• 9702 November 2007 Paper 4 Worked Solutions | A-Level Physics

Paper 4

SECTION A

Question 1

(a)

(i)

A radian is the angle subtended at the centre of a circle such that the arc is equal in length to the radius.

(ii)

Why 1 complete revolution is equivalent to angular displacement of 2π rad:

Length of arc = rθ and for one revolution, the length of arc formed (= circumference) = 2πr. So, θ = 2πr / r = 2π

(b)

Elastic cord has unextended length of 13.0cm. One end of cord attached to fixed point C. Small mass of weight 5.0N is hung from free end of cord. Cord extends to length of 14.8cm, as shown.

Cord and mass are now made to rotate at constant angular speed ω in vertical plane about point C. When cord is vertical and above C, its length is the unextended length of 13.0cm, as shown.

(i)

Show that angular speed ω of cord and mass = 8.7rads^-1:

EITHER The weight provides / equals the centripetal force OR The acceleration of free fall is the centripetal acceleration.

(a = rω²)

9.8 = 0.13ω²

ω = 8.7 rads^-1

(ii)

Cord and mass rotate so that cord is vertically below C, as shown. Length L of cord, assuming it obeys Hooke’s law:

{When the cord and mass is vertically below C, the centripetal force {F_C} is provided by the resultant force from the force in the cord [tension in the cord {T} – which is upwards] and the weight [W - which is downwards]. (T should be greater than W so that the mass remains attached to that cord) So, F_C = T – W giving T = W + F_C}

Force in the cord = Weight + Centripetal force

{To know the force in the cord, the spring constant, which is always constant should be found first. Hooke’s law: Force, F = kx. Extension, x = L – 13 and spring constant, k = F /x. When F = 5.0N, x = 14.8 – 13 = 1.8cm, k = spring constant [this always constant] = 5.0 / 1.8} 

Force in the cord = (L – 13) x 5/1.8    or force constant = 5.0 / 1.8

{Force constant mentioned above = spring constant}

{Weight = 5.0N, Centripetal force = m rω². Since only the weight is given, we need to find the mass to calculate the centripetal force. Mass, m = weight / g = 5.0 / 9.8}

{Force in the cord = Weight + Centripetal force}

(L – 13) x 5/1.8 = 5.0 + (5/9.8) (L x 10^-2) (8.7²)

Length of cord, L = 17.2cm

Question 2

(a)

Amount of 1.00mol of Helium-4 gas contained in cylinder at pressure of 1.02x10⁵Pa and temperature of 27^oC.

(i)

Volume of gas in cylinder:

pV = nRT

(T = 273 + 27 = 300)

V = (nRT/p =) (8.31 x 300) / (1.02x10⁵) = 0.0244m³

(ii)

Show that average separation of gas atoms in cylinder is approximately 3.4x10^-9m:

Volume occupied by 1 atom = 0.0244 / (6.02x10²³) = 4.06x10^-26m³

Separation ≈ (4.06x10^-26)^1/3 = 3.44x10^-9m

(b)

(i)

Gravitational force between 2 Helium-4 atoms that are separated by distance of 3.4x10^-9m:

Force = GMm / r² = [(6.67x10^-11) (4{1.66x10^-27})²] / (3.44x10^-9)² = 2.49x10^-46N

(ii)

Ratio of weight of a Helium-4 atom to gravitational force between 2 Helium-4 atoms that with separation 3.4x10^-9m:

Ratio = [4(1.66x10^-27) x 9.8] / (2.49x10^-46) = 2.6x10²⁰

(c)

Comment on answer to (b)(ii) with reference to 1of the assumptions of kinetic theory of gases:

One assumption of the kinetic theory of gases is that the forces between the atoms are negligible.

Comment: e.g

The ratio shows the gravitational force to be very small

The force is very much less than the weight

If there are forces, they are not gravitational forces.

Question 3

{Detailed explanations for this question is available as Solution 1125 at Physics 9702 Doubts | Help Page 243 - http://physics-ref.blogspot.com/2016/10/physics-9702-doubts-help-page-243.html}

 

Question 4

{Detailed explanations for this question is available as Solution 714 at Physics 9702 Doubts | Help Page 144 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-144.html}
 
 

Question 5

A capacitor is charged to a potential difference of 15 V and then connected in series with a switch, a resistor of resistance 12 kΩ and a sensitive ammeter, as shown in Fig. 5.1.

Question 6

A straight conductor carrying a current I is at an angle θ to a uniform magnetic field of flux density B, as shown in Fig.6.1. ... 

  

Question 7

{Detailed explanations for this question is available as Solution 1029 at Physics 9702 Doubts | Help Page 215 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-215.html}

SECTION B

Question 8

Fig. 8.1 shows a circuit incorporating an ideal operational amplifier (op-amp).

 

Question 9

(a)

Acoustic impedance is the product of density (of the medium) and the speed of sound (in the medium).

(b)

Why acoustic impedance is important when considering reflection of ultrasound at boundary between 2 media:

The difference in acoustic impedance determines the fraction of incident intensity that is reflected / amount of reflection.

(c)

Principles behind use of ultrasound to obtain diagnostic information about structures within body:

A pulse of ultrasound (which is directed into the body) is reflected at the boundary (between the tissues). (The reflected pulse is) detected and processed. The time for return of the echo gives (information on the) depth and the amount of reflection gives information on the tissue structures.

Question 10

Fig. 10.1 shows the variation with frequency f of the power P of a radio signal.

  

Question 11

In cellular phone network, country is divided into a number of cells, each with its own base station. Fig shows a number of these base stations and their connection to cellular exchange.

(a)

Explain why country is divided into a number of cells:

The carrier frequencies can be re-used (simultaneously without interference) so that the number of handsets possible is increased.

OR

Any sensible e.g. UHF used, so ‘line of sight’

(b)

What happens at base station and cellular exchange when mobile phone handset is switched on, before call is made?:

The handset sends out an (identifying) signal which is communicated by the base stations to (the computer at) the exchange. The computer selects a base station with the strongest signal and allocates a (carrier) frequency.