S5 Geography Practice Paper: River Processes, Coastal Landforms, Weather, Plate Tectonics And Map Skills

S5 Geography Practice Paper: River Processes, Coastal Landforms, Weather, Plate Tectonics and Map Skills

2026-11-12

S5 Geography

S5 geography practice paper: river discharge, coastal erosion, lapse rate and humidity, plate motion and earthquake energy, catchment runoff and map scale.

Download　Question paper PDF　|　Answer key PDF

Paper details　S5 geography practice paper with worked answers on river discharge, coastal erosion, lapse rate, earthquake energy, runoff and map scale.

Part 1: River discharge and channel characteristics (20 marks)

A river reach is 24 m wide with a mean depth of 2.5 m and a mean velocity of 1.2 m per second. Over the 8 km upstream of the reach the bed falls 12 m. The channel itself measures 11.2 km long, while the valley it occupies is 8.0 km long.

#	Question	Answer and marking notes
1	Calculate the cross-sectional area and the discharge of the reach.	Answer: Area = 24 x 2.5 = 60.0 square metres; discharge = 60.0 x 1.2 = 72.0 cubic metres per second.
2	Estimate the hydraulic radius, taking wetted perimeter as width plus twice the mean depth.	Answer: Wetted perimeter = 24 + 2 x 2.5 = 29.0 m; hydraulic radius = 60.0 / 29.0 = 2.07 m. A larger hydraulic radius means less frictional resistance.
3	Calculate the channel gradient and express it in the form 1 : N.	Answer: Gradient = 12 / 8,000 = 0.0015, that is 1 : 667.
4	Calculate the sinuosity and describe the channel form.	Answer: Sinuosity = 11.2 / 8.0 = 1.4. Above 1.3 the channel is meandering, with slower flow, deposition on the inner bank and erosion on the outer bank.

Part 2: Coastal erosion and longshore drift (20 marks)

A sea cliff is 35 m high and loses 42 cubic metres of rock per metre of coast each year. Mean wave height there rises from 1.5 m to 3.0 m. Longshore drift moves 45,000 cubic metres of sand past a section each year, and the beach is 5 km long and 30 m wide. Sediment input to the coast is 62,000 cubic metres a year against an output of 78,500 cubic metres.

#	Question	Answer and marking notes
5	Calculate the annual retreat of the cliff.	Answer: Retreat = 42 / 35 = 1.2 m per year. For the same volume lost, a lower cliff retreats faster.
6	Wave energy is proportional to the square of wave height. If height rises from 1.5 m to 3.0 m, by what factor does energy increase?	Answer: Height ratio = 3.0 / 1.5 = 2; energy ratio = 2 squared = 4.0 times. Winter swell can therefore multiply erosive power.
7	Assuming all drifted sand stays on the beach, calculate the average annual thickening of the sand layer.	Answer: Beach area = 5,000 x 30 = 150,000 square metres; thickening = 45,000 / 150,000 = 0.3 m per year.
8	Calculate the annual sediment budget and state the coastal trend.	Answer: Net change = 62,000 – 78,500 = -16,500 cubic metres. Output exceeds input, so the beach is narrowing and erosion dominates.

Part 3: Lapse rate and atmospheric humidity (20 marks)

Sea-level temperature is 28.4 degrees C and the lapse rate is 6.5 degrees C per km. At a station at 20 degrees C the actual vapour pressure is 16.8 hPa and the saturation vapour pressure at that temperature is 23.4 hPa. Saturation values are 16.0 hPa at 14 degrees C and 17.1 hPa at 15 degrees C. Annual rainfall at the foot of the windward slope is 1,180 mm; the air then rises a further 500 m, giving 22 mm per 100 m. Leeward rainfall is 72 per cent of the windward total.

#	Question	Answer and marking notes
9	Calculate the temperature at the top of a 1,600 m peak.	Answer: Lapse = 6.5 x 1.6 = 10.4 degrees C; summit temperature = 28.4 – 10.4 = 18.0 degrees C.
10	Calculate the relative humidity at the station.	Answer: Relative humidity = 16.8 / 23.4 = 71.8 per cent. The air is unsaturated and can still take up more water vapour.
11	Use the saturation values given to estimate the dew point.	Answer: The actual vapour pressure of 16.8 hPa lies between 16.0 at 14 degrees C and 17.1 at 15 degrees C, so the dew point is about 14.7 degrees C, close to saturation.
12	Calculate the annual rainfall on the leeward slope.	Answer: Windward total = 1,180 + 5 x 22 = 1,290 mm; leeward = 1,290 x 72 per cent = 928.8 mm, giving a rain shadow.

Part 4: Plate motion and earthquake energy (16 marks)

A plate moves 6.5 cm a year for 2.4 million years. One earthquake measured 5.2 and another 7.2; each unit of magnitude multiplies energy by about 31.6. P waves travel at 6.0 km per second and S waves at 3.5 km per second, and the epicentral distance is 84 km. Earthquake energy can be estimated from log E = 4.8 + 1.5 M, with E in joules.

#	Question	Answer and marking notes
13	Calculate how far the plate moves in 2.4 million years, in km.	Answer: 6.5 x 2,400,000 = 15,600,000 cm = 156,000 m = 156 km.
14	Calculate how many times more energy a magnitude 7.2 event releases than a 5.2 event.	Answer: Magnitude difference = 2.0; energy ratio = 31.6 squared which is about 1,000 times, that is roughly one thousand.
15	Calculate the difference in arrival time between the P and S waves.	Answer: P wave = 84 / 6.0 = 14.0 s; S wave = 84 / 3.5 = 24.0 s; difference = 10.0 s. A larger gap means a more distant epicentre.
16	Use the formula to estimate the energy released by a magnitude 6.0 earthquake.	Answer: log E = 4.8 + 1.5 x 6.0 = 13.8; E is about 6.3 x 10^13 joules, comparable to roughly 15,000 tonnes of TNT.

Part 5: Catchment runoff and water resources (12 marks)

A catchment covers 48 square km. A storm delivers 165 mm of rain and the runoff coefficient is 0.35. The reservoir can capture 60 per cent of the runoff, and daily supply is 180,000 cubic metres.

#	Question	Answer and marking notes
17	Calculate the total volume of rain falling on the catchment.	Answer: 48 x 165 x 1,000 = 7,920,000 cubic metres. One square km receiving 1 mm of rain gains 1,000 cubic metres.
18	Calculate the runoff generated by the storm.	Answer: 7,920,000 x 0.35 = 2,772,000 cubic metres.
19	Calculate the combined volume lost to evaporation, infiltration and groundwater recharge.	Answer: 7,920,000 x 0.65 = 5,148,000 cubic metres.
20	Calculate the volume the reservoir can capture and how many days of supply it supports.	Answer: Captured = 2,772,000 x 0.60 = 1,663,200 cubic metres; days = 1,663,200 / 180,000 which is about 9.2 days, under a fortnight.

Part 6: Map scale, gradient and area (12 marks)

A map has a scale of 1 : 20,000. Two points lie 6.4 cm apart on the map. The horizontal distance between the 120 m and 240 m contours on the map is 1.2 km. A rectangular area measures 5 cm by 4 cm on the map. A walking route is 4.8 km long and walking speed is 4 km per hour.

#	Question	Answer and marking notes
21	Calculate the real distance between the two points.	Answer: 6.4 x 20,000 = 128,000 cm = 1,280 m = 1.28 km.
22	Calculate the slope percentage of that hillside.	Answer: Height difference = 240 – 120 = 120.0 m; horizontal distance = 1,200 m; slope = 120.0 / 1,200 = 10.0 per cent.
23	Calculate the real area of the rectangular region.	Answer: 1 cm represents 200 m, so 1 square cm represents 40,000 square metres; 20 x 40,000 = 800,000 square metres, that is 0.8 square km.
24	Calculate the time needed to walk the whole route.	Answer: 4.8 / 4 = 1.2 hours, that is 72 minutes. Add rest time for uphill sections.

Find a tutor — free Tutor Registration →

A TutorZone practice paper; original questions and passages, not publisher material.