CCEA GCSE Physics · Web view2018. 4. 26. · CCEA GCSE Physics. Unit 1. Unit 2. This Answers...

WORKBOOK ANSWERSCCEA GCSE Physics Unit 1

Unit 2

This Answers document provides suggestions for some of the possible answers that might be given for the questions asked in the Workbook. They are not exhaustive and other answers may be acceptable – they are intended as a guide to give teachers and students feedback.

Exam-style question answers

The answers given for exam-style questions of an explanatory, or evaluative, type set out what is called ‘indicative content guidance’. Just showing the examiner you are familiar with some or most of the content will not suffice. You need to demonstrate that you understand it and are willing and able to use it in a way that directly addresses the question. The indicative content shown for each question is not exhaustive. Questions may be approached in a number of different ways. The choice of approach is yours. Just make sure your approach answers the question.

Assessing whether or not a question has been answered is where ‘levels marking’ comes in. For questions with maximum marks of 6, 12 or 15, three levels of attainment are recognised. Where the maximum mark is 20, four levels are defined. The table below gives an idea of the descriptors that are used to determine the quality of an answer and the mark to be awarded.

Level DescriptorMark (depending on maximum)

1 Demonstrates isolated elements of knowledge and understanding.Question not addressed.

1–2 (max. 6)1–4 (max. 12)1–5 (max. 15)

2 Demonstrates knowledge and understanding — not always relevant or accurate.Shows some awareness of the question.

3–4 (max. 6)5–8 (max. 12)6–10 (max. 15)

3 Demonstrates knowledge and understanding that is mostly relevant and accurate.Addresses the question directly.

5–6 (max. 6)9–12 (max. 12)11–15 (max. 15)

CCEA GCSE Physics Workbook

© Roy White 2018 Hodder Education

UNIT 1

Unit 1

Motion

There are often several ways to answer these questions. Students doing GCSE Further Maths will be aware of Newton’s equations of motion, which can make the work quicker. The answers here use only the equations required by your specification.

1 a Vectors have magnitude and direction, scalars have magnitude only.

b Velocity and acceleration are vectors (V). Speed, distance and rate of change of speed are scalars (S).

2 a Distance = 150 + 105 + 150 + 105 = 510 m

b Average speed =

c Average velocity =

Kevin’s displacement is zero, so his average velocity is 0.

3 a Average speed =

Average speed =

Now , so v = 4 m/s

b

c The acceleration down the ramp is uniform; OR friction is constant.

d The reaction time when using a stopwatch is so long that it is unsatisfactory when measuring short times. It would be better to use a light gate and electronic timer.



UNIT 1

4 a

b

c A sloped deck means that the aircraft is already rising when it leaves the carrier, so it gains height quicker. It can also mean that the aircraft experiences higher resistance on landing, so it comes to rest quicker.

Note that questions requiring you to ‘suggest’ a reason mean that some creative thinking is required.

5 Indicative content:

Hypothesis: average speed increases with height of runway, but in a non-linear way (1)

Apparatus: runway, trolley (or ball bearing etc.), stopwatch (or timer with light gate), ruler (or tape), pencil (2)

Description to include (3):

A runway is raised to create a ramp

Vertical height of the ramp is measured

Horizontal pencil lines at start and end of runway

Measure the distance between the lines

Allow the trolley etc. to move from upper line from rest

Time from the start to the ball (or trolley wheel) crossing finishing line

Carry out repetitions for reliability

Carry out repetitions for various ramp heights

Calculation: average speed = (1)

To confirm the hypothesis, plot a graph of average speed against runway height. A rising graph confirms the hypothesis; a decreasing graph refutes it. (1)

QWC: to obtain this mark your spelling, punctuation and grammar must be accurate and you present your work logically using appropriate scientific terms. (1)

Tip: Always give as much detail as you can in QWC questions.



UNIT 1

Exam-style questions

1 a i Average speed =

ii Average speed =

initial speed = 2 × 12 = 24 cm/s

iii

iv 24 cm/s

b i Average speed =

ii Average velocity =

2 a The graph should have:

the axes labelled exactly as in the table

a scale chosen to use as much of the grid as possible, consistent with ease of use

the five points plotted accurately (to within a small square)

a ruled line of best fit passing through the points

b The initial speed is the intercept on the vertical axis, 0.3 m/s

c Gradient =

So (say) gradient =

3 a Final speed v = u + at = 0 + (10 × 2) = 20 m/s

Average speed = ½(u + v) = ½(0 + 20) = 10 m/s

Distance = average speed × time = 10 m/s × 2 s = 20 m



UNIT 1

b Over the three seconds of the motion:

Final speed v = u + at = 0 + (10 × 3) = 30 m/s

Average speed = ½(u + v) = ½(0 + 30) = 15 m/s

Distance = average speed × time = 15 m/s × 3 s = 45 m

Distance travelled in third second = distance travelled in 3 s − distance travelled in 2 s

= 45 m − 20 m = 25 m

Motion graphs

1 a Gnasher is at rest (not moving)

b Gnasher is moving at a steady speed between A and B

Speed =

c Straight line from (0 s, 20 m) to (100 s, 120 m)

d The rabbits are at same position where the graphs cross: at 20 s, 40 s and 100 s.

2 a Distance = area under graph up to 20 s = (15 × 3) + ½(3 + 1.4) × 5 = 56 m

b Acceleration = gradient =

c Total distance travelled = 56 m (up to 20 s) + 14 m (from 20 s to 30 s)

Average speed =


1 a Rate of change of speed = gradient =

b Horizontal line from (0 s, 15 m/s) to (60 s, 15 m/s)

c Distance = area under the graph

First bus travels ½(60 + 45) × 20 = 1050 m

Second bus travels 15 × 60 = 900 m



UNIT 1

d No. The buses may not be travelling in same direction; nor may they have started from the same position

2 a 10 s (between t = 80 s and t = 90 s only)

b Between t = 70 s and t = 80 s (the graph has a negative slope and is the steepest)

c Retardation = change∈velocitytimetaken =

−12.4m /s10 s = −1.24 m/s2

d Its velocity is always positive

e Distance = area under the graph up to 50 s = m = 556 m

Newton’s first and second laws

1 a Every object remains at rest or moves with a constant velocity unless compelled to do otherwise by a force

b i 0 N

ii 650 N

2 a Friction (or drag)

b Resultant force = 55 − 15 = 40 N

c

d Mass of bicycle = combined mass − mass of cyclist = 80 kg − 60 kg = 20 kg

3 Resultant force = ma = 0.125 kg × 25 m/s2 = 3.125 N

Upward force = weight of rocket + resultant force = 1.25 N + 3.125 N = 4.375 N

4 a Final speed after 4 s = 2 × average speed = 2 × 16 m/s = 32 m/s

a =

Resultant force = ma = 60 kg × 8 m/s2 = 480 N

b Drag force = weight − resultant force = 600 N − 480 N = 120 N



UNIT 1


1 a a =

b Increase in speed = a × t = 30 m/s2 × 130 s = 3900 m/s

c As the burn continues, fuel is used up, so the mass being accelerated decreases

2 There are two ways to solve this problem

First: The additional 500 N causes an increase of 1 m/s2 in the acceleration

Using F = ma gives:

500 = m × 1 and m = 500 kg

Then using Fengine − friction = ma gives:

700 − friction = 500 × 1, which leads to friction = 200 N

Second: This requires solving simultaneous equations. Using F = ma twice gives:

1200 − friction = 2m

700 − friction = 1m

Subtracting gives:

500 = m (the same as in the first way above)

So as above m = 500 kg and friction = 200 N

Mass, weight and Hooke’s law

1 Mass is the amount of matter in an object; weight is the force due to the pull of gravity on an object

2Earth Moon Mars

Value of g / m/s2 10 1.6 3.8Weight of object / N 50 8 19Mass of object / kg 5 5 5

3 a The extension of a helical spring is directly proportional to the applied force, provided the limit of proportionality is not exceeded

b F = ke (or F = kx)



UNIT 1

4 a 6 N stretches the spring 9 cm

2 N stretches the spring 3 cm

So, the total length with 2 N = 9 cm + 3 cm = 12 cm

b

c The spring has been stretched beyond its elastic limit (or proportionality limit) and is now permanently deformed

5 Hypothesis: up to a certain limit the extension is directly proportional to the stretching force

Apparatus: iron stand, boss head, clamp, helical spring, mass carrier, selection of 100 g masses, ruler

Dependent variable: the extension of the spring

Independent variable; the stretching force applied to the spring

What to do and measure:

Attach the spring and mass carrier to the clamp, boss and iron stand

Measure the original length of the spring using the ruler

Attach a 100 g mass (1 N force) to the carrier and measure the new length of the spring

Repeat for additional 100 g masses up to about 600 g

Confirming the hypothesis:

Subtract the original length of the spring from each of the spring length measurements to work out the extension for each load

Plot a graph of load against extension

If the line of best fit is straight (up to a limit) and goes through the (0, 0) origin, the hypothesis is confirmed; otherwise it is rejected


1 a The graph should be labelled as in the table, with a linear scale on axes covering at least half of the paper available. The graph curves after (19 cm, 12 N).

b 13 cm; the intercept on the total length axis



UNIT 1

c

d 12 N (approximately)

e A straight-line graph passing through (9 cm, 0 N) and (17 cm, 12 N)

Pressure

1 Pascal (Pa), N/cm2, N/mm2

2

3 Weight = m × g = 180 × 10 = 1800 N

Area = 0.6 m × 0.5 m = 0.3 m2

Pressure

4 a i The raft has a large area, so the total weight of the house exerts only a small pressure on the ground; this means the house is less likely to sink into the soft earth

ii The pin’s point has a very small area so even a small force (from the thumb) allows it to exert a huge pressure on the notice board


1 a A = 0.30 m × 0.25 m = 0.075 m2

b F = P × A = 70 000 × 0.075 = 5250 N

c Force from outside = 5250 N − 3750 N = 1500 N

Pressure = 20 kPa

2 a W = mg = 0.5 × 10 = 5 N



UNIT 1

b F = P × A = 100 000 × 0.005 = 500 N

c The upward force (500 N) is greater than the force due to the cup and water on the card plus the weight of card itself (6 N); so the card is likely to remain in place

Moments

1 80 N cm

2 a The weight on the pan plus weight of pan itself must be 2 N (because the weights are same distance from pivot)

So, the weights in pan must add to 1.8 N; and there must be nine weights (each of 0.2 N)

b The clockwise moment from 3 N is now greater than the anticlockwise moment from the pan’s weights

So the clockwise moment must be increased; so move the pivot to the left

c Suppose the distance moved by the pivot to the left is d. Then, applying the principle of moments:

3 × (50 − d) = 2 × (50 + d)

150 − 3d = 100 + 2d

So 5d = 50; and d = 10 cm

3 a W acts vertically downwards from the centre of the beam (100 cm from each end)

F acts vertically upwards through the pivot

b Anticlockwise moment = clockwise moment

30 kN × 40 cm = W × 60 cm

W = 20 kN

Forces upwards = forces downwards;

so F = 30 kN + 20 kN = 50 kN


1 a At A the anticlockwise moment = the clockwise moment

600 × (3 − 1) = RB × 5



UNIT 1

RB = 240 N

b At B the anticlockwise moment = the clockwise moment

600 × 3 = RA × 5

RA = 360 N

c The sum of the upward reactions is 600 N; which is the weight of the plank

Centre of gravity

1 The centre of gravity of an object is the point through which the entire weight of the body can be thought to act.

2

3 Tumbler A is likely to be less stable than tumbler B because it has a higher centre of gravity; both tumblers are in stable equilibrium; a stationary ball on the table is likely to be in neutral equilibrium


1 It has a wide, heavy base; which gives it a low centre of mass

It is in stable equilibrium, so it requires considerable force to topple

2 a Gravity

b The tractor topples only if the weight vector is outside the base, producing a resultant moment; making the area between the four wheels as big as possible increases the area of the base, making this less likely

Density

1 D = M/V and kg/m3



UNIT 1

2 a

b

3 V = 76 − 65 = 11 cm3

The material is probably lead because its density is closest to 11 g/cm3



Use a ruler (or digital calipers) to measure the length, breadth and height of the solid in cm

The product of these three numbers is the solid’s volume in cm3

Measure the mass of the solid, in grams, on a top-pan balance

Divide the mass by the volume to find the density in g/cm3

2 Repeat for reliability and average the values calculated to obtain the most reliable value for the density

3 a Mass of salt water = density × volume = 1.06 g/cm3 × 500 cm3 = 530 g

Mass of water = 500 g

Mass of salt = 530 − 500 = 30 g

b The mass of salt per unit volume is halved; so the technician must add 500 cm3 of pure water to 500 cm3 of salt water

Energy forms and resources

1 HEat, ELectricity, SOund, LIght, MAgnetism, STrain, KInetic, GRavitational

2 a The law of conservation of energy states that energy can be neither created nor destroyed, but it can change its form.

b Y, 600 J (the useful output energy can never be greater than the input energy)



UNIT 1


1Oil Wind Nuclear Solar Hydroelectric Coal Tidal

Renewable (R) or Non-renewable (N) N R N R R N R

2 a i Gas, peat

ii They burn to produce carbon dioxide, a polluting greenhouse gas

b i Nuclear

ii They are very radioactive

Energy flow and efficiency

1

Input energy Device Useful output energySound Microphone electricalChemical Car engine kineticElectrical Toaster heatElectrical Loudspeaker soundChemical Battery electrical

2 a Kinetic energy (useful form) = 1250 − (740 + 260) = 1250 − 1000 = 250 kJ

Useful fraction =

b 0.2 (the same as part a)

3 Efficiency =


1 Input energy =

2 Total input energy = total output energy = 750 kJ + 1350 kJ + 900 kJ = 3000 kJ

Useful output energy = 900 kJ (kinetic) + (⅓ × 1350 kJ) = 900 + 450 = 1350 kJ



UNIT 1

Efficiency =

Work and power

1 Energy = power × time = 24 kW × (3 × 60 s) = 4320 kJ = 4.32 MJ

2 a The weightlifter was applying an upward force on the mass; and it was rising upwards

b Gravity

3 Weight lifted = 6 N; Work done = 9 J; Power = 300 mW (0.3 W)


1 a W = force × distance = 120 N × 60 m = 7200 J

b Power =

2 Force =

3 Measure the weight, W, of the student in newtons using bathroom scales

Use a ruler to measure the heights of about five risers in a staircase

Hence work out the average height of a riser

Count the number of stairs in the staircase

Multiply the number of stairs by the average height of a riser to find the vertical height, H, of the staircase

Use a stopwatch to measure the time taken, t, for the student to run from the bottom of the staircase to the top

Calculate the power, P, of the student using the equation

For greater reliability, repeat the experiment three times and calculate the student’s average power

4 a 650 N

b W = Fd = 650 N × (120 × 0.45)m = 35 100 J



UNIT 1

Kinetic energy and gravitational potential energy

1 KE staying the same; GPE decreasing

2 a GPE = mgh = 0.050 kg × 10 N/kg × 280 m = 140 J

b 140 J

3Height above ground / m GPE / J KE / J Total energy / J Speed / m/s

2.5 100 0 100 02.0 80 20 100 3.21.6 64 36 100 4.20.9 36 64 100 5.70 0 100 100 7.1


1 a Energy lost = 15 000 − 7200 = 7800 J

b KE = ½mv2

7200 = ½ × 64 × v2

v = 15 m/s

c Work done against friction = friction × distance = energy lost

friction × 195 = 7800 J

So friction =

2 KE = original energy − GPE − energy lost as heat and sound = 120 J − 72 J − 28 J = 20 J

3 Useful work = KE of car = ½ × 800 × 42 = 6400 J

Work done by pushers = F × d = (2 × 250) × 20 = 10 000 J

Work done against friction = 10 000 − 6400 = 3600 J



UNIT 1

Friction =

Atomic structure

1Particle Relative mass Relative charge LocationProton 1 +1 Inside nucleusElectron 1/1840 −1 Orbiting nucleusNeutron 1 0 Inside nucleus

2 Atomic number — Number of protons in the nucleus — Symbol, Z

Mass number — Number of particles in the nucleus — Symbol, A

3 a Neutrons 143, protons 92, electrons 92

b

c Isotopes are different forms of the same element having the same atomic number, but different mass number. This means their nuclei have the same number of protons, but different numbers of neutrons

4 a Rutherford–Bohr model

b Plum pudding model

cPlum pudding model Rutherford–Bohr modelElectrons are distributed throughout the atom like currents in a bun

Discrete electrons orbit a central nucleus in circular paths

Positive charge is smeared throughout the atom, like dough in a bun

Positive charge exists as discrete particles within a central nucleus

There is no central nucleus Every atom has a tiny particle at its centre, called a nucleus, which contains almost all of the atom’s mass

5 , ,


1 Another element

2 11 protons, 12 neutrons, 0 electrons (in nucleus!)



UNIT 1

3 A and C are isotopes. Each has 11 protons. A has 12 neutrons, B has 13 neutrons


It was carried out in a highly evacuated chamber

Alpha-particles were directed at a very thin gold foil

The scattered alpha-particles hit a zinc sulfide screen mounted on a microscope lens

These impacts caused the emission of flashes of light (scintillations)

The light flashes allow the experimenter to ‘see’ the alpha-particle impacts

These flashes were observed using a moveable microscope

5 a Atoms are mostly empty space; alpha-particles were deflected when they passed close to a nucleus

b The alpha particles were back-scattered only when they came very close to a nucleus; providing evidence that a nucleus is small, relatively massive and positively charged

Radioactivity

1 Radioactivity is the random, spontaneous disintegration of an unstable nucleus by the emission of an alpha particle, a beta particle or a gamma wave

2Alpha Beta Gamma

Relative charge +2 −1 0Comes from a disintegrating

Nucleus nucleus nucleus

Relative mass 4 1/1840 0Nature particle particle EM waveRange 2–4 cm

in air0.5 cm in aluminium

many cm in lead

Ionisation ability (high/medium/low)

high medium low

Consists of 2 protons and 2 neutrons

fast electron high-energy EM waves

3 Half-life is the time taken for the activity of a radioactive source to fall to half of its original activity.

4 a The time to fall from (say) 80 Bq to 40 Bq = one half-life = 2 hours

b 4 hours = 2 half-lives



UNIT 1

2 hours before the experiment, the activity was 2 × 80 = 160 Bq

4 hours before the experiment, the activity = 2 × 160 = 320 Bq


1 a 8192 → 4096 → 2048 → 1024; so 3 half-lives are involved

3 half-lives = 12 hours; 1 half-life = 12 ÷ 3 = 4 hours

b 24 hours represents a further 3 half-lives; 1024 → 512 → 256 → 128

So the activity is 128 Bq

2 The radiation must penetrate paper but be partially absorbed; so it must be beta

The source must have a long half-life to avoid frequent replacement/recalibration; so the source must be D.

Nuclear fission and nuclear fusion

1 Nuclear fission is the splitting of a uranium nucleus by a slow neutron into two highly radioactive nuclei and two or three neutrons

Nuclear fusion is the joining together of light hydrogen nuclei to make a much heavier helium nucleus

2

3 Achieving a temperature of about 1 000 000°C and maintaining it long enough to initiate fusion;

Containment of the reactants and products at this enormous temperature

4 In stars (such as our Sun)


Uranium (nuclei) in the fuel rods in the reactor…

… absorbs a slow neutron, initiating the fission process

The nucleus splits (fissions) into two lighter, radioactive nuclei; such as barium and krypton …

… with the emission of two or three more neutrons …



UNIT 1

… and a considerable amount of energy

These fission neutrons go on to cause further fission; producing a chain reaction


1 Fission does not produce carbon dioxide, a major greenhouse gas that causes climate change

A fission reactor would provide many highly paid jobs for local people

2 a International Thermonuclear Experimental Reactor

b Controlled thermonuclear fusion; in which there is a net production of useful energy



UNIT 2

Unit 2

Waves

1 In a longitudinal wave, the particles vibrate parallel to the direction in which the wave is moving

In a transverse wave, the particles vibrate perpendicular to the direction in which the wave is moving

2 The wavelength of a transverse wave is the distance between consecutive crests.

The frequency of a wave is the number of waves that pass a fixed point in one second.

A frequency of 1 kHz means 60 000 waves pass a fixed point every minute.

The amplitude of a wave is its maximum displacement from its undisturbed position.

3 v = fλ = 300 Hz × 4 cm = 1200 cm/s

4 Gamma waves, X-rays, ultraviolet light, visible light, infrared light, microwaves, radio waves

5 Audible sound ranges in frequency from 20 Hz to 20 kHz; ultrasound has a frequency above 20 kHz

6 Distance from trawler to sea-bed = (time for echo to reach trawler) × (speed of ultrasound); = (½ × 0.4 s) × 1500 m/s; = 300 m

7 Sonar uses ultrasound; radar uses microwaves, which are electromagnetic waves


1 Wavelength = 10 m; frequency = 0.25 Hz; speed = 2.5 m/s

2

3 Distance = (time for wave to reach Y) × (speed of radar wave)

= (½ × 0.0005 s) × (3 × 108 m/s) = 75 000 m = 75 km



UNIT 2

4 Medicine: for example ultrasound scan of foetus to check development in womb;

Industry: for example checking for cracks in railway lines

Reflection and refraction of waves

1

2

3Water waves reflecting (increases / decreases / stays the same)

Water waves refracting (increases / decreases / stays the same)

Frequency stays the same stays the same

Speed stays the same decrease



UNIT 2


1 a, b, d

c 90°

2 a

b

Reflection and refraction of light

1 The image is same size as the object, laterally inverted, erect and virtual.

2 a 40°

b 50°

3



UNIT 2

4 Angle of reflection at A = 65°; the angle between the reflected ray at A and mirror A = 25°;

The angle between the incident ray at B and mirror B = 25°; the angle of reflection at B = 65°


1 Angle of incidence = 45°; angle of refraction = 34°

2 a 0o

b

3 a

b Dispersion

c The different colours in white light travel at different speeds in glass; so they refract by slightly different amounts



UNIT 2

Critical angle and total internal reflection

1 The critical angle for some glass is 42°. This means that when light, travelling in glass towards air, has an angle of incidence at the glass/air boundary of 42 degrees, the angle of refraction in the air is 90 degrees.

2 The light must be travelling in a dense medium towards a boundary with a rarer medium; and the angle of incidence in the rare medium must be larger than the critical angle.

3 a For example endoscopy, keyhole surgery

b For example optical fibres

4


1 • Put a semicircular glass block on sheet of white paper

Draw round the outline of the block with a pencil

Mark the centre, X, of the straight diameter

Direct a ray of light towards X with light entering the glass from the curved side

Move the ray box to increase the angle of incidence until the refracted ray at X just emerges along the diameter

Draw two dots on the incident ray and join them to point X

Draw the normal at X

Measure the critical angle

Repeat and find an average value for the critical angle



UNIT 2

2 If the optical fibre is bent too much, the angle of incidence in the core falls below the critical angle and light refracts out of the core.

3 Slower; total internal reflection can only occur if the light is slower in the core

Lenses

1 If the lens is broader at the centre than at the edges it is converging; if the lens is narrower at the centre than at the edges it is converging.

2 a, b, c

d 12 cm

e For example a projector

f 2


1 a A virtual image is one which cannot be projected on a screen

b i, ii, iii



UNIT 2

2 Real images in a convex lens are always inverted

3 a i Long sight (hypermetropia)

ii Convex

iii

b i Short sight (myopia)

ii Concave lens

iii

Conductors and insulators

1 Conductors have free electrons; insulators do not

2 a 1.4 V



UNIT 2

b 4.2 V

c


1 Q = It = 1.60 A × 60 s = 96 C

Number of electrons =

Ohm’s law and resistance

1

2

3 a

b



UNIT 2

4 a X

b Y

c Z


1

Resistance of wire Y is higher by 3 Ω

2

Q = It = 2 A × 15 s = 30 C

3 Except when actually recording measurements or changing the current, ensure that the power supply is switched off

The filament lamp

1 a, b



UNIT 2

c

d As the current in the lamp increases, the resistance of the lamp increases.


1 a R = V/I, but when V = 0, I = 0 and division by 0 is not possible mathematically; However, the filament does have a resistance even when no current flows through it; this can be found using a different technique

bVoltage / V 0.0 0.2 0.8 1.8 3.2 5.0 7.2Current / A 0.0 0.2 0.4 0.6 0.8 1.0 1.2Resistance / Ω 1.0 2.0 3.0 4.0 5.0 6.0

c Quadrupling the voltage does not cause the resistance to quadruple

d The atoms are vibrating with greater amplitude as they absorb more kinetic energy from the drifting electrons

Series and parallel resistors

1 24 Ω



UNIT 2

2 a R = 3 Ω + 6 Ω = 9 Ω

b so R = 2 Ω

3 a 2 V

b Because the largest resistance is 24 Ω, this has the voltage of 6 V across it; so, by proportion, the other resistors are 8 Ω and 16 Ω

4

Switch Resistance between X and Y / Ω CommentA B

Open Open 36 24 Ω in series with 2 × 24 Ω in parallelOpen Closed 32 24 Ω in series with 3 × 24 Ω in parallelClosed Open 12 First resistor is shorted out, so only 2 × 24 Ω in parallelClosed Closed 8 First resistor is shorted out, so only 3 × 24 Ω in parallel


1 The current in the 2.0 Ω resistor is 1.5 A; so the voltage across the parallel combination is 3.0 V; so the resistance of the unknown resistor is 1.0 Ω

2 a

b

c Total resistance of new combination

R2 = 6 Ω



UNIT 2

Factors affecting resistance

1 a A

b If X is (1, 6), the graph is a curve passing through (1, 6), (2, 3), (3, 2) and (6, 1)

2 a Length and temperature

b As the cross-sectional area increases, the resistance decreases

c Resistance (vertical axis) against the reciprocal of the cross-sectional area (horizontal axis)

d (say) k = RA = 12 Ω × 1 mm2 = 12 Ω mm2

e


1 aVoltage, V / V 0.24 0.24 0.24 0.24 0.24Current, I / mA 300 200 150 120 100Resistance, R / Ω 0.8 1.2 1.6 2.0 2.4Length, L / mm 100 150 200 250 300

b In every case, doubling the length causes the resistance to double

c Resistance (vertical axis) against length (horizontal axis)

d (say)

e R = kL = (0.008 Ω/mm) × (225 mm) = 1.8 Ω

2 Doubling the length doubles the resistance; halving the cross-sectional area also doubles the resistance; so the resistance of the cut from the second reel is quadrupled to 2.4 Ω

a 0.6 Ω and 2.4 Ω connected in series gives a total resistance of 3.0 Ω

b 0.6 Ω and 2.4 Ω connected in parallel gives a total resistance of



UNIT 2

Electrical energy and power

1 P = V2R

2 A

3 P = IV = 0.015 A × 3 V = 0.045 W = 0.045 J/s

The energy produced every minute is 0.045 × 60 = 2.7 J

4 A kilowatt-hour is the amount of electrical energy; used in one hour; by a device that takes in 1000 joules; of electrical energy every second


1 a Current in 60 W bulb

Current in 6 W LED

Excess current = 0.25 A − 0.025 A = 0.225 A

b Energy saved = 54 J/s = 54 × 3600 J/hour = 194 400 J

2 a Time spent in the shower per day = (5 + 7 + 8 + 10) = 30 minutes = 0.5 hours

Time per week = 0.5 × 7 = 3.5 hours

Energy used in 1 week = 3 kW × 3.5 hours = 10.5 kWh

Cost = 10.5 kWh × 16 p/kWh = 10.5 × 16 p = 168 p = £1.68

b John’s father is right; energy used in 1 day = 3 kW × 0.5 h = 1.5 kWh = 1500 × 60 × 60 J = 5400 000 J

Magnetism and electromagnetism

1 a i



UNIT 2

ii

b


1 a (soft) iron

b, c



UNIT 2

d It is insulated; if it is not insulated, the conducting iron core would short-circuit the coil

2 Increase the current in the coil; increase the number of turns (per unit length) in the coil

Force on a current-carrying conductor in a magnetic field

1 a

b Fleming’s left-hand (motor) rule

2 a When the coil is horizontal, with A to the left of B, there is maximum turning effect on the coil. The coil turns clockwise. When the coil is vertical, there is no force on the coil, but its own momentum makes the coil overshoot the vertical. As it does this the direction of the current is reversed, so the coil continues to rotate clockwise.

b Increase the current in the coil; use a stronger magnet; increase the number of turns in the coil


1



UNIT 2

2 Electromagnets can be made to be much stronger than permanent magnets; this, in turn, makes the motors more powerful

Electromagnetic induction and transmission of electricity

1 a The production of a voltage across the ends of a conductor when the magnetic flux linked with that conductor has changed (or flux lines are cut)

b The taut wire is cutting through field lines; this induces a voltage across the ends of the wire

Because the wire is part of a closed circuit, a current will flow through it and it will be detected by the ammeter

c A current would be induced as before; but because it will flow in the opposite direction, the needle on the ammeter will flick to the other side of zero; before returning to the centre

d No current is induced because no flux lines are being cut

2 a The needle would deflect to one side of zero and then return to the centre again

b The needle would deflect to the other side of zero and then return to the centre again

c Soft iron is an electrical conductor; if the coils were made of non-insulated wire, the iron would cause the coils to be short-circuited

d The current is alternating; with a frequency of 1 Hz


1 a Step-down

b

c Is × Vs = Ip × Vp so



UNIT 2

2 a i Step-up

ii The voltage increases

iii The current decreases

b i Step-down

ii The voltage decreases

iii The current increases

The Solar System

1 There are eight planets in our Solar System. They all orbit our Sun in elliptical paths. The planets closest to the Sun are called rocky planets because their surfaces are solid and we can land spacecraft on them.

The planets furthest from the Sun are called gas planets. There are four of them. These planets are much larger than the planets closer to the Sun.

The Sun is an ordinary star. It is the source of almost all of our energy.

Asteroids and comets also orbit the Sun. Asteroids are large rocks. Most asteroids are found between the planets Mars and Jupiter.

2 Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune

3 a Gravity

b A satellite is an object that goes around another

c A natural satellite is a heavenly body that orbits another; for example Earth is a natural satellite of the Sun

An artificial satellite is put into space by humans and orbits another object; for example GAMBIT was a spy satellite that orbited the Earth

d Two from: astronomy, communications, Earth observation, weather monitoring


1 a Mars

b Jupiter

c A cloud of dust and gas (usually hydrogen)

2 Gravity



UNIT 2

3 a Ice and rocks (silicates)

b Some of the ice melts as a comet approaches the Sun; comets contain water, which might contain life-forms

4 a Spacecraft might collide with an asteroid

b There are many more asteroids than planets so a collision with an asteroid is more likely

The orbits of the planets are known and highly predictable; asteroids often collide with other asteroids; their orbit paths are not predictable

Stars and their life cycle

1 Asteroid, moon, planet, star, solar system, galaxy

25 The temperature is now high enough for nuclear fusion to begin2 The pressure and density of the nebula starts to rise1 Hydrogen and dust in a nebula come together due to gravity3 The nebula becomes so hot that it emits light and infrared radiation4 As more and more hydrogen is captured, the core temperature eventually

exceeds 13 million degrees6 A star is born

3 The outward force, called radiation pressure, has been in equilibrium with the inward force of gravity

4

protostar → main sequence → red

giant → white dwarf → black

dwarf

5


1 A main sequence star is one in the main phase of its life



UNIT 2

2 a

b A neutron

3 Black holes are incredibly dense; and so have such enormous gravitational fields that nothing can escape from them, not even light

Evidence for Big Bang model

1 14 000 000 000 (14 billion) years

2 Expanded and cooled

3 C

4 Expanding

5 Electrons


1 The distance between Andromeda and the Milky Way is decreasing

2 Microwave

Space travel and life on other planets

1 When a planet is very close to its star, the temperature would be so high that life could not be sustained; when the planet is very far from its star, the temperature would be so low that life could not be sustained


1 a A planet that orbits a star other than our Sun; it is not part of our Solar System

b Carbon, nitrogen and water are essential for life like our own; water can be formed from oxygen and hydrogen

2 a The distance light travels in one year

b Distance = speed × time = (3 × 105) × (4.22 × 365 × 24 × 60 × 60) = 4 × 1013 km

c Time



UNIT 2



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