Free lesson plan 8 of 8 · Key Stage 4

Energy and crater size: does doubling the energy double the crater?

GCSE (Years 10 and 11)60 minutesScience, National Curriculum for England

Students drop a ball into sand from measured heights, calculate its gravitational potential energy and impact speed, and find that crater diameter grows only with roughly the fourth root of energy, before comparing their tiny impacts with real ones.

Printable worksheet

Learning objectives
  • Calculate gravitational potential energy using Ep = mgh, and use conservation of energy to find the impact speed.
  • Carry out an investigation, identifying the independent, dependent and control variables, and plot crater diameter against energy.
  • Describe a non-linear relationship from a graph, and compare a laboratory impact with a real one using kinetic energy = ½mv².

National Curriculum links

Key vocabulary

Resources

Background for the teacher

As the ball falls, energy is transferred from its gravitational potential energy store to its kinetic energy store. If air resistance is small, as it is for a dense ball over a metre or so, the kinetic energy at impact equals mgh. On impact, that energy is transferred to the sand: moving grains thrown out as ejecta, heating, and sound.

The key result is that crater diameter does not rise in proportion to energy. Experiments with balls dropped into sand show that diameter grows roughly with the fourth root of the impact energy. Double the energy and the crater is only about 19% wider (the fourth root of 2 is about 1.19); to double the diameter you need about 16 times the energy. That makes sense: a crater twice as wide is also about twice as deep, so it has roughly eight times the volume of sand to move, and that sand has to be lifted higher too.

Real impacts are vastly more energetic. A meteoroid hits the atmosphere at between 11 and 72 km/s, so even a 1 kg rock at 20 km/s carries ½ × 1 × 20,000² = 2 × 10⁸ J, hundreds of millions of times the energy of the classroom ball. The diameters of large craters on planets also grow much more slowly than the energy that made them. At such speeds the impactor and some of the target vaporise in an explosion-like release of energy, which is why most real craters are round even when the impactor struck at a slant.

Illustrative results show the pattern you can expect, though your values will depend on the ball and the sand: a 0.050 kg ball might make a crater about 4.0 cm wide from 0.10 m, 4.8 cm from 0.20 m, 5.7 cm from 0.40 m, 6.7 cm from 0.80 m and 7.4 cm from 1.20 m.

Lesson plan

Starter 10 minutes

Ask: if a falling rock had twice as much energy, would its crater be twice as wide? Take a vote and record the result.

Recap Ep = mgh, with g = 9.8 N/kg, and work through one example together: a 0.050 kg ball at 1.20 m has Ep = 0.050 × 9.8 × 1.20 = 0.588 J, about 0.59 J. Ask students to identify the independent, dependent and control variables for the investigation.

Main 35 minutes

Groups weigh their ball, level the sand with the card, and drop the ball from heights of 0.10, 0.20, 0.40, 0.80 and 1.20 m, measured from the sand surface to the bottom of the ball. They measure each crater rim to rim in two directions at right angles and take the mean, then lift the ball out, stir the sand lightly and re-level it before the next drop. Each height is repeated at least twice.

Students calculate Ep for each height, complete the table and plot mean crater diameter (y-axis) against energy (x-axis), drawing a smooth curve of best fit.

Plenary 15 minutes

Return to the starter vote. What happened to the diameter when the energy went up eight times, from 0.10 m to 0.80 m? Students should find it rose by a factor of well under two, close to the fourth root of 8 (about 1.7).

Discuss the evaluation: sources of error (judging the rim, packing of the sand, dropping from exactly the same point), and the limits of the model compared with a real impact. Show the site's impact calculator and try doubling an asteroid's diameter, which multiplies its mass and energy by eight.

Differentiation

Support

Provide the results table with the Ep formula filled in for the first row, and graph axes already scaled.

Core

Complete the investigation, calculations and graph, and describe the relationship in words.

Challenge

Calculate the impact speed for each height using v = √(2gh), and test the fourth-root idea by plotting diameter against the fourth root of energy, which should give a straight line through the origin.

Safety and risk assessment

Check these points against your employer's risk assessment and your school's policies before the lesson.

HazardHow to control it
Eye injury from flying sand, flour or a rebounding ballWear eye protection; drop, never throw; keep faces back from the tray at impact.
Slipping on spilt sand, flour or a loose steel ballKeep balls on the tray, retrieve them at once, and sweep up spills straight away.
Flour dust and allergies or asthmaCheck for allergies; use sand if anyone is affected; keep flour away from naked flames.
Falls from climbing to drop the ballMaximum drop height 1.20 m from a clamped metre rule; nobody stands on stools or benches.
Back strain lifting heavy traysLeave sand trays on the bench, or lift with two people.

Assessment questions with answers

  1. Calculate the gravitational potential energy of a 0.050 kg ball held 1.20 m above the sand. (g = 9.8 N/kg)Answer: Ep = mgh = 0.050 × 9.8 × 1.20 = 0.588 J, or 0.59 J to 2 significant figures.
  2. Assuming no energy is transferred to the surroundings during the fall, calculate the ball's speed just before it hits the sand.Answer: ½mv² = 0.588 J, so v = √(2 × 0.588 ÷ 0.050) = √23.52 = 4.8 m/s.
  3. Name the independent variable, the dependent variable and two control variables.Answer: Independent: drop height (and so the energy). Dependent: crater diameter. Control: the same ball, the same sand, levelled and loosened the same way each time, and the same way of measuring the rim.
  4. When the energy increased eight times, a group's crater diameter rose from 4.0 cm to 6.7 cm. Is crater diameter proportional to energy? Explain.Answer: No. If it were proportional, the diameter would also have risen eight times, to 32 cm. It rose by a factor of about 1.7, close to the fourth root of 8 (1.68).
  5. A 1.0 kg meteoroid enters the atmosphere at 20 km/s. Calculate its kinetic energy and compare it with the ball's 0.59 J.Answer: Ek = ½ × 1.0 × (20,000)² = 2.0 × 10⁸ J. That is about 340 million times the ball's energy (2.0 × 10⁸ ÷ 0.59 ≈ 3.4 × 10⁸).

Homework

Use the site's impact calculator. Keep the speed, angle and density the same and record the crater size for an impactor 10 m, 20 m and 40 m across. Each doubling of diameter multiplies the mass and energy by eight. By what factor does the crater diameter change? Compare this with your classroom result.

Answer guidance: Each doubling of impactor diameter multiplies the energy by eight, but the crater grows by a factor of well under eight, typically around one and a half to two. Note that small impactors may break up in the air before reaching the ground. Like the classroom result, crater size grows much more slowly than energy.

Worksheet

Energy and crater size: a one-page worksheet for pupils that prints cleanly, with answers for the teacher.

Use with these pages

The full set

These eight lessons are a free starter pack, written specially for this site. The full set of 25 lesson plans, with worksheets, end-of-unit tests and mark schemes, is in the Corvanelle book Teaching Meteorites. The lessons in the book are different lessons, so the starter pack and the book work together rather than repeating each other.

Teaching Meteorites: Lesson Plans and Activities