y12 mock

y12 work, and springs

y13 questions on pendulum bobsleigh and geostationary orbit

y13 capacitors in Series and Parallel answers

y13 Shm questions

Y12 Questions on Energy and work

I have put in ^ to indicate a superscript (power)

1. (a) P.E. at top = 80 × 9.8(1) × 150 = 118 000 (J) (1)


K.E. at bottom and at top = 0 (1)

Elastic P.E. at top = 0, at bottom = P.E. at top for ecf = 118 000 J (1) 3

(b) 24 N m–1 × 100 m = 2400 N 1

(c) elastic P.E. is area under F-x graph (1)

graph is a straight line so energy is area of triangle (1)

elastic P.E. = ½ × kx × x = (½kx2) (1) 2

(d) loss of P.E. = 100 × 9.8(1) × 150 = 147 000 J (1)

gain of elastic P.E. = ½ × 26.7 × 1052 = 147 000 J (1) 2

(e) idea that a given (unit) extension for a shorter rope requires a greater force 1

[9]



2. (a) (i) speed = d / t C1

= 24 / 55

= 0.436 (m s–1) allow 0.44 A1

do not allow one sf

(ii) kinetic energy = ½ m v2 C1

= 0.5 x 20 x (0.436)^2

= 1.9 (J) note ecf from (a)(i) A1

(iii) potential energy = mg h C1

= 20 x 9.8 x 4

= 784 (J) A1

penalise the use of g = 10

(b) (i) power = energy / time or work done / time C1

= (15 x 784) / 55

note ecf from (a)(iii)

= 214 (W) A1

(ii) needs to supply children with kinetic energy B1

air resistance B1

friction in the bearings of the rollers / belt B1

total mass of children gives an average mass of greater than 20 kg B1

Max B2

[10]

y12 Past questions on Hooke's Law and Young's Modulus

Note 1.2 x 10^4 means 1.2 times ten to the power of 4 (etc)
 
1. (a) (i) Stress = force / area C1


force = stress x area

= 180 x 10 ^ 6 x 1.5 x 10 ^ –4

= 27000 (N) A1

(ii) Y M = stress / strain C1

= 180 x 10^6 / 1.2 x 10^–3 or using the gradient C1

= 1.5 x 1011 N m–2 A1

(b) brittle

elastic/ graph shown up to elastic limit

obeys Hooke’s law / force α extension / stress α strain

no plastic region B3

MAX 3

[8]



2. (a) One reading from the graph e.g. 1.0 N causes 7 mm C1

Hence 5.0 (N) causes 35 +/- 0.5 (mm) A1

(allow one mark for 35 +/- 1 (mm)

(b) (i) Force on each spring is 2.5 (N) C1

extension = 17.5 (mm) allow 18 (mm) or reading from graph A1

[allow ecf from (a)]

(ii) strain energy = area under graph / ½ F x e C1

= 2 x 0.5 x 2.5 x 17.5 x 10 ^–3

= 0.044 (J) A1

[allow ecf from (b)(i)]

(c) E = stress / strain C1

Stress = force / area and strain = extension / length C1

extension = (F x L) / (A x E)

= (5 x 0.4) / (2 x10^–7 x 2 x 10^11)

= 5.(0) ^ 10–5 (m) A1

(d) strain energy is larger in the spring B1

extension is (very much larger) (for the same force) for the spring B1

[11]





3. (a) (i) F = kx / k is the gradient of the graph C1

k = 2.0 / 250 x 10^–3 = 8.0 A1

Correct unit for value given in (a)(i)

i.e. 0.008 or 8 x 10^–3 requires N mm–1.

Allow N m–1 / kg s–2 if no working in (a)(i).

Do not allow unit mark if incorrect physics in part (a)(i) B1

(ii) W = ½ (F x extension) / area under the graph C1

= ½ x 2.0 x 0.250

= 0.25 (J) A1

(b) (i) F = 8 x 0.15 = 1.2 (N) A1

(ii) Hooke’s law continues to be obeyed / graph continues as a straight

line / k is constant / elastic limit has not been reached B1

(c) (i) 1. correct time marked on the graph with a V (t = 0.75 s or 1.75 s) B1

2. tangent in the correct place for downward velocity or implied

by values B1

value between 0.95 to 1.1(m s–1) A1

(ii) 1. X marked in a correct place (maximum or minimum on graph) M1

2. relates the extension / compression to F = kx to explain why the

force is a maximum or maximum extension gives max force or

maximum extension gives max acceleration A1

[12]



4. cast iron: brittle

brittle explained as having no plastic region

elastic

elastic explained as returning to original length when

the load is removed / linear graph / Hooke’s law obeyed

or equivalent words MAX 3

copper: ductile

ductile explained as can be formed into a wire

initially elastic

plastic where it stretches more and more with little

increase in stress

plastic explained as does not return to its original length

when the load is removed

reference to necking at the end MAX 3

polythene: easy to deform / deformed with a small force

plastic

ductile

polymeric MAX 2

MAX 8

QWC: spelling, punctuation and grammar B1

organisation and logic B1

[10]



5. (a) The extension of a spring is directly proportional to the applied force M1

as long as the elastic limit is not exceeded) A1



(b) (i) Correct pair of values read from the graph

force constant = 12/0.080 C1

force constant = 150 (N m–1) A1

(ii) extension, x = × 80 (= 133.33) (mm) C1

(E = ½ Fx)

energy = ½ × 2/ × 133.33 × 10–3

energy = 1.33 (J) A1

(iii) The spring has not exceeded its elastic limit B1

(iv) (elastic potential energy = kinetic energy)

M1

m and k are constant, therefore x prop v. M1

[9]

Answers to questions on Young's modulus

Questions on Young's modulus

Calculations on stress, strain and the Young modulus




Practice questions

These are provided so that you become more confident with the quantities involved, and with the large and small numbers.



Try these

A strip of rubber originally 75 mm long is stretched until it is 100 mm long.

1. What is the tensile strain?

2. Why has the answer no units?

3. The greatest tensile stress which steel of a particular sort can withstand without breaking is about 109 N m-2. A wire of cross-sectional area 0.01 mm2 is made of this steel. What is the greatest force that it can withstand?

4. Find the minimum diameter of an alloy cable, tensile strength 75 MPa, needed to support a load of 15 kN.

5. Calculate the tensile stress in a suspension bridge supporting cable, of diameter of 50 mm, which pulls up on the roadway with a force of 4 kN.

6. Calculate the tensile stress in a nylon fishing line of diameter 0.36 mm which a fish is pulling with a force of 20 N

7. A large crane has a steel lifting cable of diameter 36 mm. The steel used has a Young modulus of 200 GPa. When the crane is used to lift 20 kN, the unstretched cable length is 25.0 m. Calculate the extension of the cable.


Stress, strain and the Young modulus




1. A long strip of rubber whose cross section measures 12 mm by 0.25 mm is pulled with a force of 3.0 N. What is the tensile stress in the rubber?



2. Another strip of rubber originally 90 mm long is stretched until it is 120 mm long. What is the tensile strain?



3. The marble column in a temple has dimensions 140 mm by 180 mm.

I. What is its cross-sectional area in mm2?

II.

III. Now change each of the initial dimensions to metres – what is the cross-sectional area in m2?

IV. If the temple column supports a load of 10 kN, what is the compressive stress, in N m–2?



V. The column is 5.0 m tall, and is compressed by 0.1 mm. What is the compressive strain when this happens?



VI.

VII. Use your answers to parts 5 and 6 to calculate the Young modulus for marble.





4. A 3.0 m length of copper wire of diameter 0.4 mm is suspended from the ceiling. When a 0.5 kg mass is suspended from the bottom of the wire it extends by 0.9 mm.

I. Calculate the strain of the wire.







II. Calculate the stress in the wire.







III. Calculate the value of the Young modulus for copper.

Y12 Homework on Sankey Diagrams

Answers to calculations only.

Constructing Sankey Diagrams

1. An energy efficient light bulb is rated at 20W. It produces 5W of light. Calculate its efficiency and draw a Sankey Diagram to scale.

Efficiency = Useful power out/ power in x 100%

Eff= (5/20) x 100 = 25%

2. Paula transfers 40 000J of chemical energy during a race. She transfers 32 000J of heat energy to the surroundings during the race. Calculate her efficiency and draw a Sankey diagram to scale.

Efficiency = useful energy out/ energy in x 100%

Eff = 40 000 -32 000) /40000 = 8000/40000 = 0.2 =20%

3. Bradley does 1600J of work turning the pedals on his bike. 1577.6J is transferred to the rear sprockets. How much heat is lost and what is the efficiency of Bradley’s bike. Draw a Sankey diagram of Bradley’s chain

Efficiency = useful energy out/ energy in x 100%

Eff = 1577.6 / 1600 = 0.986 = 98.6%

4. The coal in Thomas’ boiler contains 40 kJ of energy. He loses 25.76kJ as heat as he puffs along a branch line. How efficient is Thomas and draw a Sankey diagram.

Efficiency = useful energy out/ energy in x 100%

Eff = (40 - 25.76)/40 = 0.356 = 35.6%

5. The fuel in Diesel’s tank contains 60kJ of chemical energy. He does 20.1kJ of work on the mainline. Calculate Diesel’s efficiency and draw a Sankey diagram.

Efficiency = useful energy out/ energy in x 100%

Eff = 20.1/60 = 0335 = 33.5%

6. The electric engine in Jeremy’s car can develop 3kW. If the car develops 2.658kW what is its efficiency.

 Efficiency = Useful power out/ power in x 100%

Eff = 2.658/3 = 0.886 = 88.6%

7. Jeremy dreams of a Ferrari Enzo which can develop a maximum of 700 bhp. (1 bhp = 750 Watts). Sadly for Jeremy cars are not very efficient. Typically, only about 30% of the energy that is available from the combustion of the petrol actually ends up overcoming friction to move the car forwards. Of the 70% of energy is that is not usefully converted, 55% may heat the cooling water that surrounds the engine block whilst  15% may be in the hot exhaust gases. To make car engines more efficient  the fuel has to burn at a higher temperature and the exhaust must be kept cooler. Draw a Sankey diagram of Jeremy’s dinosaur

y13 Homework Gravity

y13 test on Gravity

Y13 Gravity

Elastic collisons q1

Y12 JAn 2006

(a) Voltmeter connected in parallel with X B1
(b) Same reading / no effect / no change B1
(c)(i) LDR / light-dependent resistor B1
(c)(ii) The resistance decreases (as the intensity of light increases) B1
(c)(iii) 3.5 – 4.0 × 10-7 (m) (to) 6.5 – 7.5 × 10-7 (m) B1
(d)(i) IVR= / )10(8.48.13−×=R C1
resistance = 375 ≈ 380 (Ω) A1
(d)(ii)1 Q = It (Allow with or without the Δ notation) C1
Q = 4.8 × 10-3 × 30 C1
charge = 0.144 ≈ 0.14 (C) A1
(d)(ii)2 W = VQ / W = VIt C1
W = 1.8 × 0.144 / W = 1.8 × 4.8 × 10-3 × 30
energy = 0.259 ≈ 0.26 (Possible ecf) A1
unit: joule / J / VC /VAs B1
(Allow 1/3 if power is 0.0086 (W))
[Total: 13]
2
(a) Kirchhoff’s second B1
(b) Ohm’s B1
(c) Resistance B1
(d) Electronvolt (Allow eV) B1
[Total: 4]
3
(a)(i) 21111RRR+= / 2121RRRRR+= C1
3012011+=R / 30203020+×=R
resistance = 12 (Ω) A1
(a)(ii) R = 10 + 12
resistance = 22 (Ω) (Possible ecf) B1
(b) R = 10 (Ω) / Resistance between B and C = 0 M1
100.5=I
reading = 0.5 (A) A1
[Total: 5]
5
Any four from: B1 × 4

1 (As temperature increases) the resistance of the thermistor / T decreases

2 The total resistance decreases (Possible ecf)

3 The current increases (in the circuit) (Possible ecf)

4 The (voltmeter) reading increases / voltage across R increases (Possible ecf)

5 The voltage across the thermistor / T decreases (Possible ecf)

6 Correct use of the potential divider equation / comment on the ‘sharing’
of voltage / correct use of V = IR

[Total: 4]
6
(a) ALRρ= (Allow any subject) B1
(b) The resistance decreases M1
by a factor of four (because resistance is inversely proportional to radius2) A1
(c)(i) A25103.1105.32200−−×××= / RLAρ= C1
2200103.1105.3)(25−−×××=A C1
(A = ) 2.07 × 10-10 (m2) ≈ 2 × 10-10 (m2) A0
(c)(ii) RIP2= / VIP= and V = IR C1
0.50 = I2 × 2200 C1
current = 0.015 (A) A1
(2.23 × 10-4 scores 2/3 – answer not square rooted)
[Total: 8]

y13

Work

Work

Take g as 9.8 ms-2 or 9.8 Nkg-1

How much work is done if you push a shopping trolley with a constant force of 60N and it moves 5m in a direction parallel to the force?

A delivery driver lifts a mass of 6.5kg onto the back of a lorry 1.5m from the ground. How much work is done in this energy transfer?

How much work is done lifting a 5kg bag 1.2m and place it on the table. In your calculation you assume all the work is done against what?

How much work is done pulling a bag of rubbish 10m across a field by pulling on a string at 400 to the horizontal with a force of 250N?

A toy car has a mass of 110g and its clockwork engine exerts a force of 0.12N. Unfortunately it is not well made and its wheels are at a 130 angle to its direction of motion. What work does it do in travelling 25cm along a heavily carpeted floor?

A pyramid builder is organising his gang of acolytes to pull a large stone block up a ramp. The stone weighs 2.5t and the ramp has a height of 7m. The ramp is 15m long and the acolytes exert a force of 1.2 kN. How much work to they do? How much work is done on the stone to lift it through a height of 7m? How much work is done against friction?

An 80kg baseball player slides to a halt from a speed of 8 ms-1 in a distance of 4m. What is the average stopping force exerted on him by the ground? How much work is done on him? Where does the energy come from?

A 50g ball bearing is dropped from a height of 50cm into a tray of fine sand and embeds itself to a depth of 1.5cm. What was the average vertical force exerted on the ball bearing by the sand? How much work is done on the ball bearing?

y13 Homework

y13

Y11 Homework

Gravitational Potential Energy & Kinetic Energy

acceleration due to gravity = 9.8 m/s2

1. A car has a mass of 750 kg. Calculate its kinetic energy at the following velocities
(a) 10 m/s (b) 15 m/s (c) 20 m/s (d) 30 m/s (e) 35 m/s {NB 1 m/s =2 m p h}

2. (a) A car has a mass of 1000 kg. Calculate its KE at a velocity of 25 m/s.
(b) A train has a mass of 37 tonnes {1 tonne = 1000 kg}. If it also has a velocity of 25 m/s what is its kinetic energy ?

3. A woman has a mass of 65 kg.What is her GPE at the top of a 12 metre diving board ?

4. A cat has a mass of 6 kg. What is its GPE at the top of a tree that is 4.2 m above the ground ?

5. The Eiffel Tower is 300 m high. What is the GPE of a 100 g bird perched on the top of it ?

6. The Great Pyramid of Khufu (Cheops to the Greeks) is 146 m high. It is made of stones of mass 250 t. What is the GPE of the topmost stone ?

7. The Empire State Building is 449 m high(but this includes a 68 m TV mast).What is the GPE of a ball of mass 400 g at the top of the building (not the TV mast) ? If the ball is dropped over the side what will happen to the GPE ? What will be its speed at the instant before it hits the ground ? (Ignoring air resistance)

8. Olympus Mons is a volcano on Mars. It is 25 km high. What would be the GPE of a human of mass 50 kg on its summit ? (surface gravity = 0.38 that of Earth) Compare this with the same human standing on the summit of Everest 8534 m high.

9. The space shuttle has a mass of 80 t. Escape velocity (the velocity an object has to travel at in order to escape the earth's gravitational field) is 11 km/s. What is its KE at this speed? If all this KE is turned into GPE how high would it be ? What assumption do you have to make in your calculation.

How much energy would it take to accelerate the space shuttle to light speed (300 000 km/s). {This is not possible, also the equation you will use to calculate KE is not valid near the speed of light but it will do for this purpose} A nuclear power station produces around 500 MW
(500 000 000 J per second). How long would 100 power stations take to produce this amount of energy?
How feasible is travelling close to the speed of light?
The nearest star is about 4 light years away. How long would it take the space shuttle to get there at 11 km/s (22 000 mph)? {1 light year is the distance light will travel in 1 year}
How feasible is interstellar travel?

Y11 Homework on Forces

1. A light bulb converts 500 J of electrical energy into heat and light in
5 seconds. What is its power rating?

2. An electric kettle converts 210,000 J of electrical energy into heat energy in 1 minute. Calculate its power rating?

3. A person of weight 500 N climbs a 5m flight of stairs in 100 s. Calculate the power produced.

4. A car of mass 500 kg is driven up an incline of vertical height 5m in 5s. Calculate the power output of the engine.
[Gravitational field strength = 10 N/kg]

5. Calculate the energy transferred by a 300 W TV in 25 seconds.

6. Calculate the energy transferred by a 0.0002 W calculator in 5 minutes.

7. A hoist motor develops 500 W. It does 1000 J of work in lifting a mass. How long does it take to lift the mass?

8. A car develops 44 kW. It does 1980 kJ of work in travelling a certain
distance. How long does this take?

Hw on Velocity Time graphs

phy2 2003

6732 Unit Test PHY2

1. Resistance of lamps

V2/R OR I = 60/12 = (5 A) 1

R=(12 V x 12 V) /60 W 1

R =2.4ohm 1 total 3

Resistance variation

Lamp A: resistance of A decreases with current increase 1

Lamp B: resistance of B increases with current increase 1

Dim filament

Lamps are dim because p.d. across each bulb is less than 12 V 1

Why filament of lamp A is brighter

Bulbs have the same current 1

p.d. across A > p.d. across B/resistance A> Resistance B 1

OR

power in A > power in B 2

total 8

2 Table

Physical Quantity - Typical value

Resistance of a voltmeter -10 M ohm

Internal resistance of a car battery - 0.05 ohm

Internal resistance of an EHT supply -10 M ohm

Resistivity of an insulator - 2.0 x 10¹⁵ ohm m

Drift velocity of electrons in a metallic conductor - 0.3 mm s^-1

Temperature of a working filament bulb - 3000 K

[Mark is lost if 2 or more values are put into one box] total 6

3. Current in heating element

P= VI 1

I = 500 W / 230 V 1

I = 2.2 A 1

Or

P=V²/R 1

R= 230 x 230/105.8 1

I = 2.2A 1

Drift velocity

Drift velocity greater in the thinner wire 1

Explanation

Quality of written communication. 1

See I = nAQv 1

is the same (at all points) 1

(probably) n (and Q) is the same in both wires 1 total 8

4 Resistance of films

R= pl/A 1

R = pl/wt or A = wt 1

Resistance calculation

R = (6.0 x 10^-5) (8 x 10^-3 m) / (3 x 10^-3m) (0.001 x 10^-3 m) 1+1

(Correct substitution but values in mm 1)

R= 160 ohm 1 ecf if in mm

Resistance of square film

L=w 1

R= pl/lt=p/t 1

R= pw/wt=p/t 1

Total 7

5. Definition of specific heat capacity

energy (needed) 1

(per) unit mass/kg and per unit temperature change K or C 1

OR

Correct formula [does not need to be rearranged] 1

with correctly defined symbols 1

Circuit diagrams (see end)

Accept voltmeter across heater and ammeter as well as voltmeter across heater only

Means of varying p.d./current 1

Voltmeter in parallel with a resistor symbol 1

Ammeter in series with any representionof a heater. 1

Other apparatus

(Top pan) balance / scales 1

Stopwatch / timer / clock 1

Explanation

Energy/heat loss to surroundings/air/bench 1

OR

Mc delta T +delta,Q = VIt or equivalent in words (e.g. student ignores energy loss in calculations) 1

Modifications

Any two from

Use of insulation around block

Ensure all of heater is within block

Grease heater/thermometer

Total 10

Specific Heat Capacity Calculation

C = delta Q/ m delta T = (860 x 10³)/ (1.4 kg) (750 – 22) = 844 (J kg ^–1 K^-1)

Conversion of kJ to J 1

Subtraction of temp 1

Answer 1

Energy transfers diagram

Label 2 (energy to) (warm) water (and trough) 1

Label 3 (energy used to) evaporate water / cause evaporation/latent heat/change of state 1

Total 5

Gas equation

PV = nRT [Accept symbols or words] 1

Molar gas constant unit

R= PV/nT

P – kgm^-1 s 1

V – m³ T– K n – mol all three for 1

Kinetic energy of molecule

Nm =M

density = Nm²>/3V correctly combined the 2 equations 1

Nm²> = 3nRT density = any mass –: volume 1

Show that

Kinetic energy =m²>/2=(3/2)n(RT/N) 1

Sketch graph

PV on y axis Temperature/’C on x axis

[accept axes reversed and correct graph] 1

Straight line graph with negative intercept 1

Gradient R 1

Intercept at – 273 ’C 1

[All these marks can be scored on graph)

total 10

8. Definition of e.m.f. of a cell

work/energy (conversion) per unit charge 1

for the whole circuit / refer to total energy 1

OR

Work/energy per unit charge 1

converted from chemical to electrical (energy) 1

OR

E = W/Q for whole circuit 1

All symbols defined 1

OR

E = P/I for whole circuit 1

All symbols defined 1

[Terminal p.d, when no current drawn scores 1 mark only] 2 max

Circuit diagram

See top

R 1

A in series 1

R (can be variable) 1 A and V correct 1 V as shown

Or across R+ A

Or across battery

[2nd mark is consequent on R(fixed, variable )or lamp]

Sketch graph

See top

Graph correctly drawn with axes appropriately labelled and consistent with the circuit drawn 1

Intercept on R axes equivalent to( – )r and Gradient equivalent to ( – )r [Gradient mark consequent on graph mark] 1

[Gradient may be indicated on graph]

Grade boundaries

A 41 B 37 C 33 D 26 E 25

phy1 2003

6731 Unit Test PHY1

1. Magnitude of resultant force

4 cm line S / 1.7 cm line N 1

8 cm line NE / 8N resolved into two perp. components (5.7E & 1.7N or 5.7N) 1

Correct construction for vector sum 1

5.7 – 6.1 N 1

Name of physical quantities

Vectors 1

Two other examples

Any two named vectors other than force (if >2, must all be vectors) 1

2. Calculation of average velocity

Use of v = s/t 1

v = 1.86 m s -1 / 1.9 m s -1 1

Acceleration of trolley

Selecting u = u + 2as 1

Correct substitutions 1

2.87 m s /2.9 m s /3.0 m s-1 1

Tension in string

Use of F = ma 1

2.73 N / 2.76 N / 2.85 N 1

Assuming no action other horizontal force/table smooth/light string/inextensible string 1

Explanation

Suspended mass/system is accelerating 1

Idea of resultant force on the 0.4 kg mass 1

4. Addition to diagram

Downwards arrow Y through middle third of left leg 1

Downward arrow Z with correct line of action 1

[Ignore lengths of arrows and point of action] [Must have at least one correct label to get 2 marks; no labels gets max 1 out of 2] [One correct label can get 2 marks]

Explanation

Quality of written communication

Clockwise moments = Anticlockwise when balanced

Y is smaller than X but acts further from P

Moment of XP /Moment of Y = F x YP

Z has little or no moment about P/Z acts through P

Gravitational potential energy

Use of mgh 1

Vertical drop per second = (8.4 m) sin (3) 1

-.9 x 10 J/Js -1/W 1

What happens to this lost gpe

Becomes internal energy/used to do work against friction and or heat energy. 1[mention of K.E. loses the mark]

Estimate of rate at which cyclist does work

Rate of working = 2 x 3.9 x 10 W 1

=7.8 to 10 W 1

[3.9 x 10 W earns 1 out of 2]

6. Momentum and its unit

Momentum = mass x velocity 1

Kgms-1 or N s 1

Momentum of thorium nucleus before the decay

Zero 1

Speed of alpha particle/radium nucleus and directions of travel

Alpha particle because its mass is smaller/lighter 1

So higher speed for the same (magnitude of) momentum OR Newton 3rd Law argument 1

Opposite directions/along a line 1

Nuclear equation

Correct symbol and numbers for tin OR beta 1

Correct symbols and numbers for the other two 1

Decay constant

Use of lambda = 0.69/ half life 1~

1.57 x 10^-15 y-1 OR 4.99 x 10 s-1 1

Activity of source and comparison with normal background count

rate

Use of A = lambda N 1

0.11/0.12 (Bq) 1

Lower (than background) [Allow ecf – assume background = 0.3 to 0.5] 1

8. Radiation tests

Alpha:

Test 2 or 2 and 1 1

Count drops when alphas have been stopped by the air / alphas have a definite range / (only) alpha have a short range (in air) 1

Beta:

Test 3/3 and 1, because 1 mm aluminium stops (some) beta/does not stop any gamma rays 1

Gamma:

Test 4 or 4 and 1, because 5 mm aluminium will stop all the betas, (so there must be gamma too)/gamma can penetrate 5 mm of aluminium

Table

Target for Alpha scattering

Gold atoms/gold foil gold leaf/gold film/very thin sheets of gold/metal foil etc. [NOT thin gold sheet] 1

Target for Deep inelastic scattering Protons/neutrons/nucleons /liquid hydrogen/nuclei 1

Conclusions

(i) Atom mainly empty space/nucleus is very small 1

Nucleus dense/massive 1

(ii) Nucleons have a substructure 1

Made of quarks 1