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13+ · Science

CE.S-T11 Science: Full Practice Paper 11

ISEB COMMON ENTRANCE CE.S-T11 · Calculator allowed · about 60 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Which chemical reagent is used to test a food sample for starch, turning from orange-brown to blue-black if starch is present?

A) Benedict's solution
B) Iodine solution
C) Biuret solution
D) Universal indicator

(Total for Question 1 is 1 mark)

Which of these is a way the human body responds to becoming too hot, to help cool it back down?

A) Shivering
B) Sweating
C) Vasoconstriction (blood vessels near the skin narrowing)
D) Hairs on the skin standing up

(Total for Question 2 is 1 mark)

Complete the two missing words in these sentences about how plants disperse their seeds.

(a)Seeds with hooks or spines that catch onto the fur of a passing animal are dispersed by ______.(1)

(b)Seeds such as those of the sycamore tree, which have wing-like structures that make them spin as they fall, are dispersed by ______.(1)

(Total for Question 3 is 2 marks)

Which term describes a random change to an organism's genes (DNA) that can create new variation within a species?

A) Mutation
B) Adaptation
C) Selection
D) Fertilisation

(Total for Question 4 is 1 mark)

Nadia is investigating osmosis using potato chips. She cuts four identical potato chips, each with a starting mass of 5.0 g, and places one chip into each of four beakers, containing 50 cm3 of sucrose (sugar) solution at a different concentration. After 30 minutes, she removes each chip, gently blots it dry, and re-weighs it, recording the percentage change in mass.

Sucrose concentration (mol/dm3): 0.0	0.2	0.4	0.6
Percentage change in mass (%): +12	+4	-5	-13

(a)Identify the independent variable in Nadia's investigation.(1)

(b)State one variable that Nadia kept the same to make this a fair test.(1)

(c)Describe the pattern shown by the results.(1)

(d)Explain, in terms of osmosis, why the potato chip gained mass in the most dilute solution but lost mass in the more concentrated solutions.(1)

(Total for Question 5 is 4 marks)

Describe the role of each of these three parts of the body's immune response in fighting a pathogen (such as bacteria) that has entered the body: white blood cells, antibodies, and memory cells.

(Total for Question 6 is 3 marks)

A pupil looks at a plant cell under a light microscope and makes a scale drawing of it. The actual width of the real cell is 0.05 mm, and the width of the cell as drawn in the pupil's diagram is 20 mm. Calculate the magnification of the pupil's drawing, using magnification = image size / actual size.

(Total for Question 7 is 2 marks)

Grace is investigating how the concentration of caffeine solution affects the heart rate of Daphnia (water fleas), small transparent animals whose heartbeat can be seen and counted under a microscope. She places a Daphnia in a drop of pond water on a microscope slide, adds a drop of caffeine solution of a different concentration each time, waits two minutes, then counts the number of heartbeats in 15 seconds.

Caffeine concentration (%): 0	0.5	1.0	1.5
Heartbeats counted in 15 seconds: 32	40	51	53

(a)Identify the independent variable in Grace's investigation.(1)

(b)State one variable that Grace kept the same to make this a fair test.(1)

(c)Describe the pattern shown by the results.(1)

(d)Suggest why increasing the concentration of caffeine increases the Daphnia's heart rate.(1)

(Total for Question 8 is 4 marks)

Peppered moths exist in two colour forms: pale (speckled) moths and dark moths. Before the Industrial Revolution, tree trunks in many parts of Britain were pale, often covered in pale lichen. During the Industrial Revolution, soot and pollution from factories blackened tree trunks in many industrial areas. Over the following decades, the proportion of dark moths in these industrial areas increased greatly, while pale moths remained common in cleaner, rural areas.

(a)State the original source of the variation between pale and dark peppered moths, before the tree trunks changed colour.(1)

(b)Explain why dark moths became better camouflaged than pale moths on the sooty, blackened tree trunks of industrial areas.(1)

(c)Explain why this improved camouflage led to more dark moths surviving to reproduce than pale moths, in industrial areas.(1)

(d)Explain how the repetition of this process over many generations caused the overall colour of the moth population in industrial areas to change.(1)

(Total for Question 9 is 4 marks)

In a food chain, only a small percentage of the energy available at one trophic (feeding) level is transferred to the next trophic level up. Explain why food chains rarely contain more than four or five organisms (trophic levels), referring to energy losses between each level.

(Total for Question 10 is 6 marks)

Which test result would confirm that a gas collected during an experiment is oxygen?

A) A glowing splint held in the gas relights
B) A lit splint held in the gas is extinguished with a squeaky pop
C) The gas turns damp red litmus paper blue
D) The gas turns limewater cloudy (milky)

(Total for Question 11 is 1 mark)

Which group (family) of the periodic table contains reactive metals, such as lithium, sodium and potassium, which react vigorously with water?

A) Group 1 (the alkali metals)
B) Group 7 (the halogens)
C) Group 0 (the noble gases)
D) The transition metals

(Total for Question 12 is 1 mark)

Complete the two missing words in these sentences about copper sulfate crystals.

(a)Blue copper sulfate crystals, which contain water molecules built into their crystal structure, are described as ______ crystals.(1)

(b)When heated strongly, blue hydrated copper sulfate crystals lose their water and turn into a white powder, called ______ copper sulfate.(1)

(Total for Question 13 is 2 marks)

Which piece of equipment is used to add a liquid, such as an acid, drop by drop while accurately measuring the volume added, for example during a titration?

A) A beaker
B) A burette
C) A boiling tube
D) A filter funnel

(Total for Question 14 is 1 mark)

Priti reacts an excess of magnesium ribbon with 30 cm3 of dilute hydrochloric acid at four different concentrations, and uses a gas syringe to measure the volume of hydrogen gas produced in the first 30 seconds of each reaction.

Concentration of acid (mol/dm3): 0.5	1.0	1.5	2.0
Volume of gas in 30 seconds (cm3): 18	33	45	50

(a)Identify the independent variable in Priti's investigation.(1)

(b)State one variable that Priti kept the same to make this a fair test.(1)

(c)Describe the pattern shown by the results.(1)

(d)Explain, in terms of particles, why increasing the concentration of the acid increases the rate of this reaction.(1)

(Total for Question 15 is 4 marks)

Calculate the relative formula mass (Mr) of carbon dioxide, CO2. (Relative atomic masses: C = 12, O = 16)

(Total for Question 16 is 2 marks)

Describe a test, including the positive result, that would show a gas produced in an experiment is: carbon dioxide, hydrogen, and chlorine.

(a)Carbon dioxide.(1)

(b)Hydrogen.(1)

(c)Chlorine.(1)

(Total for Question 17 is 3 marks)

When colourless chlorine gas is bubbled through a colourless solution of potassium bromide, the solution turns orange.

(a)Name the two products formed in this reaction, and write a word equation for the reaction.(2)

(b)Explain, in terms of reactivity, why chlorine is able to displace bromine from potassium bromide solution.(1)

(c)Predict, giving a reason, whether bromine would displace chlorine from a solution of potassium chloride.(1)

(Total for Question 18 is 4 marks)

Complete the missing word to name the salt produced in each acid-alkali neutralisation reaction.

(a)Hydrochloric acid + sodium hydroxide -> sodium ______ + water(1)

(b)Sulfuric acid + potassium hydroxide -> potassium ______ + water(1)

(c)Nitric acid + calcium hydroxide -> calcium ______ + water(1)

(Total for Question 19 is 3 marks)

Diamond and graphite are both forms of the element carbon (they are allotropes of carbon), yet diamond is extremely hard and does not conduct electricity, while graphite is soft and slippery and does conduct electricity. Explain, in terms of their structure and bonding, why diamond and graphite have such different physical properties, even though both are made only from carbon atoms.

(Total for Question 20 is 6 marks)

Which quantity is measured in units of pascals (Pa)?

A) Force
B) Pressure
C) Energy
D) Power

(Total for Question 21 is 1 mark)

Which instrument is used to measure atmospheric (air) pressure?

A) Thermometer
B) Barometer
C) Ammeter
D) Voltmeter

(Total for Question 22 is 1 mark)

Complete the two missing words in these sentences about pressure.

(a)Pressure is calculated using the formula: pressure = force / ______.(1)

(b)For the same force, pressure is greater when that force acts over a ______ area.(1)

(Total for Question 23 is 2 marks)

A pupil investigates how the current flowing through a fixed resistor changes as the potential difference (voltage) across it is increased, using a variable power supply, an ammeter and a voltmeter connected in a circuit with the resistor.

Potential difference (V): 2	4	6	8
Current (A): 0.4	0.8	1.2	1.6

(a)Identify the independent variable in this investigation.(1)

(b)State one variable that was kept the same to make this a fair test.(1)

(c)Describe the pattern shown by the results.(1)

(d)Using the readings at 6 V, calculate the resistance of the resistor, using resistance = potential difference / current.(1)

(Total for Question 24 is 4 marks)

A cyclist accelerates steadily from a velocity of 2 m/s to a velocity of 8 m/s over 4 seconds.

(a)Calculate the change in velocity of the cyclist.(1)

(b)Calculate the cyclist's acceleration, using acceleration = change in velocity / time taken.(2)

(Total for Question 25 is 3 marks)

Describe how each of these three methods reduces the amount of heat energy wasted from a house: loft insulation, cavity wall insulation, and double glazing.

(Total for Question 26 is 3 marks)

A swimmer pushes backwards against the wall of a swimming pool with her feet, which causes her to accelerate forwards through the water.

(a)State Newton's third law of motion.(1)

(b)Identify the pair of forces involved when the swimmer pushes off the wall.(1)

(c)The force the wall exerts on the swimmer and the force the swimmer exerts on the wall are equal in size. Explain why the swimmer noticeably accelerates forwards, while the wall (and the pool it is attached to) does not visibly move.(2)

(Total for Question 27 is 4 marks)

A pupil investigates how the length of a guitar string affects the pitch (frequency) of the note produced when it is plucked. She uses the same string, tuned to the same tension each time, but presses her finger down on a fret to change the length of string that is free to vibrate, and uses a tuning app to measure the frequency of the note produced.

Length of vibrating string (cm): 60	45	30	15
Frequency of note (Hz): 220	290	440	880

(a)Identify the independent variable in this investigation.(1)

(b)State one variable that was kept the same to make this a fair test.(1)

(c)Describe the pattern shown by the results.(1)

(d)Explain, in terms of vibrations, why a shorter length of string produces a note with a higher pitch.(1)

(Total for Question 28 is 4 marks)

A 2 kW electric heater is switched on for 3 hours. Electricity costs 20p per kilowatt-hour (kWh). Calculate the total cost of running the heater for these 3 hours.

(Total for Question 29 is 2 marks)

Planets, such as the Earth, travel around the Sun in a roughly circular path (orbit), rather than moving off into space in a straight line. Explain, using ideas about forces, why gravity keeps the planets in orbit around the Sun.

(Total for Question 30 is 6 marks)

Mark scheme · CE.S-T11 Science: Full Practice Paper 11

Question 1

B1 B cao
Answer: B) Iodine solution

Question 2

B1 B cao
Answer: B) Sweating

Question 3

(a) B1 animal (accept animal dispersal) cao
(a) Answer: Animal
(b) B1 wind cao
(b) Answer: Wind

Question 4

B1 A cao
Answer: A) Mutation

Question 5

(a) B1 the concentration of the sucrose (sugar) solution oe
(a) Answer: The concentration of the sucrose solution.
(b) B1 any one of: the starting mass/size of each potato chip; the volume of solution (50 cm3); the type/variety of potato used; the time left in the solution (30 minutes); the temperature oe
(b) Answer: For example, the starting mass of each potato chip.
(c) B1 as the concentration of sucrose solution increases, the percentage change in mass decreases, changing from a gain in mass at low concentrations to a loss in mass at higher concentrations oe
(c) Answer: As sucrose concentration increases, the percentage change in mass decreases, changing from a mass gain at low concentrations to a mass loss at higher concentrations.
(d) B1 in the dilute solution, the solution has a higher water content (higher water potential) than the potato cells, so water moves into the chip by osmosis, increasing its mass; in the more concentrated solutions, the solution has a lower water content (lower water potential) than the potato cells, so water moves out of the chip by osmosis, decreasing its mass oe
(d) Answer: In the dilute solution, water moves into the chip by osmosis because the solution has a higher water content than the potato cells, increasing its mass; in the more concentrated solutions, water moves out of the chip by osmosis because the solution has a lower water content than the cells, decreasing its mass.

Question 6

B1 white blood cells detect and engulf (digest) pathogens (phagocytosis) oe
B1 antibodies are produced by white blood cells and bind specifically to antigens on the surface of the pathogen, helping to destroy it (or mark it for destruction) oe
B1 memory cells remain in the body after the infection has cleared and allow a faster, stronger antibody response if the same pathogen infects the body again oe
Answer: White blood cells engulf pathogens. Antibodies bind to antigens on the pathogen's surface to help destroy it. Memory cells remain afterwards, allowing a faster response if the same pathogen returns.

Question 7

M1 correct substitution: magnification = 20 / 0.05 oe
A1 x400 (400) cao
Answer: x400

Question 8

(a) B1 the concentration of caffeine solution oe
(a) Answer: The concentration of caffeine solution.
(b) B1 any one of: the same Daphnia (or Daphnia of the same size/species); the time waited before counting (2 minutes); the time over which heartbeats are counted (15 seconds); the volume of solution added; the temperature oe
(b) Answer: For example, the time waited before counting heartbeats.
(c) B1 as the concentration of caffeine solution increases, the number of heartbeats counted (the heart rate) increases, although the increase becomes smaller between 1.0% and 1.5% oe
(c) Answer: As caffeine concentration increases, the number of heartbeats counted increases, although the increase is smaller between 1.0% and 1.5%.
(d) B1 caffeine is a stimulant that speeds up the activity of the heart (or nervous system), so a higher concentration of caffeine has a greater stimulant effect, further increasing heart rate oe
(d) Answer: Caffeine is a stimulant that speeds up the heart's activity, so a higher concentration has a greater effect, increasing heart rate further.

Question 9

(a) B1 random mutations in the moths' genes (DNA), which already existed in the population before the Industrial Revolution oe
(a) Answer: Random mutations in the moths' genes, present in the population before the trees changed colour.
(b) B1 dark moths blended in with the dark, sooty bark, making them harder for predators (such as birds) to see, whereas pale moths stood out clearly against the dark bark oe
(b) Answer: Dark moths blended in with the dark, sooty bark, making them harder for predators to spot, while pale moths stood out.
(c) B1 because dark moths were less likely to be seen and eaten by predators, more of them survived long enough to breed and pass on the allele for dark colour to their offspring, compared with the more visible pale moths oe
(c) Answer: Because dark moths were less likely to be eaten, more of them survived to breed and pass on the allele for dark colour to their offspring.
(d) B1 because dark moths bred successfully more often each generation, the allele for dark colour became more common (was passed on to more offspring) with each generation, so the proportion of dark moths in the population increased generation after generation, a result of natural selection oe
(d) Answer: Because dark moths bred successfully more often each generation, the allele for dark colour became more common with each generation, so the population became mostly dark moths over time.

Question 10

Level 1 (1-2): Basic, isolated statement(s) about energy or food chains, with little or no reference to specific energy losses or why chains are short.
Level 2 (3-4): Some relevant ideas about energy being lost between levels (e.g. as heat, or not all being eaten) are given, but the explanation of why this limits the length of food chains is incomplete or not fully linked.
Level 3 (5-6): A clear, logically linked explanation of how energy is lost at each trophic level (e.g. through respiration/heat, movement, or parts not eaten/egested), so only a small percentage passes to the next level, meaning there is too little energy left to support another feeding level after four or five stages.
Indicative content:
- Not all the energy taken in by an organism at one trophic level is passed on to the organism that eats it.
- Much of the energy is lost as heat to the surroundings, released during respiration to power life processes (movement, growth, keeping warm).
- Some energy is lost in materials that are not eaten (e.g. bones, roots, leaves) or are egested as waste (faeces), rather than being digested and absorbed.
- Only the energy stored in the parts of the organism's body that are actually eaten and digested by the next consumer can be passed on to the next trophic level.
- As a result, typically only about 10% of the energy at one trophic level is transferred to the next trophic level (a large majority, roughly 90%, is lost at each stage).
- Because such a large proportion of energy is lost at every stage, the total amount of energy remaining becomes too small to support a further trophic level after about four or five stages, so food chains rarely extend beyond this length.

Question 11

B1 A cao
Answer: A) A glowing splint held in the gas relights

Question 12

B1 A cao
Answer: A) Group 1 (the alkali metals)

Question 13

(a) B1 hydrated cao
(a) Answer: Hydrated
(b) B1 anhydrous cao
(b) Answer: Anhydrous

Question 14

B1 B cao
Answer: B) A burette

Question 15

(a) B1 the concentration of the hydrochloric acid oe
(a) Answer: The concentration of the hydrochloric acid.
(b) B1 any one of: the volume of acid (30 cm3); the amount (length/mass) of magnesium ribbon (in excess each time); the temperature; the time over which gas is collected (30 seconds) oe
(b) Answer: For example, the volume of acid used.
(c) B1 as the concentration of the acid increases, the volume of gas produced in 30 seconds increases, although the increase becomes smaller between 1.5 and 2.0 mol/dm3 oe
(c) Answer: As acid concentration increases, the volume of gas produced in 30 seconds increases, although the increase is smaller between 1.5 and 2.0 mol/dm3.
(d) B1 a more concentrated acid has more acid particles in the same volume, so there are more frequent collisions between acid particles and magnesium atoms, increasing the rate at which hydrogen gas is produced oe
(d) Answer: A more concentrated acid has more acid particles in the same volume, causing more frequent collisions with the magnesium and increasing the rate of reaction.

Question 16

M1 correct substitution: Mr = 12 + (16 x 2) oe
A1 44 cao
Answer: 44

Question 17

(a) B1 bubble the gas through limewater; the limewater turns cloudy (milky) oe
(a) Answer: Bubble the gas through limewater; it turns cloudy (milky).
(b) B1 hold a lit splint at the mouth of the test tube; a squeaky pop is heard oe
(b) Answer: Hold a lit splint at the mouth of the tube; a squeaky pop is heard.
(c) B1 hold a piece of damp (litmus) paper at the mouth of the test tube; the paper is bleached white (turns from coloured/blue to white) oe
(c) Answer: Hold damp litmus paper at the mouth of the tube; it is bleached white.

Question 18

(a) B1 potassium chloride and bromine oe
(a) B1 chlorine + potassium bromide -> potassium chloride + bromine oe
(a) Answer: Potassium chloride and bromine are formed: chlorine + potassium bromide -> potassium chloride + bromine.
(b) B1 chlorine is more reactive than bromine (higher up in Group 7, the halogens), so it can displace the less reactive bromine from its compound (potassium bromide) oe
(b) Answer: Chlorine is more reactive than bromine, so it can displace the less reactive bromine from potassium bromide.
(c) B1 no; bromine would not displace chlorine, because bromine is less reactive than chlorine, and a less reactive halogen cannot displace a more reactive halogen from its compound oe
(c) Answer: No, bromine would not displace chlorine, because bromine is less reactive than chlorine.

Question 19

(a) B1 chloride cao
(a) Answer: Chloride
(b) B1 sulfate cao
(b) Answer: Sulfate
(c) B1 nitrate cao
(c) Answer: Nitrate

Question 20

Level 1 (1-2): Basic, isolated statement(s) about diamond or graphite, with little or no reference to bonding or the number of bonds per carbon atom.
Level 2 (3-4): Some relevant ideas about the arrangement of carbon atoms or the number of covalent bonds are given, but the explanation linking structure to hardness, slipperiness or conductivity is incomplete or not fully linked.
Level 3 (5-6): A clear, logically linked explanation of diamond's rigid structure with four strong covalent bonds per carbon atom (giving extreme hardness and no free electrons) compared with graphite's layered structure with three covalent bonds per carbon atom, weak forces between layers, and delocalised electrons.
Indicative content:
- Diamond and graphite are both giant covalent (macromolecular) structures made entirely of carbon atoms, but the carbon atoms are arranged and bonded differently in each.
- In diamond, each carbon atom forms four strong covalent bonds to four other carbon atoms, in a rigid, tetrahedral, three-dimensional lattice.
- Because every atom is held in place by four strong covalent bonds in all directions, a huge amount of energy/force is needed to break the structure apart, making diamond extremely hard.
- Diamond has no free (delocalised) electrons or ions able to move, so it cannot conduct electricity.
- In graphite, each carbon atom forms only three strong covalent bonds, arranged in flat hexagonal layers (sheets).
- The layers in graphite are held together only by weak forces between the layers (not covalent bonds), so the layers can slide over each other easily, which is why graphite is soft and slippery, and can be used as a lubricant or in pencils.
- Each carbon atom in graphite has one outer (spare) electron that is not used in bonding; these electrons become delocalised and free to move along the layers, which allows graphite to conduct electricity, unlike diamond.

Question 21

B1 B cao
Answer: B) Pressure

Question 22

B1 B cao
Answer: B) Barometer

Question 23

(a) B1 area cao
(a) Answer: Area
(b) B1 smaller cao
(b) Answer: Smaller

Question 24

(a) B1 the potential difference (voltage) across the resistor oe
(a) Answer: The potential difference (voltage) across the resistor.
(b) B1 the resistor used (its resistance); the temperature oe
(b) Answer: For example, the same resistor (and its resistance) was used each time.
(c) B1 as the potential difference increases, the current also increases; the current is directly proportional to the potential difference (doubling the pd doubles the current) oe
(c) Answer: As the potential difference increases, the current increases in direct proportion to it.
(d) B1 5 ohm cao (working: 6 / 1.2 = 5)
(d) Answer: 5 ohms

Question 25

(a) B1 6 m/s cao (working: 8 - 2)
(a) Answer: 6 m/s
(b) M1 correct substitution: acceleration = 6 / 4 oe (ft from (a))
(b) A1 1.5 m/s2 cao
(b) Answer: 1.5 m/s2

Question 26

B1 loft insulation (e.g. thick fibreglass/mineral wool) traps a layer of air in the loft space, reducing heat loss (by conduction/convection) through the roof oe
B1 cavity wall insulation fills the gap between the inner and outer walls with an insulating material, reducing heat loss (by conduction/convection) through the walls oe
B1 double glazing traps a layer of air (or gas) between two panes of glass, reducing heat loss (by conduction) through the windows oe
Answer: Loft insulation traps air in the loft, reducing heat loss through the roof. Cavity wall insulation fills the gap between the walls, reducing heat loss through the walls. Double glazing traps air between two panes of glass, reducing heat loss through the windows.

Question 27

(a) B1 whenever one object exerts a force on a second object, the second object exerts an equal and opposite force back on the first object oe
(a) Answer: Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force back on the first.
(b) B1 the swimmer's feet push backwards on the wall, and the wall pushes forwards on the swimmer's feet, with these two forces equal in size and opposite in direction oe
(b) Answer: The swimmer pushes backwards on the wall, and the wall pushes forwards on the swimmer, with the two forces equal in size and opposite in direction.
(c) B1 for the same size of force, a smaller mass gains a greater acceleration than a larger mass (acceleration = force / mass) oe
(c) B1 the swimmer has a far smaller mass than the wall/pool/ground it is attached to, so the same force produces a large, noticeable acceleration on the swimmer but only an immeasurably tiny (negligible) acceleration on the wall oe
(c) Answer: For the same size of force, a smaller mass gains a greater acceleration; since the swimmer has a far smaller mass than the wall and pool, the same force gives her a noticeable acceleration while the wall's acceleration is immeasurably tiny.

Question 28

(a) B1 the length of the vibrating string oe
(a) Answer: The length of the vibrating string.
(b) B1 any one of: the tension of the string; the string used (its thickness/material); how hard/where the string was plucked oe
(b) Answer: For example, the tension of the string.
(c) B1 as the length of the vibrating string decreases, the frequency of the note produced increases oe
(c) Answer: As the length of the vibrating string decreases, the frequency of the note produced increases.
(d) B1 a shorter length of string vibrates back and forth more quickly (completes more vibrations per second), producing a sound wave with a higher frequency, which is heard as a higher-pitched note oe
(d) Answer: A shorter length of string vibrates more quickly, producing a higher-frequency sound wave, which is heard as a higher-pitched note.

Question 29

M1 correct substitution: energy used = power x time = 2 x 3 (= 6 kWh); cost = 6 x 20p oe
A1 120p (accept GBP 1.20) cao
Answer: 120p (GBP 1.20)

Question 30

Level 1 (1-2): Basic, isolated statement(s) about gravity or orbits, with little or no reference to a constantly changing direction or an unbalanced force.
Level 2 (3-4): Some relevant ideas about gravitational attraction pulling the planet towards the Sun are given, but the explanation of why this keeps the planet in a curved orbit, rather than pulling it straight into the Sun, is incomplete or not fully linked.
Level 3 (5-6): A clear, logically linked explanation that the Sun's gravity provides a constant, unbalanced pulling force towards the Sun, which continuously changes the direction (not the speed) of the planet's motion, curving its path into a closed, roughly circular/elliptical orbit rather than a straight line or a direct collision course.
Indicative content:
- Without any force acting on it, an object (such as a planet) would travel in a straight line at a constant speed (Newton's first law of motion).
- The Sun exerts a gravitational force of attraction on each planet, pulling it towards the centre of the Sun.
- This gravitational force is unbalanced (there is no equal and opposite force acting on the planet in the plane of its orbit), so it causes the planet's direction of travel to continuously change.
- Because gravity constantly pulls the planet sideways/towards the Sun as it moves forward, the planet's straight-line path is continuously curved, resulting in it moving in a roughly circular/elliptical orbit around the Sun, instead of flying off in a straight line.
- The planet does not fall directly into the Sun because it also has a large sideways (tangential) velocity/speed; the combination of this forward motion and the constant pull of gravity results in a stable, curved orbital path rather than a straight-line collision course or a straight-line escape.
- Every planet orbits the Sun in this way because every planet experiences the Sun's gravitational pull, which is what holds the whole Solar System together.

Science: Full Practice Paper 11 - Worksheets, Questions and Revision

CE.S-T11 Science: Full Practice Paper 11

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28

Question 29

Question 30