• 9702 November 2013 Paper 21 & 22 Worked Solutions | A-Level Physics

Paper 21 and Paper 22 are similar.

Question 1

(a)        Choose any 2 :

Kelvin (K), Ampere (amp/A), Mole (mol), Candela (Cd)

(b) 

(i) Work = Force x Distance = (Mass x Acceleration) x Distance : kg x ms^-2 x m

Work : kg m² s^-2

Alternatively:

Energy = ½ mv² or mc² : kg (ms^-1)² = kg m² s^-2

(ii) Density = mass/ volume : kgm^-3

Acceleration of free fall, g : ms^-2

Cross-sectional area of wire, A : m²

Original length, l₀ : m

Therefore, C : kg m² s^-2 / (kg m^-3)² (ms^-2)² m² (m)³ = kg^-1 ms^-2 

Question 2

(a)

Time-base setting = 0.20μscm^-1

That is, 1cm (1 cell) represents 0.20μs
Speed of pulse, v = 3 x 10⁸ ms^-1

Number of cells between peaks of the 2 pulses = 4

Speed, v = distance, d / time, t

Therefore, distance travelled by pulse, d = v x t

t = 4 x 0.20 = 0.80μs

d = 3 x 10⁸ x 0.80 x 10^-6 = 240 m

Since the pulse travels from the source to the reflector, and from the reflector back to the source,

Distance between source and reflector = 240/2 = 120m  

(b)

Speed of sound = 300ms^-1

Speed of light = 3 x 10⁸ ms^-1

Time taken by sound wave = 240 / 300 = 0.8s

This time corresponds to 4 cells (4cm) since the distance 240m corresponds to 4 cells.

Therefore,

4 cells represent 0.8s

1 cell represent 0.8/4 = 0.2s

Time-base setting is 0.20scm^-1            [unit required]

Question 3

{Detailed explanations for this question is available as Solution 673 at Physics 9702 Doubts | Help Page 135 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-135.html}

Question 4

(a)

The torque of a couple is the product of one of the forces multiplied by the perpendicular distance between the forces.

(b)

(i)

Weight at P – vertically down

Choose any 1:

Normal reaction - vertically up

Contact force at P – at point of contact with the pin

(ii)

Torque = 35 x 0.25 x 2 = 17.5 ≈ 18 Nm

(iii)

The two 35N forces are equal and opposite and the weight and the upward / contact / reaction force are equal and opposite.

(iv)

The wheel is not in equilibrium since the (resultant) torque is not zero.

Question 5

{Detailed explanations for this question is available as Solution 540 at Physics 9702 Doubts | Help Page 106 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-106.html}

Question 6

{Detailed explanations for this question is available as Solution 389 at Physics 9702 Doubts | Help Page 72 - http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-72.html}

Question 7

(a)

(i)

Choose any 1:

The direction of the fields is the same.

Fields are uniform.

Constant electric field strength

E = V/d where E is the uniform electric field between the 2 parallel plates, separated by the distance d and V is the electric potential. 

(ii)

Reduce the potential difference across the plates

Increase the separation of the plates

(iii)

α-particles have opposite charge to β-particles (deflection in opposite direction)

β-particles have a range of velocities OR energies (different deflections) while α-particles all have the same velocity OR energy (constant deflection)

α-particles are more massive (do not use ‘heavier’) (deflection is less for greater field strength)

(b)

W = 234, X = 90, Y = 4 and Z = 2

(c)

A = 32, B = 16, C = 0 and D = -1

 

• Linear Algebra: #2 Basic Definitions

Definition.
Let X and Y be sets. The Cartesian product X × Y , of X with Y is the set of all possible pairs (x, y) such that x ∈ X and y ∈ Y .


Definition.
A group is a non-empty set G, together with an operation (The operation is usually called “multiplication” in abstract group theory, but the sets we will deal with are also groups under “addition”.), which is a mapping ‘ · ’ : G × G → G, such that the following conditions are satisfied.

1. For all a, b, c ∈ G, we have (a · b) · c = a · (b · c),
2. There exists a particular element (the “neutral” element), often called e in group theory, such that e · g = g · e = g, for all g ∈ G.
3. For each g ∈ G, there exists an inverse element g^-1 ∈ G such that g · g^-1 = g^-1· g = e. 

If, in addition, we have a · b = b · a for all a, b ∈ G, then G is called an “Abelian” group.


Definition.
A field is a non-empty set F, having two arithmetical operations, denoted by ‘+’ and ‘·’, that is, addition and multiplication (Of course, when writing a multiplication, it is usual to simply leave the ‘·’ out, so that the expression a · b is simplified to ab). Under addition, F is an Abelian group with a neutral element denoted by ‘0’. Furthermore, there is another element, denoted by ‘1’, with 1 ≠ 0, such that F \ {0} (that is, the set F, with the single element 0 removed) is an Abelian group, with neutral element 1, under multiplication. In addition, the distributive property holds:

 a · (b + c) = a · b + a · c and (a + b) · c = a · c + b · c,

 for all a, b, c ∈ F.


The simplest example of a field is the set consisting of just two elements {0, 1} with the obvious multiplication. This is the field ℤ/2ℤ (Here ℤ is the set of integers). Also, as we have seen in the analysis lectures, for any prime number p ∈ N, the set ℤ/pℤ of residues modulo p is a field. The following theorem, which should be familiar from the analysis lectures, gives some elementary general properties of fields.


Theorem 1
Let F be a field. Then for all a, b ∈ F, we have:

1. a · 0 = 0 · a = 0,
2. a · (−b) = −(a · b) = (−a) · b,
3. −(−a) = a,
4. (a⁻¹)⁻¹ = a, if a ≠ 0,
5. (−1) · a = −a,
6. (−a) · (−b) = a · b,
7. a · b = 0 ⇒ a = 0 or b = 0. 

Proof. An exercise (dealt with in the analysis lectures).

So the theory of abstract vector spaces starts with the idea of a field as the underlying arithmetical system. But in physics, and in most of mathematics (at least the analysis part of it), we do not get carried away with such generalities. Instead we will usually be confining our attention to one of two very particular fields, namely either the field of real numbers ℜ, or else the field of complex numbers ℂ.

Despite this, let us adopt the usual generality in the definition of a vector space.


Definition.
A vector space V over a field F is an Abelian group — with vector addition denoted by v + w, for vectors v, w ∈ V. The neutral element is the “zero vector” 0. Furthermore, there is a scalar multiplication F × V → V satisfying (for arbitrary a, b ∈ F and v, w ∈ V):

1. a · (v + w) = a · v + a · w, 
2. (a + b) · v = a · v + b · v, 
3. (a · b) · v = a · (b · v), and 
4. 1 · v = v for all v ∈ V. 

Examples
Given any field F, then we can say that F is a vector space over itself. The vectors are just the elements of F. Vector addition is the addition in the field. Scalar multiplication is multiplication in the field.

Let ℜⁿ be the set of n-tuples, for some n ∈ N. That is, the set of ordered lists of n real numbers. One can also say that this is

ℜⁿ = ℜ × ℜ × · · · × ℜ (n times), 

the Cartesian product, defined recursively. Given two elements

 (x₁, · · · , x_n) and (y₁, · · · , y_n)

in ℜⁿ, then the vector sum is simply the new vector

(x₁+ y₁, . . . , x_n+ y_n) 

Scalar multiplication is

a ·  (x₁, · · · , x_n) =  (a ·x₁, · · · , a ·x_n)

It is a trivial matter to verify that ℜⁿ, with these operations, is a vector space over ℜ.


Let C₀([0, 1], ℜ) be the set of all continuous functions f : [0, 1] → ℜ. This is a vector space with vector addition

(f + g)(x) = f(x) + g(x), 

for all x ∈ [0, 1], defining the new function (f + g) ∈ C₀([0, 1], ℜ), for all f, g ∈ C₀([0, 1], ℜ). Scalar multiplication is given by

(a · f)(x) = a · f(x) 

for all x ∈ [0, 1].

• Linear Algebra: #1 Introduction

The mathematical idea of a vector plays an important role in many areas of physics.

• Thinking about a particle traveling through space, we imagine that its speed and direction of travel can be represented by a vector v in 3-dimensional Euclidean space ℜ³. Its path in time t might be given by a continuously varying line — perhaps with self-intersections — at each point of which we have the velocity vector v(t).


• A static structure such as a bridge has loads which must be calculated at various points. These are also vectors, giving the direction and magnitude of the force at those isolated points.


• In the theory of electromagnetism, Maxwell’s equations deal with vector fields in 3-dimensional space which can change with time. Thus at each point of space and time, two vectors are specified, giving the electrical and the magnetic fields at that point. 


• Given two different frames of reference in the theory of relativity, the transformation of the distances and times from one to the other is given by a linear mapping of vector spaces. 


• In quantum mechanics, a given experiment is characterized by an abstract space of complex functions. Each function is thought of as being itself a kind of vector. So we have a vector space of functions, and the methods of linear algebra are used to analyze the experiment. 

Looking at these five examples where linear algebra comes up in physics, we see that for the first three, involving “classical physics”, we have vectors placed at different points in space and time. On the other hand, the fifth example is a vector space where the vectors are not to be thought of as being simple arrows in the normal, classical space of everyday life. In any case, it is clear that the theory of linear algebra is very basic to any study of physics.

 But rather than thinking in terms of vectors as representing physical processes, it is best to begin these lectures by looking at things in a more mathematical, abstract way. Once we have gotten a feeling for the techniques involved, then we can apply them to the simple picture of vectors as being arrows located at different points of the classical 3-dimensional space.

• Complex Analysis: #31 Functions of Finite Order with Zeros

Theorem 53
Let f be an entire function of strict order α. Then there exists a C > 0 such that, for all R > 0 and D_R = {z ∈ ℂ : |z| ≤ r}, we have

v_f(R) ≤ CR^α, 

where v_f(R) is the number of zeros of f in D_R.

Proof
Assume first that f(0) ≠ 0. According to Jensen’s inequality (leaving out the part where we integrate from 0, and remembering that |f(z)| ≤ Ke^|z|α , for some K > 0), we have, for R > 1

Finally, if we do have f(0) = 0, then take g(z) = f(z)/z^m, where m is the order of the zero at 0. Then the theorem will apply to g, and since m is then fixed, to f as well.


Theorem 54
Again, f is an entire function of strict order α with zeros {z_n}, listed (with multiplicity) in order of increasing absolute value. We assume that f(0) ≠ 0. Then for every δ > 0 we have the series

∑ |z_n|^−α−δ

converging.

Proof
Using partial summation and the previous theorem, where we assume that R ∈ ℕ, we have

and this last sum is convergent.

This, combined with the discussion concerning the genus of a canonical product of discrete elements of ℂ (see section 24), leads to:

Corollary (Hadamard)
Let f be an entire function of finite order α with zeros {z_n}_n∈ℕ. Then

IMPORTANT NOTE:
This series on Complex Analysis has been taken from the lecture notes prepared by Geoffrey Hemion.  The document can be found at his homepage.

• Complex Analysis: #30 Jensen`s Formula

But before doing that, we look at the easier Jensen’s inequality.


Theorem 51 (Jensen’s Inequality)
Let R > 0 be given and let the (non-constant) analytic function f be defined in a region containing the closed disc D_R = {z ∈ ℂ : |z| ≤ R}. Assume f(0) ≠ 0 and also f(z) ≠ 0 for all z with |z| = R. Let the zeros of f in D_R be z₁, . . ., z_n . (Here a zero is listed m times if it is a zero of order m.) We assume the zeros are ordered according to their increasing absolute value. Let ||f||_R = max{|f(z)| : |z| = R}. Then we have

More generally, thinking about various values of R, let v_f(R) = n be the number of zeros of f in D_R, where R, thus n, is allowed to vary. Then we have Jensen’ inequality:

It is obviously analytic in D_R. Furthermore, we have |g(z)| = |f(z)| when |z| = R. This implies that |g(w)| ≤ ||f||_R for all w ∈ D_R. [If we had a point w_∗ in the interior of D_R with |g(w_∗)| > ||f||_R, then we can assume that it is maximal with respect to this property. However, that would contradict theorem 19.] Therefore

Theorem 52 (Jensen’s Formula)
The same assumptions as in the previous theorem. Then we have

Proof
In this proof, we will first look at two very simple cases:
1. We first prove Jensen’s formula in the simple case that there are no zeros of f in D_R. Then again, as in exercise 12.1, we have an analytic function g, defined in a neighborhood of D_R, with f = e^g. Or put another way, g = log f. (To be definite, we could specify that log f(0) should be in the principle branch of the logarithm.) Then Cauchy’s formula is simply

which establishes the theorem in the first case.

2. The second case is even simpler. Namely, let ζ be a complex number with |ζ| < R. That is, ζ is some point in the interior of D_R. This second case is that the function f is simply f(z) = z − ζ. We then define a new function, namely

Complex Analysis: #29 Order of Entire Function

• Complex Analysis: #29 The Order of an Entire Function

Theorem 49
Let f : ℂ → ℂ be an entire function. Assume that there exist real constants C > 0, λ > 0, such that Re(f(z)) ≤ C(1 + |z|^λ), for all z ∈ ℂ. Then f is a polynomial, at most of degree [λ].

Proof
To begin with, we know that, for k, l ∈ Z, we have

Therefore, taking r → ∞, we see that if k > λ then a_k = 0. An analogous argument shows also that b_k = 0.

Definition 17
An entire function f is said to have finite order if there exists some real number ρ > 0 and a constant C > 0, such that

|f(z)| ≤ Ce^|z|ρ, 

for all z ∈ ℂ. The infimum over all such ρ is the order of f. That is to say, α is the order of f if |f(z)| ≤ Ce^|z|α+∈ for all ∈ > 0 and z ∈ ℂ. If |f(z)| ≤ Ce^|z|α for all z ∈ ℂ then α is the strict order of f.


Theorem 50
Let f be an entire function of finite order with no zeros. Then f = e^g, where g is a polynomial whose degree is the order of f.

Proof
According to exercise 12.1, given f, then there exists an entire function g with f = e^g. But then we must have

So Re(g(z)) ≤ |z|^α if α is the order of f, and therefore the result follows from theorem 49.

Going beyond this, we would like to think about entire functions of finite order, but with zeros This leads us to Hadamard’s theorem. But before we arrive there, let us think about Jensen’s formula.