physics examination problems solutions and...

PHYSICS EXAMINATION PROBLEMSSOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHY1021

Name of module Vector Mechanics

Date of examination January 2016


Module Code PHY1021




Module Code PHY1021



Question 1. [25]

ri = (5, 6)

The angle (⇡ 14.7

�) you can find from the dot product.

At first sight, there appears to be three unknowns: k, ↵ and �. You can find an equation with these

quantities in if you look at one component of the vector. Now, consider the possible values of ↵ and

� given the conditions on them. You will find two possibilities but only one combination works. The

result is that k = 2 so the total number of moves is 6.

The x-axis reversal means that

�x

=

�1 0

0 1

!

(could have written this as a 3x3 matrix including z, but no need.)

By taking the determinant, you can prove that this transformation is improper.

The y-axis mirror will look like the transpose of the x-axis mirror. Both transformations require you

to multiply the matrices (in this case, in either order). You can then take the determinant of the

result to show that it is proper.

1


Module Code PHY1021




Module Code PHY1021




Module Code PHY1021



Question 2. [25]

J = �p

where J is the impulse experienced by a body and �p is the change in its momentum as a result of

the impulse.

(a) V = �51.8m s

�1ˆı

(b) �K = �3.7 · 106 J

(c) J = �8.2 · 104 kgm s

�1ˆı. Careful of the sign here.

(d) F = 82MN

ˆı. Again, careful of the sign.

You will need the Galilean transformation to get U 0. To get the speed, you then need to take the

square root of the dot product of this vector with itself:

U

0=

pU

2+ u

2+ 2uU cos↵

The key is that the vertical speed is una↵ected by the frame motion. As a result

U

0sin � = U sin↵

By looking at U

0you can see that it is minimised if u = 0.

2


Module Code PHY1021




Module Code PHY1021




Module Code PHY1021



Question 3. [25]

As the points lie in a plane, take the cross product of the position vectors of any two of the points.

You will find that any choice will result in a vector with one component, which is in the

ˆk direction.

Particle 1 accelerates uniformly toward the centre as the tension is proportional to the free-fall

acceleration due to gravity. As a result, use the standard equations of motion to find:

t =

s2m1b

m2g

To remain at a fixed distance, the tension must only supply the centripetal force required to keep

the particle moving in a circle:

u = 1.98m s

�1

The angular momentum

L = �0.0396 kgm

2s

�1ˆk

Be careful of the sign: it is defined by the direction of rotation in the question.

Particle 3 will have a position vector x

ˆı + b

ˆ|. Use this in the formula for angular momentum and

you will find it to be equal to the original angular momentum of particle 1.

3


Module Code PHY1021




Module Code PHY1021




Module Code PHY1021



Question 4. [25]

u ⇡ 0.93c

ct

x

light line light line

1 · 103

c3.6µs

Be careful to show that the world line connecting the events has a steeper gradient than the light

line.

t

0 � T = T (� � 1) ⇡ 5.9µs

(Note that the question asks for how much longer the journey takes.)

You can show that the intervals are the same without substituting any numbers. Note that �x

0=

��uT .

n

p

e

⌫̄

The conservation of four-momentum

Pi = Pf

Pn = Pp + Pe + P⌫̄

4


Module Code PHY1021




Module Code PHY1021




Module Code PHY1021



(a) By looking separately at the x and y components of the relativistic momentum, you will find

that pn = 0 and pp = pe = p

⌫̄

.

(b)

Kn = 0

Kp = 2 · 10�4MeV

Ke = 0.25MeV

5

PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHY1022

Name of module Introduction to Astrophysics


1. (i) The first part of this question is bookwork.

p = h f / c.

The wavelength is 830 nm.

(ii) Equate the centrifugal force (me υ2 / r) to the Coulomb force (ke e2 / r2) and solve for υ.

The next part is bookwork.

Insert the velocity given above into Bohr’s quantized angular momentum and solve for rn.

Use E = KE + U, inserting the velocity into ½ me υ2 for KE and using the electrostatic potential –ke e2 / rn for U.

The wavelength of the transition is 487 nm.

The final part is bookwork.

2. (i) The first part is bookwork.

The minimum uncertainty in position is 5.78 × 10−7 m.

λ = h / p.


(ii) The first part is bookwork.

(a) yes, all conservation laws obeyed.

(b) No, violates baryon number conservation.

(c) No, violates baryon number and e-like lepton number conservation.

3. (i) The first part is bookwork.

E.g., axions and WIMPS.


Ωm = 0.3 and ΩΛ = 0.7.

(ii) The first part is bookwork.

Temperature in center of sun of ~107 K gives particles energies of ~kev, but fusion requires ~Gev.


~1% of hydrogen burned is converted to energy.

Hydrogen burning can power the sun for 1.48 × 1010 years.

4. The first part is bookwork.

(a) 1.6 AU.

(b) 8.4 × 1024 kg.

(c) 7.0 × 106 m.

(d) 5.8 × 103 kg/m3, made of rock or rock and iron.

(e) 5.0.

(f) 100 pc.

(g) 7.9 m.

(h) 9.7 microns.

(i) This is bookwork.

(j) 1.4 × 109.

PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHY2021

Name of module Electromagnetism I


1. (i) Coursework for Gauss's law.

Charge per unit length: QL= 2πra

Electrostatic field inside: E = a

εr̂ and outside is:

E = aR

ε0rr̂

Show that ∇⋅E = ρ = 0

The potential difference is 141 V. 2. (i) Coursework. Coursework. Coursework. Coursework. The monopole moment is -e (i.e. the negative of the elemental charge). The important thing here is to compute the calculation of the dipole moment in vector form. The

answer can be written p = (9.6×10−29, 0, 0) Cm. You can also use Cartesian unit vectors to express the answer.

The potential is the sum of the monopole and the dipole contributions. Final answer: -0.256 V, with the monopole contribution being -0.267 V and the dipole contribution being +0.011 V.

3. (i) Coursework. The particle will travel in a helical path, spiraling around the z-axis. t=0.05 seconds Number of circles is 153.03, so the passes through the y-z plane is 306. (The radius of gyration

is 0.0104 m, and the time to make one circle is 3.267×10−4 seconds.) (ii) Coursework (self-study pack 3). The current is 1 A. The emfs are 7 V and 35 V. The resistance R is 1 Ohm. 4. (i) Coursework. Coursework. Coursework. Derivations from coursework. Compute

B =∇×

A

The field strength is 0.0004 T.


Module Code PHY2022Name of module Quantum Mechanics 1Date of examination January 2016

1. The state of a [non-relativistic quantum] particle is described by a [continuous, non-singular, complex] wave function which can be normalized so that its square modulus is equal to the probability density for [the results of] a position measurement[, and from which all possible predictions about the physical properties of the system can be obtained].

[] indicate things that I was very lenient about if missed.

To normalize let Ψnorm = CΨ , then C 2 = 1

Ψ 2 dVall space∫

. This makes Ψ 2 equal to, not

just proportional to, the probability density for a position measurement.

SOLUTION TO DEGREE EXAMINATION QUESTIONfor Coordinator

Name of setter Usher Paper Question

Name of module Quantum Mechanics 1

Year of examination 2016 PHY2022 1 (cont)Initials of checker

tt

Trz

a

.L

:--\.*'--:--

I-b

L

rr2

2t gw (K<

: 2U Sr,..f (/c r )

(^(sr

(4)

Use this symmetry when normalizing, so normalization condition is:

K = πL

.

2. Wavefunction continuous and single valued (otherwise, probability not single valued) Normalizable: wavefunction and 1st derivative go to zero at infinity and has to be

square-integrable. 1st derivative continuous (so that 2nd derivative successfully mimics the potential for

(almost) all realistic potentials. Exception to this when there is an infinite discontinuity in the potential. WF zero in finite region of infinite potential.




Year of examination 2016 PHY2022 2Initials of checker

(rJ

sz Derztfalud e^'l.ntaous a il?,

eXcqzt&^t fuGh,t€€ O I rc^rrttJ^fry oJ Pd7,artrlY.

oy rlF

tre m€n9T7?7arc ilZ

I aec+uJ

€ eyp /-

The symmetric well requires symmetric probability density which requires either symmetric or antisymmetric wavefunctions.





(rJ

sz Derztfalud e^'l.ntaous a il?,

eXcqzt&^t fuGh,t€€ O I rc^rrttJ^fry oJ Pd7,artrlY.

oy rlF

tre m€n9T7?7arc ilZ

I aec+uJ

€ eyp /-

For the antisymmetric solutions k cot ka( ) = −q





S'/nn,zatc C=o

E =O

c4t7tNk6a3 G/U€C

ccls hut F t?xp

7ot cz ol"o(r

ntoT6ZQt C STaT D= o

t= €xP

+^/b €

_ gzab

( rzfcm,'(+ppl(fr4oN7

l{ T*tJ( u^\ SmztlaWwn

FSott h!fttu lt cor(tz^\ = a f-seq'YaD,

3. Set of functions φn{ } orthogonal if φm*φn dV

all space∫ = 0, for m ≠ n

Proof that Hermitian operators are orthogonal: see pages 7 to 9 of lecture notes for

Section 9.

N = 114

.





P rzotla

Frto4pragyll 7 olrY

+ 2r2 3v I

7t4

t4

atcTculG-3

t4 t+

I+ i6,4

NlTtt P,

C (wpt,rrt oA

h6rae,r,rze x/z*r7 TtEta \ D --t LJ . F /c N6\)

7sg(ra7)

.

A = 187

Wavefunction after measurement yielding A = 2 is φ2 . Immediate re-measurement will give same outcome as previous measurement, with

probability 1. Alternate measurements of incompatible variables will never be deterministic - there

will always be a range of possible values with no ability to predict the outcome in advance.

4.





a^/

70€

l< Yn

P /- 20.

P r→ r + dr( ) = u 2 4πr2dr

Coulomb potential binding electron to proton in hydrogen is −e2 4πε0r . The

Quarkonium potential is –a/r (in this approximation). Reduced mass of Quarkonium is

mQ 2 . Therefore

a0 =

2!2

mQa and

E0n = −

mQ

4!2a2

n2.

E1n = u0n*Vu0n dV

all space∫ .

E1n =

32ba0 quarkonium( ) = 3b!

2

mQa.


Module Code PHY1021




Module Code PHY1021




Module Code PHY1021



Question 1. [25]

ri = (5, 6)

The angle (⇡ 14.7

�) you can find from the dot product.

At first sight, there appears to be three unknowns: k, ↵ and �. You can find an equation with these

quantities in if you look at one component of the vector. Now, consider the possible values of ↵ and

� given the conditions on them. You will find two possibilities but only one combination works. The

result is that k = 2 so the total number of moves is 6.

The x-axis reversal means that

�x

=

�1 0

0 1

!

(could have written this as a 3x3 matrix including z, but no need.)

By taking the determinant, you can prove that this transformation is improper.

The y-axis mirror will look like the transpose of the x-axis mirror. Both transformations require you

to multiply the matrices (in this case, in either order). You can then take the determinant of the

result to show that it is proper.

1


Module Code PHY1021




Module Code PHY1021




Module Code PHY1021



Question 2. [25]

J = �p

where J is the impulse experienced by a body and �p is the change in its momentum as a result of

the impulse.

(a) V = �51.8m s

�1ˆı

(b) �K = �3.7 · 106 J

(c) J = �8.2 · 104 kgm s

�1ˆı. Careful of the sign here.

(d) F = 82MN

ˆı. Again, careful of the sign.

You will need the Galilean transformation to get U 0. To get the speed, you then need to take the

square root of the dot product of this vector with itself:

U

0=

pU

2+ u

2+ 2uU cos↵

The key is that the vertical speed is una↵ected by the frame motion. As a result

U

0sin � = U sin↵

By looking at U

0you can see that it is minimised if u = 0.

2


Module Code PHY1021




Module Code PHY1021




Module Code PHY1021



Question 3. [25]

As the points lie in a plane, take the cross product of the position vectors of any two of the points.

You will find that any choice will result in a vector with one component, which is in the

ˆk direction.

Particle 1 accelerates uniformly toward the centre as the tension is proportional to the free-fall

acceleration due to gravity. As a result, use the standard equations of motion to find:

t =

s2m1b

m2g

To remain at a fixed distance, the tension must only supply the centripetal force required to keep

the particle moving in a circle:

u = 1.98m s

�1

The angular momentum

L = �0.0396 kgm

2s

�1ˆk

Be careful of the sign: it is defined by the direction of rotation in the question.

Particle 3 will have a position vector x

ˆı + b

ˆ|. Use this in the formula for angular momentum and

you will find it to be equal to the original angular momentum of particle 1.

3


Module Code PHY1021




Module Code PHY1021




Module Code PHY1021



Question 4. [25]

u ⇡ 0.93c

ct

x

light line light line

1 · 103

c3.6µs

Be careful to show that the world line connecting the events has a steeper gradient than the light

line.

t

0 � T = T (� � 1) ⇡ 5.9µs

(Note that the question asks for how much longer the journey takes.)

You can show that the intervals are the same without substituting any numbers. Note that �x

0=

��uT .

n

p

e

⌫̄

The conservation of four-momentum

Pi = Pf

Pn = Pp + Pe + P⌫̄

4


Module Code PHY1021




Module Code PHY1021




Module Code PHY1021



(a) By looking separately at the x and y components of the relativistic momentum, you will find

that pn = 0 and pp = pe = p

⌫̄

.

(b)

Kn = 0

Kp = 2 · 10�4MeV

Ke = 0.25MeV

5

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