S6 Combined Science Practice Paper (Set 1): Force And Motion, Circuits And Power, Mole Calculation, Titration, Genetics And Energy Flow

S6 Combined Science Practice Paper (Set 1): Force and Motion, Circuits and Power, Mole Calculation, Titration, Genetics and Energy Flow

2026-11-15

S6 Combined Science

S6 Combined Science paper (Set 1) with answers: acceleration and braking force, kettle power and cost, mole calculation, titration, genetics ratio and energy flow.

Paper details　S6 Combined Science practice paper (Set 1) with full answers covering force and braking, electrical power and cost, mole calculation, titration, genetics and energy flow in ecosystems.

Section A: Force and Motion (18 marks)

A car of mass 1200 kg travels along a straight road. The driver raises the speed from 8.0 m/s to 20.0 m/s in 4.0 s, and then stops the car with a constant braking force in 40.0 m.

#	Question	Answer and marking notes
1	(a) Calculate the average acceleration of the car during the acceleration.	Answer: a = (20.0 – 8.0) / 4.0 = 3.0 m/s2, in the direction of motion.
2	(b) Calculate the distance travelled by the car during the acceleration.	Answer: s = (u + v) / 2 x t = (8.0 + 20.0) / 2 x 4.0 = 56.0 m.
3	(c) Calculate the braking force acting on the car while it stops.	Answer: from v2 = u2 + 2as, deceleration a = 20.0^2 / (2 x 40.0) = 5.0 m/s2; braking force F = ma = 1200 x 5.0 = 6000 N.
4	(d) Calculate the kinetic energy of the car at 20.0 m/s and compare it with the work done by the braking force.	Answer: Ek = 0.5 x 1200 x 20.0^2 = 240000 J (240 kJ). Work done by the braking force = F x s = 6000 x 40.0 = 240000 J. The two values are equal, so the kinetic energy is fully removed by the work done by the braking force.

Section B: Circuits and Electrical Power (18 marks)

An electric kettle is marked 220 V, 2200 W. It is connected to a 220 V supply and heats a full kettle of water for 45 minutes. Electricity costs 1.2 dollars per kW h, and the heating time is proportional to the mass of water.

#	Question	Answer and marking notes
5	(a) Calculate the current in the kettle while it is working.	Answer: I = P / V = 2200 / 220 = 10.0 A.
6	(b) Calculate the resistance of the heating element.	Answer: R = V / I = 220 / 10.0 = 22.0 ohm.
7	(c) Express the electrical energy used in 45 minutes in both kW h and joules.	Answer: E = 2.2 kW x 0.75 h = 1.65 kW h; in joules, E = 2200 W x 2700 s = 5940000 J (5.94 x 10^6 J).
8	(d) Calculate the cost of this heating. If only half a kettle of water is heated, find the new heating time and cost.	Answer: cost = 1.65 kW h x 1.2 dollars = 1.98 dollars. Half a kettle takes 45 / 2 = 22.5 minutes and costs 0.99 dollars.

Section C: Mole Calculation and Conservation of Mass (16 marks)

25.0 g of calcium carbonate (CaCO3) is strongly heated until it decomposes completely into calcium oxide (CaO) and carbon dioxide (CO2). Relative atomic masses: Ca = 40.0, C = 12.0, O = 16.0. The molar volume of a gas at room conditions is 24.0 dm3/mol.

#	Question	Answer and marking notes
9	(a) Calculate the relative formula mass of CaCO3 and the number of moles in the sample.	Answer: Mr(CaCO3) = 40.0 + 12.0 + 3 x 16.0 = 100; moles = 25.0 / 100 = 0.250 mol.
10	(b) Calculate the mass of calcium oxide formed.	Answer: the mole ratio of CaCO3 to CaO is 1 : 1, so n(CaO) = 0.250 mol; m = n x Mr = 0.250 x 56 = 14.0 g.
11	(c) Calculate the mass of carbon dioxide formed and check it by conservation of mass.	Answer: n(CO2) = 0.250 mol; m = 0.250 x 44 = 11.0 g. Check: 14.0 + 11.0 = 25.0 g, equal to the 25.0 g of reactant, so mass is conserved.
12	(d) Calculate the volume of carbon dioxide formed at room conditions.	Answer: V = n x molar volume = 0.250 x 24.0 = 6.00 dm3.

Section D: Acid-Base Neutralisation and Titration (16 marks)

25.0 cm3 of 0.100 mol/dm3 sodium hydroxide solution is titrated with 0.125 mol/dm3 hydrochloric acid, using phenolphthalein as indicator.

#	Question	Answer and marking notes
13	(a) Calculate the number of moles of sodium hydroxide taken.	Answer: n = c x V = 0.100 x 0.0250 = 2.50 x 10^-3 mol.
14	(b) Calculate the volume of hydrochloric acid needed to reach the end point.	Answer: NaOH + HCl -> NaCl + H2O, mole ratio 1 : 1, so n(HCl) = 2.50 x 10^-3 mol; V = n / c = 2.50 x 10^-3 / 0.125 = 20.0 cm3.
15	(c) Calculate the concentration of sodium chloride in the mixed solution at the end point.	Answer: n(NaCl) = 2.50 x 10^-3 mol; total volume = 25.0 + 20.0 = 45.0 cm3 = 0.045 dm3; c = n / V = 0.0556 mol/dm3.
16	(d) Explain why the colour of phenolphthalein changes at the end point.	Answer: before the titration the solution is alkaline and phenolphthalein is pink in alkaline conditions. When the added acid has just neutralised all the sodium hydroxide, the solution becomes neutral (or slightly acidic), in which phenolphthalein is colourless, so the pink colour fades to colourless at the end point.

Section E: Inheritance and Mendel Heredity (16 marks)

In tomato plants the allele for tall stem (T) is dominant to the allele for dwarf stem (t). Two tall plants are crossed and produce 400 offspring, of which 289 are tall and 111 are dwarf.

#	Question	Answer and marking notes
17	(a) State the most likely genotypes of the two parents and the expected phenotypic ratio of the offspring.	Answer: both parents are heterozygous Tt. Each produces gametes T and t, so the Punnett square gives TT : Tt : tT : tt = 1 : 2 : 1. TT, Tt and tT are all tall, so the phenotypic ratio is tall : dwarf = 3 : 1.
18	(b) Assuming a 3 : 1 ratio, calculate the expected numbers of tall and dwarf plants among 400 offspring.	Answer: tall = 400 x 3 / 4 = 300; dwarf = 400 x 1 / 4 = 100.
19	(c) A tall plant is test-crossed with a dwarf parent. Find the percentage of tall plants among the offspring.	Answer: the tall parent in a test cross is Tt and produces T and t gametes, while the dwarf parent is tt and produces only t gametes. The offspring are half Tt (tall) and half tt (dwarf), so 50% are tall.
20	(d) Explain why the observed 289 : 111 differs from the expected 3 : 1.	Answer: the 3 : 1 ratio is a theoretical probability, and the genotype of each offspring results from random union of gametes. With a finite sample of 400 plants, random fluctuation makes the observed numbers depart from the expectation. Differences in germination rate, growing conditions and survival also affect the counts, so 289 : 111 is still consistent with 3 : 1.

Section F: Energy Flow in an Ecosystem (16 marks)

In a grassland ecosystem the producers fix 40000 kJ of energy per square metre per year, and the efficiency of energy transfer between adjacent trophic levels is about 10%.

#	Question	Answer and marking notes
21	(a) Calculate the energy available per square metre per year to the primary, secondary and tertiary consumers.	Answer: primary consumers = 40000 x 0.10 = 4000 kJ; secondary consumers = 4000 x 0.10 = 400 kJ; tertiary consumers = 400 x 0.10 = 40 kJ.
22	(b) Calculate the energy lost and the percentage lost when energy passes from the primary to the secondary consumers.	Answer: lost = 4000 – 400 = 3600 kJ; percentage lost = 3600 / 4000 x 100% = 90%.
23	(c) State three ways in which energy is largely lost between trophic levels.	Answer: first, respiration oxidises organic matter and releases energy as heat; second, not all ingested food is digested and absorbed, so some leaves the body in faeces and excretory waste; third, part of each trophic level is never eaten and is broken down by decomposers after death, so that energy does not pass to the next level.
24	(d) Explain why food chains rarely have more than four or five trophic levels.	Answer: only about 10% of the energy passes on at each step; from 4000 kJ the tertiary consumers receive just 40 kJ, and one more step leaves only 4 kJ, which cannot support the survival and reproduction of another trophic level, so the length of a food chain is limited by energy.

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