Answers to End-of-chapter questions - Cambridge OCR...

Answers to End-of-chapter questions

Chapter 11 a Linear momentum = mass × velocity [1]

b Momentum is a product of a scalar quantity (mass) and a vector quantity (velocity);hence it is a vector quantity. [1]

c v2 = u2 + 2as and u = 0 [1]v = 405.32 ×× = 16.7 m s–1 ≈ 17 m s–1 [1]

p = mv = 900 × 16.7 ≈ 1.5 × 104 [1]Unit: kg m s–1 [1]

d Initial momentum = final momentum [1](4.0 × 2.0) + (4.0 × –3.0) = 8.0v [1]8.0v = –4.0

v = 0.8

0.4− = –0.50 m s–1 [1]

The objects move to the left with a combined speed of 0.50 m s–1 [1]

2 a i In an elastic collision, both momentum and kinetic energy are conserved. [1]

ii In an inelastic collision, momentum is conserved but kinetic energy is not.Some of the kinetic energy is transformed into other forms such as heat. [1]

b Initial momentum = 0.35 × 2.8 = 0.98 kg m s–1 [1]Final momentum = –(0.35 × 2.5) = –0.875 kg m s–1 ≈ –0.88 kg m s–1 [1]Change in momentum = –0.875 – 0.98 = –1.855 kg m s–1 ≈ –1.9 kg m s–1 [1]

c Momentum is conserved. [1]The initial momentum of the ball is equal to the final momentum of the ball andthe momentum of the snooker table. [1]

3 Momentum = mass × velocity [1]Principle of conservation of momentum:The total momentum before an interaction (collision) is equal to the total momentumafter the interaction. [2]This is true provided there are no external forces acting on the system. [1]Principle of conservation of energy:The total energy before a collision is equal to the total energy after the collision. [1]Kinetic energy is only conserved in elastic collisions. [1]

4 a Momentum = mass × velocity [1]

b Total initial momentum = total final momentum0 = mAvA + mBvB [1]mAvA = –mBvB [1]

Therefore B

A

v

v =

A

B

m

m−

COAS Physics 2 1


Chapter 21 a i 1 Maximum force = 48 N 2 Contact time = 0.25 s [1]

ii Impulse of the force = area under the graph [1]Impulse = (6.5 ± 1.0) N s [1]

b i a = m

F =

50.0

48[1]

Acceleration = 96 m s–2 [1]

ii Impulse = change in momentum = ∆p [1]6.5 = (0.50v) – (0.50 × 0) (the initial velocity of the ball is zero)

v = 50.0

5.6 = 13 m s–1 [1]

iii Kinetic energy = 21 mv2 = 2

1 × 0.50 × 132 [1]

Kinetic energy = 42.25 J ≈ 42 J [1]

c F = ma =

−

t

uvm [1]

F = 0.50 ×

−−

18.0

1480.0[1]

Force = (–)61 N [1]

2 a Momentum = mv = 500 × 0.60 = 300 [1]Unit: kg m s–1 or N s [1]

b i The momentum of the steel bar increases. [1]According to Newton’s second law of motion, there must be a force on the barbecause force = rate of change of momentum. [1]OrThe bar is accelerating. [1]The bar must be experiencing a force because force = mass × acceleration. [1]

ii The force acts to the right. [1]

iii Time = speed

distance =

60.0

0.3 = 5.0 s [1]

iv F = t

p

∆∆

= t

mumv −[1]

F = 0.5

)6.0500()8.1500( ×−×[1]

Force = 120 N [1]

3 a Change in momentum = mv – mu = m(v – u) [1]Change in momentum = 1100 × (24 – 0) = 2.64 × 104 kg m s–1 ≈ 2.6 × 104 kg m s–1 [1]

b Braking force F = t

p

∆∆

= 20

1064.2 4×[1]

Braking force = 1.32 × 103 N ≈ 1.3 kN [1]

c Magnitude of deceleration = m

F =

1100

1032.1 3× = 1.2 m s–1 [1]

s = ? u = 24 m s–1 v = 0 a = 1.2 m s–1

v2 = u2 + 2as02 = 242 + (2 × –1.2 × s) [1]

s = 4.2

242

= 240 m [1]

COAS Physics 2 2


Chapter 3

1 a i Speed = time

distance =

T

rπ2

v = 501

015.0π2 ××[1]

v = 4.71 m s–1 ≈ 4.7 m s–1 [1]

ii a = r

v2

= 015.0

71.4 2

[1]

a = 1.48 × 103 m s–2 ≈ 1.5 × 103 m s–2 [1]

b The tension in the belt is not sufficient to provide the centripetal force. [1]Hence the belt will slip / will not grip the pulley. [1]

2 a The centripetal force is the net force acting on an object describing a circle.It is directed towards the centre of the circle. [1]

b i Speed = time

distance

v = 0.3

15.0π2 ××[1]

v = 0.314 m s–1 [1]

F = r

mv 2

= 15.0

314.0060.0 2×[1]

Force = 0.0394 N ≈ 0.039 N [1]

ii The centripetal force on the toy increases with its speed. [1]The toy falls off because the frictional force between the turntable and the toyis not sufficient to provide the centripetal force. [1]

3 a Change in potential energy = kinetic energy gained [1]

mgh = 21 mv2

v = gh2 = 70.081.92 ×× [1]

v = 3.71 m s–1 ≈ 3.7 m s–1 [1]

b Centripetal force = r

mv 2

= 50.1

71.3050.0 2×[1]

Centripetal force = 0.459 N [1]T – mg = 0.459 [1]T = 0.459 + (0.050 × 9.81) = 0.95 N [1]

c The weight is equal to the tension only when the ball is at rest in a vertical position. [1]The ball is not is not in equilibrium in the vertical position because it has an upward (centripetal) acceleration. [1]

Chapter 41 a Arrow at S pointing towards the centre of O. [1]

b i g = 2r

GM[1]

ii At S: g = 25

40[1]

g = 1.6 N kg–1 [1]

COAS Physics 2 3


iii At C: g = 24

40 = 2.5 N kg–1 [1]

iv Average g = 2

5.26.1 + = 2.05 N kg–1 [1]

Increase in GPE: ∆E = mgh = 3.0 × 103 × 2.05 × 2.0 × 107 [1]

≈ 1.2 × 1011 J [1]

c F = mg = r

mv 2

g = r

v2

[1]

v = gr = 8100.16.1 ×× = 1.26 × 104 m s–1 [1]

v = T

rπ2; 1.26 × 104 =

T

8100.1π2 ××[1]

Period T = 4

8

1026.1

1028.6

××

≈ 5.0 × 104 s [1]

2 a i It is the gravitational force per unit mass. [1]

ii g = 2r

GM[1]

iii g = 4.0 N kg–1 when r = 10 × 106 m (or any suitable pair from the curve) [1]

M = G

gr 2 =

11

26

1067.6

)1010(0.4−×

×× [1]

Mass = 6.0 × 1024 kg [1]

iv The straight-line graph through the origin. [1]

v At the surface, g = 9.8 N kg–1 when r = 6400 kmFor distance within the Earth (linear graph), r = 64 km [1]because the field strength is a factor of 100 times smaller. [1]For distance beyond the surface (curve), r = 64 000 km [1]

because the field strength ∝ 2

1

r. [1]

b 21

Earth

R

GM = 2

2

Moon

R

GM[1]

Moon

Earth

M

M =

2

2

1

R

R = 92 = 81 [1]

Mass of moon = MMoon = 81

100.6 24× = 7.4 × 1022 kg [1]

3 a It is the gravitational force per unit mass. [1]

b i g = 4.5 N kg–1 [1]

ii g = 2r

GM[1]

COAS Physics 2 4


iii g ∝ 2

1

r; hence as r increases by a factor of 3, the field strength decreases by

a factor of 32 = 9. [1]

Hence g = 9

40 = 4.44 N kg–1 [1]

4 a i Arrows of equal lengths [1]directed towards the centre of the orbit. [1]

ii A centripetal force is required for circular motion. [1]Hence the gravitational force must be along the line joining the stars. [1]

b F = gravitational force between the stars; G = gravitational constant;M = mass of each star; R = radius of the orbit. [1]

c i Speed = time

distance

In a time of one period, the distance travelled is equal to the circumference. [1]

Hence, v = T

Rπ2

ii F = R

Mv 2

[1]

F = R

TRM 2)2( π× =

2

2π4

T

MR[1]

iii2

2

4R

GM =

2

2π4

T

MR[1]

M = 2

222 π4

TG

RR

××××

= 16π2

2

3

GT

R[1]

d M = 211

3112

)400 86100(1067.6

)105.0(π16

××××××

− [1]

Mass = 4.0 × 1030 kg [1]

Chapter 51 a i A motion where the acceleration is directly proportional to the displacement

from the equilibrium position [1]and the acceleration is directed towards the equilibrium position. [1]

ii From the graph, period = 0.25 s [1]

f = T

1 =

25.0

1 = 4.0 Hz [1]

iii a = (2πf )2x [1]a is maximum when x = A (= 5.0 mm)Maximum acceleration = (2π × 4.0)2 × 5.0 × 10–3 [1]a = 3.16 m s–2 ≈ 3.2 m s–2 [1]

b i Any three from:Resonance occurs at the natural frequency of the oscillating system. [1]At resonance, the natural frequency is equal to the frequency of the driver. [1]At resonance, the amplitude of the oscillating system is maximum. [1]At resonance, the oscillating system absorbs maximum energy from the driver. [1]

COAS Physics 2 5


ii 1 Greater mass leads to smaller amplitude. [1]This is because resonance will occur at a lower frequency. [1]

2 Greater stiffness leads to smaller amplitude. [1]This is because the resonance will occur at higher frequency. [1]

2 a i In figure 1, the acceleration a is in the opposite direction to the displacement x. [1]In figure 2, a ∝ x. [1]

ii Gradient of line = (2πf)2 [1]

(2πf)2 = 050.0

50 = 1000 [1]

f = π2

1000 = 5.03 Hz ≈ 5 Hz [1]

iii ‘Cosine’ curve with initial amplitude 25.0 mm. [1]Correct period of 0.2 s shown on graph. [1]The amplitude decreases (exponentially) with time. [1]

b i There is an acceleration (or centripetal force) towards the centre of the circle. [1]This acceleration is constant because the speed of rotation is constant. [1]

ii From Figure 2, acceleration = 50 m s–2 when deflection is 50 mm.

a = r

v2

[1]

50 = 10

2v[1]

v = 500 = 22.4 m s–1 ≈ 22 m s–1 [1]

3 a i Cosine curve. [1]Correct period of 1.0 s shown on the graph. [1]Amplitude decreases (exponentially) with respect to time. [1]

ii The amplitude will decrease more rapidly [1]because of increased damping (or air resistance). [1]The frequency of oscillation will decrease (or the period is longer) [1]because the mass is greater. (a = (2πf )2x; a is smaller because of greater mass; hence f will decrease.) [1]

b Resonance occurs at the natural frequency of the oscillating aeroplane. [1]At resonance, the amplitude is maximum, as the oscillator absorbs greatest energy. [1]As frequency is increased from 0 to 1 Hz, the amplitude increases. [1]The amplitude us maximum at 1.0 Hz. [1]As the frequency is increased from 1.0 Hz to 2.0 Hz, the amplitude decreases. [1]

4 a i A motion where the acceleration is directly proportional to the displacementfrom the equilibrium position [1]and the acceleration is directed towards the equilibrium position. [1]

ii It is a straight-line graph through the origin (hence a ∝ x). [1]The graph has a negative gradient (hence acceleration is always directedtowards the equilibrium position). [1]

b i Amplitude = 0.050 m [1]

ii Gradient of line = (2πf )2 [1]

(2πf )2 = 050.0

5.12 = 250

COAS Physics 2 6


f = π2

250 = 2.52 Hz [1]

Period = f

1 =

52.2

1 ≈ 0.40 s [1]

c i Cosine curve. [1]Correct period of 0.40 s shown on graph. [1]Correct amplitude of 0.05 m shown. [1]

ii Displacement = 0 [1]Any suitable time, e.g. 0.1 s, 0.3 s, 0.5 s, etc. [1]

Chapter 61 E = mc∆θ [1]

E = 600 × 1.1 × 103 × (70 – 30) [1]Energy = 2.6 × 107 J [1]

2 a i The internal energy of a system is the sum of the random distribution of kineticand potential energies associated with the molecules (or atoms) that make upthe system. [2]

ii The specific heat capacity of a substance is the energy required to raise thetemperature by 1 K (or 1 °C) per unit mass of the substance. [1]

b Any four from:The material is heated electrically. [1]The energy supplied to the material is E = VIt (V = p.d., I = current and t = time) [1]Measure the mass of the material and the change in temperature. [1]The specific heat capacity c is determined using the equation VIt = mc∆θ [1]The material can be insulated (lagged) to minimise heat losses to the surroundings. [1]

c E = mc∆θ = 2.0 × 920 × (293 – 0) [1]Energy = 5.39 × 105 J ≈ 540 kJ [1]

3 a i ρ = V

m =

Av

m (where m is the mass per unit time, V is the volume per unit time,

A is the cross-sectional area and v is the velocity)m = ρ × A × v [1]0.090 = 1000 × 7.5 × 10–5 × v [1]v = 1.2 m s–1

ii The speed will be doubled because the cross-sectional area is halved.Speed = 2.4 m s–1 [1]

iii Force = rate of change of momentum [1]

F = t

p

∆∆

= 0.1

)2.1090.0()4.2090.0( ×−×[1]

F = 0.11 N [1]

iv Direction of force is opposite to the direction of flow of water from shower head. [1]

b i E = mc∆θ [1]Dividing both sides of the equation by time t gives:

t

E =

t

m× c∆θ [1]

Power = t

E = 0.090 × 4200 × (27 – 15) [1]

Power = 4.54 × 103 [1]Unit: watts (W) [1]

ii Energy losses in the pipe from the heater tot he shower head. [1]

COAS Physics 2 7


iii The change in temperature is doubled to:(27 – 15) × 2 = 24 °CMaximum temperature = 15 + 24 = 39 °C [1]

Chapter 71 a pV = nRT [1]

n = RT

pV =

)15273(31.8

101.210280 33

+×××× −

[1]

n = 0.246 mol ≈ 0.25 mol [1]

bT

pV = constant (V = constant)

15273

10280 3

+×× V

= T

V×× 310290[1]

T = 280

290× 288 = 298.3 K [1]

Temperature = 298.3 – 273 ≈ 25 °C [1]

c Internal energy = kinetic energy ∝ T [1]

Therefore ratio = 288

3.298≈ 1.04 [1]

2 a The Celsius and Kelvin scales have the same increment. [1]Adding (or subtracting) 273 to 500 000 does not make a significant difference. [1]

b i According to pV = nRT, as pressure (or volume) tends to zero, the temperaturetends to 0 K. [1]

ii Mean kinetic energy of molecules ∝ T [1]As the temperature tends to 0 K, the kinetic energy tends to zero. [1]

c iT

p = constant ⇒

3000p

= 400

p[1]

p = 300

400p0 = 1.33p0 [1]

iiT

pV = nR or n ∝ p [1]

f = 0p

p =

400

300 = 0.75 [1]

3 a When the temperature is 0 K, the pressure or volume of the gas tends to zero. [2]

bT

pV = constant [1]

17273

100.1 5

+×× V

= 43273

100.1100.1 63

−×××

[1]

V = 230100.1

290100.15

9

××××

= 1.26 × 104 m3 ≈ 1.3 × 104 m3 [1]

c i pV = nRT

n = RT

pV =

)43273(31.8

1026.1100.1 45

−××××

[1]

n = 5.23 × 105 mol ≈ 5.2 × 105 mol [1]

ii Mass = 5.23 × 105 × 4.0 × 10–3 = 2.1 × 103 kg [1]

COAS Physics 2 8


d Internal energy = total kinetic energy ∝ T [1]The temperature decreases from 17 °C (290 K) to –43 °C (230 K),

hence the kinetic energy must decrease by a factor of 230

290 = 1.261. Therefore:

kinetic energy = 1900 × 1.261 = 1500 MJ [1]

e i Weight acting downwards and greater upthrust/lift acting upwards. [1]

ii Net force = Ma1.3 × 105 – Mg = Ma [1]

M = 2781.9

103.1 5

+×

[1]

M ≈ 3.5 × 103 kg [1]

Chapter 81 a Correct field pattern as in figure. [1]

The field lines touch the plate at right angles. [1]

The field direction is shown correctly (away from the plate). [1]

b i Correct triangle of forces as shown. [1]

tan 20° = W

F = 5100.1 −×

F[1]

F = 1.0 × 10–5 × tan 20° ≈ 3.64 × 10–6 N [1]

ii E = Q

F =

9

6

102.1

1064.3−

−

××

[1]

E = 3.0 × 103 [1]Unit: V m–1 or N C–1 [1]

c F = 20πε4 r

Qq(Q = q = 1.2 × 10–9 C and F = 3.64 × 10–6 N) [1]

r2 = 612

29

1064.31085.8π4

)102.1(−−

−

×××××

= 3.56 × 10–3 m2 [1]

r ≈ 6.0 × 10–2 m [1]

COAS Physics 2 9


d Correct field pattern as shown. [1]

The field pattern for the two point charges is symmetrical about the plane midway between the charges. [1]

2 a Horizontal field lines between the plates. [1]

Field lines equally spaced (to show uniform field). [1]Correct directions of fields (from +1200 V to 0 V, and from +600 V to 0 V) [1]

b E = d

V[1]

E = 040.0

600 = 1.5 × 104 V m–1 [1]

c F = EQ = 1.5 × 104 × 1.6 × 10–19 [1]Force = 2.4 × 10–15 N [1]

d Work done = kinetic energy; Fd = 21 mv2

v2 = 31

15

1011.9

040.0104.22−

−

××××

= 2.108 × 1014 [1]

v = 1.542 × 107 m s–1 ≈ 1.5 × 107 m s–1 [1]

e VQ = 21 mv2; speed ∝ V

The p.d. is doubled; hence speed = 2 × 1.452 × 107 ≈ 2.05 × 107 m s–1 [1]

f Fewer electrons will reach the grid B or plate C. [1]Hence the current will fall. [1]

3 a Electric field strength is the force experienced per unit positive charge. [2]

COAS Physics 2 10


b i Correct field pattern. [1]

The field pattern is symmetrical (quality mark) [1]Correct direction of field lines from positive to negative. [1]

ii For each charge, E = 20πε4 r

Q; r = 4.0 × 10–10 m [1]

E = 21012

19

)100.4(1085.8π4

106.1−−

−

×××××

= 9.0 × 109 [1]

Total E = 2 × 9.0 × 109 = 1.8 × 1010 (Both fields are in the same direction.) [1]Unit: V m–1 or N C–1 [1]

c i Equal but opposite forces act on each charge. [2]

ii A: No motions of dipole as forces at each end are equal and opposite. [1]B: The dipole rotates (clockwise). [1]The dipole experiences a couple/turning effect. [1]

4 a i Correct field pattern. [1]

Correct direction of field lines from positive ion to negative ion. [1]

ii F = 20πε4 r

Qq[1]

F = 21012

1919

)100.5(1085.8π4

106.1106.1−−

−−

×××××××

[2]

Force = 9.2 × 10–10 N [1]

b FH = mHaH and FI = mIaI [1]The forces are the same. [1]Simple harmonic motion implies a ∝ –x. [1]

Therefore I

H

x

x =

I

H

a

a =

H

I

m

m = 127 [1]

COAS Physics 2 11


c i Correct sine or cosine curve. [1]

Amplitude is 8.0 × 10–12 m [1]

Correct period of 1.5 × 10–14 s shown. (Period = f

1 ≈ 1.5 × 10–14 s.) [1]

ii The molecules are resonating. [1]This happens when the frequency of the incident infrared radiation is equal tothe natural frequency of oscillation of the molecules. [1]

Chapter 91 a Arrow(s) between the north and south poles of the magnet. [1]

b Current = R

V =

0.8

5.1 = 0.1875 A [1]

F = BIL [1]F = 1.2 × 10–2 × 0.1875 × 24 = 0.054 [1]Unit: newton (N) [1]

c The force will decrease by a factor of four. [1]

This is because the resistance of the wire increases by a factor of four (R ∝ A

1)

and reduced the current by a factor of four. [1]

2 a i Equally spaced (horizontal) electric field lines between the parallel plates. [1]Correct direction of the electric field (towards the –7.0 kV plate). [1]

ii Work done = energy transformed; eV = 21 mv2 [1]

v = m

eV2 =

31

19

1011.9

7000106.12−

−

××××

= 4.96 × 107 m s–1 ≈ 5.0 × 107 m s–1 [1]

b i The arrow (force) is perpendicular tot he path and towards the centre of the arc. [1]

ii Out of the plane of the paper. [1]Fleming’s left-hand rule gives the direction of the force. [1]

iiir

mv2

= Bev [1]

r = Be

mv[1]

r = 193

731

106.1100.3

1096.41011.9−−

−

××××××

= 9.4 × 10–2 m [2]

COAS Physics 2 12


c Change in direction of magnetic field to change direction of electron beam. [1]Change magnitude of current in the coil to move the electron beam across. [1]

3 a It has a positive charge; it is repelled by the positive charges on the ‘top’ plate. [1]

b i Kinetic energy = eV [1]Kinetic energy = 1.6 × 10–19 × 300 = 4.8 × 10–17 J [1]

ii 21 mv2 = 4.7 × 10–17 [1]

v = 26

17

103.2

108.42−

−

×××

= 6.46 × 104 m s–1 ≈ 6.5 ≈ 104 m s–1 [1]

c E = d

V[1]

d = E

V = 4100.4

600

× = 1.5 × 10–2 m (1.5 cm) [1]

d i Semicircle to the right of the hole. [1]

iir

mv2

= BQv [1]

r = BQ

mv[1]

r = 19

426

106.117.0

1046.6103.2−

−

×××××

[1]

Radius = 5.46 × 10–2 m [1]Distance from hole = 2 × 5.46 × 10–2 m ≈ 0.11 m [1]

4 a Magnetic flux density = fieldin conductor oflength current

force

×[1]

The field is at right-angles to the magnetic field. [1]

b i Arrow towards the centre of the circle. [1]

ii The field is out of the plane of the paper. [1]The direction of motion of the proton gives the direction of the conventional current when using Fleming’s left-hand rule. [1]

iii F = Bev [1]

iv F = r

mv2

[1]

r

mv2

= Bev [1]

B = er

mv =

60106.1

105.11067.119

727

×××××

−

−[1]

B ≈ 2.6 × 10–3 [1]Unit: tesla (T) [1]

v B = er

mv ∝ v [1]

Therefore for protons moving at double the speed, the magnetic flux densityshould be doubled in order to contain the protons within the tube. [1]

Chapter 101 a i F is towards the open end of the tube. [1]

The direction can be determined using Fleming’s left-hand rule. [1]

COAS Physics 2 13


ii F = BIw [1]

iii F = 0.15 × 800 × 25 × 10–3 N [1]Force = 3.0 N [1]

b i According to Faraday’s law:induced e.m.f. = rate of change of magnetic flux linkage. [2]The rate of change of magnetic flux is directly proportional to the speed v. [1]

ii The flux linkage is doubled. [1]According to Faraday’s law, the induced e.m.f. is also doubled. [1]

2 a Faraday’s law: induced e.m.f. = rate of change of magnetic flux linkage [2]Magnetic flux = magnetic flux density × area × cosθ ; φ = BAcosθ [1]When field is normal to area, φ = BA [1]Magnetic flux linkage = number of turns × magnetic flux (= BAN) [1]

b A sine or cosine curve shown for the voltage against time graph. [1]Doubling the speed will:double the maximum induced voltage [1]because the rate of change of flux is doubled; [1]double the frequency. [1]Doubling the number of turns will:double the maximum induced voltage [1]because the flux linkage is doubled. [1]Removing the iron will:reduce the induced voltage because the magnetic flux density is reduced. [1]

3 a i Magnetic flux = magnetic flux density × area (normal to field) = BA [1]

ii Flux linkage = number of turns × magnetic flux [1]Flux linkage = NBx2 [1]

b Faraday’s law: E = t

N

∆∆ )( φ

= t

NBx2

[1]

E = NBx × t

x = NBxv [1]

e.m.f. = 1250 × 0.032 × 0.020 × 0.10 = 0.080 V (80 mV) [1]

Chapter 111 a i Q = VC [1]

Therefore W = 21 QV = 2

1 (VC)V = 21 CV 2 [1]

ii Parabolic shape through the origin. [1]

Accurate plot showing W = 21 × 2.2 × V 2 = 1.1V 2 [1]

b i Time constant = CR [1]Time constant = 2.2 × 6.8 × 103 = 1.5 × 104 s (5 h 9 min) [1]

ii Energy lost = 21 × 2.2 × 5.02 – 2

1 × 2.2 × 4.02 [1]

Energy lost = 9.9 J [1]

COAS Physics 2 14


iii V = V0e–t/CR

t = –RC ln

0V

V[1]

t = –1.5 × 104 × ln

0.5

0.4 ≈ 3.3 × 103 s [1]

iv Power = 3103.3

9.9

× = 3.0 × 10–3 W [1]

2 a Q0 = VC [1]Q0 = 5000 × 1.2 × 10–11 = 6.0 × 10–8 [1]Unit: coulomb (C) [1]

b i Time constant = RC = 1.2 × 1015 × 1.2 × 10–11 = 1.44 × 104 s ≈ 1.4 × 104 s [1]

ii Current = R

V = 15102.1

5000

× = 4.16 × 10–12 A ≈ 4.2 pA [1]

iii 1.44 × 104 s ≈ 1.4 × 104 s (same as time constant in b I) [1]

iv t = RC; Q = Q0e–t/CR = Q0e–1 [1]

0Q

Q = e–1 ≈ 0.37; hence the charge lost is about 63%, which is roughly 3

2 . [1]

3 a i Parallel field lines between the plates. [1]Correct field direction (away from the positive charges). [1]

ii E = d

V; V = Ed = 3.0 × 105 × 1.5 × 103 = 4.5 × 108 V = 450 MV [1]

b The capacitor plates model the base and centre regions of the clouds. [1]The battery acts as the source of energy; it models the wind. [1]The resistor is included to account for the slow rate of charging of the clouds. [1]

c i I = t

Q[1]

Current = 25

20 = 0.80 A [1]

ii R = I

V =

80.05

105.4 8

××

[1]

Resistance = 1.13 × 108 Ω ≈ 1.1 × 108 Ω [1]

iii C = V

Q = 8105.4

20

×[1]

Capacitance = 4.44 × 10–8 F ≈ 4.4 × 10–8 F [1]

iv Time constant = RC [1]Time constant = 1.13 × 108 × 4.44 × 10–8 F ≈ 5.0 s [1]

4 a Capacitance = p.d.

charge[1]

b i 1 Q = VC [1]Q = 11 × 0.47 × 10–6 = 5.2 × 10–6 C [1]

2 W = 21 CV 2 = 2

1 × 0.47 × 10–6 × 112 [1]

Energy = 2.84 × 10–5 J ≈ 2.8 × 10–5 J [1]

COAS Physics 2 15


ii 1 I = R

V =

2200

11[1]

Current = 5.0 × 10–3 A [1]

2 Time constant = CR [1]Time constant = 2200 × 0.47 × 10–6 = 1.03 × 10–3 s ≈ 1.0 ms [1]

c i The graph should have a constant-ratio property (for a given time interval). [1]

Attempt made, e.g. 8.1

70.0 ≈

70.0

30.0 ≈ 0.4 [1]

ii Time constant = 1 ms and Q = Q0e–t/CR

Charge at 2.0 ms = 5.2 × 10–6 × e–2 = 7.04 × 10–7 C [1]Charge at 1.0 ms = 5.2 × 10–6 × e–1 = 1.91 × 10–6 C [1]Charge flow = (1.91 – 0.704) × 10–6 C ≈ 1.2 × 10–6 C [1]Alternatively:area under graph = charge flow [1]attempt to find area [1]correct answer [1]

Chapter 121 Alpha scattering provided evidence for the existence of the nucleus of an atom. [1]

The nucleus contains most of the mass of the atom. [1]The nucleus must have a very small size (10–14 m), because very few α-particles werescattered through large angles. [1]The nucleus must be (positively) charged, because it repelled the positive the (positively charged) α-particles. [1]When high-speed electrons are fired at the nuclei, they show a diffraction pattern. [1]The moving electrons travel as waves in accordance with the de Broglie equation

(λ = p

h). [1]

The diffraction is significant when the wavelength of the electrons is similar tothe size of the nucleus. [1]

2 a i An arrow pointing away from N but in line with N and A. [1]

ii Arrow along a line passing perpendicular to the tangent at the point of closestapproach and directed away from the path of the α-particle. [1]

b i KE = 21 mv2 = 8.0 × 10–13 J [1]

v2 = 27

13

107.6

100.82−

−

×××

; v = 1.545 × 107 m s–1 ≈ 1.5 × 107 m s–1 [1]

ii p = mv = 6.7 × 10–27 × 1.5 × 107 ≈ 1.0 × 10–19 kg m s–1 [1]

c i KE = 21 mv2 =

m

p

2

2

so for the same KE, p ∝ m [1]

A proton has a quarter of the mass of an α-particle; hence for the same kinetic energy, the initial momentum will be half.Therefore, momentum = 0.50 × 10–19 kg m s–1. [1]

ii Any two from: average force is smaller, smaller recoil or less interaction time. [2]

COAS Physics 2 16


d F = 20πε4 r

Qq[1]

F = 2130 )103.7(πε4

792−××

× ee[1]

F = 21312

219

)1037(10858π4

)106.1(792−−

−

×××××××

.. ≈ 6.8 × 10–2 N [2]

3 a i Number of protons = 3 [1]

ii Number of neutrons = 7 – 3 = 4 [1]

iii The number of electrons for a neutral atom is 3. Since the net charge is +e,this ion must have 2 electrons. [1]

b i Isotopes are nuclei of the same element with different numbers of neutrons butthe same number of protons. [1]

ii The molar mass of iron-55 is 55g. [1]

Number of nuclei = 23

6

1002.655

1030

××× −

[1]

= 3.28 × 1017 ≈ 3.3 × 1017 [1]

iii A = 55 – 3 = 52; Z = 26 – 1 = 25 [1]

Hence the new nucleus can be represented as X5225 . [1]

4 a i Proton or neutron. [1]

ii Electron. [1]

b Proton = (u u d) [1]Neutron = (u d d) [1]

c i Any 3 from: charge, momentum, baryon number, strangeness and mass–energy. [3]

ii For proton, B = +1 [1]For neutron, B = +1 [1]Hence baryon number for π+ particle = (+1 + 1) – (+1 + 1) = 0 [1]

Chapter 131 a i These are slow-moving neutrons. [1]

ii Thermal neutrons have a greater chance of interacting with the 235U nucleus. [1]

b Binding energy of uranium nucleus = 7.6 × 235 = 1786 MeV [1]Binding energy of products = 146 × 8.2 + 87 × 8.6 = 1945.4 MeV [1]Energy released = 1945.4 – 1786 = 159.4 MeV ≈ 159 MeV [1]

2 a Number of protons = 92; number of neutrons = 143 [1]

b Radius = r = 1.41 × 10–15 × 235⅓ = 8.701 × 10–15 m [1]

Density = volume

mass = 315

25

)10701.8(π4

1089.3−

−

×××

[1]

Density ≈ 1.41 × 1017 kg m–3 [1]

c The mass of the nucleons is greater than the mass of the nucleus. [1]This difference is equivalent to the binding energy of the nucleus. [1]

COAS Physics 2 17


d Determine the total mass of the nucleons (143mn + 92mp). [1]Determine the mass defect ∆m by subtracting the mass of the nucleus from the totalmass of the nucleons. [1]Calculate the binding energy ∆E of the nucleus using ∆E = ∆mc2. [1]

Binding energy per nucleon = 235

E∆[1]

3 Mass of helium nucleus = 4 × 1.67 × 10–27 = 6.68 × 10–27 kg [1]

Number of nuclei in 1.00 kg = 271068.6

00.1−×

= 1.50 × 1026 [1]

Energy generated = 1.50 × 1026 × (28.4 × 106 × 1.6 × 10–19) ≈ 6.8 × 1014 J [1]

Chapter 14

1 a For β+: Au19279 → e0

1+ + Pt19278 + ν [1]

For β–: Au19279 → e0

1− + Hg19280 + ν [1]

b For β+: initial mass = 191.92147 u [1] final mass = 191.91824 + 0.00055 = 191.91879 u [1]The final mass is less than the initial mass, so β+ decay can occur spontaneously. [1]For β–: initial mass = 191.92147 u [1] final mass = 191.92141 + 0.00055 = 191.92196 u [1]The final mass is greater than the initial mass, so β– decay cannot occur naturally. [1]

c For β+: mass loss = 191.92147 – 191.91879 = 0.00268 u [1] = 0.00268 × 1.66 × 10–27 = 4.45 × 10–30 kg [1]

energy emitted = ∆mc2 = 4.45 × 10–30 × (3.0 × 108)2 [1] = 4.00 × 10–13 J [1]

2 a U23692 → Zr100

40 + Te13152 + 5 n1

0 [1]

b i No change in the nucleon number. Proton number increases by 1. [1]

ii Nucleon number = 100 and proton number = 44. [1]

3 a A: This is the activity of the source; it is the rate of decay of nuclei. [1]λ: This is the decay constant; it is the probability of decay of a nucleus per unit time interval. [1]N: This is the number of undecayed nuclei in the sample. [1]

b i Z = 90 and A = 234 [1]

ii Mass = 2.0 × 10–6 × 7.0 × 106 = 14 kg [1]

iii Number of nuclei = 238

000 14 × 6.02 × 1023 = 3.54 × 1025 [1]

iv λ = 79 102.3105.4

693.0

××× = 4.81 × 10–18 s–1 [1]

A = λN = 4.81 × 10–18 × 3.54 × 1025 ≈ 1.7 × 108 [1]Unit: Bq or s–1 [1]

c There is a change of state, with water changing into steam (water vapour). [1]This change of state requires latent heat. [1]In order to calculate the energy required to heat the steam beyond 393 K, the specific heat capacity of the steam must be known. [1]

4 a i 29 protons [1]

COAS Physics 2 18


ii 34 neutrons [1]

b λ = 7102.3120

693.0

×× = 1.805 × 10–10 s–1 ≈ 1.8 × 10–10 s–1 [1]

c i Q = VC = 90 × 1.2 × 10–12 [1]Q = 1.08 × 10–10 C ≈ 1.1 × 10–10 C [1]

ii Number = e

Q =

19

10

106.1

1008.1−

−

××

[1]

Number = 6.75 × 108 ≈ 7 × 108 [1]

iii A = λN [1]

N = λA

= 10

8

10805.1

1075.6−×

×[1]

Number ≈ 3.7 × 1018 [1]

iv 1 year is less then 1% of 120 years, so expect it to be within 1%. [1]

5 a i Bi21283 → Po212

84 + e01− + ν [2]

ii Po21284 → Pb208

82 + He42 [2]

b Place a radiation detector close to the source to determine the count rate. [1]Place a thin sheet of paper between the detector and source. If the source emitted α radiation only, the count rate would drop dramatically to almost the background rate. [1]If the source emitted β radiation only, the paper would have little effect, but a sheet of aluminium at least a couple of millimetres thick would reduce the count rate to close to the background rate. [1]If the source emits both α and β radiation, the paper will partially reduce the count rate; the aluminium will further reduce it to close to the background rate. [1]

c i The molar mass of bismuth is 212 g. 1 mole contains 6.02 × 1023 nuclei. [1]

Number of nuclei = N = 212

100.1 9−× × 6.02 × 1023 = 2.84 × 1012 [1]

A = λN = 0.0115 × 2.84 × 1012 [1]A ≈ 3.27 × 1010 min–1 ≈ 3 × 1010 min–1 [1]

ii t½ = 0115.0

693.0 ≈ 60 min [1]

iii Smooth curve passing through (0, 33 × 103 min–1), (60 min, 16 × 103 min–1) and (120 min, 8 × 103 min–1). [1]

Chapter 151 a Any seven from:

X-ray source and detectors round patient. [1]X-ray source is rotated around patient. [1]X-rays are absorbed by tissue and bone of patient. [1]X-rays are absorbed more by high-Z matter such as bone (or absorbed less by low-Z matter such as soft tissues). [1]The attenuation mechanism is mainly the photoelectric effect. [1]There is a possibility of using a contrast medium (e.g. barium or iodine). [1]A computer is used to process the image. [1]A 3-D image is produced. [1]

COAS Physics 2 19


b The patients are exposed to ionising radiation. [1]Ionising radiation could cause cancer / damage cells. [1]It is expensive or time-consuming. [1]

2 Any seven from:C (cathode) is the negative heater and A (anode) is the positive target (e.g. tungsten). [1]Electrons are accelerated from C to A. [1]The high-speed electrons hit the target metal and produce X-ray photons. [1](Some of) the kinetic energy of the electrons is transferred into X-ray photons. [1]About 1% of the incident electron energy is converted into X-rays; the rest is transferred into heat in the target metal. [1]The graph of intensity against X-ray photon energy shows characteristic spectral linesand ‘broad-background’ braking or Bremsstrahlung radiation. [1]

The minimum wavelength λ of the X-rays can be determined using eV = λhc

. [1]

3 a I = I0e–µx [1]347 = I0e–250 × 0.025 [1]

I0 = 31093.1

347−×

= 1.78 × 105 W m–2 [1]

b intensity = area sectional-cross

power

power = intensity × cross-sectional area = 347 × π × (0.10 × 10–2) 2 [1]power = 1.09 × 10–3 W ≈ 1.1 mW [1]

c i power of electron beam = 18 × 15.0

100[1]

= 1.2 × 104 W [1]

ii 21 mv2 × 7.5 × 1017 = 1.2 × 104 [1]

v = 1731

4

105.71011.9

102.12

×××××

− = 1.87 × 108 m s–1 ≈ 1.9 × 108 m s–1 [1]

4 a Graph as shown in figure. [1]Six points plotted correctly. [1]Smooth curve drawn through the data points. [1]Additional (seventh) point plotted correctly. [1]

b i Thickness = (1.0 ± 0.1) mm (see example dashed lines on figure) [1]

ii I = I0e–µx [1]0.5 = e–µ × 1.0 [1]–µ × 1.0 = ln 0.5 [1]

µ = 0.1

5.0ln

− ≈ 0.69 [1]

Unit: mm–1 (Answer is 690 if unit is m–1.) [1]

5 Formation of image – any three from:The X-rays are detected by a film. [1]High Z (or proton number) means greater attenuation. [1]

COAS Physics 2 20


Reference to the photoelectric effect or energy range around 1 keV – 100 keV [1]Attenuation (absorption) coefficient ∝ Z3 [1]

Explanation of contrast material:X-rays do not show soft tissues well. [1]Low Z (or proton number) of tissues means less attenuation. [1]Contrast material has high Z or example given, e.g. barium or iodine. [1]Taken orally or as an enema, or can be injected. [1]

Example of type of structure that can be imaged:Intestines or throat or stomach. [1]

Chapter 161 a Any one from:

Bone growth, blood circulation in lung/brain/liver, function of the heart/brain. [1]

b i Mo9942 → m99

43Tc + e01− + ν [1]

m9943Tc → Tc99

43 + γ [1]

ii Technetium-99m emits γ-ray photons and these will pass through the patient. [1]Technetium-99m has a short half-life (of 6 h) and hence does not stay inside the patient for a long time. [1]

iii 1 λ = 21

693.0

t = 360067

693.0

× = 2.87 × 10–6 s–1 ≈ 2.9 × 10–6 s–1 [1]

2 A = λN

N = λA

= 6

6

1087.2

10600−×

× = 2.09 × 1014 [2]

Mass = 23

14

1002.6

1009.2

××

× 99 ≈ 3.4 × 10–8 g [1]

3 Activity = A0e–λ t = 600 × 3600301087.2 6

e ×××− −[1]

Activity = 440 MBq [1]

2 a i When a strong magnetic field is applied to nuclei, they start to precess. [1]The natural angular frequency of precession of the nuclei is known asthe Larmor frequency. [1]

ii The nuclei resonate when radio frequency waves are directed at the nuclei. [1]Then these radio waves are switched off, the nuclei take some time to go back to their lower energy state – this time is referred to as the relaxation time. [1]The relaxation time depends on the environment of the nuclei. [1]

b It does not use ionising radiation and therefore is not dangerous. [1]MRI gives better soft-tissue contrast than a CAT scan. [1]

c A large superconducting magnet, which produces the strong external magnetic field. [1]A set of gradient coils, which produce an additional external magnetic field that varies across the length, width and depth of the patient’s body. [1]A radio-frequency (RF) coil that transmits radio-wave pulses into the patient. [1]A radio-frequency (RF) coil that detects the signal emitted by the relaxing nuclei. [1]A computer that controls the gradient coils and RF pulses and which produces anddisplays images. [1]

3 a A is the collimator. [1]This is made of lead tubes. It ensures that only γ-ray photons travelling along theaxis of the tubes are detected, and thus ensures a sharp image. [1]B is the scintillation crystal. [1]

COAS Physics 2 21


This changes the γ-ray photons into visible light photons. [1]C is the computer. [1]This processes the signals from the photomultiplier tubes to form an image. [1]

b i Number = 39 [1]Number = 19 683 ≈ 2.0 × 104 electrons [1]

ii Current = t

Q =

t

Ne[1]

Current = 9

194

100.2

106.11097.1−

−

××××

[1]

Current ≈ 1.6 × 10–6 A

Chapter 171 a Acoustic impedance Z = ρc, where ρ is the density of the material and

c is the speed of ultrasound in the material. [1]Unit: Z → [kg m–3 × m s–1] → [kg m–2 s–1] [1]

b i Air–skin boundary:

0

r

I

I = 2

12

212

)(

)(

ZZ

ZZ

+−

= 26

26

)430106.1(

)430106.1(

+×−×

[1]

Fraction reflected = 0.9989 [1]Fraction transmitted = 1 – 0.9989 ≈ 0.0011 (0.11%) [1]

Gel–skin boundary:

0

r

I

I = 2

12

212

)(

)(

ZZ

ZZ

+−

= 266

266

)106.1107.1(

)106.1107.1(

×+××−×

[1]

Fraction reflected = 0.00092 [1]Fraction transmitted = 1 – 0.00092 ≈ 0.999 (99.9%) [1]

ii With gel on the skin, there is acoustic matching. [1]Only a small amount of ultrasound is reflected: most enters the body.Hence a useful image is produced. [1]

2 a Any five from:A voltage is applied to the piezoelectric crystal. [1]This makes the crystal change its shape. [1]The crystal material is named, e.g. quartz. [1]An alternating voltage makes the crystal vibrate. [1]The crystal resonates when the applied voltage has the same frequency asthe natural frequency of the crystal. [1]The crystal is damped to stop it vibrating when the voltage is switched off. [1]This is achieved be a backing material/epoxy resin. [1]

b i Distance 5.4 cm ± 0.1 cm read from graph. [1]Time interval = 5.4 × 20 µs = 108 µs [1]Distance = speed × time = 1500 × 108 × 10–6 = 0.162 m [1]

Size of fetal head = 2

162.0 = 0.081 m = 8.1 cm [1]

ii There is greater intensity of reflected ultrasound at the skin. [1]This is because of the large difference between the acoustic impedancesof air and skin. [1]The peaks reflected from the fetal head will have very small ‘heights’. [1]

3 a The two factors are: speed of ultrasound and density of the material. [2]

COAS Physics 2 22


b i0

r

I

I = 2

12

212

)(

)(

ZZ

ZZ

+−

0

r

I

I = 4.42 × 10–4 =

2

12

12

+−

ZZ

ZZ[1]

+−

12

12

ZZ

ZZ = 41042.4 −× = 0.0210 [1]

The values of

+−

12

12

ZZ

ZZ worked out from the table are:

soft tissue–blood: 0.0124; soft tissue–water: 0.0415; soft tissue–brain: 0.0156soft tissue–bone: 0.654; soft tissue–muscle: 0.0210 [1]The unknown medium is muscle. [1]

ii ∆t = 26.5 × 10–6 s c = 1.54 × 103 m s–1

Depth = 2

tc∆ =

2

1054.1105.26 36 ××× − = 0.0204 m [1]

Depth = 2.04 cm [1]

iii c = fλ

λ = f

c =

6

3

105.3

1054.1

××

[1]

Wavelength = 4.4 × 10–4 m [1]

Chapter 181 a There are fusion reactions within the core of the star; these release photons. [1]

The radiation pressure from the photons prevents gravitational collapse. [1]

b A red giant is a star that increases in size when its hydrogen fuel becomes exhausted.[1]A red giant has enormous size; for a star the mass of the Sun, up to the radius ofthe Earth’s orbit. [1]It appears red because its surface is cooler than that of our Sun. [1]It is brighter than our Sun because it has a greater surface area than our Sun, and so emits greater power. [1]

c It first evolves into a super red giant. [1]As the nuclear fuel runs out, the star rapidly collapses against its neutron core. [1]This generates a shockwave that explodes the stellar material into space. [1]This explosion is known as a supernova. [1]The remnant is either a neutron star or a black hole. [1]

d There are no fusion reactions within a white dwarf; hence it cannot be a star. [1]It simply leaks away radiation and energy from past fusion reactions. [1]

2 a The wavelengths have been redshifted. [1]Hence, the star must be moving away from us. [1]

bλλ∆

= c

v[1]

c

v =

5.119

4.1 = 0.0117 [1]

v = 0.0117 × 3.0 × 108 [1]Speed ≈ 3.5 × 106 m s–1 [1]

3 a The composition of a star can be determined by examining the spectrum obtained from it. [1]The spectrum is compared with the spectra of known elements in the laboratory. [1]

COAS Physics 2 23


b i Parallax is the apparent change in the position of a star against the backgroundof distant stars. [1]The parallax is obtained from measurements taken when a star is viewed from different positions (e.g. different positions of the Earth in its orbit). [1]

ii [1]

The parsec is the distance at which the radius of the Earth’s orbit around the Sun [1]subtends an angle of 1 arc second. [1]

4 a Olber’s paradox: In an infinite and static universe [1]the night sky ought to be bright. [1]This is because the line of sight always ends up on a star. [1]Hubble’s law: speed of recession of galaxy ∝ distance of the galaxy from us [1]All galaxies in the universe are receding from each other. [1]This implies that the universe must have had zero size / been a singularityin the finite past. [1]

b i 1 pc = 3.1 × 1016 m [1]

H0 = 166 101.310

000 70

××[1]

Hubble constant = 2.26 × 10–18 s–1 ≈ 2.3 × 10–18 s–1 [1]

ii Age = 0

1

H = 181026.2

1−×

Age = 4.43 × 1017 s ≈ 4.4 × 1017 s [1]

iii Size of observable universe = speed of light × ageSize = 3.0 × 108 × 4.43 × 1017 [1]Size ≈ 1.3 × 1026 m [1]

c The universe has expanded at a constant rate. [1]

5 a i Distance (pc) = seconds) (arcparallax

1[1]

Distance = seconds arc 314.0

1

Distance = 3.18 pc [1]

ii 1 pc = 3.1 × 1016 mDistance = 3.18 × 3.1× 1016 ≈ 9.9 × 1016 m [1]

Chapter 191 a Any six from:

Very high temperatures at the Big Bang. [1]Quarks and leptons formed. [1]

COAS Physics 2 24


Temperature of the universe decreases as it expands. [1]Strong nuclear force takes effect. [1]Protons, neutrons and pions are formed. [1]Helium nuclei are formed by fusion. [1]25% of the mass is helium. [1]Lots of γ radiation. [1]The universe becomes transparent to electromagnetic radiation. [1]

b Temperature of 3 K (or 2.7 K). [1]Agrees with expected cooling of the universe. [1]

2 a The universe is homogeneous and isotropic. [1]

b i H0 = 166

3

101.310

105.7

××

= 2.419 26 × 10–18 s–1 [1]

Age = 0

1

H = 1810419.2

1−×

= 4.133 × 1017 s [1]

Age = 360024365

10133.4 17

×××

= 1.3 × 1010 years [1]

ii The universe is flat. [1]The density of the universe is equal to the critical density. [1]

iii Curve passing through point P. [1]The curve returns to the time axis. [1]

iv The universe has a finite lifetime. It ends up in a Big Crunch. [1]

3 a [1]

Open: universe expands for all time; gradient of the graph is always > 0, even after infinite time. [1]

Closed: universe collapses back. [1]

Flat: universe expands for all (finite) time; gradient of the graph is zero after infinite time. [1]

Some correct reference to critical density (e.g. density > critical density for an open universe). [1](Can score the marks on a fully labelled diagram.)

b ρ0 = G

H

π8

3 20

20H =

3

1011067.6π8 2611 −− ××××[1]

H0 = 2.36 × 10–18 s–1 ≈ 2.4 × 10–18 s–1 [1]

COAS Physics 2 25

Answers to End-of-chapter questions - Cambridge OCR...

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