• 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].