A Level Physics Worksheets 2pq3k

18 Marking scheme: Worksheet (A2) 1

2

3

30 π × 2π = ≈ 0.52 rad 360 6 210 b θ= × 2π ≈ 3.7 rad 360 0.05 c θ= × 2π ≈ 8.7 × 10−4 rad 360

a

θ=

[1] [1] [1]

1 .0 × 360 = 57.3° ≈ 57° 2π 4 .0 b θ= × 360 ≈ 230° 2π 0.15 c θ= × 360 ≈ 8.6° 2π

a

θ=

[1] [1] [1]

88 days is equivalent to 2π radians. 44 θ= × 2π = π rad 88 1 b θ= × 2π ≈ 0.071 rad (4.1°) 88

a

[1] [1]

4

a b c d

Friction between the tyres and the road. Gravitational force acting on the planet due to the Sun. Electrical force acting on the electron due to the positive nucleus. The (inward) force between the clothes and the rotating drum.

5

a

ω=

[1] [1] [1] [1]

v 150 = r 20000

[1]

ω = 7.5 × 105 rad s−1 v r 150 2 a= 20 000 a = 1.125 m s–2 ≈ 1.1 m s−2 F = ma = 80 × 1.125 F = 90 N

b a=

c

6

a

[1]

2

[1]

[1] [1] [1] [1]

8 .2 = 0.82 s 10 ii Distance = circumference of circle = 2π × 0.80 = 5.03 m ≈ 5.0 m distance 5.03 iii speed = = time 0.82 −1 speed, v = 6.13 m s ≈ 6.1 m s−1 v2 iv a = r 6.13 2 a= 0.80 a = 47 m s–2 i

Time =

AS and A Level Physics

Original material © Cambridge University Press 2010

[1] [1] [1] [1] [1] [1] [1]

1

18 Marking scheme: Worksheet (A2)

F = ma = 0.090 × 47 F ≈ 4.2 N b The tension in the string. c The stone describes a circle, therefore the angle between the velocity and the acceleration (or centripetal force) must be 90°.

[1] [1] [1]

v

7

distance time 2πr 2π × 0.12 speed v = = 1.6 T v = 0.471 m s−1 ≈ 0.47 m s−1 mv 2 ii F = ma = r 0.300 × 0.4712 F= 0.12 frictional force ≈ 0.55 N b Frictional force = 0.7mg mv 2 0.7 mg = r v = 0.7 gr = 0.7 × 9.81 × 0.12

a

i

speed =

−1

8

[1]

a

[1] [1] [1] [1] [1] [1] [1]

[1]

−1

speed = 0.908 m s ≈ 0.91 m s

[1]

Kinetic energy at B = loss of gravitational potential energy from A to B 1 2 mv = mgh or v = 2 gh 2 v = 2 × 9.81 × 5.2 =10.1 m s–1 ≈ 10 m s−1

[1] [1]

2

v r 10.12 a= 16 a = 6.38 m s−2 ≈ 6.4 m s−2 ii Net force = ma R − mg = ma R = mg + ma = m(a + g) = 70(6.38 + 9.81) R ≈ 1.1 × 103 N

b i

a=


[1] [1] [1] [1] [1] [1]


2


9

a

R cos 20° = W = 840 × 9.8 840 × 9.8 R= = 8760 N cos 20°

b R sin 20° = centripetal force =

[1] [1]

mv 2 r

[1]

mv 2 R 840 × 32 2 r= 8760 sin 20° r = 287 m ≈ 290 m r=

[1]

[1] [1]

mv 2 r 0.120 × 4.0 2 = 2.4 N net force = 0.80 weight W of stone = mg = 0.120 × 9.81 = 1.18 N ≈ 1.2 N At the top: W + TB = 2.4 so TB = 2.4 − 1.2 = 1.2 N At the bottom: TA − W = 2.4 so TA = 2.4 + 1.2 = 3.6 N T 3 .6 = 3.0 ratio = A = 1 .2 TB

10 net force =



[2] [1] [1] [1] [1]

3


Gravitational field strength at a point, g, is the force experienced per unit mass at that point.

2

F= −

GMm r2

Therefore: G =

3

4

[1]

⎡ N m2 ⎤ Fr 2 → [N m2 kg–2] → ⎢ 2 ⎥ Mm ⎣ kg ⎦

[1]

F m F = mg = 80 × 1.6 F = 128 N ≈ 130 N (F is the ‘weight’ of the astronaut.)

g=

a

F= F=

GMm r2 6.67 × 10 −11 × 1.7 × 10 −27 × 1.7 × 10 −27

[1]

(5.0 ×10 )

−14 2

(

a b

c

g= −

[1]

)

[1] [1] [1] [1]

1 ). r2 Doubling the distance decreases the field strength by a factor of four. 2 GM (5 R ) ratio = 2 GM (59 R ) The field strength obeys an inverse square law with distance (g ∝

[1] [1] [1]

2

GM r2 6.67 × 10 −11 × 1.0 × 10 26

[1] [1]

g= − g=

7

[1]

GM r2

59 2 ⎛ 59 ⎞ ratio = 2 = ⎜ ⎟ 5 ⎝ 5 ⎠ ratio ≈ 140 6

[1] [1] [1]

F ≈ 7.7 × 10–38 N 6.67 × 10 −11 × 5.0 ×10 28 × 5.0 ×10 28 b F= 2 8.0 × 1012 F = 2.61 × 1021 N ≈ 2.6 × 1021 N 6.67 × 10 −11 × 1500 2 c F= 2.0 2 F = 6.00 × 10–6 N ≈ 6.00 × 10–6 N 5

[1]

(2.2 × 10 )

7 2

[1] (magnitude only)

[1]

g = 13.8 N kg–1 ≈ 14 N kg–1

[1]

GM r2 6.67 × 10 −11 × 5.0 × 10 23 GM r2 = = = 8.34 × 1012 m2 4.0 g

[1]

g= −

r=

8.34 × 1012 ≈ 2.9 × 106 m


[1]

[1]


1


8 a

F= − F=

GMm r2

[1]

⎞ 6.67 × 10 −11 × 1800 × 6.0 × 10 24 ⎛ 7.8 × 1010 ⎜r = = 3.9 × 1010 m ⎟⎟ 2 ⎜ 2 3.9 × 1010 ⎠ ⎝

(

)

[1]

F = 4.74 × 10–4 N ≈ 4.7 × 10–4 N GMm b F= − 2 r 6.67 × 10 −11 × 1800 × 6.4 × 10 23 F= 2 3.9 × 1010

[1]

F = 5.05 × 10–5 N ≈ 5.1 × 10–5 N F a= (F is the net force.) m

[1]

(

c

9 a

[1]

)

[1]

4.74 × 10−4 − 5.05 × 10−5 a= 1800

[1]

a ≈ 2.4 × 10–7 m s–2

[1]

(towards the centre of the Earth)

GMm r2 6.67 × 10 −11 × 5000 × 6.0 × 10 24

F= − F=

(6800 × 10 )

3 2

[1] (r = 6400 + 400 = 6800 km)

F = 4.33 × 104 N ≈ 4.3 × 104 N F 4.33 × 10 4 b a= = m 5000 a = 8.66 ≈ 8.7 m s–2 v2 c a= r 2 v = ar = 8.66 × 6800 × 103 v = 7.67 × 103 m s–1 ≈ 7.7 km s–1

[1]

[1] [1] [1] [1] [1]

The work done in bringing unit mass from infinity to the point b 0J 6.67 × 10 −11 × 6.0 × 10 24 ([1] mark only if minus sign missed) c Ep =− 6.4 × 10 6 = −6.25 × 106 J d 6.25 × 106 J

10 a

11 a

Gravitational force on planet = Centripetal force =

[1]

GMm r2

[1] [1] [1] [2] [1] [1] [1]

mv 2 r

[1] mv 2 GMm = r r2 GM r

Equating these two forces, we have:

[1]

GM r

[1]

Therefore: v2 =


or v =


2


6.67 × 10 −11 × 2.0 × 10 30 GM = r 1.5 × 1011 v = 2.98 × 104 m s–1 ≈ 30 km s–1

b v=

[1] [1]

12 The field strengths are the same at point P.

GM M

=

GM E

(R − x )2

[1]

R−x=x×

ME MM

[1]

R−x=x×

81

x

2

10x = R

so


so R − x = 9x R x= 10

[1] [1]


3


2

3

4

5

6

a The period of an oscillator is the time for one complete oscillation. b The frequency of an oscillator is the number of oscillations completed per unit time (or per second). The gradient of a displacement against time graph is equal to velocity. The magnitude of the velocity (speed) is a maximum at 0 s or 0.4 s or 0.8 s. b For s.h.m., acceleration ∝ – displacement. The magnitude of the acceleration is maximum when the displacement is equal to the amplitude of the motion. The magnitude of the acceleration is a maximum at 0.2 s or 0.6 s or 1.0 s.

a

13.2 12 T = 1.1 s 1 1 b f= = T 1.1 f = 0.909 ≈ 0.91 Hz

a

T=

[1] [1] [1] [1]

[1] [1] [1] [1]

[1] [1]

a Amplitude = 0.10 m b Period = 4.0 × 10–2 s 1 1 c f= = T 0.04 f = 25 Hz d ω = 2πf = 2π × 25 ω = 157 rad s–1 ≈ 160 rad s–1 e Maximum speed = ωA = 157 × 0.10 maximum speed = 15.7 m s−1 ≈ 16 m s−1

[1] [1] [1] [1] [1] [1] [1] [1]

⎛t⎞ Phase difference = 2π × ⎜ ⎟ ⎝T ⎠ where T is the period and t is the time lag between the motions of the two objects. ⎛t⎞ ⎛ 2.5 ⎞ phase difference = 2π × ⎜ ⎟ = 2π × ⎜ ⎟ T ⎝ ⎠ ⎝ 10 ⎠ π phase difference = ≈ 1.6 rad 2 ⎛t⎞ ⎛ 5 .0 ⎞ b Phase difference = 2π × ⎜ ⎟ = 2π × ⎜ ⎟ ⎝T ⎠ ⎝ 10 ⎠ phase difference = π ≈ 3.1 rad

a

a

A = 16 cm

[1] [1] [1] [1]

2π 2π = T 2.8 ω = 2.24 rads–1 ≈ 2.2 rads–1 c a = (2πf )2x (magnitude only) For maximum acceleration, the displacement x must be 16 cm. 2 1 ⎞ ⎛ −2 a = ⎜ 2π × ⎟ × 16 × 10 2.8 ⎠ ⎝ a = 0.806 m s–2 ≈ 0.81 m s–2 d Maximum speed = ωA = 2.24 × 0.16 maximum speed = 0.358 m s−1 ≈ 0.36 m s−1

b ω = 2πf =


[1]


[1] [1] [1]

[1] [1] [1] [1]

1


2π 2π = T 2.0 ω = 3.14 rad s–1 ≈ 3.1 rad s–1 b a = –(2πf )2x or a = –ω2x a = 3.142 × 3.0 × 10–2 a ≈ 0.30 m s–2 c x = A cos (2πft) = A cos (ωt) x = 3.0 × 10–2 cos (3.14 × 6.7) x ≈ –1.7 × 10–2 m

7 a

8 a

ω = 2πf =

[1] [1] [1] [1] [1] [1] [1] [1]

Gradient of x–t graph = velocity

[2] b Gradient of v–t graph = acceleration (for s.h.m. acceleration ∝ −displacement)

[2] c

Kinetic energy =

1 2 mv ∝ v2 2

[2] d Potential energy = total energy − kinetic energy

[2]



2


9 a

a = −(2πf )2x Therefore (2πf )2 = 6.4 × 105

[1] [1]

6.4 × 10 5 = 127 Hz ≈ 130 Hz 2π b F = ma Acceleration is maximum at maximum displacement, so magnitude of maximum force is given by: F = ma = 0.700 × (6.4 × 105 × 0.08) F = 3.58 × 104 N ≈ 3.6 × 104 N f=

According to Hooke’s law, F = –kx (The minus sign shows that the force is directed towards the equilibrium position.) From Newton’s second law: F = ma Equating, we have: ma = –kx ⎛k⎞ Hence: a = − ⎜ ⎟ x ⎝m⎠ b For s.h.m. we have a = –(2πf )2x k Hence (2πf )2 = m 1 k Therefore f = 2π m 1 1 = c f= = 2.5 Hz T 0 .4 1 k 2.5 = 2π 850 k = (2π × 2.5)2 × 850 ≈ 2.1 × 105 N m–1

10 a



[1]

[1] [1] [1] [1] [1]

[1] [1]

[1] [1] [1]

3


The atoms in a solid are arranged in a three-dimensional structure. There are strong attractive forces between the atoms. The atoms vibrate about their equilibrium positions. b The atoms in a liquid are more disordered than those in a solid. There are still attractive electrical forces between molecules but these are weaker than those between similar atoms in a solid. The atoms in a liquid are free to move around. c The atoms in a gas move around randomly. There are virtually no forces between the molecules (except during collisions) because they are much further apart than similar molecules in a liquid. The atoms of a gas move at high speeds (but no faster than those in a liquid at the same temperature).

2

The atoms move faster because their mean kinetic energy increases as the temperature is increased. The atoms still have a random motion.

3

a

4

[1] [1] [1] [1]

a

[1] [1] [1] [1] [1] [1] [1] [1]

The internal energy of a substance is the sum (of the random distribution) of the kinetic and potential energies of its particles (atoms or molecules). b There is an increase in the average kinetic energy of the aluminium atoms as they vibrate with larger amplitudes about their equilibrium positions. The potential energy remains the same because the mean separation between the atoms does not change significantly. Hence, the internal energy increases because there is an increase in the kinetic energy of the atoms. c As the metal melts, the mean separation between the atoms increases. Hence, the electrical potential energy of the atoms increases. There is no change in the kinetic energy of the atoms because the temperature remains the same. The internal energy of the metal increases because there is an increase in the electrical potential energy of the atoms.

[1] [1] [1] [1] [1] [1] [1] [1]

Change in thermal energy = mass × specific heat capacity × change in temperature

[1]

5 The specific heat capacity refers to the energy required to change the temperature of a substance. Specific latent heat of fusion is the energy required to melt a substance; there is no change in temperature as the substance melts.

[1] [1]

6

E = mc∆θ E = 6.0 × 105 × 4200 × (24 – 21) E = 7.56 × 109 J ≈ 7.6 × 109 J

[1] [1] [1]

7

E = mc∆θ E = 300 × 10–3 × 490 × (20 – 300) E = –4.1 × 104 J (The minus sign implies energy is released by the cooling metal.)

[1] [1] [1]

8

E = mLf = 200 × 10−3 ×3.4 × 105

[1]

= 6.8 × 104 J


[1]


1


9

a

i ii iii b i ii iii

T = 273 + 0 = 273 K T = 273 + 80 = 353 K T = 273 – 120 = 153 K θ = 400 – 273 = 127 °C θ = 272 – 273 = –1 °C θ = 3 – 273 = –270 °C

[1] [1] [1] [1] [1] [1]

The thermal energy E supplied and the specific heat capacity c remain constant. The mass m is larger by a factor of 3. Since E = mc∆θ, we have: E 1 ∆θ = ; ∆θ ∝ m mc 15 = 5.0 °C Therefore ∆θ = 3 b The thermal energy E supplied is halved but the specific heat capacity c and the mass m remain constant. Since E = mc∆θ, we have: E ∆θ = ; ∆θ ∝ E mc 15 Therefore ∆θ = = 7.5 °C 2

10 a

[1]

[1] [1] [1] [1] [1]

Melting point = 600 °C (There is no change in temperature during change of state.) b The lead is being heated at a steady rate and therefore the temperature also increases at a steady rate. c The energy supplied to the lead is used to break the atomic bonds and increase the separation between the atoms of lead (and hence their potential energy increases). d E = mc∆θ E = 200 × 10–3 × 130 × (600 – 0) E = 1.56 × 104 J ≈ 1.6 × 104 J e In a time of 300 s, 1.56 × 104 J of energy is supplied to the lead. Rate of heating = power 1.56 × 10 4 power = 300 power = 52 W f Energy supplied = 52 × 100 = 5200 J

[1]

11 a

Lf =

[1] [1] [1] [1] [1]

[1] [1] [1]

∆E 5200 = ∆m 0.2

[1]

= 26 000 J kg−1

[1]

12 The energy supplied per second is equal to the power of the heater. In a time of 1 s, water of mass 0.015 kg has its temperature changed from 15 °C to 42 °C. E = mc∆θ (where E is the energy supplied in 1 s) E = 0.015 × 4200 × (42 – 15) E = 1.7 × 103 J The power of the heater is therefore 1.7 kW. m (You may use P = ( )c∆θ ) t

[1] [1] [1] [1] [1]

13 The gas does work against atmospheric pressure. Energy to do this work is taken from the internal energy of the gas.



[1] [1]

2


14 Heat ‘lost’ by hot water = heat ‘gained’ by cold water. 0.3 × c × (90 – θ) = 0.2 × c × (θ – 10) where c is the specific heat capacity of the water and θ is the final temperature. The actual value of c is not required, since it cancels on both sides of the equation. Hence: 0.3 × (90 – θ) = 0.2 × (θ – 10) 27 – 0.3θ = 0.2θ – 2.0 0.5θ = 29 so θ = 58 °C

[1] [1]

15 Heat ‘lost’ by metal = heat ‘gained’ by cold water 0.075 × 500 × (θ – 48) = 0.2 × 4200 × (48 – 18) (θ is the initial temperature of the metal.) 0.2 × 4200 × 30 θ – 48 = 0.075 × 500 θ – 48 = 672 θ = 720 °C

[1] [1]


[1] [1] [1]

[1] [1] [1]


3


2

3

number of atoms = number of moles × NA number of atoms = 1.0 × 6.02 × 1023 ≈ 6.0 × 1023 b Number of molecules = 3.6 × 6.02 × 1023 ≈ 2.2 × 1024 c Number of atoms = 0.26 × 6.02 × 1023 ≈ 1.6 × 1023

a

There are 6.02 × 1023 atoms in 4.0 g of helium. 0.004 = 6.645 × 10–27 kg ≈ 6.6 × 10–27 kg mass of atom = 23 6.02 × 10

There are 6.02 × 1023 atoms in 0.238 kg of uranium. 0.238 = 3.95 × 10−25 kg ≈ 4.0 × 10−25 kg mass of atom = 23 6.02 × 10 mass of uranium b i number of moles = molar mass of uranium a

number of moles =

0.12 = 5.04 × 10–4 ≈ 5.0 × 10–4 238

ii number of atoms = number of moles × NA number of atoms = 5.04 × 10–4 × 6.02 × 1023 = 3.06 × 1020 ≈ 3.1 × 1020 4

The absolute zero of temperature is –273.15 °C or 0 K. This is the lowest temperature any substance can have. At absolute zero of temperature, the substance has minimum internal energy.

5

a

6

Pressure × volume = number of moles × universal gas constant × thermodynamic temperature b PV = nRT 1.0 × 8.31 × 293 nRT P= = V 0.020 5 P = 1.22 × 10 Pa ≈ 1.2 × 105 Pa (120 kPa) a

PV = nRT Comparing this equation with y = mx, we have: y = PV, x = T, gradient, m = nR A graph of PV against T is a straight line through the origin.

Correct graph gradient n= R

b PV = nRT At a constant temperature, the product PV is a constant. Hence a graph of PV against P is a straight horizontal line.


[1] [1] [1] [1]

[1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1]

[1] [1]

[1] [1] [1]


1


PV = nRT 4 .0 n= = 0.138 moles 29 nRT 0.138 × 8.31× (273 + 34) P= = V 0.030 4 P = 1.17 × 10 Pa ≈ 1.2 × 104 Pa (12 kPa) P b is constant when the volume of the gas is constant. T The pressure is doubled, hence the absolute temperature of the gas is also doubled. Therefore: temperature = 2 × (273 + 34) = 614 K temperature in °C = 614 – 273 = 341 °C ≈ 340 °C

7 a

PV RT 180 × 10 3 × 2.0 × 10 −2 300 × 10 3 × 2.0 × 10 −2 + n= 8.31 × (273 − 13) 8.31 × ( 273 − 13) n = 4.44 moles ≈ 4.4 moles b Total volume, V = 4.0 × 10–2 m3, T = 273 – 13 = 260 K nRT P= V 4.44 × 8.31 × 260 P= 4.0 × 10 − 2 P ≈ 2.4 × 105 Pa (240 kPa)

8 a

n=

400 F = A 1.6 × 10 −3 P = 2.5 × 105 Pa PV b n= RT 2.5 × 10 5 × 2.4 × 10 −4 n= 8.31 × (273 + 5.0) n = 2.6 × 10–2 moles c i mass = number of moles × molar mass mass = 2.6 × 10–2 × 29 = 0.754 g ≈ 0.75 g mass ii density = volume 0.754 × 10 −3 density = 2.4 × 10 − 4 density = 3.14 kg m–3 ≈ 3.1 kg m–3

9 a

P=


[1] [1] [1] [1] [1] [1] [1] [1] [1]

[1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1]

[1] [1]


2


10 Mean kinetic energy of atom ∝ absolute temperature 1 mv 2 ∝ T or v2 ∝ 2T 2 m Since the mass m of the atom is constant, we have: v ∝ T The temperature of 0 °C in kelvin is T = 273 K 10 000 The absolute temperature increases by a factor of (= 36.6) 273 10 000 = 6.05 Hence the speed will increase by a factor of 273 The speed of the atoms at 10 000 K = 1.3 × 6.05 ≈ 7.9 km s–1

The particles have a range of speeds and travel in different directions. 3 3 b i Mean kinetic energy = kT = × 1.38 × 10 − 23 × 5400 2 2 = 1.118 × 10–19 J ≈ 1.1 × 10–19 J 1 2 3 ii mv = kT 2 2

11 a

3kT 3 × 1.38 × 10 −23 × 5400 = m 1.7 × 10 − 27 4 speed = 1.147 × 10 m s–1 ≈ 11 km s–1 v=

[1] [1] [1] [1] [1] [1] [1] [1]

[1]

Mean kinetic energy =

2 × 5.65 × 10 −21 7.31 × 10 − 26 speed = 393 m s–1 ≈ 390 m s–1 v=

c

[1]

[1]

3 3 kT = × 1.38 × 10 − 23 × 273 2 2 = 5.65 × 10–21 J ≈ 5.7 × 10–21 J b There are 6.02 × 1023 molecules of carbon dioxide. 0.044 = 7.31 × 10 − 26 kg mass of molecule = 6.02 × 10 23 1 mv 2 = 5.65 × 10 − 21 J 2

12 a

[1]

[1] [1] [1] [1] [1]

[1] [1]

3 3 Total kinetic energy of one mole of gas = kT × NA = RT (Note: R = k × NA) 2 2 For an ideal gas, the change in internal energy is entirely kinetic energy. 3 3 Change in internal energy = R × (373 − 273) = × 8.31 × 100 2 2 change in internal energy = 1.2465 kJ ≈ 1.2 kJ

[1]

[1] [1]

The molecule has 3 degrees of freedom for translational motion and 2 degrees of freedom for rotation – making a total of 5. [1] 1 5 Therefore, mean energy = 5 × kT = kT [1] 2 2 ii The molecule has 3 degrees of freedom for translational motion and 3 degrees of freedom for rotation – making a total of 6. [1] 1 Therefore, mean energy = 6 × kT = 3kT [1] 2 b Internal energy = 3kT × NA = 3RT (Note: R = k × NA) [1] internal energy per unit kelvin = 3R [1] –1 = 3 × 8.31 ≈ 25 J K [1]

13 a

i



3


2

Electric field strength is the force per unit charge at that point. The potential at a point is the work that must be done to bring unit charge from infinity to that point. a

E=

[1] [1]

V V ,d= d E

[1]

5000 = 1.25 × 10–2 m ≈ 1.3 cm 400 000 b F = EQ = 400 000 × 1.6 × 10–19 N F = 6.4 × 10–14 N

d=

3

E=

Q 4 πε 0 r

2

so k =

[1] [1] [1]

1 1 = 4 πε0 4π × 8.85 × 10 −12

k = 8.99 × 109 m F–1 ≈ 9.0 × 109 m F–1 4

[1]

The force between the charges obeys an inverse square law with distance; that is, F ∝

1 r2

Point B: The distance is the same. The force between the charges is F. Point C: The distance is doubled, so the force decreases by a factor of 4. F The force between the charges is . 4 Point D: The distance is trebled, so the force decreases by a factor of 32 = 9. F The force between the charges is . 9 Point E: The distance between the charges is 8 R. The force between the charges decreases by a factor of ( 8 ) 2 = 8 F . The force between the charges is 8 5

a

E=

Q

[1] [1] [1] [1] [1] [1] [1] [1] [1]

4 πε 0 r 2

2.5 × 10 −6 4 π × 8.85 × 10 −12 × 0.15 2 E = 9.99 × 105 V m–1 ≈ 1.0 × 106 V m–1 b The distance from the centre of the dome increases by a factor of 3. The electric field strength decreases by a factor of 32 = 9. 1.0 × 10 6 = 1.1 × 104 V m–1 Therefore E = 9

E=


[1]


[1] [1] [1]

[1]

1


6

a

i

E=

Q

[1]

4 πε 0 r 2

20 × 10 −6 80 (r = = 40 cm) −12 2 2 4 π × 8.85 × 10 × 0.40 E = 1.124 × 106 V m–1 ≈ 1.1 × 106 V m–1 40 × 10 −6 ii E = 4 π × 8.85 × 10 −12 × 0.40 2 E = 2.248 × 106 V m–1 ≈ 2.2 × 106 V m–1 (The electric field doubles because the charge is doubled, E ∝ Q.) E=

7

8

9

[1] [1] [1] [1]

b Net field strength, E = 2.2 × 106 – 1.1 × 106 = 1.1 × 106 V m–1 The direction of the electric field at X is to the left.

[1] [1]

Q = V × 4πεor = 20 000 × 4 × π × 8.85 × 10−12 × 0.15 Q = 3.3 × 10–7 C 9.0 ×10 9 × 3.3 ×10 −7 kQ b E = 2 = r 0.15 2 5 = 1.32 × 10 V m–1 ≈ 1.3 × 105 V m–1 c F = eV = 1.6 × 10−19 × 1.32 × 105 F = 2.11 × 10−14 N ≈ 2.1 × 10−14 N

[1] [1]

a

a

V=

9.0 × 109 × − 2000 × 10−9 kQ Q = = 4 πε 0 r r 5 × 10− 2

([1] mark only if minus sign omitted)

[1] [1] [1] [1]

[2]

V = 3.6 × 105 J C–1 =360 kV

[1]

Similarities • Both produce radial fields.

[1]

1 . r2 • The field strengths are defined as force per unit (positive) charge or mass. • Both produce action at a distance. Differences • Electrical forces can be either attractive or repulsive, whereas gravitational forces are always attractive. • Gravitational forces act between masses, whereas electrical forces act between charges. •

Both obey an inverse square law with distance; that is, F ∝



[1] [1] [1]

[1] [1]

2


10 The electric field strength due to the charge +Q is equal in magnitude but opposite in direction to the electric field strength due to the charge +3Q. [1] Therefore: Q 3Q = (where R is the distance between the charges = 10 cm) [1] 2 4 πε 0 x 4 πε 0 ( R − x) 2 3 1 = 2 x ( R − x) 2 R−x so = 3 x

x(1 +

3)=R

[1] [1] so

x=

R

1+ 3

= 0.37 R

[1]

x = 0.37 × 10 = 3.7 cm 11 ratio =

ratio =

e 2 4πε 0 r 2 Gm 2 r 2

[1]

(where m = mass of proton and r = separation)

e2 4 πε 0 Gm 2

[1]

The r2 cancel and so this ratio is independent of the separation. (1.6 × 10−19 ) 2 ratio = 4 π × 8.85 × 10 −12 × 6.67 × 10−11 × (1.7 × 10 −27 ) 2

ratio ≈ 1.2 × 1036 F=

[1] [1] [1] [1] [1] [1]

2 × 2.3 × 10 −13 = 2.7 × 10–14 1.7 × 10 − 27

[1]

v = 2.7 × 10−14 = 1.6 × 107 m s–1


[1]

[1]

= 1.44 × 106 V c W = VQ = 1.44 × 106 × 1.6 × 10–19 = 2.3 × 10–13 J 2W 1 2 d mv = W ⇒ v2 = 2 m v2 =

[1]

[1]

Q1Q2 1.6 × 10 −19 × 1.6 × 10 −19 9 = 9 × 10 × (10 −15 ) 2 4 πε 0 r 2 = 230 N Q 1.6 × 10 −19 b V= = 9 × 109 × 10 −15 4 πε 0 r

12 a

[2]

[1]


3


a

Q = VC = 9.0 × 30 × 10–6 Q = 2.7 × 10–4 C (270 µC)

[1] [1]

Q 2.7 × 10 −4 = e 1.6 × 10 −19 15 15 number = 1.69 × 10 ≈ 1.7 × 10 a i The charge is directly proportional to the voltage across the capacitor. Hence doubling the voltage will double the charge. charge = 2 × 150 = 300 nC ii Since Q ∝ V for a given capacitor, increasing the voltage by a factor of three will increase the charge by the same factor. charge = 3 × 150 = 450 nC 150 × 10 −9 Q b C= = V 3.0 –8 C = 5.0 × 10 F

b number of excess electrons =

2

3

4

1 2 1 V C = × 9.02 × 1000 × 10–6 2 2 E = 4.05 × 10–2 J ≈ 4.1 × 10–2 J b For a given capacitor, energy stored ∝ voltage2. energy = 22 × 4.05 × 10–2 ≈ 0.16 J

a

E=

1

[1]

[1]

1 ≈ 83 µF 0.012 1 1 1 = + + C1 C2 C3

[1] [1]

C total 1 1 1 1 = = 0.13 µF−1 + + 10 50 100 C total

[1]

1 ≈ 33 µF 0.03 Total capacitance of the two capacitors in series is 83 µF (from b). Ctotal = 83 + 50 = 133 µF ≈ 130 µF

Ctotal =

Ctotal = C1 + C2 Ctotal = 100 + 500 = 600 µF b The potential difference across parallel components is the same and equal to 1.5 V. a


[1]

[1]

1 ≈ 7.7 µF 0.13 d Total capacitance of the two capacitors in parallel = 50 + 50 = 100 µF. 1 1 1 = = 0.03 µF−1 + 50 100 C total

5

[1] [1]

[1] [1]

Ctotal =

e

[1] [1]

[1] [1] [1]

Ctotal = C1 + C2 Ctotal = 20 + 40 = 60 nF 1 1 1 b = + C total C1 C2 1 1 1 = + = 0.012 µF−1 100 500 C total

c

[1]

[1]

a

Ctotal =

[1]


[1] [1] [1] [1] [1] [1] [1] [1] [1]

1


Q = VC = 1.5 × 600 × 10–6 Q = 9.0 × 10–4 C (900 µC) 1 1 d E = QV = × 9.0 × 10–4 × 1.5 2 2 E = 6.75 × 10–4 J ≈ 6.8 × 10–4 J

[1] [1]

c

[1] [1]

1 2 1 V C = × 322 × 10 000 × 10–6 2 2 E = 5.12 J ≈ 5.1 J 5.12 E b P= = t 0.300 P ≈ 17 W

6 a

E=

[1] [1] [1] [1]

Q = VC = 12 × 1000 × 10–6 Q = 1.2 × 10–2 C (12 mC) b i Ctotal = C1 + C2 Ctotal = 1000 + 500 = 1500 µF Q ii V = (The charge Q is conserved and C is the total capacitance.) C 1.2 × 10 −2 V= = 8.0 V 1500 × 10 −6

7 a

8 a

∆Q = I∆t =

I 225 ×10 −3 = = 4.5 × 10–3 C 50 f

Q 4.5 × 10 −3 = 9 .0 V = 5.0 × 10–4 F = 500 µF i The capacitors are in parallel, so the total capacitance = 2C and charge stored = 2Q; current =2I = 2 × 225 = 450 mA 1 1 ii The capacitors are in series, so the total capacitance = C and charge stored = Q; 2 2 1 1 × 225 = 113 mA current = I = 2 2

Q = CV = 200 × 10–6 × 200 = 0.040 C 1 1 b E = CV 2 = × 200 × 10–6 × 2002 = 4.0 J 2 2 c The two capacitors now make a pair of capacitors in parallel of total capacitance = 100 µF + 200 µF = 300 µF Q 0.040 The charge is conserved, therefore V = = C 300 × 10 −6 = 133 V 1 1 d E = CV 2 = × 300 × 10–6 × 1332 2 2 = 2.67 J e The energy is lost as the connected wires are heated as the current es through them.

9 a


[1] [1]

[1]

b C= c

[1] [1] [1] [1]


[1] [1] [1]

[1] [1] [1] [1] [1] [1] [1] [1] [1]

2


10 The capacitors are in parallel, so the total capacitance = 3C. The total charge Q remains constant. Q2 . The energy stored by a capacitor is given by E = 2C Q2 Q2 Einitial = and Efinal = 2C 2(3C )

Q 2 2(3C ) Efinal 1 = = 2 Einitial 3 Q 2C 1 2 Fraction of energy ‘lost’ as heat in resistor = 1 – = . 3 3 The resistance governs how long it takes for the capacitor to discharge. The final voltage across each capacitor is independent of the resistance. Hence, the energy lost as heat is independent of the actual resistance of the resistor. Fraction of energy stored =



[1] [1] [1] [1] [1] [1]

[1]

3


2

3

4

5

The direction of the magnetic field should be clockwise (and not anticlockwise). The separation between adjacent circular field lines should increase further away from the wire (see below).

a The conductor is pushed to the left. b The conductor is pushed to the left. c The conductor is pushed out of the plane of the paper. F B= Il N [B] = = N A−1 m−1 Am F = BIl F = 0.12 × 3.5 × 0.01 (length = 1.0 cm) F = 4.2 × 10–3 N

F = BIl sin θ F = 0.050 × 3.0 × 0.04 × sin 90° F = 6.0 × 10–3 N b F = 0.050 × 3.0 × 0.04 × sin 30° F = 3.0 × 10–3 N c F = 0.050 × 3.0 × 0.04 × sin 65° F = 5.44 × 10–3 N ≈ 5.4 × 10–3 N

[1] [1]

[1] [1] [1] [1]

[1] [1] [1] [1]

a

[1] [1] [1] [1] [1] [1]

6

Force experienced by PQ = force experienced by RS (but in opposite direction). No force experienced by QR and PS (since current is parallel to the field). torque = one of the forces × perpendicular distance between forces = (BIL)x torque = BI(Lx), Lx = area of loop = A torque = BIA ∝ A The torque is directly proportional to the area of the loop.

[1] [1] [1] [1] [1]

7

a

Current is at right angles to magnetic field. Left-hand rule produces force on AB towards the right. Wire leaves mercury and breaks /current stops/force stops. Weight causes AB to fall back/return and make again. b i Moment = Fd 3.5 × 10−5 = 1.0 × 10−3 N F= 0.035

[1] [1] [1] [1] [1]



[1]

1


F Bl 1.0 ×10 −3 I= = 2.38 A ≈ 2.4 A (6.0 ×10 −3 × 0.07)

ii F = BIl ⇒ I =

8

9

[1] [1]

F = BIl × number of turns F = 0.19 × 2.8 × 0.07 × 25 = 0.93 N b Torque= Fd = 0.93 × 0.03 torque = 0.028 N m c The longest side always stays at 90º to the magnetic field as the coil turns so the force is constant. The perpendicular distance between the forces changes as the coil turns so the torque (moment) changes.

a

F = mg = (103.14 – 102.00) × 10–3 × 9.81 so F = 1.12 × 10–2 N ≈ 1.1 × 10–2 N F b B= Il 1.12 × 10 −2 B= 8.2 × 0.05 B = 2.73 × 10−2 T (27 mT)

a



[1] [1] [1] [1] [1] [1] [1] [1]

[1] [1]

2


F = EQ = 5.0 × 105 × 3.2 × 10−19 F = 1.6 × 10−13 N

2

a

[1] [1]

600 V = d 3.0 × 10 − 2 E = 2.0 × 104 V m−1 The field acts towards the negative plate. b The electric field is uniform between the plates (except at the ‘edges’). The electric field is at right angles to the plate. c i Since the droplet is stationary, the electric force on the droplet must be equal and opposite to its weight. The electric force must act upwards, so the charge on the droplet must be negative. E=

ii E =

[1] [1] [1] [1] [1]

[1]

[1]

F Q

6.4 × 10 −15 F = E 2.0 × 10 4 Q = 3.2 × 10−19 C

Q=

3

F = BQv F = 0.18 × 1.6 × 10–19 × 4.0 × 106 F = 1.15 × 10–13 N ≈ 1.2 × 10−13 N

4

a

[1] [1] [1] [1] [1]

F = BQv F = 0.004 × 1.6 × 10–19 × 8.0 × 106 F = 5.12 × 10−15 N ≈ 5.1 × 10−15 N

[1] [1] [1]

−15

5.12 ×10 F = m 9.11×10−31 a = 5.63 × 1015 m s−2 ≈ 5.6 × 1015 m s−2 From circular motion, the centripetal acceleration a is given by: v2 a= r (8.0 × 10 6 ) 2 v2 r= = a 5.63 × 1015 r = 1.14 × 10−2 m ≈ 1.1 × 10−2 m (1.1 cm)

b a=

c



[1] [1]

[1] [1]

1


5

a

Both arrows at A and B are towards the centre of the circle. The force on the electron is at 90° to the velocity. Hence the path described by the electron is a circle. c The magnetic force provides the centripetal force. mv 2 Therefore: BQv = r BQr mv BQ = or v = r m −3 2.0 × 10 × 1.6 × 10 −19 × 5.0 × 10 −2 v= 9.1 × 10 −31 v = 1.76 × 107 m s–1 ≈ 1.8 × 107 m s−1 BQr d v= , so the speed v is directly proportional to the radius r. m 1.76 × 10 7 = 8.8 × 106 m s−1 Radius is halved, so v = 2

[1]

b

6

a

Ek = 15 × 103 × 1.6 × 10−19 = 2.4 × 10−15 J (1 eV = 1.6 × 10−19 J) 1 mv 2 = 2.4 × 10−15 2

2 × 2.4 × 10 −15 1.7 × 10 − 27 v = 1.68 × 106 m s−1 ≈ 1.7 × 106 m s−1 mv 2 b F = ma = r 1.7 × 10 −27 × (1.68 × 10 6 ) 2 F= 0.05 −14 F = 9.60 × 10 N ≈ 9.6 × 10−14 N c F = BQv 9.60 × 10 −14 F B= = Qv 1.6 × 10 −19 × 1.68 × 10 6 B ≈ 0.36 T distance d speed = time circumference 2π × 0.05 = time = speed 1.68 × 10 6 time = 1.87 × 10−7 s ≈ 1.9 × 10−7 s

v=


[1] [1] [1] [1] [1] [1] [1]

[1] [1]

[1] [1] [1]

[1] [1] [1] [1] [1]

[1] [1]


2


7

a

In order for the positively charged ions to emerge from the slit, the net force perpendicular to the velocity must be zero.

electrical force on ion = magnetic force on ion EQ = BQv The charge Q cancels. E = Bv V The electric field strength is E = . Therefore, the magnetic flux density is: d V d (5.0 × 10 3 ) / 0.024 E B= = = v v 6.0 × 10 6 B = 3.47 × 10–2 T ≈ 35 mT BQr mv b v= so r = m BQ ( m1 − m 2 ) v ∆r = BQ 8

The centripetal force is provided by the magnetic force. mv 2 Therefore: Bev = r mv Ber Be = or v = r m circumference 2πr T= = speed Ber m 2 πm The radius r of the orbit cancels. Hence: T = Be The time T is independent of both the radius of the orbit r and the speed v. b The faster electron travels in a circle of larger radius.

a



[1]

[1] [1] [1]

[1] [1] [1] [1] [1] [1] [1] [1]

[1] [1]

3


a

i Magnetic flux, Φ = BA ii Flux linkage = NΦ = NBA b Flux linkage = N(B cos θ)A = NBA cos θ (The component of B normal to the plane of the coil is B cos θ.)

[1] [1] [1]

2

Area A of coil = x2 flux linkage = NBA = NBx2

[1] [1]

3

a

4

a

There is no change to the magnetic flux linking the coil, hence according to Faraday’s law, there is no induced e.m.f. [1] b The magnetic flux density B increases as the magnet moves towards the coil. [1] There is an increase in the magnetic flux linking the coil, hence an e.m.f. is induced across the ends of the coil. [1] flux linkage = NBA so B =

flux linkage NA

[1]

1.4 × 10 −4 70 × 4.0 × 10 − 4 B = 5.0 × 10−3 T b flux linkage = NBA cos θ flux linkage = 70 × 0.050 × 4.0 × 10−4 × cos 60° flux linkage = 7.0 × 10−4 Wb

[1] [1] [1] [1]

magnetic flux, Φ = BA = 40 × 10–3 × (0.03 × 0.03) Φ = 3.6 × 10−5 Wb ii flux linkage = NΦ = 200 × 3.6 × 10−5 flux linkage = 7.2 × 10−3 Wb b Final flux linkage = 0, initial flux linkage = 7.2 × 10−3 Wb. Hence, change in magnetic flux linkage is 7.2 × 10−3 Wb.

[1] [1] [1] [1] [1] [1]

Initial magnetic flux = BA = 0.15 × (π × [8.0 × 10−3]2) initial magnetic flux = 3.02 × 10−5 Wb final magnetic flux = 0 average magnitude of induced e.m.f. = rate of change of magnetic flux linkage ∆Φ 0 − 3.02 × 10 −5 , so E = 1200 × E= N 0.020 ∆t E = 1.81 V ≈ 1.8 V (magnitude only) e.m.f. 1.81 = b average current = resistance 6.3 I = 0.287 A ≈ 0.29 A

[1] [1] [1]

B=

5

a

6

a

[1]

i



[1] [1] [1] [1]

1


7

8

9

a

distance = speed × time = 2.0 × 1.0 = 2.0 m

[1]

b Area swept = length × distance travelled = 0.10 × 2.0 = 0.20 m2 c Change in magnetic flux = area swept × magnetic flux density change in magnetic flux = 0.20 × 0.050 = 1.0 × 10−2 Wb d Magnitude of e.m.f. = rate of change of magnetic flux linkage ∆ ( NΦ ) E= (N = 1) ∆t 1.0 × 10 −2 = 1.0 × 10−2 V (1 Wb s−1 = 1 V) E= 1 .0 e E = Blv = 0.050 × 2.0 × 0.10 = 1.0 × 10−2 V

[1] [1] [1] [1]

Initial magnetic flux = BA = 0.060 × π × (1.2 × 10−2)2 initial magnetic flux = 2.72 × 10−5 Wb final magnetic flux = –2.72 × 10−5 Wb (since the field is reversed) average magnitude of induced e.m.f. = rate of change of magnetic flux linkage − 2.72 ×10 −5 − 2.72 ×10 −5 ∆Φ , so E = 2000 × E= N 0.030 ∆t E = 3.62 V ≈ 3.6 V (magnitude only)

[1] [1] [1]

There is a current in the primary coil when the switch is closed. This current creates a magnetic flux in the primary coil. Due to the soft iron ring, the magnetic flux created by the primary coil also links the secondary coil. With the switch closed, there is no change in the magnetic flux linkage at the secondary and hence the lamp is not lit. When the switch is opened, the magnetic flux decreases to zero in a short period. The rapid change in magnetic flux at the secondary coil creates an e.m.f. and the lamp illuminates for a short period. Eventually there is no magnetic flux at either the primary or the secondary coil and hence there is no e.m.f. induced – the lamp stays off. b Change the cell to an a.c. supply (or keep switching on and off very fast).

[1] [1]

[1] [1]

a



[1]

[1]

[1] [1] [1]

2


10

distance = speed × time = vt area swept = length × distance travelled = Lvt change in magnetic flux = area swept × magnetic flux density change in magnetic flux = (Lvt) × B = BLvt magnitude of e.m.f. = rate of change of magnetic flux linkage BLvt E= = BLv t E = BLv = 40 × 10−6 × 0.20 × 0.30 E = 2.4 × 10−6 V (2.4 µV) The magnitude of the induced e.m.f. in a circuit is directly proportional to the rate of change of magnetic flux linkage. b Lenz’s law expresses the principle of conservation of energy. c i magnetic flux = magnetic flux density × cross-sectional area of coil or Φ = BA ii Flux linkage = NBA Coil X: flux linkage = NBA = 200 × 0.10 × (π × 0.022) ≈ 2.5 × 10−2 Wb Coil Y: flux linkage = NBA = 4000 × 0.01 × (π × 0.032) ≈ 1.1 × 10−1 Wb The coil Y has greater flux linkage.

[1] [1] [1] [1] [1] [1] [1] [1]

11 a



[1] [1] [1] [1] [1] [1] [1]

3


a

T=

1 f

[1]

0.1 s b i t = 0, 0.05 and 0.15 1 ii t = of a period 4 t = 0.025 s iii t = 0.075 s 1 iv t = = sin (20πt) 2 t = 0.0125 s

[1] [1] [1] [1] [1]

[1] [1]

2

a Corresponding parts of each wave occur at the same time. b Power P always positive. Two peaks for P in time taken for one peak of I and V. Sinusoidal shape above axis.

3

a

i

P 690 = V 230 Irms = 3.0 A

I=

[1] [1] [1]

ii Ipeak = 3 2 = 4.2 A iii Vpeak = 230 2 = 325 V b Correct shape all above axis. Two cycles shown (i.e. one cycle of current waveform). Peak and average power marked.


[1] [1] [1] [1]


[1] [1] [1] [1]

1


4

5

6

7

8

a.c. Current is positive and negative OR current flows one way, then the opposite way. V2 b Voltage switches between +2 and −2 V. Power is , so it has the same value for both R positive and negative values of the voltage. c i 2V ii 2 V ⎛ 2 .5 ⎞ ⎛ 6 ⎞ Average power = ⎜ ⎟× ⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠ = 7.5 W a

a

Period = 80 ms = 0.080 s 1 = 12. 5 Hz f= T ii Peak voltage = 4.5 V 4.5 = 3.2 V r.m.s. voltage = 2 b V0 = 4.5 V ω = 2πf = 2π × 12.5 = 78.5 s−1 ≈ 79 s−1

a

[1]

[1] [1] [1] [1]

115 × 500 = 1000 turns 230

115 Vs = 0.023 A r.m.s. = R 5000 500 Ip = 0.023 × = 0.0125 A ≈ 0.013 A r.m.s. 1000

[1] [1] [1] [1] [1] [1] [1] [1] [1] [1]

b Is =

[1]

c

[1]

d Peak voltage = 115 2 = 162 V so the cables will break down. 9

[1]

[1]

I=

Number of turns on the primary =

[1] [1] [1]

[1]

i

P 24 = 6 V I = 4.0 A 1150 b Vp = 6 × 30 Vp = 230 V 30 × 4 24 or Ip = c Ip = 230 1150 Ip = 0.10 A d Maximum p.d. = 6 2 maximum p.d. = 8.5 V e Heat is still produced inside the wires even if it cannot be conducted to the outside. The wires may melt if they cannot lose the heat.

a

[1] [1]

a

I=

[1] [1]

P 1000 = = 10 A r.m.s. V 100

[1]

P = I 2R = 102 × 5 = 500 W b At high voltages the current is less for the same power. Power lost in cable = I 2R so power lost is less.



[1] [1] [1]

2


A ‘packet’, or quantum, of electromagnetic energy.

[1]

2

The energy of a photon is proportional to the frequency of the radiation. Hence a γ-ray photon has greater energy than a photon of visible light (and therefore is more harmful).

[1]

3

4

Electromagnetic radiation travels through space as waves and, as such, shows diffraction and interference effects. b Electromagnetic radiation interacts with matter as ‘particles’. The photoelectric effect provides strong evidence for the particle-like (photon) behaviour of electromagnetic radiation.

[1]

a

a

3.0 × 108 λ 6.4 × 10 −7 14 f = 4.69 × 10 Hz ≈ 4.7 × 1014 Hz

c = fλ so f =

c

=

[1]

[1]

[1] [1]

b E = hf E = 6.63 × 10−34 × 4.69 × 1014 E = 3.1 × 10−19 J

[1] [1] [1]

5

For an electron to escape from the surface of the metal, it must absorb energy from the photon that is greater than the work function. [1] The work function is the minimum energy required by the electron to escape from the surface of the metal. [1] The photon of visible light has energy less than the work function of the metal, whereas the photon of ultraviolet radiation has energy greater than the work function. [1]

6

a

7

The electron loses energy. This energy appears as a photon of electromagnetic radiation. b Energy of photon = E1 − E2 Therefore: E − E2 hf = E1 − E2 or f = 1 h c The change in energy ∆E is greater. Hence the frequency of the radiation is greater (f ∝ ∆E). The spectral line will be the right side of the line shown on the spectrum diagram.

[1] [1] [1]

There are six spectral lines.

[1]

Correct transitions shown on the energy level diagram.

[1]



[1] [1] [1]

1


8

∆E = hf =

hc

[1]

λ

6.63 ×10 −34 × 3.0 ×10 8 670 ×10 −9 ∆E = 2.97 × 10−19 J ≈ 3.0 × 10−19 J ∆E =

9

10 11 12

[1] [1]

a Continuous spectrum b Emission spectrum c Absorption spectrum

[1] [1] [1]

Electrons travel through space as waves. Evidence for this is provided by the diffraction of electrons by matter (e.g. graphite).

[1]

The electronvolt is the energy gained by an electron travelling through a potential difference of one volt.

[1]

The kinetic energy Ee of the electron is: Ee = VQ = 6.0 × 1.6 × 10–19 Ee = 9.6 × 10–19 J The energy EUV of the ultraviolet photon is: hc EUV = hf =

[1] [1] [1]

λ

6.63 × 10 −34 × 3.0 × 108 2.5 × 10 −7 EUV = 7.96 × 10−19 J ≈ 8.0 × 10−19 J The energy of the ultraviolet photon is less than the kinetic energy of the electron. (The student is correct.)

EUV =

The threshold frequency is the minimum frequency of electromagnetic radiation that just removes electrons from the surface of the metal. b At the threshold frequency, the energy of the photon is equal to the work function φ of the metal. Hence: φ = hf0 (f0 = threshold frequency) 1.9 × 1.6 × 10 −19 f0 = 6.63 × 10 −34 f0 = 4.6 × 1014 Hz

[1] [1]

13 a

14 a

E = hf =

hc

[1] [1] [1] [1]

λ

6.63 × 10 −34 × 3.0 × 108 550 × 10 −9 E = 3.62 × 10–19 J ≈ 3.6 × 10–19 J b Power emitted as light = 0.05 × 60 = 3.0 W E=

[1] [1] [1]

3 .0 3.62 × 10 −19 = 8.3 × 1018

Number of photons emitted per second =

15 φ = 4.3 × 1.6 × 10–19 = 6.88 × 10–19 J hc 6.63 × 10 −34 × 3.0 × 108 Energy of photon = = = 9.47 × 10−19 J −7 λ 2.1 × 10 energy of photon = work function + maximum kinetic energy of electron maximum kinetic energy of electron = (9.47 − 6.88) × 10−19 ≈ 2.6 × 10−19 J


[1]


[1] [1] [1] [1] [1] [1]

2


h mv h 6.63 × 10 −34 ν= = mλ 1.7 × 10 −27 × 2.0 × 10 −11

16 λ =

[1] [1]

v = 1.95 × 104 m s−1 (20 km s−1)

[1]

17 Energy lost by a single electron = energy of photon (The energy lost by a single electron travelling through the LED reappears as the energy of a single photon.) Therefore:

eV =

hc

[1]

λ

hc 6.63 ×10 −34 × 3.0 ×10 8 = λe 5.8 ×10 −7 ×1.6 ×10 −19 V = 2.14 V ≈ 2.1 V V=

18 a

[1] [1]

Kinetic energy of electron = VQ = Ve 1 p2 mev2 = Ve or = Ve (where p = mev) 2 2me p=

λ=

[1]

[1] [1]

2meVe h h = (de Broglie equation) me v p

Therefore, λ = b λ=

[1]

h 2meVe

h2 h or V = 2me λ2 e 2meVe

[1]

(6.63 × 10 −34 ) 2 2 × 9.1 × 10 −31 × (4.0 × 10 −11 ) 2 × 1.6 × 10 −19 V ≈ 940 V V=

19 Using f =

c

λ

[1] [1]

and Einstein’s photoelectric equation (hf = φ +

Red light

⇒ ⇒

Blue light

⇒ ⇒

h × 3.0 × 108 640 × 10 −9 4.688 × 1014 h h × 3.0 × 108 420 × 10 −9 7.143 × 1014 h

1 2 mv max ): 2

=

φ + (0.9 × 1.6 × 10−19)

=

φ + 1.440 × 10−19

=

φ + (1.9 × 1.6 × 10−19)

=

φ + 3.040 × 10−19

[1]

(equation 1)

[1]

(equation 2)

[1]

Equations 1 and 2 are two simultaneous equations. (7.143 – 4.688) × 1014 h = (3.040 − 1.440) × 10−19 (3.040 − 1.440) × 10 −19 h= ≈ 6.5 × 10−34 J s (7.143 − 4.688) × 1014


[1] [1]


3


External energy has to be supplied to excite or free an electron. (Allow: The electrons are trapped in an energy well.) b An energy level of 0 eV means the electron is free from the atom. The minimum energy is equal to 3.00 eV. Energy needed to free electron = 3.00 × 1.6 × 10−19 Energy needed to free electron = 4.80 × 10−19 J c i The difference between the energy levels −3.00 eV and −1.59 eV is equal to 1.41 eV. Hence, an electron jumps from −3.00 eV energy level to −1.59 eV energy level. hc ii ∆E = hf =

20 a

λ

hc 6.63 ×10 −34 × 3.0 ×10 8 = ∆E 1.41×1.6 ×10 −19 λ = 8.82 × 10−7 m

λ=

[1]

[1] [1] [1] [1] [1] [1] [1] [1]

21 a This is the lowest energy level occupied by an electron in an atom. [1] b The shortest wavelength corresponds to the change in energy between the two most widely separated energy levels. Hence, ∆E = 10.43 eV [1] hc ∆E = hf = [1]

λ

hc 6.63 ×10 −34 × 3.0 ×10 8 = ∆E 10.43 ×1.6 ×10 −19 λ = 1.19 × 10−7 m

λ=

[1] [1]

2.18 ×10 −18 = –2.18 × 10−18 J 12 2.18 ×10 −18 E2 = − = –5.45 × 10−19 J 2 2 hc b E2 − E1 = ∆E =

22 a

E1 = −

[1] [1] [1]

λ

hc 6.63 × 10 −34 × 3.0 × 10 8 = ∆E (21.8 − 5.45) × 10 −19 λ = 1.22 × 10−7 m This spectral line lies in the ultraviolet region of the spectrum.

λ=

c

1 ⎞ ⎛ 1 ∆E = 2.18 × 10−18 ⎜ 2 − 2 ⎟ = 1.607 × 10−20 J 7 ⎠ ⎝6 −34 hc 6.63 × 10 × 3.0 × 108 λ= = 1.238 × 10−5 m = ∆E 1.607 × 10− 20 λ ≈ 1.24 × 10−5 m (12.4 µm) This spectral line lies in the infrared region of the electromagnetic spectrum.



[1] [1] [1] [1] [1] [1] [1]

4

30 Marking scheme: Worksheet (A2) 1 a change in energy = change in mass × (speed of light)2 or b i ∆E = ∆mc2 ∆E = 0.001 × (3.0 × 108)2 ∆E = 9.0 × 1013 J ii ∆E = ∆mc2 ∆E = 9.1 × 10−31 × (3.0 × 108)2 = 8.19 × 10−14 J ∆E ≈ 8.2 × 10−14 J

∆E = ∆mc2

6.65 × 10 −27 = 4.01 u 1.66 × 10 − 27 2.16 × 10 −26 ii Mass = = 13.01 u 1.66 × 10 − 27 b i Mass = 1.01 × 1.66 × 10−27 = 1.68 × 10−27 kg ii Mass = 234.99 × 1.66 × 10−27 = 3.90 × 10−25 kg

2 a

i

Mass =

3 In all nuclear reactions the following quantities are conserved: • charge (or proton number) • nucleon number • mass–energy • momentum. Any three of the above. 4 a The nucleons within the nucleus are held tightly together by the strong nuclear force. b The binding energy of a nucleus is the minimum energy required to separate the nucleus into its constituent protons and neutrons. binding energy c binding energy per nucleon = number of nucleons 128 binding energy per nucleon = 16 binding energy per nucleon = 8.0 MeV

The half-life of a radioactive isotope is the mean time taken for the number of nuclei of the isotope to decrease to half the initial number. N b i 20 minutes is 1 half-life, so number of nuclei left = 0 2 ii 1.0 hour is 3 half-lives. 3 N ⎛1⎞ Number of nuclei left = ⎜ ⎟ N 0 = 0 8 ⎝2⎠

[1] [1] [1] [1] [1] [1]

[1] [1] [1]

[3] [1] [1]

[1] [1]

5 a

Activity is equal to the number of emissions (or decays of nuclei) per second. Hence, there are 540 α-particles emitted in 1 second. ii Number of α-particles emitted in 1 h = 540 × 3600 ≈ 1.9 × 106 b Energy released in 1 s = number of α-particles emitted in 1 s × energy of each α-particle energy released in 1 s = 540 × 8.6 × 10−14 energy released in 1 s = 4.64 × 10−11 J ≈ 4.6 × 10−11 J c Rate of emission of energy = energy released per second rate of emission of energy = 4.6 × 10−11 J s−1 = 4.6 × 10−11 W

6 a

[1] [1] [1]

[1]

i



[1] [1] [1] [1] [1] [1]

1


7 a

The nuclide

56 26 Fe

is the most stable.

[1]

It has the maximum value for the binding energy per nucleon. b Binding energy = binding energy per nucleon × number of nucleons binding energy ≈ 12.3 × 10−13 × 12 binding energy ≈ 1.5 × 10−11 J c From the graph, the binding energies per nucleon of 21 H and 42 He are approximately

[1]

1.0 × 10−13 J and 11.2 × 10−13 J. energy released = difference in binding energy per nucleon × number of nucleons energy released = [11.2 × 10−13 – 1.0 × 10−13] × 4 energy released = 4.08 × 10−12 J ≈ 4.1 × 10−12 J d High temperatures (~108 K) and pressures.

[1] [1] [1] [1] [2]

8 92 11 proton + 143 01 neutron →

235 92 uranium

[1]

mass defect = [(143 × 1.009) + (92 × 1.007)]u – (234.992)u mass defect = 1.939 u = 1.939 × 1.66 × 10−27 kg mass defect = 3.219 × 10−27 kg binding energy = mass defect × (speed of light)2 binding energy = 3.219 × 10−27 × (3.0 × 108)2 = 2.897 × 10−10 J binding energy binding energy per nucleon = number of nucleons 2.897 × 10 −10 = 1.233 × 10−12 ≈ 1.2 × 10−12 J binding energy per nucleon = 235 9 a

Fission is the splitting of a heavy nucleus like

[1] [1]

235 92 U

[1] [1] [1] [1] [1]

[1]

into two approximately equal fragments.

The splitting occurs when the heavy nucleus absorbs a neutron. b i All particles identified on the diagram.

[1] [1]

ii In the reaction above, there is a decrease in the mass of the particles. [1] 2 According to ∆E = ∆mc , a decrease in mass implies that energy is released in the process. [1] iii The change in mass is ∆m given by: ∆m = [1.575 × 10−25 + 2.306 × 10−25 + 2(1.675 × 10−27)] – [3.902 × 10−25 + 1.675 × 10−27][1] [1] ∆m = –4.250 × 10−28 kg (The minus sign means a decrease in mass and hence energy is released in this reaction.) ∆E = ∆mc2 [1] [1] ∆E = 4.250 × 10−28 × (3.0 × 108)2 −11 −11 ∆E = 3.83 × 10 J ≈ 3.8 × 10 J [1]



2


10 Binding energy of ‘reactant’ = 236 × 7.59 = 1791 MeV (binding energy of neutron = 0) Total binding energy of ‘products’ = (146 × 8.41) + (87 × 8.59) ≈ 1975 MeV Therefore energy released = 1975 – 1791 = 184 MeV 11 a

t1/2 =

0.693

λ

so λ =

0.693 t1/2

[1]

0.693 56 λ = 1.238 × 10−2 s−1 ≈ 1.2 × 10−2 s−1 b A = λN 0.693 A= × 6.0 × 1010 56 A ≈ 7.4 × 108 Bq

λ=

12 a

A = λN

so λ =

[1] [1] [1] [1] [1]

A

[1]

N

5.0 × 10 9 8.0 × 1014 λ = 6.25 × 10−6 s−1 ≈ 6.3 × 10−6 s−1 0.693 b t1/2 = λ=

c

[1] [1] [1]

[1] [1] [1]

λ 0.693 t1/2 = 6.25 × 10 −6 t1/2 = 1.11 × 105 s ≈ 1.1 × 105 s N = N0 e−λt

[1] [1] [1]

−6

N = 8.0 × 1014 e − (6.25×10 × 40×3600 ) N = 3.25 × 1014 ≈ 3.3 × 1014

[1] [1]

The decay constant is the probability that an individual nucleus will decay per unit time. 0.693 0.693 so λ = b i t1/2 = λ t1/2 0.693 λ= 18 × 24 × 3600 λ = 4.46 × 10−7 s−1 ≈ 4.5 × 10−7 s−1 ii A = λN A = 4.46 × 10−7 × 4.0 × 1012 A = 1.78 × 106 Bq ≈ 1.8 × 106 Bq iii 36 days is equal to 2 half-lives.

13 a

[1] [1] [1] [1] [1] [1] [1] [1]

2

⎛1⎞ activity = ⎜ ⎟ × 1.78 × 106 = 4.45 × 105 Bq ≈ 4.5 × 105 Bq ⎝2⎠

number of nuclei = number of moles × NA 1.0 × 10 −6 number of nuclei = × 6.02 × 1023 226 number of nuclei = 2.66 × 1015 ≈ 2.7 × 1015 b A = λN ⎛ 0.693 ⎞ 0.693 × 2.66 × 1015 ⎟⎟ × N = A = ⎜⎜ 1600 × 365 × 24 × 3600 ⎝ t1/2 ⎠

[1]

14 a

[1] [1] [1] [1]

A ≈ 3.7 × 104 Bq


[1]


3


15 According to Einstein’s equation: ∆E = ∆mc2 In this case, ∆E is the energy of two photons and ∆m is the mass of two protons. Hence:

2×

λ=

hc

λ

= (2 × mp) c2

[1]

hc h 6.63 × 10 −34 = = mp c 2 mp c 1.7 × 10 − 27 × 3.0 × 10 8

[1]

λ = 1.3 × 10−15 m

[1]

16 For fusion, we have: energy released per kg = number of ‘pairs’ of 21 H in 1 kg × 4.08 × 10−12 J (from 7 c)

⎛ 1 1000 ⎞ energy per kg = ⎜ × × 6.02 × 10 23 ⎟ × 4.08 × 10−12 2 2 ⎝ ⎠ energy per kg = 6.14 × 1014 J ≈ 6.1 × 1014 J For fission, we have: energy released per kg = number of nuclei in 1 kg × 3.83 × 10−11 J (from 9 b) ⎛ 1000 ⎞ × 6.02 × 10 23 ⎟ × 3.83 × 10−11 energy per kg = ⎜ ⎝ 235 ⎠ 13 energy per kg = 9.8 × 10 J There is less energy released per fusion than per fission. However, there are many more nuclei per kg for fusion. Hence fusion produces more energy per kg than fission. 17 N = N0 e−λt

[1] [1] [1] [1] [1] [1] [1]

N = e −λt = e −( 0.693t / t1/2 ) N0

[1]

9

fraction left = e − ( 0.693×5.0×10 / 4.5×10 fraction left = 0.463 ≈ 0.46


[1]

0.693 t1/2

and λ =

fraction left =

[1] [1]

9

)

[1] [1]


4


Processor (or processing device) Output device

[1] [1]

2

a Axis labels Graph correct shape

[1] [1]

b Thermistor, strain gauge, microphone, etc.

[1]

Length of fine/thin wire sealed in a plastic case ρl ρl 5.0 × 10−7 × 0.15 b i R= so A = = 150 A R −10 2 A = 5.0 × 10 m 1 1 ii Extends by so resistance increases by a factor of , which is 1 Ω. 150 150

[1] [1]

Fraction of the output of a device is fed back to the input reducing any changes in the input to the system. b Increased bandwidth, less distortion, greater stability in gain (less affected by temperature changes).

[1] [1]

Inverting amplifier When Vin is positive then Vout is negative, or vice versa. b The output voltage reaches the power supply voltage and cannot exceed this value. c 10 V d i Gain becomes −10 rather than −5, i.e. sloping line has twice the gradient. Vout is +10 V for Vin below −1 V or Vout is −10 V for Vin above +1 V. ii Gain becomes −2.5 rather than −5, i.e. sloping line has half the gradient. iii The sloping line increases in length before flattening off. e Output voltage peaks at ±5 V. Input and output voltage out of phase.

[1] [1] [1] [1] [1] [1] [1] [1] [1] [1]

3

4

5

a

a

a

f

When the input signal reaches above 2 V or below −2 V the output voltage remains at −10 V or +10 V.



[1] [1] [1]

[2]

[1] [1]

1


6

The output is not saturated so the potential at both the inverting and non-inverting inputs is very nearly the same. Since the non-inverting input is earthed, this value is 0 V. b 1.0 V V 1 = 5.0 × 10−4 A c I= = R 2000 d Negligible current goes into the (–) terminal of the op-amp since it has a very high input resistance (impedance). R e Size of gain = f = 5 so Rf = 5Rin Rin

a

Rf = 5 × 2 = 10 kΩ 7

8

[1] [1] [1] [1] [1]

[1]

Any two from: infinite open-loop voltage gain, infinite input resistance (or impedance), infinite bandwidth, zero output resistance (or impedance) V 10 −3 b i I= = 4 R 10 I = 1.0 × 10−7 A ii V = IR = −1.0 × 10−7 × 100 × 103 = −1.0 × 10−2 V V − 1.0 ×10 −2 iii Gain = out = = −10 Vin 1.0 ×10 −3 iv 1.0 × 10−2 V (with Q being more positive)

a

[2] [1] [1] [1] [1] [1]

For [2] marks, any two from the following points. [2] With an inverting amplifier the output is half a cycle out of phase with the input, whereas with a non-inverting amplifier the input and output are in phase. The input goes to the (−) terminal on the op-amp for an inverting amplifier and to the (+) terminal for the non-inverting amplifier. The input resistance (impedance ) is higher for an op-amp used as a non-inverting amplifier. − Rf but for an inverting amplifier For an inverting amplifier the gain is Rin R the gain is 1 + 1 . R2 R 40 b G = 1+ 1 = 1+ [1] 20 R2

a

G=3 V V 8.0 c G = out ⇒ Vin = out = = 2.67 V ≈ 2.7 V G 3 Vin V 8.0 d I= = R (20 + 40 )×10 3 e

[1] [1] [1]

I = 1.33 × 10−4 A ≈ 1.3 × 10−4 A The voltage is the same as for part c; voltage across the 20 kΩ resistor = 2.7 V



[1] [1]

2


The output is either +9 V (+Vs) or −9 V (−Vs) [1] + – – + [1] depending on whether V > V or V > V b −9 V (−Vs) [1] – + c i If the resistances are all the same, then V and V are equal, so it is unclear which of the two is larger. The comparator is at its switching point and could go either way. [1] – ii V stays the same (at 4.5 V). [1] V + falls (due to larger resistance of X and larger potential drop across X). [1] Vout is −9 V (it may have already been −9 V). [1]

9 a

10 a

Two LEDs and series resistor shown Correct direction shown for LEDs

[1] [1]

output of op-amp

red

green

b When output of op-amp is +9 V then the green LED is forward biased with enough p.d. across it. The red LED is reverse biased. When output of op-amp is −9 V then the red LED is forward biased with enough p.d. across it. The green LED is reverse biased.



[1] [1]

3

32 Marking scheme: Worksheet (A2) 1 It is an electromagnetic wave. Wavelength in the range 10−8 m to 10−13 m

[1] [1]

Energy = 50 × 103 × 1.6 × 10−19 = 8.0 × 10−15 J hc E=

[1] [1]

2 a b

λ

6.63 × 10 −34 × 3.0 × 10 8 8.0 × 10 −15 wavelength = 2.49 × 10−11 m ≈ 2.5 × 10−11 m

λ=

[1] [1]

3 Compton scattering and pair production.

[2]

4 A CAT scan produces a three-dimensional image of the body.

[1]

5 A technique that does not involve cutting the body, e.g. CAT scan.

[1]

6 It is a wave similar to sound (a longitudinal wave) but with a frequency above the audible range/above about 20 kHz.

[1] [1]

7 a

v = fλ

1.5 × 10 3 1.8 × 10 6 = 8.33 × 10−4 m ≈ 0.83 mm b High-frequency ultrasound implies shorter wavelengths. Hence, smaller details can be seen on an ultrasound scan. wavelength =

8 Acoustic impedance = density of medium × speed of ultrasound (Z = ρc) 9 a

Z = ρc 6.40 × 106 = ρ × 4000 6.40 × 10 6 ρ= = 1600 kg m−3 4000

⎛ Z − Z1 ⎞ ⎟⎟ b fraction of intensity reflected = ⎜⎜ 2 ⎝ Z 2 + Z1 ⎠

[1] [1] [1] [1] [1]

[1] 2

[1] 2

c

⎛ 1.66 − 1.63 ⎞ −5 =⎜ ⎟ = 8.3 × 10 ⎝ 1.66 + 1.63 ⎠ percentage of intensity reflected = 0.0083% There is only a very small amount of ultrasound reflected at the boundary between these two materials. This is because their acoustic impedances are very similar.

[1] [1] [1] [1]

10 Superconducting magnet, RF transmission coil, RF receiver coil, gradient coils and computer. [5] 11 a The rotation of the spin/magnetic axis about the direction of the external magnetic field. b 40 MHz c 3.0 × 10 8 c wavelength = = f 40 × 10 6 = 7.5 m



[1] [1] [1] [1]

1


12 High-energy electrons made to bombard a metal target. When the electrons are accelerated and/or slowed down, electromagnetic radiation is produced. For large accelerations, radiation is in the X-ray region. Many different values of acceleration so many different wavelengths giving a continuous spectrum. Incident electron may excite an orbiting electron in a target atom. De-excitation of these electrons gives rise to an emission line spectrum. power cross-sectional area b Power = 250 × π(2.0 × 10−3)2 ≈ 3.1 × 10−3 W

13 a

intensity =

[1] [1] [1]

14 A material (e.g. barium) with high-atomic number/attenuation coefficient used to absorb X-rays. It is used to image the edges of soft tissues, such as the intestines (following a barium meal). 15 Maximum energy of X-ray photon = 80 keV hc = 80 × 10 3 × 1.6 × 10 −19

[1] [1] [1] [1]

λ min

λmin =

[1] [1] [1] [1] [1] [1] [1]

6.63 × 10 −34 × 3.0 × 10 8 = 1.554 × 10 −11 m ≈ 1.6 × 10 −11 m 80 × 10 3 × 1.6 × 10 −19

[1]

16 The energy of the incident X-ray photon is used to remove an electron close to the nucleus of an atom. [1] 3 The attenuation coefficient is proportional to the cube of the atomic (proton) number (µ ∝ Z ). [1] The attenuation coefficient is inversely proportional to the cube of the photon energy (µ ∝ E −3). [1] 17 a

Sketch graph showing: labelled axes correct shape of line intensity showing an exponential decrease

[1] [1] [1]

b I = I0e−µx where µ is the (linear) attenuation/absorption coefficient. c 0.10 = e−0.693x ln 0.10 = –0.693x –2.3 = −0.693x x = 3.32 mm ≈ 3.3 mm

[1] [1]

X-ray image produced of target area. Image is stored in a computer. X-ray beam rotated about target so that many images produced from different angles. Computer stores and processes images to give image of slice through target. Process repeated for different slices. b X-ray image is flat shadow image of whole structure. CAT scan gives two-dimensional image of a slice through structure. Images of slices can be processed to give image of structure in any plane.

[1] [1] [1] [1] [1] [1] [1] [1]

18 a



[1] [1] [1]

2


19 Pulse of ultrasound transmitted into body where it is reflected at the boundary between different tissues. Reflected wave is detected and processed. Time for echo to reach detector indicates depth of tissue boundary. Intensity of echo gives information about tissue boundary.

[1] [1] [1] [1] [1]

20 Amount of reflection of a wave at a boundary depends on the difference in acoustic impedance of the two media. Acoustic impedance is speed of wave × density of medium. Acoustic impedance of air is very different from that for skin so very little ultrasound energy would into skin at the skin–air boundary. Coupling medium has acoustic impedance close to that of skin (and probe) so greatly reduces reflection at skin surface.

[1] [1] [1] [1] [1] [1]

2 × thickness = 4000 × (13 × 10−6) 2 × thickness = 5.2 × 10−2 5.2 × 10 −2 = 2.6 × 10 −2 m (2.6 cm) thickness = 2 b In a B-scan, the transducer is moved around the body. This produces a two-dimensional outline of an organ.

21 a

22 a

[1] [1]

[1] [1] [1]

Nuclei of hydrogen (and certain other nuclides) have spin/have a magnetic axis. In a strong magnetic field, the nuclei precess about the direction of the field with a frequency known as the Larmor frequency. Radio wave pulse at Larmor frequency causes resonance and nuclei precess in high-energy state. After pulse has ceased, nuclei relax, emitting an RF signal.

[1] [1] [1] [1] [1] [1]

Diagram showing:

[1] [1] [1] [1] [1] [1] [1] [1] [1]

b

magnet producing large field magnets producing non-uniform field RF coils. Patient subject to large magnetic field and calibrated non-uniform field. RF pulse at Larmor frequency transmitted to patient. RF emissions from patient are detected and processed. Hydrogen nuclei within patient have Larmor frequency dependent on magnetic field strength so that location (and concentration) of hydrogen nuclei can be detected. Total image built up by varying the non-uniform field to give specific field strength at different positions within the patient.



[1]

3


23 X-rays:

Ultrasound: MRI:


diagnosis of broken bones relatively cheap, can detect flaws within the bone structure. fetal development can distinguish between soft tissues; it is immediate and relatively safe. spinal/back problems fine detail provided of soft and hard tissues.


[1] [1] [1] [1] [1] [1]

4


2

3

The audio signals have the same frequency. The amplitude of the signal/trace/carrier rises and falls with the same number of cycles (two in the time of the trace). The audio signals have different volumes/loudness/amplitudes. The amplitude of the trace rises and falls more in the bottom trace. b The time (or horizontal distance along each trace) for one rapid variation is the same in both traces. a

In amplitude modulation the amplitude of the carrier wave is altered to carry the signal (the frequency remains the same). In frequency modulation the frequency of the carrier wave is altered to carry the signal (the amplitude remains the same). b i 30 × 2 = 60 kHz ii 800 − 60 = 740 kHz iii Alters from 740 kHz to 860 kHz 6000 times a second. c More transmitters may be needed as the range of FM is less than that of AM. Equipment to transmit and receive FM is more expensive.

[1] [1] [1] [1] [1]

a

a

i ii iii b i ii iii

Carrier wave Sidebands 5 kHz 2.5 × 10−5 s 2.0 × 10−4 s Correct amplitude-modulated shape 8 carrier wave oscillations per oscillation of the amplitude Correct times marked

[1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1]

time for one wave of audio signal, b ii

time for one wave of carrier, b i

c

Correct shape Sidebands extending from 40 to 55 kHz and from 40 to 25 kHz

[1] [1] 25


40

55 f / kHz


1


4

5

Analogue signal can have any value (within limits). Diagram to show analogue signal. Digital can have only a few values, e.g. two, and nothing in-between these values. Diagram to show digital signal.

[1] [1] [1] [1]

b The value of the signal is measured at regular intervals of time. The value obtained is converted into a binary number (with a certain number of bits). The binary numbers obtained are placed one after the other and form the digital signal.

[1] [1] [1]

a 0101 and 1000 b Values shown as horizontal lines of 0.5 ms duration Values plotted: 5, 8, 8, 6, 2, 1, 2, 6, 8, 8 Graph correct with labels

[2] [1] [1] [1]

a

V / mV

8 6 4 2 0 0

c

1

2

3

4

5 t / ms

Any variation in the signal that occurs between sampling is not detected Increasing sampling frequency decreases the time between samples Frequency at least 2 × signal, i.e. 600 Hz (frequency of signal ≈ 300 Hz) ii The variation in voltage can use more voltage levels (28 levels rather than 24 levels). The signal voltage at every sample is closer to the actual value. i

6

⎛ 6.0 × 10 −3 ⎞ ⎟ = 164.7 dB ≈ 165 dB Signal-to-noise ratio (in dB) = 10 lg ⎜⎜ −19 ⎟ 2 . 0 10 × ⎝ ⎠

7

a

1.0 × 10 −3 = 30 dB 0.001× 10 −3 b Signal becomes 0.001 mW or signal-to-noise ratio is 1 0 dB

Signal-to-noise ratio = 10 lg



[1] [1] [1] [1] [1]

[1]

[1] [1] [1]

2


8

Any two reasons and explanation. Less attenuation so fewer repeater/regeneration amplifiers needed. More bandwidth so more data can be sent per second/more telephone calls made at once. Less interference/noise so fewer regeneration amplifiers needed. Lower diameter/weight so easier to handle/cheaper.

9 a

i Any value less than 10 m (and more than 1 mm) ii Any value between 10 and 100 m b The sky wave uses reflection by the ionosphere for transmission. The ionosphere fluctuates in its ability to reflect.

Orbits around the Earth’s equator. Takes one day for a complete orbit. Stays over one point on the Earth OR height of orbit 36 000 km above Earth’s surface. b First satellites used wavelengths of about 5 cm; typically now between 1 mm and 1 cm. c Advantage: permanent link with ground station/dishes do not have to be moved. Disadvantage: greater time delay for signal OR further away so signal weaker.

[2] [2] [2] [2] [1] [1] [1] [1]

10 a

[1] [1] [1] [1] [1] [1]

11 The public switched telephone network connects every telephone through exchanges. Without exchanges too many telephones and interconnecting wires are needed. One cable can handle many telephone conversations at once. Sampling places a series of digital bits from many telephone conversations on one cable.

[1] [1] [1] [1]



3

18 Worksheet (A2) Data needed to answer questions can be found in the Data, formulae and relationships sheet. 1

Convert the following angles into radians. a 30° b 210° c 0.05°

2

[1] [1] [1]

Convert the following angles from radians into degrees. a 1.0 rad b 4.0 rad c 0.15 rad

3

[1] [1] [1]

The planet Mercury takes 88 days to orbit once round the Sun. Calculate its angular displacement in radians during a time interval of: a 44 days b 1 day.

4

In each case below, state what provides the centripetal force on the object. a b c d

5

[1] [1]

A car travels at a high speed round a sharp corner. A planet orbits the Sun. An electron orbits the positive nucleus of an atom. Clothes spin round in the drum of a washing machine.

An aeroplane is circling in the sky at a speed of 150 m s−1. The aeroplane describes a circle of radius 20 km. For a enger of mass 80 kg inside this aeroplane, calculate: a her angular velocity b her centripetal acceleration c the centripetal force acting on her.

6

[1] [1] [1] [1]

[2] [3] [2]

The diagram shows a stone tied to the end of a length of string. It is whirled round in a horizontal circle of radius 80 cm.

The stone has a mass of 90 g and it completes 10 revolutions in a time of 8.2 s. a Calculate: i the time taken for one revolution ii the distance travelled by the stone during one revolution (this distance is equal to the circumference of the circle) iii the speed of the stone as it travels in the circle iv the centripetal acceleration of the stone v the centripetal force on the stone. b What provides the centripetal force on the stone? c What is the angle between the acceleration of the stone and its velocity?



[1] [1] [2] [3] [2] [1] [1]

1

18 Worksheet (A2)

7

A lump of clay of mass 300 g is placed close to the edge of a spinning turntable. The centre of mass of the lump of clay travels in a circle of radius 12 cm.

a

The lump of clay takes 1.6 s to complete one revolution. i Calculate the rotational speed of the clay. ii Calculate the frictional force between the clay and the turntable.

b The maximum magnitude of the frictional force F between the clay and the turntable is 70% of the weight of the clay. The speed of rotation of clay is slowly increased. Determine the speed of the clay when it just starts to slip off the turntable. 8

[4]

The diagram shows a skateboarder of mass 70 kg who drops through a vertical height of 5.2 m.

The dip has a radius of curvature of 16 m. a Assuming no energy losses due to air resistance or friction, calculate the speed of the skateboarder at the bottom of the dip at point B. You may assume that the speed of the skateboarder at point A is zero. b i Calculate the centripetal acceleration of the skateboarder at point B. ii Calculate the force R acting on the skateboarder at point B. 9

[2] [3]

[2] [3] [3]

A car of mass 820 kg travels at a constant speed of 32 m s−1 along a banked track. The track is banked at an angle of 20° to the horizontal.

The net vertical force on the car is zero. Use this to show that the force R on the car is 8.56 kN. b Use the answer from a to calculate the radius of the circle described by the car.

a



[2] [4]

2

18 Worksheet (A2)

10 A stone of mass 120 g is fixed to one end of a light rigid rod.

The stone is whirled at a constant speed of 4.0 m s−1 in a vertical circle of radius 80 cm. Calculate the ratio:

tension in the rod at A tension in the rod at B

[6]

Total:


59

Score:


%

3


Define gravitational field strength at a point in space.

[1]

2

Show that the gravitational constant G has the unit N m2 kg−2.

[2]

3

The gravitational field strength on the surface of the Moon is 1.6 N kg−1. What is the weight of an astronaut of mass 80 kg standing on the surface of the Moon?

[2]

4

Calculate the magnitude of the gravitational force between the objects described below. You may assume that the objects are ‘point masses’. a

two protons separated by a distance of 5.0 × 10−14 m (mass of a proton = 1.7 × 10−27 kg)

b two binary stars, each of mass 5.0 × 1028 kg, with a separation of 8.0 × 1012 m

c

5

[3]

[2]

two 1500 kg elephants separated by a distance of 5.0 m

[2]

The diagram shows the Moon and an artificial satellite orbiting round the Earth. The radius of the Earth is R.

Write an equation for the gravitational field strength g at a distance r from the centre of an isolated object of mass M. b By what factor would the gravitational field decrease if the distance from the centre of the mass were doubled? c The satellite orbits at a distance of 5R from the Earth’s centre and the Moon is at a distance of 59R. Calculate the ratio: gravitational field strength at position of satellite gravitational field strength at position of Moon a

6 The planet Neptune has a mass of 1.0 × 1026 kg and a radius of 2.2 × 107 m. Calculate the surface gravitational field strength of Neptune.



[1] [2]

[3]

[3]

1

19 Worksheet (A2)

7 Calculate the radius of Pluto, given its mass is 5.0 × 1023 kg and its surface gravitational field strength has been estimated to be 4.0 N kg−1.

[3]

8 A space probe of mass 1800 kg is travelling from Earth to the planet Mars. The space probe is midway between the planets. Use the data given to calculate: a the gravitational force on the space probe due to the Earth b the gravitational force on the space probe due to Mars c the acceleration of the probe due to the gravitational force acting on it.

[3] [2] [3]

Data

separation between Earth and Mars = 7.8 × 1010 m mass of Earth = 6.0 × 1024 kg

mass of Mars = 6.4 × 1023 kg

9 An artificial satellite orbits the Earth at a height of 400 km above its surface. The satellite has a mass 5000 kg, the radius of the Earth is 6400 km and the mass of the Earth is 6.0 × 1024 kg. For this satellite, calculate: a the gravitational force experienced b its centripetal acceleration c its orbital speed. 10 a Explain what is meant by the term gravitational potential at a point. b Write down the gravitational potential energy of a body of mass 1 kg when it is at an infinite distance from another body. c The radius of the Earth is 6.4 × 106 m and the mass of the Earth = 6.0 × 1024 kg. Calculate the potential energy of the 1 kg mass at the Earth’s surface. d Write down the minimum energy required to remove the body totally from the Earth’s gravitational field. 11 The planets in our solar system orbit the Sun in almost circular orbits. a Show that the orbital speed v of a planet at a distance r from the centre of the Sun is given by: GM v= r b The mean distance between the Sun and the Earth is 1.5 × 1011 m and the mass of the Sun is 2.0 × 1030 kg. Calculate the orbital speed of the Earth as it travels round the Sun.

[3] [2] [3] [2] [1] [3] [1]

[4]

[2]

12 There is a point between the Earth and the Moon where the net gravitational field strength is zero. At this point the Earth’s gravitational field strength is equal in magnitude but opposite in direction to the gravitational field strength of the Moon. Given that: mass of Earth = 81 mass of Moon calculate how far this point is from the centre of the Moon in of R, where R is the separation between the centres of the Earth and the Moon. Total:


57

Score:


[4] %

2

20 Worksheet (A2) 1

2

For an oscillating mass, define: a the period b the frequency.

[1] [1]

The graph of displacement x against time t for an object executing simple harmonic motion (s.h.m.) is shown here.

a State a time at which the object has maximum speed. Explain your answer. b State a time at which the magnitude of the object’s acceleration is a maximum. Explain your answer.

[2] [2]

3

An apple is hung vertically from a length of string to form a simple pendulum. The apple is pulled to one side and then released. It executes 12 oscillations in a time of 13.2 s. a Calculate the period of the oscillations. [2] b Calculate the frequency of the oscillations. [2]

4

This is the graph of displacement x against time t for an oscillating object.

Use the graph to determine the following: a the amplitude of the oscillation b the period c the frequency in hertz (Hz) d the angular frequency in radians per second (rad s−1). e the maximum speed of the oscillating mass.



[1] [1] [2] [2] [2]

1

20 Worksheet (A2)

5

6

7

Two objects A and B have the same period of oscillation. In each case a and b below, determine the phase difference between the motions of the objects A and B. a

[2]

b

[2]

A mass at the end of a spring oscillates with a period of 2.8 s. The maximum displacement of the mass from its equilibrium position is 16 cm. a What is the amplitude of the oscillations? b Calculate the angular frequency of the oscillations. c Determine the maximum acceleration of the mass. d Determine the maximum speed of the mass.

[1] [2] [3] [2]

A small toy boat is floating on the water’s surface. It is gently pushed down and then released. The toy executes simple harmonic motion. Its displacement–time graph is shown here.

For this oscillating toy boat, calculate: a its angular frequency b its maximum acceleration c its displacement after a time of 6.7 s, assuming that the effect of damping on the boat is negligible.



[2] [3] [3]

2

20 Worksheet (A2)

8 The diagram shows the displacement–time graph for a particle executing simple harmonic motion.

Sketch the following graphs for the oscillating particle: a velocity–time graph b acceleration–time graph c kinetic energy–time graph d potential energy–time graph.

[2] [2] [2] [2]

9 A piston in a car engine executes simple harmonic motion. The acceleration a of the piston is related to its displacement x by the equation: a = −6.4 × 105x a Calculate the frequency of the motion. b The piston has a mass of 700 g and a maximum displacement of 8.0 cm. Calculate the maximum force on the piston.

[3] [2]

10 The diagram shows a trolley of mass m attached to a spring of force constant k. When the trolley is displaced to one side and then released, the trolley executes simple harmonic motion.

Show that the acceleration a of the trolley is given by the expression: ⎛k⎞ a = − ⎜ ⎟x ⎝m⎠ where x is the displacement of the trolley from its equilibrium position. b Use the expression in a to show that the frequency f of the motion is given by: 1 k f= 2π m c The springs in a car’s suspension act in a similar way to the springs on the trolley. For a car of mass 850 kg, the natural frequency of oscillation is 0.40 s. Determine the force constant k of the car’s suspension.

a

Total:


59

Score:


[3] [2]

[3] %

3

21 Worksheet (A2) Specific heat capacity of water = 4200 J kg−1 K−1 Specific latent heat of fusion of water = 3.4 × 105 J kg−1 1

2 3

Describe the arrangement of atoms, the forces between the atoms and the motion of the atoms in: a a solid b a liquid c a gas.

[3] [3] [3]

A small amount of gas is trapped inside a container. Describe the motion of the gas atoms as the temperature of the gas within the container is increased.

[3]

a Define the internal energy of a substance. b The temperature of an aluminium block increases when it is placed in the flame of a Bunsen burner. Explain why this causes an increase in its internal energy.

c 4 5 6

7 8

A lump of metal is melting in a hot oven at a temperature of 600 °C. Explain whether its internal energy is increasing or decreasing as it melts.

[1] [3]

[4]

Write a word equation for the change in the thermal energy of a substance in of its mass, the specific heat capacity of the substance and its change of temperature.

[1]

The specific heat capacity of a substance is measured in the units J kg−1 K−1, whereas its specific latent heat of fusion is measured in J kg−1. Explain why the units are different.

[2]

During a hot summer’s day, the temperature of 6.0 × 105 kg of water in a swimming pool increases from 21 °C to 24 °C. Calculate the change in the internal energy of the water.

[3]

A 300 g block of iron cools from 300 °C to room temperature at 20 °C. The specific heat capacity of iron is 490 J kg−1 K−1. Calculate the heat released by the block of iron.

[3]

Calculate the energy that must be removed from 200 g of water at 0 °C to convert it all into ice at 0 °C.

[2]



1

21 Worksheet (A2)

Change the following temperatures from degrees Celsius into kelvin. i 0 °C ii 80 °C iii −120 °C b Change the following temperatures from kelvin into degrees Celsius. i 400 K ii 272 K iii 3 K

9 a

[3]

[3]

10 An electrical heater is used to heat 100 g of water in a well-insulated container at a steady rate. The temperature of the water increases by 15 °C when the heater is operated for a period of 5.0 minutes. Determine the change of temperature of the water when the same heater and container are individually used to heat: a 300 g of water for the same period of time b 100 g of water for a time of 2.5 minutes.

[3] [3]

11 The graph below shows the variation of the temperature of 200 g of lead as it is heated at a steady rate.

a Use the graph to state the melting point of lead. b Explain why the graph is a straight line at the start. c Explain what happens to the energy supplied to the lead as it melts at a constant temperature. d The initial temperature of the lead is 0 °C. Use the graph to determine the total energy supplied to the lead before it starts to melt. (The specific heat capacity of lead is 130 J kg−1 K−1.) e Use your answer to d to determine the rate of heating of the lead. f Assuming that energy continues to be supplied at the same rate, calculate the specific latent heat of fusion of lead.



[1] [1] [1] [3] [2] [3]

2

21 Worksheet (A2)

12 The diagram shows piped water being heated by an electrical heater.

The water flows through the heater at a rate of 0.015 kg s−1. The heater warms the water from 15 °C to 42 °C. Assuming that all the energy from the heater is transferred to heating the water, calculate the power of the heater.

[5]

13 A gas is held in a cylinder by a friction-free piston. When the force holding the piston in place is removed, the gas expands and pushes the piston outwards. Explain why the temperature of the gas falls.

[2]

14 Hot water of mass 300 g and at a temperature of 90 °C is added to 200 g of cold water at 10 °C. What is the final temperature of the mixture? You may assume there are no losses to the environment and all heat transfer takes place between the hot water and the cold water.

[5]

15 A metal cube of mass 75 g is heated in a naked flame until it is red hot. The metal block is quickly transferred to 200 g of cold water. The water is well stirred. The graph shows the variation of the temperature of the water recorded by a datalogger during the experiment.

The metal has a specific heat capacity of 500 J kg−1 K−1. Use the additional information provided in the graph to determine the initial temperature of the metal cube. You may assume there are no losses to the environment and all heat transfer takes place between the metal block and the water.

Total:


71

Score:


[5]

%

3


2

Determine the number of atoms or molecules in each of the following. a 1.0 mole of carbon b 3.6 moles of water c 0.26 moles of helium

[1] [1] [1]

The molar mass of helium is 4.0 g. Determine the mass of a single atom of helium in kilograms.

[2]

The molar mass of uranium is 238 g. a Calculate the mass of one atom of uranium. b A small rock contains 0.12 g of uranium. For this rock, calculate the number of: i moles of uranium ii atoms of uranium.

[2] [1]

4

Explain what is meant by the absolute zero of temperature.

[3]

5

a Write the ideal gas equation in words. b One mole of an ideal gas is trapped inside a rigid container of volume 0.020 m3. Calculate the pressure exerted by the gas when the temperature within the container is 293 K.

[1]

3

6

7

A fixed amount of an ideal gas is trapped in a container of volume V. The pressure exerted by the gas is P and its absolute temperature is T. a Using a sketch of PV against T, explain how you can determine the number of moles of gas within the container. b Sketch a graph of PV against P when the gas is kept at a constant temperature. Explain the shape of the graph. A rigid cylinder of volume 0.030 m3 holds 4.0 g of air. The molar mass of air is about 29 g. a Calculate the pressure exerted by the air when its temperature is 34 °C. b What is the temperature of the gas in degrees Celsius when the pressure is twice your value from part a?



[2]

[3]

[4] [3] [4] [4]

1

22 Worksheet (A2)

8

The diagram shows two insulated containers holding gas. The containers are connected together by tubes of negligible volume.

The internal volume of each container is 2.0 × 10−2 m3. The temperature within each container is −13 °C. The gas in container A exerts a pressure of 180 kPa and the gas in container B exerts a pressure of 300 kPa. a Show that the amount of gas within the two containers is about 4.4 moles. b The valve connecting the containers is slowly opened and the gases are allowed to mix. The temperature within the containers remains the same. Calculate the new pressure exerted by the gas within the containers. 9 The diagram shows a cylinder containing air at a temperature of 5.0 °C. The piston has a cross-sectional area 1.6 × 10−3 m2 It is held stationary by applying a force of 400 N applied normally to the piston. The volume occupied by the compressed air is 2.4 × 10−4 m3. The molar mass of air is about 29 g. a Calculate the pressure exerted by the compressed air. b Determine the number of moles of air inside the cylinder. c Use your answer to b to determine: i the mass of air inside the cylinder ii the density of the air inside the cylinder. 10 The mean speed of a helium atom at a temperature of 0 °C is 1.3 km s–1. Estimate the mean speed of helium atoms on the surface of a star where the temperature is 10 000 K. 11 The surface temperature of the Sun is about 5400 K. On its surface, particles behave like the atoms of an ideal gas. The atmosphere of the Sun mainly consists of hydrogen nuclei. These nuclei move in random motion. a Explain what is meant by random motion. b i Calculate the mean translational kinetic energy of a hydrogen nucleus on the surface of the Sun. ii Estimate the mean speed of such a hydrogen nucleus. (The mass of hydrogen nucleus is 1.7 × 10−27 kg.) 12 a Calculate the mean translational kinetic energy of gas atoms at 0 °C. b Estimate the mean speed of carbon dioxide molecules at 0 °C. (The molar mass of carbon dioxide is 44 g.) c Calculate the change in the internal energy of one mole of carbon dioxide gas when its temperature changes from 0 °C to 100 °C.



[3]

[3]

[2] [3] [1] [2] [6]

[1] [2] [3] [2] [5] [3]

2

22 Worksheet (A2)

13 The diagram below shows three different types of arrangements of gas particles.

A gas whose particles consist of single atoms is referred to as monatomic – for example helium (He). A gas with two atoms to a molecule is called diatomic – for example oxygen (O2). A gas with more than two atoms to a molecule is said to be polyatomic – for example water vapour (H2O). A single atom can travel independently in the x, y and z directions: it is said to have three degrees of freedom. From the equation for the mean translational kinetic energy of the atom, 1 we can generalise that a gas particle has mean energy of kT per degree of freedom. 2 Molecules can also have additional degrees of freedom due to their rotational energy. a Use the diagram above to explain why: 5 i the mean energy of a diatomic molecule is kT 2 ii the mean energy of a polyatomic molecule is 3kT. b Calculate the internal energy of one mole of water vapour (steam) per unit kelvin.

Total:


75

Score:


[2] [2] [3]

%

3


a Explain what is meant by the electric field strength at a point. b Explain what is meant by the electric potential at a point.

[1] [1]

2

A pair of parallel metal plates has a potential difference of 5000 V across them. The electric field strength between them is 400 kN C−1. Calculate: a the separation between the plates b the force on a dust particle between the plates which carries a charge of 1.6 × 10−19 C.

[2] [2]

The electric field strength E at a distance r from a point charge Q may be written as: Q E=k 2 r What is the value for k?

[1]

3

4

The diagram shows a point charge +q placed in the electric field of a charge +Q.

The force experienced by the charge +q at point A is F. Calculate the magnitude of the force experienced by this charge when it is placed at points B, C, D and E. In each case, explain your answer. 5

A spherical metal dome of radius 15 cm is electrically charged. It has a positive charge of +2.5 µC distributed uniformly on its surface. a Calculate the electric field strength on the surface of the dome. b Explain how your answer to a would change at a distance of 30 cm from the surface of the dome.



[9]

[3] [2]

1

23 Worksheet (A2)

6

The diagram shows two point charges. The point X is midway between the charges.

Calculate the electric field strength at point X due to: i the +20 µC charge ii the +40 µC charge. b Calculate the resultant electric field strength at point X.

[3] [2] [2]

The dome of a van de Graaff generator has a diameter of 30 cm and is at a potential of +20 000 V. Calculate: a the charge on the dome b the electric field strength at the surface of the dome c the force on a proton near the surface of the dome.

[2] [2] [1]

a

7

8

9

An isolated charged sphere of diameter 10 cm carries a charge of −2000 nC. Calculate the potential at its surface. b Calculate the work that must be done to bring an electron from infinity to the surface of the dome.

a

[3] [2]

Describe some of the similarities and differences between the electrical force due to a point charge and the gravitational force due to a point mass.

10 The diagram shows two point charges. Calculate the distance x of point P from charge +Q where the net electric field strength is zero.

[6]

[6]

11 Show that the ratio:

electrical force between two protons gravitational force between two protons is about 1036 and is independent of the actual separation between the protons.

[6]

12 A helium nucleus consists of two protons and two neutrons. Its diameter is about 10−15 m. a Calculate the force of electrostatic repulsion between two protons at this separation. b Calculate the potential at a distance of 10–15 m from the centre of a proton. c How much work would need to be done to bring two protons this close to each other? d If one proton were stationary, at what speed would the second proton need to be fired at it to get this close? (Ignore any relativistic effects.)

Total:


66

Score:


[2] [2] [2] [3]

%

2


A 30 µF capacitor is connected to a 9.0 V battery. a Calculate the charge on the capacitor. b How many excess electrons are there on the negative plate of the capacitor?

2

The p.d. across a capacitor is 3.0 V and the charge on the capacitor is 150 nC. a

Determine the charge on the capacitor when the p.d. is: i 6.0 V ii 9.0 V.

[2] [2]

b Calculate the capacitance of the capacitor. 3

[2] [2]

[2]

A 1000 µF capacitor is charged to a potential difference of 9.0 V. a Calculate the energy stored by the capacitor. b Determine the energy stored by the capacitor when the p.d. across it is doubled.

4

For each circuit below, determine the total capacitance of the circuit.

5

The diagram shows an electrical circuit.

a b c d

Calculate the total capacitance of the two capacitors in parallel. What is the potential difference across each capacitor? Calculate the total charge stored by the circuit. Calculate the total energy stored by the capacitors.



[2] [2] [13]

[2] [1] [2] [2]

1

24 Worksheet (A2)

6

A 10 000 µF capacitor is charged to its maximum operating voltage of 32 V. The charged capacitor is discharged through a filament lamp. The flash of light from the lamp lasts for 300 ms. a Calculate the energy stored by the capacitor. b Determine the average power dissipated in the filament lamp.

7

[2] [2]

The diagram shows a 1000 µF capacitor charged to a p.d. of 12 V. a

Calculate the charge on the 1000 µF capacitor.

[2]

b The 1000 µF capacitor is connected across an uncharged 500 µF capacitor by closing the switch S. The charge initially stored by the 1000 µF capacitor is now shared with the 500 µF capacitor. i Calculate the total capacitance of the capacitors in parallel. ii Show that the p.d. across each capacitor is 8.0 V. 8

[2] [2]

The diagram shows a circuit used to measure the capacitance of a capacitor.

reed switch

mA 9.0 V

The reed switch vibrates between the two s with a frequency of 50 Hz. On each oscillation the capacitor is fully charged and totally discharged. The current through the milliammeter is 225 mA. a Calculate the charge that flows off the capacitor each time it is discharged. b Calculate the capacitance of the capacitor. c Calculate the current through the milliammeter when a second identical capacitor is connected: i in parallel with the original capacitor ii in series with the original capacitor.


[1] [2]

[1] [1]


2

24 Worksheet (A2)

9 A capacitor of capacitance 200 µF is connected across a 200 V supply. a Calculate the charge stored on the plates. b Calculate the energy stored on the capacitor. The capacitor is now disconnected from the power supply and is connected across a 100 µF capacitor.

[1] [1]

c Calculate the potential difference across the capacitors. d Calculate the total energy stored on the capacitors. e Suggest where the energy has been lost.

[3] [2] [1]

10 The diagram below shows a charged capacitor of capacitance C. When the switch S is closed, this capacitor is connected across the uncharged capacitor of capacitance 2C. Calculate the percentage of energy lost as heat in the resistor and explain why the actual resistance of the resistor is irrelevant.

Total:


64

Score:


[7]

%

3

25 Worksheet (A2) Data needed to answer questions can be found in the Data, formulae and relationships sheet. current into paper

1

The diagram shows the magnetic field pattern for a current-carrying straight wire drawn by a student in her notes. List two errors made by the student. [2]

2

A current-carrying conductor is placed in an external magnetic field. In each case below, use Fleming’s left-hand rule to predict the direction of the force on the conductor.

a

b

c

[3] 3 4 5

The unit of magnetic flux density is the tesla. Show that: 1 T = 1 N A−1 m−1

[2]

Calculate the force per centimetre length of a straight wire placed at right angles to a uniform magnetic field of magnetic flux density 0.12 T and carrying a current of 3.5 A.

[3]

A 4.0 cm long conductor carrying a current of 3.0 A is placed in a uniform magnetic field of flux density 50 mT. In each of a, b and c below, determine the size of the force acting on the conductor.

[6]

a


b

c


1

25 Worksheet (A2)

6

The diagram shows the rectangular loop PQRS of a simple electric motor placed in a uniform magnetic field of flux density B.

axis

The current in the loop is I. The lengths PQ and RS are both L and lengths QR and SP are both x. Show that the torque of the couple acting on the loop for a given current and magnetic flux density is directly proportional to the area of the loop. [5] 7

The diagram shows a rigid wire AB pivoted at the point A so that it is free to move in a vertical plane. The lower end of the wire dips into mercury. A uniform magnetic field of 6.0 × 10−3 T acts into the paper throughout the diagram. When the current is switched on, the wire continuously moves up out of the mercury and then falls back again. Explain this motion. [4] b The force on the wire due to the current may be taken to act at the midpoint of the wire. When the current is first switched on, the moment of this force about A is 3.5 × 10−5 N m. Calculate: a

i the force acting on the wire ii the current in the wire. 8

[2] [2]

The coil in the d.c. motor shown in question 6 has a length L = 7.0 × 10−2 m and width x = 3.0 × 10−2 m. There are 25 turns on the coil and it is placed in a uniform magnetic field of 0.19 T. The coil carries a current of 2.8 A. The coil is free to rotate about an axis midway between PQ and RS. a Calculate the force on the longest side of the coil. [2] b Calculate the maximum torque (moment) exerted on the coil. [2] c Explain why the force acting on the long side of the coil does not change as the coil rotates but the torque exerted on the coil varies. [2]



2

25 Worksheet (A2)

9

The diagram shows an arrangement that is used to determine the magnetic flux density between the poles of a magnet.

The magnet is placed on a sensitive top pan balance. A current-carrying wire is placed at right angles to the magnetic field between the poles of the magnet. The force experienced by the current-carrying wire is equal but opposite to the force experienced by the magnet. The magnet is pushed downward when the wire experiences an upward force. The length of the wire in the magnetic field is 5.0 cm. The balance reading is 102.00 g when there is no current in the wire. The balance reading increases to 103.14 g when the current in the wire is 8.2 A. a Show that the force experienced by the wire is equal to 1.1 × 10−2 N. [1] b Calculate the magnetic flux density of the magnetic field between the poles of the magnet. [3]

Total:


39

Score:


%

3

26 Worksheet (A2) Data needed to answer questions can be found in the Data, formulae and relationships sheet. 1 2

Calculate the force experienced by an oil droplet with a charge of 3.2 × 10−19 C due to a uniform electric field of strength 5.0 × 105 V m−1.

[2]

The diagram shows two parallel, horizontal plates separated by a vertical distance of 3.0 cm. The potential difference between the plates is 600 V.

a Calculate the magnitude and direction of the electric field between the plates. b Describe the electric field between the plates. c A charged oil droplet of weight 6.4 × 10−15 N is held stationary between the two plates. State whether the charge on the droplet is positive or negative. Explain your answer. ii Determine the charge on the oil droplet.

[3] [2]

i

3 4

Calculate the force experienced by an electron travelling at a velocity of 4.0 × 106 m s−1 at right angles to a magnetic field of magnetic flux density 0.18 T.

[2] [2] [3]

The diagram shows an electron moving at a constant speed of 8.0 × 106 m s−1 in a plane perpendicular to a uniform magnetic field of magnetic flux density 4.0 mT.

a Calculate the force acting on the electron due to the magnetic field. [3] b What is the centripetal acceleration of the electron? [2] c Use your answer to b to determine the radius of the circular path described by the electron. [2]



1

26 Worksheet (A2)

5

The diagram shows the trajectory of an electron travelling into a region of uniform magnetic field of flux density 2.0 mT. The electron enters the region of magnetic field at 90°.

a Draw the direction of the force experienced by the electron at points A and B. b Explain why the electron describes part of a circular path while in the region of the magnetic field. c The radius of curvature of the path of the electron in the magnetic field is 5.0 cm. Calculate the speed v of the electron. d Explain how your answer to c would change if the electron described a circular path of radius 2.5 cm. 6

[1] [5] [2]

A proton of kinetic energy 15 keV travelling at right angles to a magnetic field describes a circle of radius of 5.0 cm. The mass of a proton is 1.7 × 10−27 kg. a Show that the speed of the proton is 1.7 × 106 m s−1. b For this proton, calculate the centripetal force provided by the magnetic field. c Determine the magnetic flux density of the magnetic field that keeps the proton moving in its circular orbit. d How long does it take for the proton to complete one orbit?

7

[1]

[3] [3] [3] [2]

The diagram shows a velocity-selector for charged ions. Ions of speed v emerge from the slit.

The parallel plates have a separation of 2.4 cm and are connected to a 5.0 kV supply. A magnetic field is applied at right angles to the electric field between the plates such that the positively charged ions emerge from the slit of the velocity-selector at a speed of [6] 6.0 × 106 m s−1. Calculate the magnetic flux density of the magnetic field. b Ions from the velocity-selector into a mass spectrometer which contains another magnetic field, of flux density B. The ions all have charge Q but either have mass m1 or mass m2. Show that the difference in the radius of the two isotopes in the magnetic field is given by: (m1 − m 2 )v ∆r = [2] BQ a



2

26 Worksheet (A2)

8

An electron describes a circular orbit in a plane perpendicular to a uniform magnetic field. a

Show that the time T taken by an electron to complete one orbit in the magnetic field is independent of its speed and its radius, and is given by: 2 πm Be where B is the magnetic flux density of the magnetic field, e is the charge on an electron and m is the mass of an electron.

T=

[5]

b Explain in words how a faster electron takes the same time to complete one orbit as a slower electron. [1]

Total:


55

Score:


%

3

27 Worksheet (A2) 1

A flat coil of N turns and cross-sectional area A is placed in a uniform magnetic field of flux density B. The plane of the coil is normal to the magnetic field. a Write an equation for: i the magnetic flux Φ through the coil [1] ii the magnetic flux linkage for the coil. [1] b The diagram shows the coil when the magnetic field is at an angle θ to the normal of the plane of the coil. What is the flux linkage for the coil? [1]

2

A square coil of N turns is placed in a uniform magnetic field of magnetic flux density B. Each side of the coil has length x. What is the magnetic flux linkage for this coil?

3

4

The diagram shows a magnet placed close to a flat circular coil. a Explain why there is no induced e.m.f. even though there is magnetic flux linking the coil. b Explain why there is an induced e.m.f. when the magnet is pushed towards the coil.

[2]

[1]

[2]

A coil of cross-sectional area 4.0 × 10−4 m2 and 70 turns is placed in a uniform magnetic field. a The plane of the coil is at rightangles to the magnetic field. Calculate the magnetic flux density when the flux linkage for the coil is 1.4 × 10−4 Wb. [3] b The coil is placed in a magnetic field of flux density 0.50 T. The normal to the coil is at an angle of 60° to the magnetic field, as shown in the diagram. Calculate the flux linkage for the coil. [3]



1

27 Worksheet (A2)

5

A square coil is placed in a uniform magnetic field of flux density 40 mT. The plane of the coil is normal to the magnetic field. The coil has 200 turns and the length of each side of the coil is 3.0 cm. a Calculate: i the magnetic flux Φ through the coil [2] ii the magnetic flux linkage for the coil. [2] b The plane of the coil is turned through 90°. What is the change in the magnetic flux linkage for the coil? [2]

6

A flat circular coil of 1200 turns and of mean radius 8.0 mm is connected to an ammeter of negligible resistance. The coil has a resistance of 6.3 Ω. The plane of the coil is placed at right angles to a magnetic field of flux density 0.15 T from a solenoid. The current in the solenoid is switched off. It takes 20 ms for the magnetic field to decrease from its maximum value to zero. Calculate: the average magnitude of the induced e.m.f. across the ends of the coil b the average current measured by the ammeter. a

7

[5] [2]

The diagram shows a straight wire of length 10 cm moved at a constant speed of 2.0 m s−1 in a uniform magnetic field of flux density 0.050 T.

For a period of 1 second, calculate: a the distance travelled by the wire b the area swept by the wire c the change in the magnetic flux for the wire (or the magnetic flux ‘cut’ by the wire) d the e.m.f. induced across the ends of the wire using your answer to c e the e.m.f. induced across the ends of the wire using E = Blv. 8

[1] [1] [2] [2] [1]

A circular coil of radius 1.2 cm has 2000 turns. The coil is placed at right angles to a magnetic field of flux density 60 mT. Calculate the average magnitude of the induced e.m.f. across the ends of the coil when the direction of the magnetic field is reversed in a time of 30 ms. [5]



2

27 Worksheet (A2)

9 The diagram shows a step-up transformer. The ends AB of the primary coil are connected to a 1.5 V cell and a switch. The switch is initially closed and the lamp is off. The switch is suddenly opened and the lamp illuminates for a short time.

a Explain why the lamp illuminates only for a short period. b State one change to the apparatus that would allow the lamp to illuminate normally.

[4] [1]

10 A wire of length L is placed in a uniform magnetic field of flux density B. The wire is moved at a constant velocity v at right angles to the magnetic field. Use Faraday’s law of electromagnetic induction to show that the induced e.m.f. E across the ends of the wire is given by E = BLv. Hence calculate the e.m.f. induced across the ends of a 20 cm long rod rolling along a horizontal table at a speed of 0.30 m s−1. (The vertical component of the Earth’s magnetic flux density is about 40 µT.)

[8]

11 a State Faraday’s law of electromagnetic induction. b Lenz’s law expresses an important conservation law. Name this conservation law. c i Define magnetic flux for a coil placed at right angles to a magnetic field. ii Determine for which of the two coils X and Y, each placed at right angles to the magnetic field, is the magnetic flux linkage the greatest.

Total:


59

Score:


[1] [1] [1] [4]

%

3

28 Worksheet (A2) 1

2

3

4

5

An alternating voltage is given by the equation V = V0 sin 2πft where V0 = 5.0 V and f = 10 Hz. a Calculate the period of the alternating voltage. b Calculate the values of t during the first cycle of the voltage (from t = 0) for which the value of V is: i 0 ii +V0 iii −V0 iv +Vrms

[2]

[1] [2] [1] [2]

The graph shows how an alternating voltage V and an alternating current I change with time t.

a V and I are in phase with each other. Explain what is meant by in phase. b Copy the graph and add a waveform to show how the power dissipated varies with t.

[1] [3]

An electric drill is marked 230 V r.m.s., 690 W. Calculate: a i the r.m.s. current in the wire connecting the drill to the mains ii the peak current in the wire connecting the drill to the mains iii the peak value of the potential difference across the drill. b Sketch a graph of the power drawn by the drill over one cycle of the current. Mark on the graph the values of peak power and average power.

[2] [1] [1]

The diagram shows the variation of voltage with time across a resistor. a State and explain whether the current in the resistor is a.c. or d.c. [2] +2 V b Explain why the power dissipated in the resistor is the same as the power produced by a steady voltage of 2 V. [1] 0 c For the voltage variation shown, state: i the peak value [1] −2 V ii the r.m.s. value. [1] Calculate the value of the average power dissipated in a resistor when the alternating supply to the resistor has a peak current of 2.5A and a peak voltage of 6.0 V.



[3]

t/s

[2]

1

28 Worksheet (A2)

6

7

The diagram shows the trace obtained on the screen of an oscilloscope connected to a signal generator. The time-base of the oscilloscope is set at 20 ms per division and the Y-gain at 1.5 V per division. a For the signal generator, calculate: i the frequency [2] ii the r.m.s. voltage. [2] b The equation of the waveform can be written in the form V = V0 sin (ωt). Determine the values of V0 and ω. [2]

1 division

The diagram shows a step-down transformer. The primary coil has 1150 turns and the secondary coil has 30 turns. The ends of the secondary coil are connected to a lamp labelled ‘6.0 V, 24 W’. The ends AB of the primary coil are connected to an alternating voltage supply. The potential difference across the lamp is 6.0 V. a b c d e

Calculate the current in the lamp. Calculate the input voltage to the primary coil. Calculate the current in the primary coil, assuming the transformer is 100% efficient. Calculate the maximum p.d. across the lamp during one cycle of the a.c. A student suggests that to avoid the production of heat in the transformer the wires should be coated in a material that is a poor conductor of heat. Explain why this is not a sensible suggestion.

[2] [2] [2] [2]

[1]

8

An electrician uses a transformer to step the 230 V r.m.s. mains voltage down to 115 V r.m.s. The secondary coil has 500 turns and is connected to a resistor of 5000 Ω. a Calculate the number of turns on the primary coil. [1] b Calculate the current in the secondary coil. [1] c Calculate the current in the primary coil. Assume that the transformer is 100 % efficient. [1] d The electrician connects cables to the secondary coil that break down when the p.d. between the wire and earth is larger than 130 V. Explain whether the cables will break down when the transformer is switched on. [2]

9

A consumer receives 1000 W of power at 100 V r.m.s. through a 5.0 Ω cable. a Calculate the rate of heat production in the cable. b Explain why transmitting the same amount of power at a higher voltage produces less heat dissipation in the cable.

Total:


50

Score:


[2] [2]

%

2


What is a photon?

[1]

2

γ-rays from a radioactive material have higher frequency than visible light. Explain why this means that γ-rays are more harmful.

[2]

State one piece of evidence that electromagnetic radiation has: a wave-like properties b particle-like properties.

[1] [1]

A light-emitting diode emits red light of wavelength 6.4 × 10−7 m. Calculate: a the frequency of the red light b the energy of a photon of red light.

[2] [3]

Using the photons and work function, describe why electrons are emitted from the surface of a zinc plate when it is illuminated by ultraviolet radiation but not when it is illuminated by visible light.

[3]

3

4

5

6

The figure below shows an electron making a transition between two energy levels and the bright spectral emission line observed.

Explain why electromagnetic radiation is emitted when an electron jumps from energy level E1 to energy level E2. b Derive an expression for the frequency f of the radiation emitted. c State and explain the position of the spectral line when an electron makes a transition between energy levels E1 and E3.

a

[2] [2] [2]

7

An electron in an atom can occupy four energy levels. With the help of an energy level diagram, determine the maximum number of spectral emission lines from this atom. [2]

8

Lithium atoms emit red light of wavelength 670 nm. Calculate the difference between the energy levels responsible for this red light.



[3]

1

29 Worksheet (A2)

9 The diagram below shows a hot solid, at a temperature of 5000 K, emitting a continuous spectrum.

State the type of spectrum observed from: a position X b position Y c position Z.

[1] [1] [1]

10

What experimental evidence is there that suggests that electrons behave as waves?

[1]

11

The electronvolt is a convenient unit of energy for particles and photons. Define the electronvolt.

[1]

12

An electron is accelerated through a potential difference of 6.0 V. According to a student, this electron has kinetic energy greater than the energy of a photon of ultraviolet radiation of wavelength 2.5 × 10−7 m. With the aid of calculations, explain whether or not the student is correct. [5]

13 a Define threshold frequency for a metal. b The work function of caesium is 1.9 eV. Calculate the threshold frequency.

[1] [3]

14 A particular filament lamp of rating 60 W emits 5.0% of this power as visible light. The average wavelength of visible light is 550 nm. Calculate: a the average energy of a single photon of visible light b the number of photons of visible light emitted per second from the lamp.

[3] [3]

15 A plate of zinc is illuminated by electromagnetic radiation of wavelength 2.1 × 10−7 m. The work function of zinc is 4.3 eV. Calculate the maximum kinetic energy of a photoelectron.

[4]

16 Neutrons travelling through matter get diffracted just as electrons do when travelling through graphite. In order to show diffraction effects, the neutrons need to have a de Broglie wavelength that is comparable to the spacing between the atoms. Calculate the speed of a neutron that has a de Broglie wavelength of 2.0 × 10−11 m.

[3]

17 A yellow light-emitting diode (LED) is connected to a d.c. power supply. The output voltage from the supply is slowly increased from zero until the LED just starts to glow. The yellow light from the LED has a wavelength of about 5.8 × 10−7 m. Estimate the potential difference across the LED when it just starts to glow.

[4]



2

29 Worksheet (A2)

18 a

In an electron-diffraction experiment, electrons are accelerated through a p.d. V. Show that the de Broglie wavelength λ of an electron is given by: h λ= 2meVe

where me is the mass of the electron and e is the elementary charge. b Calculate the accelerating p.d. V that gives an electron a de Broglie wavelength of 4.0 × 10−11 m.

[3] [3]

19 In an experiment on the photoelectric effect, a metal is illuminated by visible light of different wavelengths. A photoelectron has a maximum kinetic energy of 0.9 eV when red light of wavelength 640 nm is used. With blue light of wavelength 420 nm, the maximum kinetic energy of the photoelectron is 1.9 eV. Use this information to calculate an experimental value for the Planck constant h. [5] 20 The diagram below shows the some of the energy levels for a helium atom.

a Explain the significance of the energy levels being negative. b When a helium atom is not excited, the electrons have an energy of −3.00 eV. This is known as the stable state of the electrons. Calculate the minimum energy, in joules, required to free an electron at this energy level. Explain your answer. c The helium atom absorbs a photon of energy 1.41 eV. i State the transition made by an electron. ii Calculate the wavelength of the radiation absorbed by the helium atom.

[1]

[3] [2] [3]

21 The figure below shows the energy level diagram for an atom of mercury. a Explain what is meant by the ground state. [1] b Calculate the shortest wavelength emitted by the atom. Explain your answer. [4]



3

29 Worksheet (A2)

22 For the hydrogen atom, the energy level En in joules is given by the equation 2.18 ×10 −18 En = − n2 where n is an integer, known as the principal quantum number.

Calculate the energy level of the ground state (n = 1) and the energy level of the first excited state (n = 2). b Determine the wavelength of radiation emitted when an electron makes a transition from the first excited state to the ground state. In which region of the electromagnetic spectrum would you find a spectral line with this wavelength? c In which region of the electromagnetic spectrum would you find the spectral line corresponding to an electron transition between energy levels with principal quantum numbers of 6 and 7? Justify your answer. a

Total:


90

Score:


[2]

[4]

[4]

%

4


2

a Write down Einstein’s famous equation relating mass and energy. b Determine the change in energy equivalent to a change in mass: i of 1.0 g ii equal to that of an electron (9.1 × 10−31 kg).

[1] [2] [2]

In nuclear physics, it is common practice to quote the mass of a nuclear particle in of the unified atomic mass unit, u. The unified atomic mass unit u is defined as one-twelfth of the mass of an atom of the carbon isotope 126 C .

Determine the mass of each of the following particles in of u: i an α-particle of mass 6.65 × 10−27 kg ii a carbon-13 atom of mass 2.16 × 10−26 kg. b Determine the mass of each of the following particles in kilograms: i a proton of mass 1.01 u ii a uranium-235 nucleus of mass 234.99 u. a

[1] [1] [1] [1]

3

State three quantities conserved in all nuclear reactions.

[3]

4

a Explain why external energy is required to ‘split’ a nucleus. b Define the binding energy of a nucleus. c The binding energy of the nuclide 168 O is 128 MeV. Calculate the binding energy

[1] [1]

per nucleon. 5

6

[2]

a Define the half-life of a radioactive isotope. b The half-life of a particular isotope is 20 minutes. A sample initially contains N0 nuclei of this isotope. Determine the number of nuclei of the isotope left in the sample after: i 20 minutes ii 1.0 hour.

[1] [2]

The activity of an α-source is 540 Bq. The kinetic energy of each α-particle is 8.6 × 10−14 J. The isotope in the source has a very long half-life. a Calculate the number of α-particles emitted by the source in: i 1 second ii 1 hour. b Determine the total energy released by the source in a time of 1 second. c State the rate at which energy is emitted from this α-source.

[1] [1] [3] [1]



[1]

1

30 Worksheet (A2)

7

The binding energy per nucleon against nucleon number graph for some common nuclides is shown below.

a Identify the most stable nuclide. Explain your answer. b Use the graph to estimate the binding energy for the nucleus of

12 6C

.

Use the graph to estimate the energy released in the following fusion reaction. 2 2 4 1 H + 1 H → 2 He d The fusion reaction shown in c is one of the many that occur in the interior of stars. State the conditions necessary to initiate such reactions in stars. c

8

Use the data given below to determine the binding energy and the binding energy per nucleon of the nuclide 235 92 U . mass of proton = 1.007 u mass of neutron = 1.009 u mass of uranium-235 nucleus = 234.992 u



[2] [2] [4]

[2] [7]

2

30 Worksheet (A2)

9

a Describe the process of induced nuclear fission. b The diagram shows the fission of uranium-235 in accordance with the nuclear equation: 235 1 95 139 1 92 U + 0 n → 38 Sr + 54 Xe + 2 0 n

[1]

Copy the diagram, adding labels to identify the neutrons, the strontium nuclide and the xenon nuclide. ii Explain why energy is released in the reaction above.

[1] [2]

iii Use the following data to determine the energy released in a single fission reaction 1 involving 235 92 U and 0 n .

[5]

i

mass of mass of

235 92 U 1 0n =

−25

= 3.902 × 10

−27

1.675 × 10

kg

kg

mass of mass of

95 −25 kg 38 Sr = 1.575 × 10 139 −25 kg 54 Xe = 2.306 × 10

10 One of the neutron-induced fission reactions of uranium-235 may be represented by the following nuclear equations. 235 1 236 92 U + 0 n → 92 U 236 92 U

87 1 → 146 57 La + 35 Br + 3 0 n The binding energies per nucleon for these nuclides are: 236 146 87 92 U, 7.59 MeV; 57 La, 8.41 MeV; 35 Br, 8.59 MeV.

Calculate the energy released in MeV when the 11 The half-life of the radon isotope

220 86 Rn −1

236 92 U

nucleus undergoes fission.

[3]

is 56 s.

a Determine the decay constant in s . b Calculate the activity of a sample containing 6.0 × 1010 nuclei of

220 86 Rn

[3] [3]

.

12 The activity of a radioactive source containing 8.0 × 1014 undecayed nuclei is 5.0 × 109 Bq. a Determine the decay constant in s−1. b Calculate the half-life of the nucleus. c How many undecayed nuclei will be left after 40 hours? 13 a Define the decay constant of a nucleus. b The thorium isotope 227 90 Th has a half-life of 18 days.

[1]

A particular radioactive source contains 4.0 × 1012 nuclei of the isotope i

Determine the decay constant for the thorium isotope

[3] [3] [3]

227 90 Th

227 90 Th

.

in s−1.

ii What is the initial activity of the source? iii Calculate the activity of the source after 36 days.

[3] [3] [2]

14 A sample of rock is known to contain 1.0 µg of the radioactive radium isotope 226 88 Ra .

The half-life of this particular isotope is 1600 years. The molar mass of radium-226 is 226 g. a Determine the number of nuclei of the isotope 226 88 Ra in the rock sample.

[2]

b Calculate the activity from decay of the radium-226 in the sample.

[3]



3

30 Worksheet (A2)

15 In a process referred to as ‘annihilation’, a particle interacts with its antiparticle and the entire mass of the combined particles is transformed into energy in the form of photons. The following equation represents the interaction of a proton (p) and its antiparticle, the antiproton ( p ). 1 1p

+ −11 p → γ + γ

The antiproton has the same mass as a proton – the only difference is that it has a negative charge. Determine the wavelength λ of each of the two identical photons emitted in the reaction above. (Mass of a proton = 1.7 × 10−27 kg.)

[5]

16 Does fusion or fission produce more energy per kilogram of fuel? Answer this question by considering the fusion reaction in 7 c and the fission reaction in 9 b. (The molar masses of hydrogen-2 and uranium-235 are 2 g and 235 g, respectively.)

[7]

17 Some astronomers believe that our solar system was formed 5.0 × 109 years ago. Assuming that all uranium-238 nuclei were formed before this time, what fraction of the original uranium-238 nuclei remain in the solar system today? The isotope of uranium 238 9 92 U has a half-life of 4.5 × 10 y.

[4]

Total:


100

Score:


%

4

31 Worksheet (A2) 1

The flowchart below shows the components that make up an electronic sensor. sensing device

What are the names of the missing components? 2

[2]

A thermistor is an example of a sensing device. a Sketch the temperature characteristic of a negative temperature coefficient thermistor. b State the name of one other sensing device.

3

a

Describe the structure of a metal-wire strain gauge.

[2] [1] [2]

−7

b A strain gauge contains 15 cm of wire of resistivity 5.0 × 10 Ω m. The resistance of the strain gauge is 150 Ω.

4

a

i Calculate the cross-sectional area of the wire in the strain gauge. ii Calculate the increase in resistance when the wire extends by 0.1 cm, assuming that the cross-sectional area and resistivity remain constant.

[1]

What is meant by negative ?

[2]

b State two advantages of negative in an operational amplifier. 5

[2]

[2]

The circuit shown is used to produce a graph of Vout against Vin by moving the slider on the variable resistor. The supply voltage to the op-amp is + Vs.



1

31 Worksheet

The graph obtained is shown.

State the type of amplifier drawn. Explain how the graph shows that the amplifier is of this type. b State why the graph flattens at the ends. c Suggest the value that was used for the supply voltage Vs. d State what changes occur to the graph if:

[2] [1] [1]

i Rin is halved in value but Rf is kept unchanged ii Rf is halved in value but Rin is unchanged from the initial value iii the supply voltage Vs is increased.

[2] [1] [1]

The variable resistor is removed and an a.c. signal of maximum voltage ±1.0 V is applied to the input of the amplifier circuit. Sketch the output voltage and input voltage on the same graph. Explain why the amplifier circuit produces distortion if an a.c. signal with a maximum voltage of 3.0 V is applied to the input.

[2]

a

e

f 6

In the circuit shown in question 5 the input voltage Vin is 1.0 V and Rin is 2.0 kΩ. a b c d e

7

[2]

Explain why the potential at the inverting input (−) is almost zero. State the value of the fall in potential across Rin. Calculate the current in Rin. Explain why the current in Rf is the same as the current in Rin. Determine the value of Rf. You will need to use the graph from question 5.

[2] [1] [1] [1] [1]

An electrical device generates a potential of +1.0 mV at point P.

a State two properties of an ideal operational amplifier. b Assuming that the operational amplifier is ideal, calculate: i ii iii iv

the current in the 10 kΩ resistor the potential at R the gain of the amplifier using your answer to b ii the potential difference between R and Q.



[2] [2] [1] [1] [1]

2

31 Worksheet

8

The circuit shows a non-inverting amplifier with an output voltage of 8.0 V.

State two differences between an op-amp used as an inverting amplifier and as a non-inverting amplifier. b Calculate the gain of the amplifier shown in the circuit. c Calculate the value of the input voltage Vin. d Calculate the value of the current in the 40 kΩ resistor. e Determine the voltage across the 20 kΩ resistor. a

9

[2] [2] [1] [2] [1]

The circuit shows an op-amp used as a comparator.

a Explain how the op-amp acts as a comparator. [2] b State the value of Vout when V − is larger than V +. [1] c The resistors and the thermistors are all chosen to have the same resistance, as closely as can be measured. i Explain why the value of Vout is uncertain. ii The temperature of thermistor X falls. Explain what, if anything, happens to V −, V + and Vout.

[1] [3]

10 A device is to be placed on the output of the circuit shown in question 9, such that when the output voltage is +9 V green light is emitted and when the output voltage is −9 V the light emitted is red. a Draw the circuit diagram of the device. b Explain how the device works.

[2] [2]

Total:


58

Score:


%

3

32 Worksheet (A2) Data needed to answer questions can be found in the Data, formulae and relationships sheet. 1 State the nature of X-ray radiation.

[2]

2 The energy of an X-ray photon is 50 keV. a Calculate the energy of the photon in joules. b Calculate the wavelength of the X-rays.

[2] [2]

3 One of the interaction mechanisms between X-rays and matter is the photoelectric effect. Name the two other interaction mechanisms.

[2]

4 State one main difference between the images produced by a normal X-ray machine and by a CAT scan.

[1]

5 Briefly explain what is meant by a non-invasive technique.

[1]

6 State what is meant by ultrasound.

[2]

7 The speed of ultrasound in soft tissue is 1.5 km s−1. a Calculate the wavelength of ultrasound of frequency 1.8 MHz. b Use your answer to part a to explain why high-frequency ultrasound is suitable for medical scans. 8 Define acoustic impedance.

[2] [1] [1]

9 The table below shows useful data for biological materials. Material

Density / kg m−3

soft tissue

1060

1540

1.63

muscle

1075

1590

1.71

bone

?

4000

6.40

blood

1060

1570

1.66

Speed of ultrasound / m s−1

Acoustic impedance Z / 106 kg m−2 s−1

a Calculate the density of bone. b Calculate the percentage of intensity of ultrasound reflected at the blood–soft tissue boundary. (Assume the waves are incident at right angles to the boundary.) c Explain why it would be difficult to distinguish between blood and soft tissue in an ultrasound scan. 10 Name the five main components of an MRI scanner.

[2] [3] [2] [5]

11 Protons have a precession frequency of 40 MHz in a strong uniform magnetic field. a Describe what is meant by precession. b State the frequency of the radio frequency (RF) radiation that will cause the protons to resonate. c Use your answer to b to determine the wavelength of the RF radiation.



[1] [1] [2]

1

32 Worksheet (A2)

12 Briefly describe the production of X-rays and explain why an X-ray spectrum may consist of a continuous spectrum and a line spectrum.

[7]

13 The intensity of a collimated X-ray beam is 250 W m−2. a Define intensity. b The diameter of the X-ray beam is 4.0 mm. Calculate the power transmitted by the beam.

[1] [2]

14 Describe what is meant by a contrast medium and state why it is used in X-ray scans.

[2]

15 The potential difference between the cathode and the anode of an X-ray tube is 80 kV. Calculate the minimum wavelength of the X-rays emitted from this tube.

[3]

16 The photoelectric effect is one of the attenuation mechanisms by which X-ray photons interact with the atoms in the body. Describe some of the characteristics of this mechanism.

[3]

17 A collimated X-ray beam is incident on a metal block. The incident intensity of the beam is I0. a Draw a sketch graph to show the variation with thickness x of the intensity I of the beam. b Write down an expression for the intensity I in of I0 and x. Explain any other symbol you use. c The linear absorption coefficient of a beam of 80 keV X-rays is 0.693 mm−1 in copper. Calculate the thickness of copper necessary to reduce the intensity of the beam to 0.10 I0.

[3] [2] [3]

18 a Describe the use of a CAT scanner. b Compare the image formed in X-ray diagnosis with that produced by a CAT scanner.

[5] [3]

19 Outline how ultrasound may be used in medical diagnosis.

[5]

20 Explain why, in medical diagnosis using ultrasound, a coupling medium is necessary between the ultrasound probe and the skin.

[6]

21 a

When an ultrasound pulse reflects from the front and back edges of a bone, it produces two peaks on an A-scan. The time interval between these two peaks is 13 µs. The speed of the ultrasound in bone is 4000 m s−1. Calculate the thickness of the bone. b Describe how a B-scan differs from an A-scan.

22 a

Outline the principles of magnetic resonance.

[6]

b Outline, with the aid of a sketch diagram, the use of MRI (magnetic resonance imaging) to obtain diagnostic information about internal body structures. 23 X-Rays, ultrasound and MRI are all used in medical diagnosis. State one situation in which each of these techniques is preferred and give reasons, one in each case, for the choice.

Total:


[3] [2]

104

Score:


[10]

[6]

%

2

33 Worksheet (A2) 1

Two amplitude-modulated radio waves are shown. Each wave has the same carrier frequency and is carrying an audio signal.

a b 2

a b

c 3

State and explain one similarity and one difference between the audio signals that they carry. Explain how the graphs show that the carrier frequency is the same. Describe the difference between amplitude and frequency modulation. A carrier wave has a frequency 800 kHz. It is modulated in frequency by an audio signal of frequency 6 kHz and amplitude 2.0 V. The frequency deviation of the carrier wave is 30 kHz V−1. i Determine the maximum frequency shift produced. ii Determine the minimum frequency of the modulated carrier wave. iii Describe how the frequency of the carrier wave changes. A country intends to start a new broadcasting system. State two reasons why it is more expensive to set up an FM broadcasting system rather than an AM system.

[4] [1] [2]

[1] [1] [1] [2]

The graph shows the frequency spectrum of an AM radio wave carrying an audio signal of a single frequency.

35

a

40

45

Frequency / kHz

State the name of the component with frequency 40 kHz. State the name of the components with frequency 35 and 45 kHz. Determine the frequency of the audio signal. Calculate the time for one complete carrier wave. Calculate the time for one complete wave of the audio signal. Sketch a graph of the variation of the signal with time. On your graph mark the values obtained in b i and b ii. The frequency spectrum shown above is formed from a carrier wave and an audio signal of one frequency. Draw the frequency spectrum formed at one instant if the audio signal contains a range of frequencies up to 15 kHz.

i ii iii b i ii iii c



[1] [1] [1] [1] [1] [3]

[2]

1

33 Worksheet (A2)

4

5

Data is often produced as an analogue signal and then converted into digital form for transmission. a Explain, with the aid of sketch graphs, the difference between an analogue and a digital signal. b Explain the process of sampling in which an analogue signal is turned into a digital signal.

[4] [3]

The diagram shows the analogue signal from a microphone.

For transmission, the signal is digitally sampled every 0.5 ms starting at time t = 0 s. In the analogue-to-digital (ADC) converter, 0 to 0.99 mV produces a digital output 0000 1 to 1.99 mV produces a digital output 0001 and so on. a b c

6 7

State the value of the digital out put when t = 0 s and when t = 0.5 ms. The digital signal from the ADC is eventually converted back into analogue form. Draw a sketch diagram showing the final analogue signal produced. i Explain how increasing the sampling frequency improves the final analogue signal produced and suggest a suitable maximum value for the sampling frequency. ii Telephone systems use 8-bit numbers, rather than 4-bit numbers. Explain why this improves the final analogue signal produced.

A laser provides power input of 6.0 mW into an optic fibre, where the average noise is 2.0 × 10−19 W. Calculate the signal-to-noise ratio in dB. A signal has a power of 1.0 mW and a noise of 0.001 mW. a What is the signal-to-noise ratio in dB? b The signal is attenuated by 30 dB and the noise remains constant. What is the new signal-to noise ratio in dB?



[2] [3] [3] [2] [1] [1] [2]

2

33 Worksheet (A2)

8

9

In the modern telephone system, more and more coaxial cable has been replaced for long-distance transmission of telephone signals by optic fibre. State and explain two reasons for this change. a

b 10 a b c

State a typical value of wavelength for: i space waves ii sky waves. Explain why satellite communication is more reliable than a sky wave for long-distance communication between two points on the Earth’s surface. Describe the orbit of a geostationary satellite. State a typical wavelength for communication between the Earth’s surface and a geostationary satellite. State one advantage and one disadvantage of the use of a geostationary satellite rather than a satellite in polar orbit for telephone communication.

[4]

[2] [2] [3] [1] [2]

11 In the original telephone system of 1876, every telephone was connected to every other telephone by a pair of wires. Today the telephone is used worldwide as the result of the invention of the exchange and the use of sampling using digital electronics. Describe how each of these developments has meant that many telephone conversations can take place at once.

[4]

Score:

%

Total:


61


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