• 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