edexcel a2 unit 4 physics on the move sample page gec
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Edexcel-A2-Unit-4-Physics-on-the-Move-Further-mechanics-Electric-and-magnetic-field-
and-Particle-physics
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A2-Unit-4-Physics-on-the-Move ..............................................................................................................1
1 Further Mechanics..................................................................................................................................1
Chapter 1 Momentum and impulse ...........................................................................................................1
1-1 Momentum and impulse impulse—momentum theorem.........................................................1
1-2 Principle of conservation of linear momentum...........................................................................3
1-3 14 Worked examples........................................................................................................................4
Chapter 2 Circular motion.......................................................................................................................10
2-1 Uniform circular motion ...........................................................................................................10
2-2 Centripetal force and centripetal acceleration...........................................................................10
2-3 15 Worked examples......................................................................................................................10
2 Electric and Magnetic fields.................................................................................................................10
Chapter 1 Electric field ...........................................................................................................................10
1-1 Electric charge Coulomb law.................................................................................................10
1-2 Electric field Electric field strength .....................................................................................12
1-3 Electric field lines .....................................................................................................................13
1-4 Electric potential difference Electric potential.......................................................................13
1-5 Linking the electric field and the electric potential...................................................................13
1-6 Capacitor Capacitance ...........................................................................................................13
1-7 The motions of charged particles in the uniform electric field .................................................13
1-8 Comparison of electric and gravitational fields ........................................................................13
1-9 29 Worked examples......................................................................................................................13
Chapter 2 Magnetic fields .......................................................................................................................17
2-1 Magnetic flux density ...............................................................................................................17
2-1-1 Magnets and fields.........................................................................................................17
2-1-2 Magnetic fields from currents........................................................................................17
2-1-3 Magnetic force on moving charges and flux density (B)...............................................17
2-1-4 Magnetic force on a current...........................................................................................17
2-2 Moving charges in a magnetic field..........................................................................................17
2-2-1 Charged particles in circular orbits................................................................................17
2-2-2 The cyclotron.................................................................................................................17
2-3 Electromagnetic induction phenomena.....................................................................................17
2-4 Faraday law of electromagnetic induction and Lenz’s law.......................................................17
2-5 Motional emf ............................................................................................................................17
2-6 32 Worked examples......................................................................................................................17
3 Particle physics ....................................................................................................................................23
Chapter 1 Atomic nucleus .......................................................................................................................23
1-1 Nucleus structure of atoms Atomic nucleus ..........................................................................23
1-2 Natural radiation phenomenon Decay half-life...................................................................26
1-3 Nuclear reaction Nuclear energy............................................................................................27
1-4 Fission.......................................................................................................................................29
1-5 Fusion of light nuclei ................................................................................................................29
1-6 19 Worked examples.........................................................................................................................29
Chapter 2 Particles and Radiation ...........................................................................................................29
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II
2-1 Particles and antiparticles .........................................................................................................29
2-2 Particle interactions...................................................................................................................29
2-3 Properties of particles and antiparticles ....................................................................................29
2-4 Quarks and Antiquarks..............................................................................................................29
2-5 45 Worked examples......................................................................................................................29
A2-Unit-4-Physics-on-the-Move
1 Further Mechanics
Chapter 1 Momentum and impulse
1-1 Momentum and impulse impulse—momentum theorem
(i) Definition of Momentum:
The product of an object’s mass m and velocity v is called its momentum:
momentum p mv
Momentum is measured in 1kgms . It is a vector.
(ii) The rate of change of momentum and Newton’s second law
The rate of change of momentum of an object is proportional to the resultant
force acting.
This can be written in the following form:
change in momentum
time takentanresul t force
In symbol:
mv muF
t
…… ①
Where v is final velocity, u is initial velocity of an object.
Equation ① can be rewritten ( )m v uF
t
And the acceleration is given by v ua
t
F ma
. So
…… ②
Note:
⒈ Equation ① and ② are therefore different versions of the same
principle (Newton’s second law).
⒉ when using equations ① and ②, remember that F is the resultant force
acting. For example, for the figure below, the resultant force is
to the right. The acceleration a can be worked out as follows:
26 20 6N
263
2
Fa m
m s
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(iii) Relationship of kinetic energy and momentum
From AS Unit 1, Kinetic energy 21
2kE m v for non-relativistic particle, and
momentum p mv . Thus pv
m , therefore
2 221 1
2 2 2k
p pE mv m
m m
(iv) Impulse
Impulse:
As resultant force, (m v uF
t
) can be rewritten Ft mv mu
In words force × time=change in momentum.
The quantity ‘force × time’ is called an impulse.
A given impulse always produces the same change in momentum,
irrespective of the mass. For example, if a resultant force of 6 N acts for 2s,
the impulse delivered is 6×2=12N s.
This will produce a momentum change of 112kgms
So a 4 kg mass will gain of velocity 13ms
Or a 2 kg mass will gain of velocity, and so on. 16ms
The graph below is for a uniform force of 6 N. in 2s, the impulse delivered is
12 Ns. numerically, and this is equal to the area of the graph between the 0
and 2 s points.
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(v) impulse—momentum theorem
From section (iii), we get that:
The resultant force × time = change in momentum.
And the quantity ‘force × time’ is called an impulse.
So
Impulse = change in momentum.
Thus:
When a resultant force F acts on an object, the impulse of the resultant force
is equal to the change in momentum of the object; this statement is called
impulse—momentum theorem:
Ft mv mu
mvmu
1 2 1 21 2 1 2
: The final momentum
: The initial momentum
1-2 Principle of conservation of linear momentum
(i) Principle of conservation of linear momentum:
Provided there are no external forces acting on a system, the total momentum
before collision is equal to the total momentum after collision.
For a collision involving two bodies (Figure 1.1), the conservation law can
be expressed symbolically as:
M v M v M v M v
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1v
2v
1v 2v
Note: , , 1v
2v
1v
2v
momentum mv
, are all vectors, and in your calculations, you must
be pay attention to the direction of the velocity.
(ii) Elastic and inelastic collisions
Even though momentum is always conserved in collisions, kinetic energy
may or may not be conserved. For this reason, we usually classify collisions
according to whether kinetic energy is conserved of not. If kinetic energy is
conserved during a collision, we call it an elastic collision. At other extreme
are collisions in which two objects stick together after impact. These
collisions are known as perfectly inelastic collisions. A collision between two
railroad cars that couple together upon impact is an example of a perfectly
inelastic collision. In between are the inelastic collisions that are neither
elastic nor perfectly inelastic.
1-3 14 Worked examples
1. A baseball of mass m = 0.14 kg has an initial velocity of as it
approaches the bat. We have chosen the direction of approach as the negative
direction. The bat applies a force that is much larger than the weight of the
ball, and the ball departs from the bat with a final velocity of . The
contact time between the bat and the ball is . Find
38 /u m
58 /v m
s
s34.0 10t s
(a) The initial and final momentum of the ball.
(b) The impulse produced by the bat.
(c) The average force exerted on the ball.
Solution:
(a) Strategy:
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So
0.14 ( 38) 5.32 /ip mu kg m s
0.14 ( 58) 8.12 /f
Initial momentum:
Final momentum: p mv kg m s
8.12 ( 5.32) 13.4 /Ft mv mu kg m s
13.4 /kg m s
(b) Strategy:
The equation impulse = force × time can not be used directly to calculate the
impulse, since the average force that the bat applies to the ball is not known.
However, since the weight of the ball is negligible, the force applied by the
bat is the net force, and the impulse of the net force equals the change in
momentum, according to the impulse—momentum theorem:
Impulse = change in momentum
That is
So the impulse produced by the bat is
(c) Strategy:
Now that the impulse is known, the contact is also known. So use the
equation: impulse = force × time to calculate the average force.
So
313.4 / 33504.0 10
impulse kg m sF Nt s
The force is positive, reflecting the fact that it points opposite to the velocity
of the approaching ball.
2. Two cars of unequal mass undergo a head-on collision in which they stick
together after the collision (Fig. 2.1). Car A has a mass of 1700 kg and an
initial velocity of 10 km/h. car B has a mass of 850 kg and an initial velocity
of 12 km/h toward the car A. the impact time is 21.0 10 /m s .
(a) What is the velocity of the combination immediately after collision?
(b) What is the accelerations of the two cars during the collision?
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1 10 /v km h
2 12 /v km h
2.67 /v km h
Strategy: use the Principle of conservation of linear momentum,
1 2 1 21 2 1 2M v M v M v M v
1 2 vv v
1 2 ( )A BBA
, since the two cars stick together, so
,
Thus
M vM v M v M
1 2BA
And choose the direction of the car A as the positive direction. Then
1 10 /v km
h 2 12 /v km
, hSolutions:
M v M v
1700 10 850 ( 12) 6800 /kg m s
(a) Total momentum before collision =
=
( )A BM vM
Total momentum after collision =
2550 v
=
From the Principle of conservation of linear momentum:
6800 / 2550kg m s v
2.67 /v m s
Then
(b) Strategy: the acceleration is the rate of change in velocity, v u
at
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Solution:
Car A:
6 2
2
2.67 / 10 /2.64 10 /
11.0 10
3600
A
change in velocity km h km ha k
impact time h
m h
Car B:
6 2
2
2.67 / ( 12 / )5.28 10 /
11.0 10
3600
B
change in velocity km h km ha k
impact time h
m h
3. Fig. 3.1 shows the variation with time, t, of the force, F, acting on a body.
Fig. 3.1 What physical quantity does the area X represent?
A the displacement of the body
B the acceleration of the body
C the change in momentum of the body
D the change in kinetic energy of the body
Solution:
Impulse:
As resultant force, (m v uF
t
) can be rewritten Ft mv mu
In words force × time=change in momentum.
The quantity ‘force × time’ is called an impulse.
Choose (C).
4. Water of density 1000 kgm–3 flows out of a garden hose of cross-sectional
area 7.2×10–4m2 at a rate of 2.0×10–4m3 per second. How much momentum
is carried by the water leaving the hose per second?
A 55.6 10 Ns
B 25.6 10 Ns
C 0.20Ns
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D 0.72Ns
Solution:
The product of an object’s mass m and velocity v is called its momentum:
momentum mv
Momentum is measured in 1kgms . It is a vector.
The mass of the water in one second is given by 41000 2.0 10 0.2m V k g
The water flows a distance 4 3
4 2
2.0 100.28
7.2 10
ms m
m
Thus, the speed per second is given by 0.28
0.28 /1
s mv m s
t s
0.2 0.28 / 0.056 /momentum mv kg m s kg m s
Therefore,
Choose (B).
5. Which row, A to D, in the table correctly shows the quantities conserved in
an inelastic collision? mass momentum Kinetic energy Total energy
A conserved Not conserved conserved conserved
B Not conserved conserved conserved Not conserved
C conserved conserved conserved conserved
D conserved conserved Not conserved conserved
Solution:
Even though momentum is always conserved in collisions, kinetic energy
may or may not be conserved. For this reason, we usually classify collisions
according to whether kinetic energy is conserved of not. If kinetic energy is
conserved during a collision, we call it an elastic collision. At other extreme
are collisions in which two objects stick together after impact. These
collisions are known as perfectly inelastic collisions. A collision between two
railroad cars that couple together upon impact is an example of a perfectly
inelastic collision. In between are the inelastic collisions that are neither
elastic nor perfectly inelastic.
Choose (D).
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6. Which one of the following statements is correct?
The force acting on an object is equivalent to
A its change of momentum.
B the impulse it receives per second.
C the energy it gains per second.
D its acceleration per metre.
Solution:
The rate of change of momentum of an object is proportional to the resultant
force acting.
This can be written in the following form: change in momentum
time taken tanresul t force
In symbol:
mv muF
t
As resultant force, (m v uF
t
) can be rewritten Ft mv mu
In words force × time=change in momentum.
The quantity ‘force × time’ is called an impulse.
Thus
The impulse per second is equivalent to the force.
Choose (B).
7. Fig. 7.1 shows how the force on a glider of mass 2000 kg changes with
time as it is launched from a level track using a catapult.
Assuming the glider starts at rest what is its velocity after 40 s?
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A 2.5ms-1
B 10ms-1
C 50ms-1
D 100ms-1
Solution:
When a resultant force F acts on an object, the impulse of the resultant force
is equal to the change in momentum of the object; this statement is called
impulse—momentum theorem:
Ft mv mu
mvmu
: The final momentum
: The initial momentum
And the area of the force-time graph equals to the impulse. Thus, the impulse
is given by
3 5140 5.0 10 10
2I N s
50 10mv mu mv mv
Therefore, , gives 5 510 10
50 /2000
v m sm
. Choose (C).
Chapter 2 Circular motion
2-1 Uniform circular motion
2-2 Centripetal force and centripetal acceleration
2-3 15 Worked examples
2 Electric and Magnetic fields
Chapter 1 Electric field
1-1 Electric charge Coulomb law
1.1 Electric charge
(i) There are only two kinds of charge in nature: positive and negative charge.
Charge of the same kind repels each other while charge of different kinds
attracts each other. The amount of charge is called the quantity of electricity.
The quantity of positive charge is usually represented by a positive number,
and that of negative charge by a negative number.
A large number of facts show that charge can be neither created nor
destroyed, but only moved from one object to another, or from one part of a
body to another. The total amount of charge of the body in the transfer
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process remains the same. This conclusion is called the law of charge
conservation. It is one of the important and fundamental laws in physics.
(ii) Elementary charge
Electrons and protons carry the same amount of different charge and the
quantity of charge . Experiments show that the quantity of
charge of all charged objects is equal to either or its multiple. Therefore,
the quantity of charge is called the elementary charge.
191.60 10e
e
C
e
The precise value of the elementary charge measured so far is 19(1.60217733 0.00000049) 10e C
It can be written as 191.60 10e C
1.2 Coulomb’s law
Coulomb’s law:
The magnitude F of the electrical force exerted by one point charge on
another point charge is directly proportional to the magnitudes and
of the charges and inversely proportional to the square of the distance r
between the charges:
1q 2q
1 22
q qF k
r
9 2 28.99 10 /k N m C
0
Where k is a proportional constant whose value in SI units is
Is common practice to express k in terms of another constant , by writing
0
1
4k
0; is called the permittivity of free space and has a value of
12 2 20
18.85 10 / ( )
4C Nm
k
. The force F is often called the electrostatic
force.
In Fig. 1.1 above, each charge exerts a force of magnitude F on the other.
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Regardless of whether the forces are (a) attractive or (b) repulsive, they are
directed along the line between the charges and have equal magnitude.
1-2 Electric field Electric field strength
(i) Definition of electric field
The electric field E that exists at a point is the electrostatic force F
experienced by a small text charge placed at that point divided by the
charge itself:
0q
0
FE
q
F
0q
The electric field is a vector, and its direction is the same as the direction of
the force on a positive test charge.
SI unit of electric field: Newton per coulomb (N/C)
Note:
★It is the charges in the environment that create an electric field at a given
point. The field exists in the sense that whenever a positive or negative
charge is placed at the point, the field exerts a force on the charge.
★Any charge in the environment contributes to the electric field that exists at
a point. To determine the net electric field, it is necessary to determine the
various contributions separately and then find the vector sum of the
contributions too get the net field.
(ii) The electric field produced by a point charge q can be obtained in general
terms from Coulomb’s law. First, note that the magnitude of the force on a
test charge is 02
qqF k
r 0q
0q
and then divide this value by to obtain the
magnitude of the field. Since is eliminated algebraically from the result,
the electric field does not depend on the text charge, so the electric field of a
point charge q is given by
2
kqE
r
As in Coulomb’s law, only the magnitude of q is used in the equation
2
kqE
r , without regard to whether q is positive or negative. If q is positive,
E
E
is directed away from q. on the other hand, if q is negative, is directed
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toward q, since a negative charge attracts a positive test charge.
1-3 Electric field lines
1-4 Electric potential difference Electric potential
1-5 Linking the electric field and the electric potential
1-6 Capacitor Capacitance
1-7 The motions of charged particles in the uniform electric field
1-8 Comparison of electric and gravitational fields
1-9 29 Worked examples
1. two very small objects, whose charges are +1.0 C and －1.0 C, are
separated by 2.0 m. Calculate the magnitude of the attractive force that either
charge exerts on the other.
Solution:
From the Coulomb’s law, we use the equation 1 22
q qF k
r to calculate the
magnitude of the force.
9 2 2 91 22 2
(1 )(1 )8.99 10 / 2.25 10
(2.0 )
q q C CF k Nm C N
r m
2. In the Bohr model of the hydrogen atom, the electron is in orbit about the
nuclear proton at a radius of . 115.29 10 m
(a) Determine the force on the electron.
(b) Find the speed of the electron, assuming the orbit to be circular.
Solution:
(a) The electron experiences an electrostatic force of attraction because of the
proton, and the magnitude of this force is
19 19
9 2 2 81 22 11 2
(1.60 10 )(1.60 10 )8.99 10 / 8.22 10
(5.29 10 )
q q C Ck Nm C N
r m
eF
Note: the electron is also pulled toward the proton by a gravitational force.
However, the gravitational force due to the proton is negligible in comparison
to the electrostatic force.
(b) The electron is in circular motion, the centripetal force is provided by the
electrostatic force. Thus,
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2
emvF
r
Therefore, 8 11
631
8.22 10 5.29 10 2.2 10 /9.11 10
eF rv m sm
3. An electron (charge and mass ) is
injected into a region of uniform electric field of magnitude
191.60 10e C g319.11 10m k 51.0 10 /E N C .
What is the initial acceleration of the electron?
Solution:
From Newton’s second law, the magnitude of the acceleration is given by F
m.
To find the acceleration, we must first find the magnitude of the force on the
electron. We can do this by using the equation 0
FE
q
Thus
The magnitude of the force on the electron is
F F qE e E
The acceleration is the force divided by the mass,
19 516 2
311.60 10 1.0 10 1.8 10 /
9.11 10Fa m sm m
e E
4. Find the electric field at a distance of 0.2 m from a charge of 4 nC.
Solution:
For the point charge, 2
kqE
r
Thus 9
9 2 22 2
4 108.99 10 / 899 /
(0.2 )
kq CE Nm C N C
r m
5. A point charge, 30q C , is placed at the origin of coordinates. Find the
electric field at the point y = 5 m on the y axis.
Solution:
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For the point charge, the magnitude of the electric field is given by 2
kqE
r
Thus 6
9 2 2 42 2
( 30 10 )8.99 10 / 1.08 10 /(5 )
kq CE Nm C N Cr m
55.0 10B
The minus sign means that the direction of the electric field is from the point
y to the origin of coordinates.
6. In a uniform electric field, a test charge , is moved from
point A to point B, and of work are performed in this process.
Calculate
60 2.0 10q C
55.0 10 J
(a) The difference in the electric potential energies of the charge between the
two points.
(b) The potential difference between the two points.
Solution:
(a) The difference in potential energy between points A and B is equal to the
work done in moving the charge from A to B.
Therefore,
A ABEPE EPE W J
(b) The potential difference between A and B is the difference in potential
energy divided by the charge
60
525
2.0 105.0 10B A
B A
EPE EPEV V V
q CJ
115.29 10 m
7. In the Bohr model of the hydrogen atom, the electron is in an orbit around
the nuclear proton at a distance . Calculate
(a) The electric potential that the proton creates at this distance.
(b) The total energy of the atom.
Solution:
(a) The electric potential created by the proton charge of is 191.60 10 C
19
119 2 2 1.60 10 27.2
5.29 108.99 10 /kq CV V
r mNm C
(b) The total energy of the hydrogen atom is the sum of the electric potential
energy and the kinetic energy of the electron. The potential energy, relative to
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a zero reference value at infinity, can be found from the equation
by using the charge on the electron and the potential of + 27.2 V determined
in part (a):
EPE qV
19 181.60 10 27.2 4.35 10C V JEPE qV
From example 2 in section 4.3.1.2, we know that the speed of the electron is
. Since the mass of the electron is , its kinetic energy
is
62.2 10 /m s 319.11 10 kg
31 6 2 182 9.11 10 (2.2 10 / ) 2.2 101 12 2
kg m s JKE mv
18 18182.2 10 ( ) 2.15 104.35 10E KE EPE J JJ
The total energy E of the hydrogen atom is
191 1.60 10eV J 13.6eV
Note: the total energy in electron volts ( ) is .
8. Two metal plates 5.0 m apart have a potential difference of 100 V between,
and the spacing of the two plates is decreased, the potential difference now is
50 V. calculate the spacing between the two plates now?
Solution:
From the equation, VE
r
We first get the value E. 100
20 /5
V VE V m
r m
And the new distance now is given by
50 2.520 /
V Vr mE V m
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Chapter 2 Magnetic fields
2-1 Magnetic flux density
2-1-1 Magnets and fields
2-1-2 Magnetic fields from currents
2-1-3 Magnetic force on moving charges and flux density (B)
2-1-4 Magnetic force on a current
2-2 Moving charges in a magnetic field
2-2-1 Charged particles in circular orbits
2-2-2 The cyclotron
2-3 Electromagnetic induction phenomena
2-4 Faraday law of electromagnetic induction and Lenz’s law
2-5 Motional emf
2-6 32 Worked examples
1. A proton in a particle accelerator has a speed of . The proton
encounters a magnetic field whose magnitude is 0.4 T and whose direction
makes an angle with respect to the proton’s velocity. Calculate
65.0 10 /m s
030
(a) The magnitude of the magnetic force on the proton.
(b) The acceleration of the proton.
(c) What would be the force and acceleration if the particle were an electron
instead of a proton?
Solution:
(a) Since the positive charge on a proton is , the magnitude of
the magnetic force is
191.60 10 C
19 6 0 13sin 0.4 1.60 10 5.0 10 sin30 1.6 10F Bqv N
(b) The acceleration of the proton follows directly from Newton’s second law
as the magnetic force divided by the mass of the proton: pm
13 227
139.6 10
1.67 101.6 10
p
Fa msm kg
N
(c) The magnitude of the force does not change when the proton is replaced
by an electron, since both have the same charge magnitude. However, the
direction of the force on the electron is opposite to that on the proton, since
the charge on the electron is negative. Furthermore, the electron has a smaller
mass me and, therefore, experiences a significantly greater acceleration:
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17 231
131.76 10
9.11 101.6 10
p
Fa msm kg
N
2. A straight wire is placed between the poles of a laboratory magnet that
produces a uniform field of 1.0 T over a region that 0.25 m wide. The wire is
perpendicular to the direction of the field and, when connected to a source of
electricity, carries a current. What must the current be if the field exerts a
force of 9.81 N on the wire?
Solution:
sinF BIL We can get the current from the equation .
sinFI
lB
When we insert the numerical values of 9.81F N , 0.25l m , B = 1.0 T, and
, we find the current: 0sin sin 90 1
9.81 39.20.25 1.0 1sin
AFIlB
3. At a speed of , electrons are perpendicularly injected into a
magnetic field
61.6 10 /m s42.0 10B T . Find the orbital radius and time for electrons
moving in a circular motion. (Suppose the mass of the electron is
, the charge of the electron is ) 319.11 10 kg 191.60 10 C
Solution:
The orbital radius is given by 31 6
219 4
9.11 10 1.6 10 / 4.56 101.60 10 2.0 10
mv kg m sR mqB C T
The time period is given by
76
22 2 1.8 101.6 104.56 10RT s
v
4. Electrons are perpendicularly projected into a magnetic field 47.0 10B T ,
and have an orbital radius in a circular motion. Find their speed.
(Suppose the mass of the electron is , the charge of the electron
is )
23.0 10 m319.11 10 kg
191.60 10 C
Solution:
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From the equation mvR
qB , we can get the speed
4 19 26
317.0 10 1.60 10 3.0 10 3.7 10 /
9.11 10BqR T C mv m sm kg
5. A cyclotron operating at a magnetic field of 1.6 T is used to accelerate
protons ( and ). What must be the
frequency of oscillation of the accelerating voltage?
27mass 1.67 10 kg 19charge 1.60 10 C
Solution:
The time period of the cyclotron is given by
2 mT
Bq
Thus the frequency is given by 19
727
(1.6 )(1.60 10 )2.4 10
2 1.67 10
12
T CHz
kg
BqfT m
6. A coil, mounted on an axle, has its plane parallel to the flux lines of a
uniform magnetic field B, as shown. When a current I is switched on, and
before the coil is allowed to move,
A there are no forces due to B on the sides PQ and RS.
B there are no forces due to B on the sides SP and QR.
C sides SP and QR attract each other.
D sides PQ and RS attract each other.
Solution:
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The magnetic force on the wire is given by:
sinF BIL , where is the angle between the magnetic field and the
direction of the current.
Thus, for the sides PQ and RS, the angle is 00 and . Therefore,
the magnetic force on PQ and RS is zero.
0180
Choose (A).
7. Protons, each of mass m and charge e, follow a circular path when
travelling perpendicular to a magnetic field of uniform flux density B. What
is the time taken for one complete orbit?
A 2 eB
m
B 2
m
eB
C 2
eB
m
D 2 m
eB
Solution: 2v
eBv mr
, gives eBrv
m , and 2 2 2r r m
TeBrv eBm
Choose (D).
8. Particles of mass m carrying a charge Q travel in a circular path of radius r
in a magnetic field of flux density B with a speed v. How many of the
following quantities, if changed one at a time, would change the radius of the
path?
· m
· Q
· B
· v
A one
B two
C three
D four
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Solution:
From the outline below, moving charges in a magnetic field: mv
RqB
mv
Thus, the particle moves in a circular path whose radius R depends on the
momentum of the particle, its charge, and the magnitude of the magnetic
field.
Choose (D)
9. Fig. 9.1 shows a square coil PQRS placed in a uniform magnetic field with
the plane of the coil parallel to the lines of magnetic field. A constant current
is passed round the coil in the direction shown, causing a force to act on side
PS of the coil. Which one of the following statements about the forces acting
on the other sides of the coil is correct?
Fig. 9.1
A A force acts on each of the other sides of the coil.
B No force acts on sides PQ and RS of the coil.
C A force acts on side RS and an equal and opposite force to this force acts
on side PQ.
D A force acts on side QR in the same direction as the force that acts on PS.
Solution:
The magnetic force on the wire is given by:
sinF BIL , where is the angle between the magnetic field and the
direction of the current.
Thus, for the sides PQ and RS, the angle is 00 and . Therefore,
the magnetic force on PQ and RS is zero.
0180
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Choose (B).
10. Fig. 10.1 shows the arrangement of an apparatus for determining the
masses of ions. In an evacuated chamber, positive ions from an ion source
pass through the slit at P with the same velocity v. After passing P, the ions
enter a region over which a uniform magnetic field is applied. The ions travel
in a semicircular path of diameter d and are detected at points such as R.
(a) (i) State the direction of the applied magnetic field.
Solution:
The magnetic force points to the center of the semicircular. Thus by the
Fleming’s left-hand rule, the direction of the magnetic field is into the
plane of diagram.
(ii) Explain why the ions travel in a semicircular path whilst in the magnetic
field.
You may be awarded additional marks to those shown in brackets for the
quality of written communication in your answer.
Solution:
By the Fleming’s left-hand rule, the magnetic force is perpendicular to both
magnetic field and velocity. The force changes direction of velocity but not
its magnitude. Thus the particle undergoes a circular path, the centripetal
force is supported by the magnetic force.
(iii) By considering the force that acts on an ion of mass m and charge Q,
having velocity v, show that the diameter d of the path of the ions is given by
2mvdBQ
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where B is the flux density of the magnetic field.
Solution: 2mv
BQvr
, gives 22
mvd r
BQ
(b) In an experiment using singly-ionised magnesium ions travelling at a
velocity of , d was 110 mm when B was 0.34T. Use this result to
calculate the charge to mass ratio of these ions.
47.5 10 /m s
Solution:
The charge to mass ratio can be obtained using the equation 2mvd
BQ
47
3
2 2 7.5 100.40 10 /
0.34 110 10
Q vC kg
m Bd
(c) (i) Some ions of the same element, whilst travelling at the same velocity
as each other at P, may arrive at a point that is close to, but slightly different
from, R.
Explain why this might happen.
Solution:
From the equation: 2mvd
BQ
d is proportional to mass m, and the ions have different mass. Thus they have
different value of d.
(ii) Other ions of the same element, also travelling at the same velocity at P
as all of the others, may travel in a path whose diameter is half that of the
others.
Explain why this might happen.
Solution:
The diameter of path d is proportional to 1
Q, and the ions are doubly ionized,
thus some ions may travel in a path whose diameter is half that of the others.
3 Particle physics
Chapter 1 Atomic nucleus
1-1 Nucleus structure of atoms Atomic nucleus
1.1 The nucleus structure model of atom
In 1909-1911, the British physicist Rutherford (1871-1937) and his
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assistants did experiments using particle scattering. Fig. 1-1 shows the
experimental setup. Using particle to shine on gold foil, some particle
change direction after passing through the gold foil. This is because the tiny
charged particles inside the gold atom have Coulomb interactions with the
particles. This phenomenon is called the scattering of particle.
The result of the experiment was that most particles passed through the
gold foil and almost all moved in their previous direction. But there were a
tiny number of particles which showed a large deflection.
Rutherford made precise records of the number of particles scattered in
various directions. Based on this, he put forward the nucleus structure model
of the atom. There is a tiny nucleus at the center of the atom called the atomic
nucleus. All the positive charge and nearly all the mass of the atom are
concentrated in the atomic nucleus. The electrons carrying a negative charge
move in space outside the nucleus.
According to this model, since the atomic nucleus is tiny when it passes
through the gold foil, most of the particles are far from the nuclei and feel
a very small repulsive force. Their movement is hardly influenced. Only a
very small number of particles pass close to the atomic nucleus, and are
affected considerably by the nucleus Coulomb repulsive force and so make a
large angle deflection (Fig. 1-2)
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particles
1.2 Constituents of the atom
Inside an atom: ① a positively charged nucleus composed of proton and
neutrons, ② Electrons that surround the nucleus.
Note: inside the atom, the number of proton = the number of electron, and if
the atom do not loss or gain electrons, it is neutral (that is no charge)
Nucleon: a proton or a neutron in the nucleus.
Inside an atom: Charge ( /C ) Mass ( /Kg )
a proton +1.60×1019 1.67×10-27
a neutron 0 1.67×10-27
an electron －1.60×1019 9.11×10-31
Isotopes: atoms of the same element that has the same numbers of proton but
different numbers of neutrons.
Represent an atom: AZ X
X: chemical symbol
A: total number of protons and neutrons, sometimes called nucleon number
or mass number. Z: number of protons, sometimes called atomic numbers.
So, the number of neutrons in the nucleus = A－Z
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Specific charge:
For charged particles, specific charge is defined as its charge divided by its
mass, that is arg
argch e
specific ch emass
Unit: C/kg
1.3 atomic measurements
The unified atomic mass unit (u) is used for measuring the masses of
atomic particles. And 271 1.66 10u kg
For example
Mass of proton: 1.00728 u
Mass of neutron: 1.00867 u
Mass of electron: 0.00055 u
1-2 Natural radiation phenomenon Decay half-life
2.1 Natural radiation phenomenon
The property of a substance emitting rays is called radioactivity. Elements
with radioactivity are called radioactive elements.
The phenomenon of elements’ spontaneous radiation of rays is called
natural radioactivity phenomenon.
2.2 Three types of radiation
(i) Alpha radiation
Alpha particle is the atom ; symbol is42 He 4
2 , sometimes in symbol .
An unstable nucleus of an element X emits an alpha particle; the product
nucleus is a different element Y.
Equation below: 4 42 2
A AZ ZX Y
Note: conservation of mass number and atomic number.
(ii) Beta decay
Beta is electron, in symbol 01 or ,
Equation below: 0
1 1A AZ ZX Y
: is called antineutrino, which has no charge and no mass.
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(iii) Gamma radiation
In symbol , it is emitted by a nucleus with too much energy, following
an alpha or beta emission.
2.3 Half-life
We can use a half-life to indicate the rate of decay of radioactive elements.
The amount of time required for half of the nuclei of a radioactive
element to decay is called the half-life of the element.
Note: different radioactive elements have different half-lives, and
sometimes the difference is very big.
The measure of radioactivity is the number of radioactive disintegrations
per second, called the activity (in symbol A). The SI unit of activity is the
Becquerel (Bq), which is one disintegration per second.
And if unstable atoms are present at time zero, then the number
remaining at some time t is
0N
0tN N e , where is a proportionality
constant (called the decay constant).
Since the activity is proportional to the number of radioactive atoms
present, the activity satisfies an exponential decay law also: 0tA A e
When the time is equal to one half-life, 12
t , the number of atoms
remaining is 02
NN , thus the equation can be written as 0tN N e
12
te
We can use this relation to obtain the half-life in terms of .
12
0.693t
Thus, we can get 1
2
0.693t
0t, therefore the equation N N e can be
written as:
12
0.693
0
tt
N N e
1-3 Nuclear reaction Nuclear energy
3.1 Nuclear reaction
In nuclear physics, the process of making a new nucleus when a nucleus is
bombarded by other particles is called a nuclear reaction.
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For example, if a high-energy particle strikes and is absorbed by a
nucleus of nitrogen-14, the new nucleus immediately decays to form a
nucleus of oxygen-17 and a proton. This is an examples of a nuclear reaction.
It can be described by the following equation: 14 4 17 17 2 8N He O 1 p
Note: the mass and charge numbers are all conserved in a nuclear reaction.
There is energy change in a nuclear reaction. The energy released in a
nuclear reaction is called nuclear energy.
Einstein’s relativity theory points out that energy has mass. If an object
gains energy, it gains mass. If it loses energy, it loses mass. The change of
energy E is linked to the change of mass m by this equation: 2E mc
191 1.60 10eV J
Where c is the speed of light: 83 10 /c m s
With nuclear particles, energy is often measured in non-SI unit
eV/MeV/GeV:
131 1.60 10
MeV J 101 1.60 10GeV J
271 1.66 10u kg 2
And with nuclear particles, mass is usually measured in u
( ). by converting 1 u into kg and applying E mc , it
is possible to show that:
3.2 Mass defect & binding energy
A helium-4 nucleus is made up of 4 nucleons (2 protons and 2 neutrons).
The calculation below (Fig. 1-3) shows that the nucleus has less mass than its
four nucleons would have as free particles. The nucleus has a mass defect of
0.029267u.
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p p
n np
p
nn
Helium-4 nucleus
4.002603u
1.007276u1.007276u1.008665u1.008665u
4.03187u
Difference in mass = 0.029267u
(mass defect = 0.029267u)
Fig. 1-3 mass defect The binding energy of a nucleus is the energy equivalent of its mass
defect. (Binding energy is the energy holds the nucleus together). Therefore,
it is the energy needed to separate the nucleus into its component parts. For
, the mass defect is 42 He
42
2(2 ) 0.029267e
p n Hmm m m u
0.029267 931.5 27.3
And 1 u is equivalent to 931 MeV, the binding energy is E MeV
1-4 Fission
1-5 Fusion of light nuclei
1-6 19 Worked examples
Chapter 2 Particles and Radiation
2-1 Particles and antiparticles
2-2 Particle interactions
2-3 Properties of particles and antiparticles
2-4 Quarks and Antiquarks
2-5 45 Worked examples
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