,. assume that there is no friction, what would be his speed at the lowest point of the ramp?
If you
(b) (c) Real ramps are not frictionless, so what must the skater do to reach rt
A railway truck of mass 10 000 kg is travelling at The two trucks stay together after the collision.
Q2
I msr
"
t.p
on tt e oiher riadf
[3 marks]
i''
,
trucks.
to revise
[1 mark]
and collides with a stationary truck of mass 15 000
(a) lhat can you say about the total kinetic energy before and after the collision without calculating anything? (b) Calculate the final velocity of the two ,, (c) Calculate the total kinetic energy before and after the collision.
Warnt
[t mark]' ,,
[1
rnark]
[2 marksJ 12 marksl
rubber band at a
As my gran always tells ffie ,'What you put in is what you get oit' truth thai works equally well for - this is a universal energy, physics revision and beer. So t guess there's no aiguing with me gran (well, her umbrella always deterred me... ). U
xt 4: StcrtoN I
FuxruER
M
ECHANtcs
Uxr 4: SrcrloN I
Funrum MtcHANtcs
1)
2) 3)
'
I
'
:
',.
:
Even if the car shown js goin$ at a constant speed, its velocit, velocity is changing since its direction is changing.
Since acceleration is defined as the rate of change of velocity, the car is accelerating even though it isn't going any faster. This acceleration is called the centripetal acceleration and is always directed towards the centre of the circle.
There are two formutas for centripetal acceleration:
ffiandry
a = centripetal acceleration in ms-2 v = linear speed in ms-1 (t = angular speed in rad s-l
r = radius
in m
From Newton's laws, if there's a centripetal acceleration, there must be a centripetal force acting towards the centre of the circle. Since F - ffid, the centripetal force must be:
ffiandffi
The centripetal force is what keeps the object moving in a circle refitove the force and the object would fly off at a tangent.
-
Ql Q2 Q3 Q4
force of the centripede.
How many radians are there in a complete circle? How is angular speed defined and what is the relationship between angular speed and linear speed? Define the period and frequen'cy of circular motion. What is the relationship between period and angular speed? ln which direction does the centripetal force act, and what happens when this force is removed?
Exam Questions
Ql
(a) At what angular speed does the Earth orbit the Sun?
(l
year:3.2 x
I
107 s)
(b) calculate the Earth's linear speed. (Assume radius of orbit = 1,5 x 10, m) :
:
......'
(c) Calculate the cenhipetal force needea.io
tj"p
[2 marks]
l
trre gurth,io it, orbit. (Mass of Earth
:
6.0 x lOu kg)
[2 marks]
(d) What is providing this force? Q2
[1 mark]
t .:'- ..,,
A bucket full of water, tiea to a ,op., is ulia[ i*,rne around at the top of the swing). The radius of the cilcle is l"rn.,
marksl
12
,,
, .' :i.
.. i
, .-::li. :'1 '1"
,,., .
,
tu) P]-99nlidPrirgthe,acceleration due to.gravity at the top of the swing, what is the minimum frequency
,:
[3:
(b) The bucket is now t*Yng with a constant angular speed of 5 rads-r. What will be the tension in the rope when the bucket is at the top of the swiig if the total mass of tire uucger tO kg? , ,
anafifi;
.,
marks]
[2 marks]
l'm spinnin' around. mtoie out of my way...
'Centripetal" just means 'centre-seeking". The centripeta.l force is what actually causes circular motion. force'. Don't get the two mixed up.
llhat you feel when you're spinning, though, is the reactiop @entrifugal)
Uxtr
4: SrcroN I
:-
Funrutn MrcHANtcs
10
uNr 4:
uNr 4: SrcrpN 2
SrcrrcrN
2
Erccrntc AND MncNErtC Ftrns
Erccrntc AND MncNEnC Frcns
1)
get further apart and the field strength decreases. you need to combine the two equations on page r o
2)
E
-!9r'
k -'-l-
where
4re
ln a uniform field, the fie.ld lines are parallelso.they're always the same distance apart. This means that the field strength is the same at all points within the field i.e. a tesl charge would experience the same force wherever it was. The field strength betw.een two paraltel plates depends on the potential E can be measured in difference, v, and the distance, d, between them, according toihe equation:
-
volts per metre (Vm-l)
ln_vgstigating a Uniform
,J,n;.,r.,
Field
Atomiser_
1)
Anatomiseri'uit"'inne.i'totoitJ,op,thatareToPp|,l"M
il'
When the circuit i; off, the arop,
iril fro*
the
top'
. "
3)
-,-,---,-i- iot1o, p[ate...=-$[ glulg.!o_t!: !ot19m plate due to their weight, ,^:u"*,r",){" wh en' th e c i rc u t s ;I,' iJ in"," il; *ii=.i....:i,l "." chargJ .iirf ."" Ln field, which exerts a force on the oil drops. A negatiVety *a;:diio"i; o"iJ;.nlth". plates bv batancing the upward forca rro; the etJctr:ic ri"ra ;itn1i" "i1
4)
lryou
i
i
r*i;;
l;il;
;;;";;
;*;'**t*mf
i;*fu"ijilrffiil"
s)
Q1 Q2
inC-fcaqe
th.i p'd,, you fnqrease the field ilie,nsth
qp fhe
;fli.;ii,Ii,l,lf'ili,:
oi!drgp-lnoveiltirwaidiihe potifivefop o.,u_-,_..
lf you increase the distance between the ptates or decrease the p.d.. y9u
redtigF_ Irs_fsld,9t1e1gt]r q;d- _ .theoildropfallstbthebottomplatedueioitsweight. _1:t:,:-=.=itl":,,:i..,j.,,.-,._-.... _.--*.-,_,,,,,,
"l
Draw the electric field lines due to a positive charge, and due to a negative charge. Write down Coulomb,s law.
Exam Questions
Ql Q2
*ih:l.llbut
The diagram rholr,yo:l-.1,1ic clarrys opposite charge, Draw electric field lines to show ihe electric field in the area surrounding ihe charges. Find the electric field strength at a distance of 1.75 x l0-r0 m from a 1.6 x
e.
'
o
iQ "
.
" "
'-0 ,
[,
marks]
l0-re C ioint crrarge. 12 marksl (a) parallel Two plates are separated by an ai.r gap of 4.5 mm. The plates Q3 are connected to a tr500 v dc supply. Wh.at is the -rr-J electric field strength between the plates? Give a suitable unit ana state the
air"rrir,
(b) The plates are now pulled further apart so that the distance between them is The electric field strength remains the same. whal is the new voltage between the
"i,rre
field.
doubled.
At t
least you get
broasf
r:
,
,:
r,
:
.
), r
,
i
:
.
: .
:
,
[2 marks]
beef...
a choice here - uniform or radial, positive or tn:ra/.#t:::it*:;?i.#J.i.t;!ir'ati??v.u):rntt?;4!i?p:r'Dt
plates?
[3 marks]
negative, attractive ar repulsive, chocolate or strawberry...
Uxtr 4: stcroN
2
Ettcrntc AND M,qcNErc Ftrns
12
like a charge bucket. The capacitance of one of these Capacitors are things that store electrical charge things tel/s you how much charge the bucket can hold. Sounis simple enough..'. ha... ha, ha, ha...
-
A farad is a huge unit so you'll usually see capacitances expressed in terms of:
,,:-:: hvestigaling tne Qnarse Stoidd .oh' e'CeiEa-eitdi'' .F-'el-u^p..4'".e$3r.qtl{.-1g.LngeluI9..9.ur.l-e.$qI{pq1',9IIip!'$ff"'qlqq;"ffi. Constantly- adjust.ihe,,iariabrc resistor"tokeep- the charging curtent constant for--,--' .-..... - -, --.I,aq.lo-hg ers you can (it's impossible when the,capacitor is nearly'firlly charged)- -.*--;.T. i: 3) i Record the p,d. atrre$ular'intervals until it equals the batterj p.d. -:'"'-4)
'
"
[',Ffom theSe]esultS, you can plbt the fullowing gfaphS:
6rraient=9=C v rge stored on Blates)
" .
:f(s),
1)
2)
In this circuit, when the switch is flicked to the left, charge builds up on the plates of the capacitor. Electrical energyt provided by the battery, is stored by the capacitor. lf the switch is flicked to the right, the energy stored on the ptates will discharge through the bulb, converting electrical energy into light and heat.
3)
Work is done removing charge from one plate and depositing opposite charge onto the other one. The energy for this muit come from the electrical energy of the battery, and is given by charge x p.d. The ener[y stored by a capacitor is equal to the work done by the battery.
4)
So, you can find the energy stored by the capacitor from the area under a graph of p.d. against charge stored on the capacitor.
The p.d. across the capacitor is proportional to the charge stored on it, so the graph will be a straight line through the origin. The energy stored is given by the yellow triangle.
5)
Area of triangle = 1/z x base x height, so the energy stored by the capacitor is: for 'work done' , buL you can also use Efor energy W stands
UNr 4:
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2
Ettcrltc
AND
m
M,qcxErtc Ftttos
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2
Erccrntc
AND
M,qcNErc Ftrms
Uttg 4: Srcrpr,t
2
Erccrntc AND M,qcxETtC FtnoS
The time it takes to charge up or discharge a Capacito-i,depends on: 1) The capacitincb of the Ilri,I
2\
tll;il.:
i..ill
.The 'i:
1.
r"iirtunte
,:t';', :ui.
oF
;o;.itai tC
tt," bircuit (Ai1'tnii
When a capacitor is discharging, the amount of charge Ieft on the plates falls exponentially with time. That means it always takes the same length of time for the charge to halve, no matter how much charge you
1)
2)
stari with
-
like radioactiVb decay (see p. 42). charging
The charge left on the plates of a capacitor
l/:Vo-Voe
discharging from full is given by the equation:
d
where Qo is the charge of the capacitor when it's fully charged.
ischarging _t
V _ Voe
Rc
I
The graphs of V against f and against f for charging and discharging are also exponential.
r is the Greekletter'tau tf f = T
- RC is put into the equation
above, then Q = Qo€-'.'''So when
the time constant, is the time taken for the charge on a discharging capacitor (Q) to fall to 37% sf Qo.
2)
It's also the time taken for the charge of a charging capacitor to rise to 63 "/" of Qo.
3)
The larger the resistance in series with the capacitor, the longer it takes to charge or discharge.
4)
ln practice, the time taken for a capacitor to charge or discharge fully is taken to be about 5RC.
Q1 Q2
1 I
e
where
rft|-
e
2.718
:i '. :,., . , :' ,'i:. ,;. 'r'. _'
r,
1)
So
t-i:
,,
'.-..
.1
,, .1' .-',
x
0.37
.
0.37 Qo
Sketch graphs to show the variation of p.d. across the plates of a capacitor with time for: a) charging a capacitor, b) discharging a capacito
What two factors affect the rate of charge of a capacitor?
Exam Question ,
Ql I
.::
I
A.'ZSO
[f
,
,.
"upacitor
'r
.i::r.
:.;,.i1i,:.1
,
I l:i .'
.;,-.
:
l:
.
is ful1y charged aom a-(V.b1tlg.poOtt"n discharged through a I kQ resistor.
(a)',Calculate the time taken for the chqrgg on the iipaeitor.to fall to 37Yo of its original value.
(c) If the charging vgltage is increased to 12
!
what effect will this have
on:
,'
t
,' ,. , - ,. l;',1r: i) the total charge stored, ii) the capacitange of the capacitor, iii) the time taken to fully charge
An analogy
:-
[2 marks] [3 marks] ,:.
[3 marks]
consider the lowly bike pump...
One way to think of the charging process is like pumping air into a bike tyre. To start with, the air goes in easily, but as the tyre pressure increases, it gets harder and harder to squeeze more air in. The analogy works just as well for discharging...
UNr 4: SrcnoN
2
Etrcrntc AND Mnct'tErtc Ftrtos
16
UNr 4: Srcrpv 2
*
Etrcrntc AND MecNlrtc Ftrtos
,,
l,l
7r
3) {,
+,r
A f*rce acts if a current-carrying wire cuts rnagnetic flux Iines. Flernins's Lsft-Hand Rule lf th* currcnt is parallet ta the flux Iines, no force acts' The First, frnqrr p*ints in the directicn of the The direction af the {arce is alr,vays perpendicular to hoth the uni{orm magn*t tc Field, the srC *nd {tnger points in Lhe directicr, *t the conventianal C,;rr*nt. current directlan and the magnetic fieldThe n y*ar Lh*lvlb pcints in the dtreclion of the The Cirection rlf the force is given by Fleming's Left-Hand Rule- '.- +>" larce (i*
which &t-tticn
takes
The S,,ze of the fOrCe, F, On a CUrrent-c Z{r\,'ing r,t'ire at right angles to a rnagnetic tielcl is proportional to th* current, I, the length of wire in ihe fiekJ, I, &nd the strength of the magnetic field, B. This sii'es the equation:
li
ln th;s equation, the magnetic field strength, B, is definecl
glacel,.
i
\;i
as: You can think
:::; e rnagteLic
iield
strength as ;L z r'-,.,-ber of flux lines ?€Y Utf i*r- &r€a,S:: :reviOUS page).
field is caused by the to the rlire -- S sin 9.-3 is perpendicular wlrich component of iielclstrerrgth lio, ior .r rvire ;it .1n angle I to the fielcl, the tbrce acting on the u'ire is giverr by: The torce on a cLirrent-carrying rvire it-. a nl.rgnetic
F'
'r.,. a:.'
.f,
Ci-lrititl B
"P;l'
.
BII sin 0 lll'
Exampfes:
i.
Field --
-----'
-- p ('tllI('ll1 *> Ill.itlrrcl it- fit'ltl
i)
lf0=9C' I= 3ll
ii)
lf t) = 3C' - = 3/lxO.5
il)
lf()=O",,
i
Pracfice Quesffon$,,"
el
Describe wl-ry a current-carrying rlire .rt right angles to att cxlcrtral nragnetic field lvill erperience a force.
Q2
Sketch the mirgnclic fickls arerund .r long straight current-carrying wirc', and a solenoid. Shoiry the direction of tlre current and magnetic field on each diagrant'
Q3 -
A copper bar can rolt frccly on lwo copper suppor6, as shown in lhe diagram' When'current is applied in thc direclion shown, which way will the bar roll?
Exaur Question Qr
x A wire carrying a currcnt of 3 A runs perpendicular to a magnetic ficld of strength 2 l0-$T. 4 cm of the wire is wilhin the field. (n) Calcuhtc thc magnirude of thc forcc on the wire. (b) If the u.ire is rotate
it
[2 marks] [2 murks.l
is at 30o to the field. what would the size of the force bc?
t revised the right-hangUute
by the A69 and ended up in.Ne,wpaftl,e','
nrake sure yolt knotry ho' lo use.it .and ttnderstanc! what ir a// nrr';trts: +ve to -ve -* tlris is as Rcvrrenrber that rhe clirection of tha ma]4,nept iblc/ ri lronr N to 5 anri that the c'urrent is front pat... intpartant;.s rrsritg the correcthand. You needtogetthoseflghtor it'llallg,oO Flt nting's leit-h;tnd
,rtii
the
k,y kt rhii soctr'on
-
.so
Urul T
4; Srcr ffiN 2
#
ftrcrs/C
ztN{J Mnr;ru
rl
ff
f /t'l /::
Nt,rl;netic fields are used
a lot when dealing with particle
bearns
-
you'll be learning more about therr
uses
in Unit 5.
Electric current in a wire is caused by the flow of negatively charged electrons. These chargcd particles-are affected so a current-carrying wire experiences a force in a magnetic field (see pages 16-17). by magnetic fields
-
You meb this equ.ilion on t.he previc-tls i'.*ge
1)
2)
The equation for the force acting on a single charged particte moving perpendicular to a magnetic iield is:
Equation2,mof,lT(]regCncralIyEquation3:
F
*
&qv sin 6 ln nrany exarn rlLI€Si ons, q is tir.: size of the charile c': the .'1.:ctri'tt wlricl'r is 1.5 r lO ' cot,i.'rnbs.
wlrcre g is the charge orr the particlc, v is its velocity ar:d 0 is the angle bctween the direction of motion arrcl thc magnetic field.
Example rns I
through a unifbrnl magtretic field of stretrgth 2'f . Find the force on the electron rvhen: a) it is travelling at right angles to the nlagnetic field. b) it is travelling at iOo to the directiort of the tttttgnetic field. (The magnitude of the charge on an electrorr is 1.6 x l0-10C.)
An e lectron is travelling at 2 x l0a
a)
b)
f-Bqu
,I.6 x 10-11' x 2 x so, F = 2 x so, F*6.4x]O-rsN
10"+
So,
1)
:) 3)
+i
6) 7r
l0ti
N
Mathematicall),, that is the condition for circular motion. This effect is used in particle accelerators such as cyclotrons (see page 25) rlhich use electric and magnetic fields to accelerate particles to very high energies along circular paths. Pos itively The radius of curvature or the path of a charged particle nrcving charged through a magnetic field gil,es you information about the particle's pa rt ic le this means you can identify different particles charge and mass by studying horv they're deflected (see pages 33-35i.
B-field
5)
F=3"2 x
By Fleming's left-hand rule the force on a moving charge in a n-lagnetic field is al\vays perpendicular to its direction af travel,
fr 4)
F=BQvsin0
s0, F=2x 1.6 x 10-t" x ? x 10j x sirr30" so, F=6.,1 xl0:tx0.5
1)
Ci:.ular p::h
Force
aaa
lf a conducting rod rnoves through a magnetic field its electrons will experience a force, which means that they r,vill accumulate at one end of the rod.
mctron
2i
This induce$ an s.m.f. (efectromotive forcei across the ends of the exactly as a battery woukl.
3i
If the rod is pa rt of a cornplete circuit, then an induced current rvill
flow through it * this
is called
rr"rd
electramagnetic inductian.
An e.m.f. is induced whenever there is relative motion between a conductor and a magnet. . : you get an e.m.f. either way. The conductor can move and the magnetic field stay stiltor the other way round
-
An e.m.f. is produced whenever lines of force (flux) are cut. Ftrrx rrrtlinp alwavs indtrces e'm'f lrrrr will onlv induce a current if thr: circuit is complete'
19
1)
time
trme a
I
:
il# dJ'!fio'^ro'irtri
Liu'.
you to calculate the e.m.f . induced by the Earth's : magnetic field across the _
:=wings?an of a ptane. Think J - of it as a moving rod and r : use the equation above. /r,
Q1 Q2 Q3 Q4
_-
t
t
t / t,
r r l r \ \
r
\ \
\ -,-
What is the difference between magnetic flux density, magnetic flux and magnetic flux Iinkage? What is needed for flux cutting to induce a current? State Faraday's law.
Explain how you can find the direction of an induced e.m.f. in a copper bar moving at right angles to a magnetic field.
Exam Questions
QI (a) What is the. force on an electron,travelling at a velocity of 5 x 106 ms I through a pelpendicular magaetic fieldof0.77T? [Thechargeonaxelectron!s-1.6,x l0-teC.] :,, (b)Explain why it follows a circular path while in the field.
[2marks] [1.mark]
A coil of area 0.23 mz is placed at right angles to a magnetic field of 2 x 104T. (a) What is the magnetic flux iiassirlg thloug!,1!e- .c"olP . - 1, ,:, j ,". ,,.. : . . . :. (b) If the coil has 150 turns, what is the magnetiC flux linkage in the coil? ' (c) Over a perio d of 2,5 seconds the mugnrtid'rieta iS'reaui.a nniffiry to 1.5
Q2
.
Beware
-
physics can induce extreme confusion...
OK... I know that might have seemed a bit scary... but the more you go through it, the more it stops being a big scary monster of doom and just becomes another couple of equations you have to remember. Plus it's one of those things that ryl{
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ErccrRtc AND M,qcNEnc Ftr.rs
21
Transformers are like voltage aerobics instructors. .They say step upt the voltage goes up. They say step down, the voltage goes down. They say star jump, and the voltage does nothing because neithir o7 them are alive it'i just induction.
-
Transformers are devices that make use of electromagnetic induction to change the size of the r-:r4e*-voltage for an alternating current. They use the principle of flux linking using two coils of wire.
Laminat.ed iron core
MagnetiG,,:,,
field in Lhb t.:.1' .' '-.' , iron core '; '
2)
An alternating current flowing in the primary (or input) coil produces magnetic flux.
3)
The magnetic field is passed through the iron core to the second ary (or output) coil, where it induces an alternating voltage of the same frequency.
4)
step-up transformers increase the voltage by having more turns on the secondary coil than the primary. Step-down transformers reduce the voltage by having fewer turns on the second ary coil.
:,, . ' ,,'. '.,'.-:'
'''
1
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23
lnside every atom, there's a nucleus containing protons and neutrons. Protons and neutrons are both known as nucleons. orbiting this core are the This is the nuclear model of the atom The.d i agram,, s,h,ows.neutral oxygen,
" ,-
- 'i .-i "ii ,:+i "ii iji=* i--1r...,*-..,r.-
.,i;rrrrt
..,t,',.:
.
.
,,. -*
,
witt,'elint $iotAri anJl ei'lrrt eleitionE i : i r'.i*-i i;+*t$itiiti$i.1ffi .
.
#l;l;[I i*il;
r, .
Tprotons 41d fieuirons' iri ri:.4,.'; .:,-
,,:.:.a
.(;+ili{
Tha"elattionl in' br:6it ,r' the nuit.u!'
:::r.:
;
You have to know the properties of electrons, protons and neutrons for the exam make sure you learn this table.
-
The masses in the table are given in atomic mass units (u) . 1 u is equal to one-twelfth the mass of an atom of carb on-12 which means it's almost exactly the sarne as the mass of a
-proton or neutron.
No... really
The proton number is sometimes called the atomic number, and has the symbol Z (l'm sure !t makes sense to someone). Z is just the number of protons in the nucleus. It's the proton number that defines the element
-
no two elements will have the same number of protons.
reactions and chemical behaviour depend on the number of electrons. So the proton number telts you a lot about its chemical properties.
The nucleon number is also called the mass number, and has the symbolA (*shrug*). It tells you how many protons and neutrons are in the nucleus. Each proton o, n""rtron has a mass of (apprqximately) 1 atomic mass unit. The mass of an electron .orp"r"d with a nucleon is virtually nothing, so the number of nucleons is about the same as the atom's mass (in atomic mass units).
Ql Q2 Q3
Explain how alpha particle scattering shows that a nucleus is both small and positively charged. List the particles that make up the atom and give their charges and relativq masses
Define the proton number and nucleon number.
Exam Question
Q1
A beam of alpha,particles is,directed onto a very thin gold film (a) Explain why the,,majority of,alpha particles are not scattered. (b) Explain how alpha particles are scattered by atomic nuclei.
[2'marks] [3 marks]
The .important things to learn from these two pages are the nuclear model for the structure of the atom (i.e. a large mass nucleus surrounded by orbiting electrons) and .iow Geiger and Marsden's alpha-particle.scattering experiment giyes
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'-
1)
A cyclotron uses two-Semicircular etectrodes to accelerate protons or other charged particles across a gap
2)
Since it is circulat, dcyclotron can take up much less room than a linear accelerator.
3)
An alternating potential difference is applied between the the particles are attracted from one side to electrodes - asenergy increases (i.e. they are accelerated). the other their A magnetic field is used to keep the particles moving in a
4)
circular motion
(in the diagram on the right,the magneticfield
would be perpendicular Lo Lhe page) .
s)
The combination of the etectric and magnetic fields makes the particles spirat outwards as their energy
1)
According to special relativity, as you accelerate an object you increase its mass. The increase in mass isn't usually noticeable itt only when you get close to the
2)
'3) ir i..
-
. ', This'means that as an object, like a particle in an accelerator, travets faster and :, faster its mass gets greater and greater. As the mass of the particle increases, it gets harder to accelerate it.
"'4)
I
Q1 Q2 Q3 Q4
,
Kim knew
that
slowing down
wouldn't decrease hei mass but it was a gotod,ekcuse.
-
This effect limits how much you can accelerate a particle in a cyclotron. Protons c?o be accelerated to energies of around 20 MeV. lf you want higher energies than that, you have to use a different type of accelerator called a synchrotron.
What is meant by thermionic emission? Sketch a labelled diagram of an electron gun that could be used to accelerate electrons.
Explain how particles are accelerated in a linear accelerator.
What forces act on a particle in a cyclotron?
Exam Questions (lJ.se
Ql
electron
mass =
9,! x
=-1.6 x lTte C and h = 6.61'x 1U3a ls) differenCe of l kV
l0-31kg,electron charge
An electron is accelerated through
a
potential
:
'
:
,,,
(b) Calculate its energy, in joules.
[,1,,mark]
[1:mark] r
(c) Calculate its'speed in ms-r and express this as a percentage of the speed of light (3.0 x .
A cyclotron is used to accelerate particles to very high
Q3
Thewavelengthoftheelectronsinamicroscopeis0.15n-. A 4.4xL06msil:,,"
t {.9x:1orlnr:I
ms-l).
:.:..:
... .I :-
[3 marks]
,::
speeds.
Q2
101
l
ttevetocityoftheelectronsis:appioximately: , ,,: rl :;,;. :f:.. :,'.1
C 6.5:x 106ms-r D 1.0x 10?mrr.
Smash high-energy particles together to see what they're, made of... A famous particle accelerator (in the physics world) is the Large Hadron Cotlider (LHC) found at CERN on the Swiss-French border - this accelerator is a whopping 27 km loop... Make sure you understand how the two types of particle accelerator on this page work - and don't forget to learn the equations on page 24 either.
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27
Leptons are fundamental particles and they don't feel the strong interaction. The only way they can interact wlth other particles is via the weak interaction ='' and gravity tana itre electromagnetic torceias well if they're charge6l'
:' -
'
:
' '
i
'r
2)
two more leptons called the muon (p-) and the tau (r-) that are just like heavy
etectrons. '
Muons and taus are unstable, and decay eventually into ordinary electrons. The electron, muon and tau each come with their own neutrinoiv.r vpTandu
3)
4)
,
v is the
G
reekleLler
nu
Neutrinos have zero or almost zero mass and zero electric charge so they don't do much. - happening In fact, d neutrino can pass right through the Earth without anything to it.
s)
Each lepton is given a lepton number of *1 , but the electron, muon and tau types of lepton have to be counted separately.
'[;*:!:lx;y;,,,iii" numbe:rffi'rirrrj the
-8i€;8tffi r+qy
0-t
tifi
i:1,.,:,.t,".,
You get three different lepton
numberst L^, L.. and L-.
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ffiffi "ir.-i]fi,, : l.
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iffi
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,
The neutron is an unstable particle that decays into a proton. (But it's much more stable when it's part of a hucleus.) It's really just an example of p- decay: \\\\\\l
I
/
/
I
/,
/
/
,/
I
\
\-
\
\
\
\
\
antineutrino has l^= -1 so the tot al lepton number is ,"ro.l. Antineutrino? Yes, well I haven't mention ed antiparticles yet. ...' ., Just wait for the next ?age
The
-/./
Ql Q2 Q3 Q4
/
/
/
t
/
I
I
List the differences between a hadron and a lepton.
Which is the only stable baryon? A particle collision at CERN produces 2 protons, 3 pions and 't neutron. What is the total baryon number of these particles?
Which two particles have lepton number L. = *1
?
Exam Questions
Ql Q2
List all the degay products of the neuhon..-Explaln,wfry thjp decay-gqgr91,b9 due !g the strg.4g,intep.a9fLqp
t3 marks]
Initially the muon was incorrectly identified
[3 marksl
Go back to the to of a e
as a meson. Explain why the muon is not a meson;
,,
not pass GO. do not collec* t200
it. Co back and read it again. I promise - read these pages about 3 or 4 times and you'll start to see a pattem. There are hadrons that feel tie force, leptons that don't. tlriront are either baryons or mesons, and they're all weird elgept for those well-known baryons: protons and neutrons. There are loads of leptons, including good'old electrons.
Do
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More stuff that seems to laugh in the face of,:common sense :- but actually, antiparticles help to physics... (Oh, and if you haven't read pages 26 and 27 yet then go back and read them now -
When Paul Dirac wrote down an equation obeyed by electrons, he found a kind of mirror image solution. 1) It predicted the existence of a particle Iike the electron but with opposite electric charge - the positron. 2) The positron turned up later in a cosmic ray experiment. Positrons are antileptons so L" = -1 for them. They have identical mass to electrons but they carry a positive charge.
:rr).) jn" iJ,,i --,": ^ ;;; ll
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t
expl ai n a lot,i,n-;gailicle t"',,,,
,
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Each particle type has a corresponding antiparticle with the same mass but with opposite charge. For instance, an antiproton is a negatively charged particle with the same mass as the proton. it doesn't do much either. Even the shadowy neutrino has an antiparticle version called the antineutrino
-
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You've probably heard about the equivalence of energy and mass. lt all comes out of Einstein's special theory of relativity. Energy can turn into mass and mass can turn into all you need is one fantastic and rather famous formula. energy if you know how
-=*
-
lf you've done any chemistry, you'll know that when you carry out a reaction the mass of yogr reactants will always i.e. mass is conserved. ln nuclear reactions, the mass of the particles you start equal the mass of your products with might be more or iess than the mass of the particles you end up with. This happens when energy is converted to mass or mass to energy. This time it's the total mass and energy thatt conserved.
: :-
:,:i r, ..
..
':.
.. ,,.
.
When you're describing nuclear reactions, the Sl units of kilograms and ioules are too big to be easily use6l. lnstead, the mega-electron volt, MeV (page 24), is used for energy and atomic mass units, u (page 23), ot eY/cz '
are used for mass.
Conversion factor:
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B " 1O-3o kg -1 MeV/c.z = 1 .7 B * 10-21 kg = GeY 1 lcz . and 1.7
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'32 '---More conservation? This is starting to sound like biology...
-kily for
?f$et ^,€O
:
a Ch;
particlt
ln any particle reaction, the total charge after the
gxgl-'ks:::ro
-:3*::*:",sy-{
re
th
e
re a c t i o n
- T, - P * n is fine for charge, butcouldn,t'!,pp*nbecausethebaryonnumb9rincreasesfrom1.,_3';-
The same goes for baryon number, e.g. the reaction P
Energy conservation is a bit complicated in particle reactions. Since mass and energy i.e. the total are equivalent (see p. 28), it's really mass-enerSy that's consen'ed
:TIqI As @
-
?j l!,",
T*1:"*l
'6$d
if you convert the masses of all the particles into energy.
j ,r-, 3c< 1"he decav aecay of a fie',I. for tne tracKs tor aL oartrcle r'< a.t QarLtcte tracks with any other interaction, momentum has to be conserved. - \'3'' i.ct-, _ n-;c like m omentum !4isn'-* : looks rrr\L ita. tar a ,,-c-:.r, J'v-r:t , ?.27 ':
'rrv'rrurrvurrr
ccrtselec. Tha:'s because some momentum is car ine antineutrino, which you can't see
(se
e.next
?
Clc tha
c0l nra ^1
d.
[-{e
be
Bu
ah
The three types of lepton number have to be conserved separately.
ret
t
)
u; +v
hasL =0atthestartand L = 1 -1 =0attheend,soit'sOK. On the other hand, the reaction Y. + l-L- + e- + ue can't happen. At the start L* = 2 and L" = 0 but at theendLrl€ -0andL -2. Tr----+ -rpplt
2)
ha
For example, the reaction
Bc
l-r
ary A
p;
Ql Q2 Q3 Q4 Q5
The
What is a quark?
The
Describe how a neutron is made up from quarks.
Which type of particle is made from a quark and an antiquark? Explain why quarks are never observed on their orvn.
P
\
List five quantities that are conserved in particle reactions.
\
;, rl
r
Exam Questions
Ql Q2 Q3
A
\
A proton is composed of which combination of three quarks?
A
ddd
B
uuu Cddu
D
[1 mark]
uud
Give the quark composition of the
n
Give two reasons why the reaction p
and explain how the charges of the quarks give rise to its charge,
* n ---+ p + K*
[2 marks$
does not happen.
physical property called strangeness
-
how coot is that...
True, there's a lot of information here, butthis page really does tie up a lot oi the stuff on the lastfew pages. Learn as much as you can from these pages, then go back to page 26, and work back through to here. Don't expect to understa it att btut you will definitely find it much easier to learn when you can see how all the bits fit in t:getly
-
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Ar/U:LEAR DE:AY
41
What a radioactive source can be used for often depends on its ionising properties. 1)
Alpha particles are strongly positive
2)
'ionising an atom transfers some of the energy from the alpha particle to the atom. The alpha particle quickly ' ionises many atoms (about 10 000 ionisations per alpha particle) and loses all its energy. This'makes alphasources suitable for use in smoke alarms because they allow current to flow, but won't travet very far. The beta-minus particle has lower mass and charge than the alpha particle, but a higher speed. This means it can still knock electrons off atoms. Each beta particle will ionise about 100 atoms, losing energy at each interaction. This lower number of interactions means that beta radiation causes much less damage to body tissue than alpha radiation. This means beta radiation can be used in medicine to target and damage cancerous cells - since it' passes through healthy tissue without causing too many problems. Camma radiation is even more weakly ionising than beta radiation, so will do even tess damage to body tissue. This means it can be used for diagnostic techniques in medicine.
3)
4)
5)
-
so they can easily pull electrons off atoms, ionising them.
Put a Geiger-Mtiller tube anywhere and the counter There are many sources of background radiation: 1)
:2.\
3) .'.. 41,
i''
'i
s)
will click
'
-
it's detecting background radiation.
The The
bili. :a:
::. :t ::: i:. t :::::t::,1
ThE
cosmii ;iUiation: C9s"iiliryr ays particles,tmqilty l-,igli:",',"rgi pioi;ns) from;s: ce!:',;..,; When they collide,with phrtieIes in,the upper atmosphere,:th6y'producehuclear: Iadiation. tiving tlrings, nrr pr"ni una animaJs
ioiiili;trrn;r,'rra **n
ie'd,iMttivE'.isuto-p.d'fatb # Man;ma{e rualSiion, tn
mos-i aieaS,'radihtion
Lithis will 6" ,1,'! ,-,;';.
from medical or industriil s6urces makes up
When you take a reading from a radioactive source, you the background radiation separately and subtract it from you r measurement.
Ql Q2 Q3 Q4
What makes an atom radioactive? Name three types of nuclear radiation and give three properties of each. Describe the differences in ionising and penetrating powers of the three main types of nuclear radiation. Give three sources of background radiation.
Exam Question
Q1
Briefly
describ-elan absorption experiment to distinguish
You may wish to include a sketch in your answer.
,,
Radioactive emissions
-
as easy as a, 0.
-y...
You need to learn the different types of radiation and their properties. Remember that alpha particles are by far the most ionising and so cause more damage to body tissue than the same dose of any other radiation - which is one reason we don't use alpha sources as medical tracers. Learn this atl really well, then go and have a brew and a bickie...
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43
Q1 Q2 Q3 Q4
Define radioactive activity. What units is it measur.ed in? Sketch a general radioactive decay graph showing the number of undecayed particles against time.
What is meant by the term 'half-life'? Describe how radiocarbon dating works.
ExamQuestions r ,. r Qt
r
.
iRadioactive decay ig a rb-ndom prgcess,i This statement means
,
Q2
,
C You can't predict when an:atom will
that;'
1
decay --1 you.cau only,,estimate the proportion that
You take a reading of 750 Bq from a pure radioactive source. and background activity in your lab is measured as 50 Bq.
will
' ':.
:
decay in a given time.
fhe radioactive source initially contains 50 000 atoms,
(a) Calc,qlate the decay. constant,for your:sample,, ;. , ,: ,,. ,. , (b) What is the half-life of this sample? (c) Approximately how many atoms of tne,. radioactive source will there be after 300
[3 marks] [2 marks]
:
,
U
seconds?
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-
Os ctLLArtoNs
3
OsctLLArlolvs
45
1)
From maximum positive displacement (e.g. maximum displacement to the right) to maximum negative displacement (e.g. maximum dlsplacement to thb teft) and batk again is called a cycle of oscillation. '
2)
The frequenn, f, of the SHM is the number of cycles per second (measured in Hz).
3)
The period,
I,
is the
rt'Sn
time taken for a complete cycle (in seconds).
fi't}ffi'$'[ri'dil:
so'a"pendul,ufit,bldck,:*iI
l,
k
tA ifi depehdehr,of 'the' ampI itude',
tick'ihg
in"'r'e$u
1
i
.e.
ionstant rfor
a,
given. o-scl'l lation);
Iar:,time ihter:valS even if 'its swing uocciffii've
rffii
li. .'ir"'-
I know it looks like there are loads of complicated equations to learn here, but don't panic it's bad really. - how notthat You'll be given these formulas in the exam, so just make sure you know what they mean and to use them.
1)
According to the definition of SHM, the acceleration, a, is directly proportional to the disptacement, x. The constant of proportionality depends on the frequency, and the acceleration is always in the opposite direction from the displacement (so there's a minus sign in the equation). \ \ \ \ I I t I / /.ubon't forgel,A is the maximum disPlacemenl
- it's not/l acceleration. \ \ \ \\ / tt 2)
-
The velocity is positive when the object's moving in one direction, and negative when it's moving in the opposite direction. For example,.a pendulum's velocity is positive when it's moving from left to right and negative when it's moving from right to left. v
x
t
t
tt
t
swinging as a form of simple harmonic motion,
- -Aus sin(cuf)
The displacement varies with time according to the equation on the right. To use this equation you need to start timing when the pendulum is at its maximum displacement i.e. when f - 0, x - A. :
Jer emy was investigating
-
Acos(c..rf)
'
-
Q1 Q2 Q3
Sketch.4 graph of how the velocity of an object oscillating with SHM varies with time.
What is the special relationship between the acceleration and the displacement in SHM? Given the amplitude and the frequency, how would you work out the maximum acceleration?
Exam Questions
Ql
(a) Define simple harmonic motion. (b) Explain *hy'tii6 motion of
Q2
a
ball bouncin$ offthe giound is not SHM., ' "" I ':
' :i' :i
:,'il
A pendulum is pulled a distance 0.05 m from its midpoint and released. It oscillates witlisimnl harmonic motion with a freq:uen"y of 1,5 Hz. Calculate: ': i (a) itsmaximurniveiociiy:' '
"'-'
,
.
(b) its displacement 0.1 s after it is released (c) the time it takes to fall to 0.01 m from the midpoint after it is released
"Simple" harmonic motion
-
hmmm. l'm not convinced.,.
The basic concept of SHM is simple enough (no pun intended). Make sure you can remember the shapes of all the graphs on page 44 and the equations from this page, then just get as much practice at using the equations as you can. U
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UNr 5: SrcnoN 3
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47
lf you set up a simple pendulum attached to an angle sensor and computer like change the length, /, the mass of the bob, mt, and the amplitude, A, -youthen get the following results:
thisq eJ ,
*', au*
q;f..$i :*itir
U
,:,*,
ffi
hung around waiting for the
Pob
experiment
to start.
The formula for the period of a pendulum is: (The derivation" quite hard, so you don't need to know it.) This formula only works for small angles of oscillation u? to about 10' from the equilibrium point.
-
Ql Q2 Q3
Write down the formulae for the period of a mass on a spring and the period of a pendulum. Describe a method you could use to measure the period of an oscillator. For a mass-spring system, what graphs could you plot to find out how the period depends on: a) the mass, b) the spring constant, and c) the amplitude? What would they look like?
Exam Questions
,, A mass of 0.l0 kg is attached to the ttom and the spring extends to a total lenct-h of.0,20in. r, .;.::.,, , , ' -- .-l-.,_ :-., ,,.r,iit, ,.t:. :. ... -..., (a) Calculate the spring constant qf the spring'in Nmr. (g = 9.8t Nkgt) Thl;n1nO isn'L.morling at this point, l
,, .,
,.
(b) The mass is pulled down a furth"r 2 cm and then released. Assuming tt or"illut., with simple harmonic motion, calculate the period of the subsequent oscillations.
"'ilis
(c) What
Q2
,"^r *"11,0:" r":U.$ * *ake thepenod of oscillation
twice as lon*Z
Two pendulums of diffefent lengths were ieleased fromrest at ihe titp of their swing. ' It took exactly the same time for the shorter pendulum to make five iomplete oscillations as it tgok the longerpendulum to make three complete oscillations. . , i r,.
^
ihe'sholterpendUlumnada.tengttrof0.20m.
Go on
-
,
Showthatthelengthofthelongeronewas0.56-."'
'
SHO the examiners what 'ol)'re made of...
The most important things to remember on these pages are those two period equations. You'll be given them in your exam, but you need to know what they mean and be happy using them. U
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OssuarloNs
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OsutmroNs
49
1)
The degree of damping can vary from light damping (where the damping force is small) to overdamping.
2)
Damping reduccs the amplitude of the oscillation over time. The heavier the damping, the quicker the amplitude is reduced to zero.
3)
Critical damping reduces the amplitude
'
(i.e. stops the system oscillating) in the shortest possible time. 4)
Car suspension systems and moving coil meters are critically darnped so that they don't oscillate
s)
Systems with even heavier damping are
6)
equilibrium than a critically dalmped system. Plastic deformation of ductile materials reduces the amplitude of oscillations in the same way as damping. As the material changes shape, it absorbs energy, So the oscillation will be smaller.
.
;
1)
tightly damped systems have a very sharp resonance peak. Their amplitude only increases ' dramatically when the driying frequencv is very close to the natural.frequeney.
2)
Heavily-damped systems have a flatter response. Their amplitude doesn't increase very much near the natural frequency and they aren't as sensitive to the driving frequency. Experiment to show how damping affects resonance How dam oina affects reson ance
(
amplitude
sharp
,,,
r,l,Stru ctu rgs', ar e',.'.,,,1,1't;'
d,amped increasin g degree Disc to increase air resistance
mi ltto hive'as flat!
at the resonant afree oscillator
flat the
Ql Q2
natural
,
response as possible .; so that,th ey ,don't.i, t' 'colb' r' 1uhe soiJnd ,i,',
frequency than
Sets driving frequency
.,,
avoid being
damaged by resonan ce. Loudspeakers are also
of damping
Oscillates at a smaller amplitude
to
driving frequency
frequency
What is a free vibration? What is a forced vibration? Draw diagrams to show how a..damped system oscillates with time when the system is lightly damped and when the system is critically damped.
Exam Questions
Ql (a) Whatisresonance? (b)
(c)
r, : ..,.. ,.
i:,,,,r:.:t,,.,
Draw a diagram to show how the amplitude of a lightly damped system varies with drivirg nelu"rcy, On the same diagram, show how the amplitude of the system varies with driving frequency when it
[2 marks]
, [2 marks] ',,
,[1'mark],' :
[1
*?tL]
[[ rnzirk]
A2 Physics
-
it can really put a dampe;r on your social life.,. , oioi22ylr1t :i{) o, *ry, viry bad... ;{
Resonance can be really useful (radios,
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AsrRoPHYStcs AND CosuoLocY
AsrBopHystcs AND CosvoLocY
51
1
) ' I Both gravitational
2)
and electric field strengths are,unit forc€s. Cravitational field strength, g, is force per'unit,mhps. Electric field strength, f, is force per unit positive charge.
Newton's law of gravitation for the force between two point masses is an inverse square law. Coulomb's law for the electric force between two point charges is also an inverse square law. So the strength of both forces goes down,rapidly as you get further away from the source of the field.
1)
2) 3)
Take a look
at
pages 10-11
for more on electric fields.
Cravitational forces are always attractive. Electric forces can be either attractive or repulsive. Objects can be shielded from electric fields, but not from gravitational fields. The size of an electric force depends on the medium between the charges, e.g. plastic or air. For gravitational forces, this makes no difference.
ln both gravitational and electrical fields, the energy change of a particle depends only on where the particle starts and finishes. lt doesn't matter what path it takes to get there.
Example in a Gravitational Field
Example in an Electric Field
UNr 5: SrcnoN 4
AsrnopHystcs AND CosuoLocy
Distances in astronomy are... well, astronomical. Astronomers use some pretty clever techniques ,
to
measure
1)
Radio telescopes can be used to send short pulses of radio waves towards a planet or asteroid (a rock flying about the Solar System) which reflect off the surface and bounce back.
2)
The telescope picks up the reflected radio waves, and the time taken (t) for them to return is measured.
3)
Since we know the speed of the radio waves (speed
of !igh.! c) we can work out the distance, d to the object using: It's 2d, not just d, because distanc e to the object
-
travels twice the there and back again.
thte pulse
lf two short pulses are sent a certain time interval apart, you can measure the distance an object has moved in that time. From this time and distance, you can calculate the average speed of the object relative to Earth. More accurate measurements can be made using Doppler shifts (see p.5B).
4)
From Copernicus onwards, astronomers were able to work out the distance the planets are from the Sun relative to the Earth, using astronomical units (AU). But they could not work out the actual distances.
t'6'ne
rbffimicti'
ii ir'{Au)
distance between
when it was carefully measured The size of the AU wasn't accurately known until 1769 during a transit of Venus (when Venus passed between the Earth and the Sun).
2)
-
All electromagnetic waves trave[ at the speed of light, c, in a vacuum (c = 3.00 x 10s ms{).
2)
lf we see the light from a star that is, say, l0light-years away then we are actually seeing it as it was 10 years ago. The further away the object is, the further back in time we are actually seeing it.
3)
1 ly is equivalent to about 63 000 AU.
1)
You experience parallax every duy. Imagine you're in a moving car. You see that (stationary)
objects in the foreground seem to be moving faster than objects in the distance.
U
2)
This apparent change in position is called parallax and is measured in terms of the angle of parallax. The greater the angle, the nearer the object is to you.
3)
The distance to nearby stars can be calculated by observing how they move relative to very distant stars when the Earth is in different parts of its orbit. This gives a unit of distance called a parsec (pc).
4)
By knowing the distance between the Sun and the Earth, and measuring the parallax angle, you can use trigonometry to work out the distance to a star.
S)
Pa.rsecs are the distance unit of choice for astronomers, so if you want to be a starry-eyed astrophysicist you'll be seeing a lot more of them.
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AsrRoPHvs/cs AND CosuoLoGY
53
1)
The luminosity of a star is the total energy it emits per sec6nd., lt's:'related.to the temperature of the star and its surface area.
2)
The luminosity is proportional to the fourth power of the star's temperature and is directly proportional to the surface area. This is Stefan's law:
ffi
where L is the luminosity of the star (in W), A is its surFace area (in m2), T is its surfacetemperature (in K) and o (a little Greek"rigma") is Stefan's constant.
- 5.67 x 10-B Wm-2K-4.
3)
Measurements give Stefan's constant as o
4)
To find the luminosity of a star, remember that stars are spheres.
You can find the luminosity of a star by substituting the surface area of a sphere, 4TEr2, into Stefan's law:
where r is
the radius of the star (or
sphere)
.
From Earth, we can measure the radiation flux that reaches us from a star. The flux is the power of radiation per square metre, so as the radiation spreads out and becomes diluted, the flux decreases. If the energy has been emitted from a point or a sphere (like a star, for example) then it obeys the inverse square law:
ffi
wherel is the luminosity of the star (in w),
The distances to most stars and objects in space are too big to measure using parallax (see previous page). For these objects you can use standard candles to find their distance. t
)
Standard candles are objects, such as supernovae and Cepheid
variables,,
.
that you can calculate the luminosity of dir'ectly.
2)
Because you know the luminosity of the standard candle, you can measure its flux on Earth and use the inverse square law to
3)
find its distance.
So, if you find a standard candle within a galaxy, you can work out how far that galaxy is from us. This is how Hubble's constant was worked out (see p. 5B).
UNr 5: SrcrroN 4
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55
lf you plot luminosity against tem.perature, you don't just get'a random collection of stars. The stars appear to group in distinct areas on the plot. The distinct areas show the main stages of a star's life cycle: the main sequence, red giants and white dwarfs. (see the next page for the details). tliis is called the Hertzsprung-Russetl Diagram.
\\\\tlttt// .; ur)i O
hotler to cooler. / / tIII I\ \\\
.F o.ot F J
5
::
,,:1
Temperature goes the "wrong way" along the x-axis from
-
.:":i::ti ' tt,'.',..
:
';
The reason you can see these areas is because stars exist in these stable stages of their life cycle for long periods of time. You don't see groups of stars in any transitional period on the H-R diagram because they are unstable and happen quickly (compared with the life of the star).
Q1 Q2 Q3
What
is Wien's displacement law and what is it used for? What is an H-R diagram a plot of?
What causes dark lines to appear in what would otherwise be a continuous spectrum from the
Sun?
Exam Questions Q
1
The star Procyon A produces a black body, spectly1 with, a peak
What is the surface temperature of
PffionA
to 3
S
f,,?,, .. .,
l
a
Q2
A star has a surface temperature of 4000 K and the luminosity
as
the Sun (3.9 x
1026
W).
(a) Which radiation curve represents this Explain your answer.
same
Eo {-,
CO
,
star
-
1, (0
X, Y or Z?
l-{
t{ o B
o
o<
(b) Calculate the star's surface
-t....1..
il,j;i :V:iit#ffi$
W
li ^r-l
..r! '..:.
ta .-7.,-.t-.. t.-.
trl
*i..
Itt
1.0
area.
2.0 Wavelength (trm)
,':":..I,.,''.'.',':
Q3
if
Sketch the basic features of an H-R diagram, indicating where you would find main sequence stars, red giants and white dwarfs.
Wavelength to the max... Who'd have thought something as boring sounding as the H-R diagram could be about the life and times (and,deaths)
stars. lknowwhatyou'rethinking,'lwanttoknowmoreaboutthestarlifecyclel
of
Wellgoonthen...justforyou...
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AsrRopHyslcs AND CosuoLocy
UNn
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AsrRopHygcs AND CosuoLocY
57
1)
2) 3)
4) s)
When the core of a star runs out of fuel, it starts to contract If the star is massive enough though, it keeps contracting.
-
forming a white dwarf core.
The electrons get squashet onto the atomic nuclei, combining with protons to form neutrons and neutrinos. The core suddenly collapses to become a NEUTRON,STAR, which the outeq l4yers fall 1
onto.
.,
.
., ,.,
Whgn the outer layers hit the surface oi the neutron star they rebound, setting up huge shoc-lryavss,i"' ' . r . : ripping the star apart and causing a supernova. The Iight from a supernova can briefly outshine ap entire galaxy. ,
.
ij;i
fait'(upttd 600-'ti
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r;
I
;iili::
i,it.
:
tEt;.
col'a)i
[n:€J.{ffii'fi.d"i6:iff.6v.i
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AsrnopHystcs AND CosuoLocy
5B
Everyone's heard otf the Big Bang theory
-
well here's some evidence for it.
I You'll haVe experienced the Doppler effect loads of times with sound waves. tmaging ar1 ambulance driving past you. As it moves towards you its siren sounds higher-pitched, but as it mories awayr its pitch is lowe'r-- This change in frequency and wavelength is called the Doppler shift. The fre'quenc1, and the wavelength change because the waves bunch together in front of the source and stretch out behind it. The amoqnt of stretching or bunching together depends on the velocity of the source. When'a iiChtsouice,move! away from us, the wavelengths of its light become longer and the frequencies become !ower.: --'': ghifts the light towards the red end of the spectrum and is called redshift. ''' '': -" this :
1)
2) 3)
4)
--.'-_" -
,
=,
s) 6)
The amount of redshift or blueshift is determined by the following
formula:
.
,
.\ is the emitted wavelength, f is the emitted frequen cy, A,\ and Af are the differences between the observed and emitted wavelengths/frequencies, v is the velocity of the source in the observer's direction and c is the'speed of Iight. 7)
B)
(v << c means "v is much less than c'
.)
The way cosmologists tend to look at this stuff, the galaxies aren't actually moving through space away from us. lnstead, space itself is expanding and the light waves are being stretched along with it. This is called .
cosmological redshift to distingJish it from"redshift producedLy sources that ire moving through spaie. The same formula works for both types of r:edshift as long as v is much less than c. lf v is close to the speed of light, you need to use a nasty, relativistic formula instead (you don't need to know that one).
The spectra from galaxies (apart from a few very close ones) all show redshift. how fast the galaxy is moving away. The amount of redshift gives the recessional velocity
-
2)
Hubble realised that the speed that galaxies moved away from us depended on how far they were away. s€€ p. 53) showed that they A plot of recessional veloiity against distance (found using Cepheid variables is expanding. were proportional, which suggests that the Universe
3)
This gives rise to Hubble's taw:
where v - recessional velocity in kms-', d - distance in Mpc and Ho = Hubble's constant in kmr'Mpc-'. 4)
Since distance is very difficult to measure, astronomers disagree on the value of Ho. It's generally accepted that Ho lies somewhere between 50 kmsrMpc-l and 100 kmsrMpcr.
s)
TheSlunitforHoissr. TogetHoinSl units,youneedvinmrlanddinm(1 Mpc=3.09x,10?2m).
The Universe is expanding and cooling down. So further back in time it rnust have been smaller and hotter. lf you trace time back far enough, you get a Hot Big Bang:
2) 3)
U
The absolute size of the Universe is unknown but there is a limit on the size of the observable Universe.This is simply a sphere (with the Earth at its centre) with a radius equal to the maximum distance that light can travel during its age.
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AsrnoPHYstcs AN D CosuoLoGY
Leroythought the Univ erse sfr.artedwith a big gianthead that shot flaming arrows fro*lts neck. All this b\gbang eiidenceis just circumstantial.
59
lf the Universe has been expanding at the Sam'e'rate for its,wholeilife, the age bf the Universe is f =
1)
1/4
(time
Unfortunately, since no one knows the exact value of Ho *" can only guess the Universe's age. lf Ho= 75 kmsrMpc-l, then the age of the Unive rse x 1/(2.4 x 10-r8 s-l) = 4.1 x 10r7 s = 13 biltion years.
2)
AII the mass in the Universe is attracted together by gravity. This attraction tends to slow down the rate of expansion of the Universe. The criticat density is the density of mass in the Universe that means gravity is just strong enough to stop the expansion at f = With a bit of mathematical jiggery-pokery you can get an equation for the critical density of the Universe in terms of the Hubble constant:
-;
lf the density of the Universe is less than the critical density, gravity is too weak to stop the expansion. The Universe would just keep expanding for ever. lf the density is greater than the critical density, gravity would be strong enough to stop the expansion and start the Universe contracting again (ending up with a Big Crunch). ffi.!; ln fact if you look at the graphs of size against time, the expansion : 5, rate is slowing down in allthree cases. So all three models suggest the Universe was expanding faster in the past than it If that's true, then we/ve overestimated the time it's taken for Universe to Bet,to the size it is now. The more dense the
6) 7)
P,
lt's no small matier w;rking out the density of the Universe. Even if you manage to count up the total mass that you can see, like stars and galaxies, there's a lot of evidence for the existence of dark matter, which can't be observed directly.
B) .
e)
Po--
lf you include all the dark matter that's been detected indirectly, current estimates of the actual density aren't far off the critical density. But even if we knew the exact density of the Universe, we could still be none the wiser about its future. ,Astronomers in the late 90s threw a spanner in the works they found evidence that the expansion is now accelerating. That means the simple models on this page might be a long way from the true picture...
-
Ql
Why does the calculated age of the Universe depend on its density?
Exam Questions
As.sume
H0: +50 kms-rMpc{ (l Mpc
:
3.09 x
1022
m). :
Calculate an estimate of the age of the Universe, and hence the size of the observable Universe. :
An object has (TakeHo
a redshift of 0.37. Estimate the speed at which
=,2,4x
10-ls's-1,
I ly:9.5 x
1015m.) '
:,,,,.;.;:,1,
,i
it is moving away from
,.r-.:
us.
'.;
;,:;::,r,;;r
;-1.;r::..
With teference to:'theispe'ed ofthe object, explain why your 4nswers to a) and b) are esiimsles..i
It's the end
'
of the world as
Eeeshk... these pages are filled with questions that could leave anyone feeling critically dense. How did the Universe start? .How old is it? Will it ever end? Just keep going through these pages and soon you'll feel like the Universe's personal psychic.
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AsrnopHystcs AND CosuoLocy
U
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AsrnoPHYStCS AND CosuoLoCY
61
Two light nuclei can combine to create a larger nucleus and release a load of energy. This is called nuclear fusion.
1)
2)
All-nuclei are positively charged so there will be an - repulsion electrostatic (or Coulomb) force of between them. Nuclei can only fuse if they overcome this electrostatic force and get close enough for the attractive force of the strong interaction to hold them together; , , , , : . ,. Typicalty they need about 1 MeV of kinetic energy and that's a lot of ene rgy.
W
d;efla.etad.6l"el
;'.i
ct bEt5tit'
r""t'i;p;,iSi6
6,'
i.itffi
l
:r
3)
j
,
-,
The energy emitted by the Sun and other stars comes from nuclear fusion reactions. Fusion can happen because the temperature in the core of stars is so high the core of the Sun is about 101 K. At these temperatures, atoms don't exist the negatively charged electrons are stripped' away t leaving positively charged nuclei and free electrons. The resulting mixture is called a plasma. A lot of energy is released during nuclear fusion because the new, heavier nuclei have a much higher binding energy per nucleon. This helps to maintain the temperature for further fusion reactions to happen.
1)
2) 3)
-
-
4)
Q1 Q2 Q3 Q4 Q5
What is the binding energy per
'
nucleon?
'
Which element has the highest value of binding energy per nucleon? What is spontaneous fission? How can fission be induced in 23sU? Describe the conditions in.the core of a star. Why are these conditions necessary for fusion?
Exam Questions Q
I
The following equation represents a nuclear reaction
]p
* lp --+ 1H *
Jp * energy released
.
'
(a) State the type of nuclear reaction gtiqn. (b) Civen t t the binding enetgy per nucleon [2 marks]
(a) Calculate'ft totrl mass Aeficit for this reaction. (b) How much energy is released in this reaction, if
a mass
deficit of .l u releases 93 I MeV of energy?
lf anyone asks. l've gone fission... that joke never gets old... Phew, until that last subsection I was wondering what all this smashing nuclei has to do with astrophysics and cosmology. Make sure you're down with the mass deficit and bezzers with the binding energy calculations they're exam faves, U
Nr 5; SrcnoN 4
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AsrnopHystcs AND CosuoLocy
ANswER/Nc ExnM
ANswrRlNG Exnu Qu rsrtoNs
a uEsrloAvs
ANswtRlNc Exnu Qu rsrtoNs
64
lJnit 4: Section 1 - Further MechaniCs 5 - Momentum
Pase
momentum before = momentum after
t)
(0.2x0.3)!0=l.lv A.2l = l.lv [t mark for workingl +
fi markl
v = 0.19
3)
fi
mst
mark]
+Q
-Q
Recognisable pattern around the charges (not just in between) mark], lines equally spaced around the charges and joined to the charges, and general symmetry of the diagram fi markl, arrows along
Reso/ve the velocitlT vector into components parallel and perpendicular to the line of the collision:
[l
Parallel: vt= 5 x cos(60") = 2.5 ms-t U markl Perpendicular: v2= 5 x sin(60") = 4.3 ms-t fi markl Find the momentum parallel to the line of the collision, (0.6 x 2.5) + (2 x A) = (a.6 x 0) + (2 x v) I.5 = 2v +v = 0.75 ms-t fi markl
field lines between the charges with arrows pointing away from the positive and towards the negative char1e
2) E---g
4rre
3) a)
[t mark for each vector]
Pase
a)
7-
E
Force and Enersv
t/zvz - th - mgh [t mark]. t/emvz = myh + +v2=2gh-2xg.Bt x2=39.24 [t mark].r/=6.26ms-t fi markl
KE
-
t/zmv2 and PE
no friction means the kinetic energy will all change back inta potential so he'll rise back to the same height as he started. fi markl Put in some more energy by actively 'skatingi fi markl The kinetic energy will be /ess after the collision. fi markl total momentum before = total momentum after t000AxI *25000v fi marklsov=0.4ms-t fi markl Before: KE = 0.5 x l0 000 x 12 = 5000 tl markl After: KE = 0.5 x 25 0Q0 x 0.42 = 2000 [t mark]
Pase t
)
a)
b) 2 m -
c) 2) a) b) c)
U 2)a)
I I
I
- 2 x (4.5 x t0-3) = 9.0 x 1U3 m [t markJ +V - Ed = lt500/(4.5 x 101)]x 9 x 101 -
l3
x
N tt markl The answers to b) and c) use the roun ded value of u calculated in parta) - if you didn'tround,you answers will be slightly differenl. d) The gravitational force between the Sun and the Earth fi markl 1022
ft mark]''r
Vm-',
3000V
= V/d
fi
'
markl
Capacitors
Capacitance:?
-
gradient of line
-ry
So the correct answer is B [l mark] Charge stored = Q = are& = I 5 x lU6 x 66
-
220 p,F.
990 lt'C.
:36 I tl markl *_;'1 'v' [l mark] -l*0.5xl22 2 mark]
markl:@=0.01s03s
3)a) w -#,,
= 3.6
.6x10-te
b) Q- CVtt markl=a.5x12-6Cft
Paseg-CircularMotion 2tr 0 t ) a), - 2.0x l0-7 rad s-t [1 mark] - ; ft mark] so Q : dl,, b) v = ru) [t mark] - 1.5 x 10tt x 2.0 x l0-7 = 30 kms'| fi markl c) F = fitu2r fi markl = 6.0 x l02a x (2,0 x lU7)2 x 1.5 x l7tt
mark].
4r x B.B5 xt 0-" x (l .7 5 xl7-to )2 :4,69sx\0t0 [t mark] Vm-'or NC-'fi markl E = V/d - t 500/(4.5 x tU3) - j.3 x l}s [1 mark]
b) d
/:
[l
^12
_
0.7 5 ms-l
4.3
1)
Electric Fields
1)
momentum before collision = momentum after collision (0.4 x 0.A48) = (0.6 x 0.A32) fi markl A.0192 = 0.0192, so mamentum is conserved fi markl
2)
ll
Pase
b) -/ v -9
t, -a'Dxt?-] 470x10-o C\"'.-..r markl
=9v [t
1x0.02
ttt
markl
mark]
l
Page 15 t
)
a)
-
Charging and Dischai,iging
37% after RC seco nds [1 mark], t = loo,o x 2.5 x 10a = 0 25 seconds fi markl
The charge falls to so
o.z
2)
a)
Cravity pulling down on the water at the top of the swing gives a centripetal acceleration of 9,Bl ms-z fi markl. lf the circular motion of the water-needs a centripetal acceleration of less than 9.Bl ffi{2, gravity witt pull it in too tight a circle. The water will fall out of the bucket. Since
b)
a=
u)2f, ,rs'
, so u) =
3.1
rads-l
This force is provided
by both the tension in the
T+ (l0x9.Bl)=250. 5o T=250-(l0x9.Bl)
ANswrns
markl.
l52N
mark], so after.0.7 seconds:
There is 6%
0.7 seconds [1 markl
fi markl iii)None [1 mark]
rope, T, and gravity:
-
ft mark]. ii) None
[t mark]
? o.Slrev s-t [t mark] force{m.'r = l0 x 52 x t - 250 NI/
u = 2rf , so f = Centripetal
=^r=:9'Bl
b) e - er*-E [!
fi
markl.
of the initial charge
Q: left
Qoe-azs
-Q0 x0.06
on the capacitor after
65
i:
Page
l7 - Magnetic Fields and Force
Unit 4: Section 3
I)A) F-BII
:
Page
2 x' l}-s x 3 x 0.04
[t mark] . : -2.4xtU6N [lmauk] b) F-BIlsin0 : 2.4 x 10-6 x sin30" - 2.4 x 10-6 x 0.5 [l markl 1 ) y, lu t.zt0-6 /Y N !llt mark] lltutttl --
:
I
a)
)
9
Charged Particles
-
in
l)a) F-Bqv-0.77x1.6xtTte x5x x
Magnetic Fields 106
[tmark]
l0-t3 N [l mark] The force acting on the electron is always at right angles to its velocity and the speed of the electron is constant. This is the
= 6.16
b)
Page 1
2)
a)
BA [t mark] = 2 x lA3 x 0.23 = 4.6 x t 0-4 Wb [t mark]
b)
iD BAN [t mark] = 2 x l0-3 x 0.23 x 150 - 0.A69 Wb tl markl
)
Q=
a) b) c)
-
dtt,'
2) 2.5
marks available
for correct answer, otherwise
one mark
for
each
correct stage of working.l
Page
2l -
,+.
a
a
Atom
[t mark] t.6x lT-ts l/eV= 1.6 x l7ts I [t m2rk] Kinetic energy = t/zmv2 - 1.6 x l0-t6 I tt markl Yz - (2 x 1.6 x l}-'u) + (9.1 x t0-31) = 3.5 x 1014 i v = I .9 x 107 ms-t [t mark] Divide by 3.0 x 108: 6.3% of the speed of tight [t mark] 1000
"V x l00A eV
P = trlv' so v - p +
63
x
10-34
= 4.42
x
l0-24
ffi
-
x
lA2a + 9.1 x lT3t Il mark]
6.
4.42
+ 0.1 5 x t/.-s kg ms-t
-
4.86
x
l}G ms-t
So the correct answer is B
I
a
,,
Particle Accelerators
-
.\-fi +p, - h +,\=
Page
n s
a
25
so p
I)a) V=Blvfi markl = 60 x lA6 x 30 x 100 - 0.1B V [t mark] b)a a
2)
3)
Electromagnetic lnduction
a
:- The Nuclear
Physics
The alternating electric fietd accelerates the particles from one side of the cyclotron to the other, increasing their energy [t mark]. The magnetic field keeps the particles moving in a circular path [t mark].
(zxto-' -r.sxro-,)(o.z:xrso) _6.ei10_, v [3
Particle
The majority of alpha particles are not scattered because the nucleus is a very small part of the whole atom and so the probability of an alpha particle getting near it is smatt [t mark]. Most alpha particles pass undeflected through the empty space around the nucleus [l mark]. Alpha particles and atomic nuclei are both positively charged fi markl. lf an alpha particle travels c/ose to a nucleus, there will be a significant electrostatic force of repulsion between them fi markl. This force causes the alpha particle to be deflected from its original path [t mark].
b) Page t
23
-
)
\ -s
o =
27
-
Classification
of
Paiticles
Proton, electron and electron antineutrino [t mark] The electron and the electron antineutrino are leptons [t mark]. Leptons are not affected by the strong interaction, so the decay can't be due to the strong interaction ft mark].
,e,
resista nce
[1 mark]
2)
Mesons are hadrons but the muon is a lepton fr mark]. , The muon is a fundamental particle but mesons are not ft mark]. Mesons feel the strong interaction butthe muon does not [t mark].
graph shou;ld have three steps, with the last step twice the height with the opposite sign to the gradient of the flux cha"nge with time. So the correct answer is B It mirk]. The
of
the others,
Axswrns
66
29
Page
-
Antiparticles
t) e*+e--)T+yftmark] This is called annihilatio,n It 2)
3)
4)
mark].
The protons, neutrons and electrons which make up the iron atoms wauld need to annihilate with their antiparticles fi markl No antiparticles are available in the iron block fl mark].
t
)
Electrical energy supplied: AE = VIAI l2 x 7.5 tB0 = t 6200 tl markl The temperature rise rs 12.7
notconserved. / The creation of a particle of matter requires the creation of its antiparticle. ln this case no antineutron has been produced [l mark].
- ener7y
*
fl
mark]
2)a) pV =
10,
t
)
-
ft mark]
p:;: NkT
Quarks
7y:
-du fi
[l
Page
- -l /3 unit.
I)a)
Charge of anti-up quark = -2/3 unit. unit fi markl Total charge =
-l
b)
is
:
t)
35
c)
Detecting Particles
-
xl
0-23 x 26A
-
35 500
Pa
U markl
-
39
Mass
-
Internal Energy
of I molecule =
fftass
of I mole
= 2.8
for knock-on electrons OR by applying Fleming's le{t-hand
l*" :ry
x
6.02
l0-2
x 1023
markl
Rearranging gives:
? - 3kT fi markl
xl0-" x 30A 7_ 3 xl .iB 4.65 xl0-'6
1.67 xl
Typical speed = r.tn.s. speed
=
Cas moiecules move at different
Os
^IZ.OZ spee
'm2
s-2 U markl
x las = 517 ms-t fl
mark]
ds because the)/ have different
amounts of energy fi markl. The molecules have different amounts of energy because they are constantly colliding and transferring energy between themselves [l mark].
Charged particles follow curved tracks in a magnetic field fi markl. +ve and -ve particle tracks curve in opposite directions [1 mark] You can identify the direction of curvature for negative particles by looking
.83
ed Lhevalue of N in this calculation so it'd fit on fhe page this in you calculator or you'll gethuge rounding errors.
- 4.65 x I A26 kg tl
number changes from 2 to I , so baryon number is not conserved tl markl. The charge changes from +l to +2, so charge not conserved [t mark]
Pase
xl0'6 xl
NA
The baryon
3)
t .9
mark]
markl
Charge of down quark
txtasfto rN: ,. - EY'-t.9xlo26 kT I B,xto-u x2gt3-
markl, so
keep numbers like
The correct answer is D, uud
2)
fi
J kg-t 6-r wouid be righttoo.
-
ft mark]. Now use the value of N to find p atthe new height:
We round
32
Page
oC
2x8.2
markl - 601 MeV Totat energy of the two photons = 601 MeV 5o the energy of photon 601 + 2 = 300.5 MeV
-
8.2
You need ilhe righl- unit for Lhe thrrd mark
2 x (300 MeV + 0.5 MeV)
t
- 4.5 -
c- # fi markl = t6200 - 9BB I k{'"C-I fi markl
- 8.2 x l0-t4 I = 0.5 MeV fi
E,o,=
l
Specificheatcapacityr
after
total energy for each particle before annihilation -- E"r, Eki,*ti" Eresl^= m-C - 9.1 I x l0-3t x (3.0 x l}u)' e
x
-
The baryon number is
Energy before
Heat and Temperature
37 -
Page
:
2)
- distance so
a)
Speed
b)
Atthough the particles are moving at an average of 400 ffis't, they are frequently colliding with other particles. [1 mark] This means their motion in any one direction is limited and so they [,] mark] only slowly move from one end of the ro?m to tl1 .oth17. , At 30 oC the average speed of the particles would be slightly faster [t mark] since the absolute temperature would have ris;en from 293 K to 30i K and the temperature determines the average speed'[l mark]. This means the speed of diffusion would also be faster fi markl.
time
= distance
rule. fi markl
2)
3)
Antineutrinos are neutral and so will not leave tracks in many standard detectors. Beta particles are charged and so will ionise particles and leave a track, and so are more easily detected. fl mark] The proton and the positive pion give
neutral pion
tracks butthe neutron and the do not. So the correct answer rs C [1 mark].
4)
I
mark for not mark for two tracks going in opposite directions, showing a track for the photon, I mark for tracks spiralling inwards.l
[t
s)
P
-
rBQ
= 3.2 - 9.2
[l
x x
ANswrns
mark] x 10-6x 1.6
l.B
x
l}-te
l0-2s kgms-l [1 mark]
400 ms-l
c)
''"'.'
Unit 5: Section 2 Page 1
)
4t = Place
'.
:
: '. :r' '
"
.
Nuclear Decay
-
Page
Radioactive Emissions
t
)
a)
Extension of spring
b)
: 9.BNm- [t mtark]. :W m, so k T : 2*E + T : 2rr"W : 2nxJo.ol :0.63 s [t mark].
c)
mx
different materials between the source and detector and
measure the amount of radiation getting through
fi
47 ;:simptle Harmoi,nic Osiillators'
k
:
T2
so
gives
markl:
-
= 0.20
fi
8.g.,
t
)
Oi.tO'm
[l mark]. Hooke's Law
if
T is
doubled,
=
STrh*tp"ndutum
is quadrupled and m is quadrupled = 4 x 0.1 0 - 0.4a kg tl markl.
T2
3T,onrpunctutumr
and f : 2*[
[1
mark].
L-etlengtlt
mark for each material stopping correct radiationl of tong pendutum
,
Page
-
t
[t mark| 5o mass needed 2)
0.10
43
-
Exponential Law of
Decay i
Dividing by
r
C [t mark].
The correct answerrs
gives 5
2n
- t. so 5lrn ^W)
gives
- slzn
I !s !W:3x^E
5x.
fi markl.
. Squaring and simplifying '
- 9l so length of long pendulum - 5/9 * 0.56 m fi markl. a
2)a) Activity, A= A = .\N
b) c)
T,
:ln2
[t
mark
+
ffteasured - background - 750 - 50 - 700 Bq fi markl 700 = 50 000 fi markl 5o .\ = A.014 s-t fi markl
^
0'693
; '.\ - 0.014 -
4g.5seconds
for the hatf-tife equation,
I
Page
/:
mark
for the correct
- 750 mark for the decay equation, number of atoms remaining after 300 secondsl N = Noe-tu = 50 0A0 x
[2 marks available
t
) a) When a
Free and Forced Vibrations
is forced to vibrate at a frequency that's
close to, or markl and oscillates with a much markl. larger than usual amplitude b) See graph below. fi markl for showing a peak at the natural frequency, [t mark] for a sharp peak. system
the same as its natural frequency
e-0'0t4x30a
I
-
hatf-life]
49 -
tt
I mark for the
*Unit 5.' Section
t
)
a)
b)
3
sharP response
amplitude
Oscillations
01
Simple harmonic motion is an oscillation in which an object always accelerates towards a fixed point fi markl with an acceleration
natural
2) a) Maximum velocity = (2nf)A = 2rx I.5 x 0.05 = 0.47 mrt [t markJ. b) Stopclock started when object released, so x = Acos(2rft) fi markl. x = 0.05 x cos(2r x 1.5 x}.l ) * Q.05 xcos(0.94) = 0.029 m
b)
driving frequency
frequency
directly proportional to is displacement from that point [1 mark]. tThe SHM equation would getyou the marks if you defined all the variables.l The acceleration of a falling bouncy batt is due to gravity. This acceleration is constant, so the motion is not SHM. fi markJ.
fi markl.
fi
c)
fi markl for a smaller peak atthe natural frequency witt actually be slightly to the left of the natural frequency due to the damping, but you'll get the mark if the peak is at the same See
graph.
[the
peak
frequency 2)
the diagraml.
:t:
if it returns to rest in the shortest time mark] when it's displaced from equilibrium and released. e.g. suspension in a car fi markl
a) A
system is criticatty damped
possible
b)
in
[t
:
c) x=Acos(2rrft) +0.01 =A.05xcos(2rxl .5t).
cos(3rt) + cas-t(0.2) = 3rt. irt = l.i7'* tt mark for working, I mark for correct answerl
So 0.2 =
f = A.l5
s.
Don't forgeLto put yaur calculator in radian modewhen you'resolving questions on circular motion it's an easy mistake to make.
-
ANswrns
5B
5Z
Paee
Unit 5: Section 4
5l
Page
-
-
A,strophysi-gs and Cos.mqlogy.
)
9.81x (6400 x 1000)2
o, :9Y r' -
x l02a kg tl
6-67 x10-11 x7.3511022.
-
$740x l ooo)2
So the correct answer is A
fi
Stars
It mark]. The rest of the star expands, and the star forms a red giant [t mark]. Red giants have a higher luminosity and a lower surface temperature fi markl than main sequence stars.
markl
1.62
of
until the core becomes hot and dense enough to fuse helium
[1 mark]
6.67 x 10-1r
= 6.02
The Life Cycle
Stars on the main sequence are in a highly stable phase where they fuse hydrogen in their core U markl. Once the hydrogen in the Coire runs out, the'star starts to collapse
Gravitational Fields
r) ,:ff+M:* 2)
t
-
Eventually the helium in the.core will run out and the star will once again begin to collapse fi, markl The core will not becoime dense enough for further fusio,n to [ake place. The core will collaptse to form a white dwarf and the statr's outer layers will be ejected fi markl. White dwarfs have a lower luminosity and a higher surface mark] temperature than main sequence stars
Nktr,
markl
ll
?age 53 1
-
Measuring Astronomical Djstances
) a) A,standard candle is an objiect in space with a known luminosity. tt markl By knowing its luminosity and measuring the ftux [t mark], you can use the inverse square.law to work outthe distance to the
markl
b)
t
)
object. fi You could .use the observed parallax of the star and the distance between the Earth and the Sun to make a right-angle:d triangle. [t mark] The dixance to the star could then be found using trigonometry. fi markl
2)
a
The star is
L
-
Aola
law
fi
c)
3.9 x 10?6
:7.5x
101um' (to 2
s.f.) [t mark]
Hubble and the Big Bang
v = Hod [t mark] where v is recessional velocity (in kms-t ), d is distance (in Mpc) and H, is Hubble's constant (in kms-tMpc-').
markl Hubble's law suggests that the l.Jniverse originated with the Big Bang fi markl and has been expanding ever since. [1 mark] Ho=v + d = 50 kms-t + t Mpc 50 kms-l = 50 x 1 03 ms-t and t Mpc - 3.a9 x 1022 m So, Ho= 50
is:
x 103 ms-t +3.09 x
2.898x10-3 -u"'u= . ,T- 436 x 10-' -6650Kto3s.f.
t
the star U markl
- oATa, so 3.9 x 1026 - 5.67 x l0-B x Ax 400An [1 mark], which gives A = 2.7 x 17te m2 [1 mark].
50 000 K nn n -ti
n.3.o3.?t[
[5 marks maxirnum, I mark each for correctly labelled
axes,
each for 'Main Sequencei'White Dwarfs' and'Red Ciants'.}
Arusvvrns
-
energy per nucleon is about 0.56 MeV in 2H, so the increase in binding so about 1.7 MeV is released (ignoring energy is about 1.72 MeV the positron) lt mark].
-
2)
a)
mass
deficit
=
rnass before
-
mass after
[t
mark]
= (2.013553 + 3.015501) - (4.A01 505 + 1.008665) =0.01BBB4ufi markl b) 0.Al BBB4 x 931 = 17.6 MeV fi markl
L
(
mark].
Nuclear Fission and Fusion
[t mark]. There are 2 nucleons
T = 2.898 x :10-3, so for this starl,u* - 2.898 x 1U3 + 40A0 - 7.25 x l0-7 m fi markl. Curve Y peaks at around 0.7 ltm (= 7 x lt7 m), so could represent
er atur.
61
b) The increase in binding
A [t markl
Decreasi ng temp
s-l
) a) Fusion [t mark]
a) According to Wien's displacement law ),^"rx
, b)
x l}-tg
,: Page
., 2)
1.62
-- 4.6 x lA2s / 9,5 x l7ts ly = 4.9 billion ly tl markl c) z = v/c is only valid if v <
55 - Luminosity and the Hertzsprung-Russell
So the correct answer is
=
2)a) zxv/c[t mark] sovx0.37x3.0x l}sx 1.1 x 108 ms-t[t mark] b) d -v/Hox !.1 x l7s / 2.4 x 1At8 --4.6x l02s m fl mark]
Diagram
t)
1A22 rt't
[t mark for the correct value, I mark for the correct unitJ t - I /Ho [t mark] t - l/t.62 x 1a-ts = 6.lB x l0t7 s x 20 billion years fi markl The observable tJniverse has a radius of Z0 billion light-years. [1 mark]
markl
5.67 x10-8 x55004
-
fi b)
where A is the surface area of the star.
o-4ol'
Page
a)
'
sphere. So the luminosity,of the star using Stefan's
59
Page
I mark
-.--
conservation in particte reactions 32 of charge 32, 33 of energy 6, 7, 28, 32, 33
tempylture 37, 39 zero 37 absorption spectra 54
\ '
qI,'
F
acceteration 6 centripetal 9 I sHM 44, 45
controlled experiments cosmic rays 26,
3
41
activity 42
cosmological redshift 58
atpha
Coulomb's
law 10,
51
critical
scatter.ing 22 alternating current (AC) 2A, alternators 20
21
,
damping 49 density 59 cyclotrons 18, 25
24
:
19, 20
field
lines 10, 11, 16
of momentum 4-7, 32, 33 continuous spectra 54
law
Faraday's
strength 1A, 11, 16, 17,50,
51
fields
electric 10, 11, 24,51 gravitational 50, 51 magnetic 16-21 , 25, 33 fission 60 Fteming's lefrhand rule 17, 18, 20 flux magnetic 16, 18, 19, 21 radiation from a star 53 forces
amplitude of oscillations 44-49
centripetal 9
angles
dark matter 59
atom stability 40
(u) 23, 28
atomic number 23
re
22
,
t 42
ltant
51
6
forced vibrations 48
:
free vibrations 48
density of the Universe 59
astronomical units (AU) 52
atom ic structu
resu
de Broglie equation 24 decay constan
51
gravitational 50, restoring 44
damping 48, 49
astronomical distances 52, 53
atomic mass units
electric 1 0,
D
radians B of parallax 52 angular speed B, g annihilation 29 antiparticles 2B-3O arc-tength B
frequency
differential equations 43
circular motion
displacement, SHM 4,4, 45
SHM
45
B .
Doppler effect 58
fundamental partictes 27
driving frequency 48, 49
fusion 56, 57,
dynamo
s
61
21
23
atoms 43 average kinetic energy 39
C
l.
B background radiation
41
, 42
baryon number 26
baryons 26, 31 becquerels 42
27, 40,
beta radiation
gamma radiation 40,
elastic collisions 7
generators
21
elastic potential energy 44
glass smash
ing 48
electric
gradients 62 gravitational
charges 10-1 energy 12
5
fields 1 0, force 10
, 24,
11
constant 50
induction 1 B-21 radiation 52,54 electromotive force (e.m.f.) 1B-21
black holes 57
electron microscopes 24
Boltzmann's constant 37
electron-positron pairs 29
bubble chambers 33, 34
electrons 22,23,27,28 electronvolts (eVs) 24 electrostatic force
C
61
energy
capacitance 12 capacitors 12-15
conservation
charging and d ischarging centripetal force 9
1
4,
1
5
11
iapacitor 12, 14,
charges in a magnetic
15
field 18, 25, 33
circular motion B, 9
cloud chambers 33
coils 1 6,
20
transformers
51
:
H hadrons 26,30 Hertzsprung-Russell diagrarp 55 hot big bang theory 58 Hubble constant 53, 59 Hubble's
law
5B
hydrogen fusion 56
of 6, 7, 28, 32, 33
electronvolts 24 internal 38,39 kinetic 6, 7, 24, 38, 39, 44 potential 38, 44, 51
Cepheid variables 53 stored on a
fields 50, force 50
51
electromagnetic 61
black body radiation 54
charge 10,
4i
electrodes, particle accelerators 24, 25
41
big giant head 5B binding energy 60,
E
,
I ideal gases 37 -39
stored by u capacitor 12, 13
inelastic collisions 7
thermal 36
internal energy 38, 39
equation of state, ideal gases 37
inverse square
equilibrium position, SHM 46, 49 evidence 2, 3 exp losions 4 exponential retations 42, 43, 62
ionisation 33
laws 10, 50, 51, 53
ionising properties of radiation 40,
41
isotopes 42, 43
21
collisions 4, 7, 24 ideal gases 38 components, resolving 5
conclusions
3
lNotx
s
K
P
kaons 26,
pair production 29,
31
science and society
31
kelvin 37
3
simple harrribnic motion (SHM) 44-47
kinetic energy 6, 7, 24, 44 ideal gases 3 B, 3 9
smoke alarms 43
particle accelerators 1 B, 24, 25 'detectors 33-35
,'
L
3
s 2,
scientific proces
parallax 52
:
peak wavelength 54
solenoids sPectra 54
sPeed B,
law 2A leptons 2 7, 32
6
1
gPecific heat caPacitY 36
Lenz's
g
'
period
distribution of gas particles 38, 39 of light 52 spontaneous fission 60 spring constant 46 standard cand les 53
linear speed B, 9
permittivity 10
stars 53-57
logarithm rules 62
photons 29 pions 26, 31 ptastic deformation 49
cycle 55-57 Stefan's law 53
penetration properties of radiation
light dampi ng 49 lighryears (ly) 52 I
circular motion I of oscillation, SHM 45-47
inear particte acceterators (l inacs) 24
luminosity 53,55,56
M;
potential energy 44, ideal gases 38 power laws 63
,
16-21 25, 33 16-21 21
fields
flux
,
main sequence
55, 56
predictions
MASS
number 23 relativistic 25 mass-energy equivalence 28, 60
oscillator
46
taus 27 temperature 36-39
2
of stars 53-56
protons 23, 26, 27, 30,
theories
31
thermionic emission 24 time constant 15
pulsars 57
transformers
mechanical energy 44
26, 29, 31 momentum 4-7, 32, 33,
2
thermal energy 36
protostars 56
n'resons
61
53, 56, 57
T
proton number 23
:
strong interaction 26, 27, supernovae
51
pressure 37
deficit 60
mass-spring
::[[::::i'jil' ,o,u
capacitors 12, 14
field strength 17
, 61
life
potential difference accelerating charges 24, 25
magnetic
40
21
a
35
quark confinement
muons 27
31
U
quarks 30-32
.uniform fields
10, 11, 51
:_1
,
1:
Universe
N
R
natural frequency 48, 49
radar 52
neutrinos 27
radial fields
neutron stars 57 neutron
s 23,
Newton's 2nd
26, 27, 30,
law
6 "
31
,
4A
activity 42 fission 60 fusion 57 ,
radiation
0, 1 1, 50
51
flux
.t.
radioactivity 40-43
vapour trails 33
radiocarbon dating 43
velocity 4,
;,
radiation 4A reactions 60
redshift 58
volume
time dilation ,34 resonance 48,49
force 44, 46 resu ltant force 6
observable Universe 58
rockets 4
oil drop experiment
Rutherford scatterin$ 22
lx orx
3
21
/
relativistic mass 25
restoring
11
45
vottage, transformers
I
\"' 1;
5
SHM 44,
41
model, atoms 22,-23
oscillations 44-48 overdamping 49
il
.
vibrations 48
o
{.
l
'll
V
velocity 58 red giants 55, 56
nucleons zzi zo
:l
.T;
53
recessional
nuclei 22, 23, 26, 60, 61 nucleon number 23, 60
+.i
f t
radio telescopes 52
radon gas 61
.tl
B
:
Newton's taw of gravitation 50, nuclear
radians
1
l
age 59 expansion 58 observable 58,59 unstable atoms 40
W weak interaction 27
webers 16 white dwarfs 55, 56 Wien's displacement law 54
