Physics 9702 · Paper 1

Conservation of Energy

Energy cannot be created or destroyed. It only moves between stores: potential, kinetic, thermal, elastic.
Every big structured question in this chapter is built on that one sentence.

The principle

The principle of conservation of energy: the total energy of a closed system is constant. Energy can be transferred between forms, but it cannot be created or destroyed (9702/12/O/N/24 Q17).
With no friction or air resistance, lost GPE = gained KE, and the reverse on the way up.
With resistance, the totals still balance, but part of the energy becomes thermal energy (and a little sound): lost GPE = gained KE + energy transferred to the surroundings.

A roller-coaster with a tall first hill and a lower second hill, a car partway down the first drop. — Energy changes between gravitational PE and kinetic energy as the car goes up and down. At a peak the car has the most GPE and the least KE; in a dip it has the most KE and the least GPE. With no friction the total is unchanged, so the car always keeps some KE and a lower later peak is fine.

Worked example

Smallest case: speed from a drop

A pendulum ball is pulled aside until it is 0.15 m above its lowest point, then released. Air resistance is negligible. How fast is it moving at the lowest point?

Lost GPE becomes KE: $mgh = \tfrac{1}{2}mv^2$ .
The mass cancels: $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.15} = 1.7 \text{ m s}^{-1}$ .

Answer

v = 1.7 \text{ m s}^{-1}

The mass cancelling is why a heavy ball and a light ball reach the same speed from the same height.

Worked example

One change: some energy is lost

Water leaves a dam outlet 20 m below the reservoir surface, moving at 16 m s⁻¹. What percentage of the lost GPE did not become kinetic energy (9702-style, from the coursebook)?

Work with 1 kg of water. GPE lost: $mgh = 1 \times 9.81 \times 20 = 196 \text{ J}$ .
KE gained: $\tfrac{1}{2}(1)(16)^2 = 128 \text{ J}$ .
Missing: $196 - 128 = 68 \text{ J}$ , so $68/196 = 35\%$ went to heat and sound.

Answer

35\% \text{ lost}

The exam chains

Worked example

Exam version: a rocket's energy totals

A rocket of mass 2.9 × 10⁶ kg rises from rest to a height of 3.2 km, reaching 320 m s⁻¹ after 20 s. Resistive forces are negligible. Find the gains in GPE and KE, then the average output power. (9702/21/O/N/25 Q1(c))

GPE: $\Delta E_p = 2.9 \times 10^{6} \times 9.81 \times 3200 = 9.1 \times 10^{10} \text{ J}$ .
KE: $\Delta E_k = \tfrac{1}{2}(2.9 \times 10^{6})(320)^2 = 1.5 \times 10^{11} \text{ J}$ .
The engines supplied both: $P = \dfrac{(0.91 + 1.5) \times 10^{11}}{20} = 1.2 \times 10^{10} \text{ W}$ .

Answer

9.1 \times 10^{10} \text{ J}, \; 1.5 \times 10^{11} \text{ J}, \; 1.2 \times 10^{10} \text{ W}

The power step is the trap: the engine paid for the KE and the GPE. Forgetting one of the two is the standard lost mark.

Worked example

One change: work against friction on a slope

A block is pushed up a slope at constant speed by a 1.8 N force along the slope, moving 8.0 m along it and rising a height h. Its weight is 4.0 N, and 4.8 J of work is done against friction. Find h. (9702/12/O/N/25 Q19)

Energy in = energy out: $1.8 \times 8.0 = 4.8 + 4.0h$ . (Constant speed, so no KE term.)
$14.4 - 4.8 = 4.0h$ , so $h = 2.4 \text{ m}$ .

Answer

h = 2.4 \text{ m}

Your turn— tap to reveal the worked answer (9702/22/M/J/23 Q2(c))

A block of weight 2.4 N leaves a machine at a height of 1.8 m moving at 3.4 m s⁻¹. Air resistance is negligible. Find the decrease in its GPE as it falls to the ground, and its kinetic energy just before landing. (9702/22/M/J/23 Q2(c))

GPE lost: $2.4 \times 1.8 = 4.3 \text{ J}$ (the weight is already a force).
Starting KE: $\tfrac{1}{2}\left(\dfrac{2.4}{9.81}\right)(3.4)^2 = 1.4 \text{ J}$ .
Final KE: $1.4 + 4.3 = 5.7 \text{ J}$ .

Answer: 4.3 J lost; 5.7 J final KE.

Two traps that repeat: an object falling at constant speed gains no KE, so its lost GPE goes to thermal energy of the air, not to motion (9702/11/M/J/25 Q20); a projectile at the top of its arc keeps its horizontal motion, so a 45° launch keeps half its kinetic energy at the peak, not zero (9702/12/F/M/25 Q18).

List where the energy starts, where it usefully goes, and what becomes heat. Every term is a number of joules, and the two totals must be equal.

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Conservation of Energy

Energy cannot be created or destroyed. It only moves between stores: potential, kinetic, thermal, elastic.
Every big structured question in this chapter is built on that one sentence.

The principle

The principle of conservation of energy: the total energy of a closed system is constant. Energy can be transferred between forms, but it cannot be created or destroyed (9702/12/O/N/24 Q17).
With no friction or air resistance, lost GPE = gained KE, and the reverse on the way up.
With resistance, the totals still balance, but part of the energy becomes thermal energy (and a little sound): lost GPE = gained KE + energy transferred to the surroundings.

Worked example

Smallest case: speed from a drop

A pendulum ball is pulled aside until it is 0.15 m above its lowest point, then released. Air resistance is negligible. How fast is it moving at the lowest point?

Lost GPE becomes KE: $mgh = \tfrac{1}{2}mv^2$ .
The mass cancels: $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.15} = 1.7 \text{ m s}^{-1}$ .

Answer

v = 1.7 \text{ m s}^{-1}

The mass cancelling is why a heavy ball and a light ball reach the same speed from the same height.

Worked example

One change: some energy is lost

Water leaves a dam outlet 20 m below the reservoir surface, moving at 16 m s⁻¹. What percentage of the lost GPE did not become kinetic energy (9702-style, from the coursebook)?

Work with 1 kg of water. GPE lost: $mgh = 1 \times 9.81 \times 20 = 196 \text{ J}$ .
KE gained: $\tfrac{1}{2}(1)(16)^2 = 128 \text{ J}$ .
Missing: $196 - 128 = 68 \text{ J}$ , so $68/196 = 35\%$ went to heat and sound.

Answer

35\% \text{ lost}

The exam chains

Worked example

Exam version: a rocket's energy totals

GPE: $\Delta E_p = 2.9 \times 10^{6} \times 9.81 \times 3200 = 9.1 \times 10^{10} \text{ J}$ .
KE: $\Delta E_k = \tfrac{1}{2}(2.9 \times 10^{6})(320)^2 = 1.5 \times 10^{11} \text{ J}$ .
The engines supplied both: $P = \dfrac{(0.91 + 1.5) \times 10^{11}}{20} = 1.2 \times 10^{10} \text{ W}$ .

Answer

9.1 \times 10^{10} \text{ J}, \; 1.5 \times 10^{11} \text{ J}, \; 1.2 \times 10^{10} \text{ W}

The power step is the trap: the engine paid for the KE and the GPE. Forgetting one of the two is the standard lost mark.

Worked example

One change: work against friction on a slope

Energy in = energy out: $1.8 \times 8.0 = 4.8 + 4.0h$ . (Constant speed, so no KE term.)
$14.4 - 4.8 = 4.0h$ , so $h = 2.4 \text{ m}$ .

Answer

h = 2.4 \text{ m}

Your turn— tap to reveal the worked answer (9702/22/M/J/23 Q2(c))

GPE lost: $2.4 \times 1.8 = 4.3 \text{ J}$ (the weight is already a force).
Starting KE: $\tfrac{1}{2}\left(\dfrac{2.4}{9.81}\right)(3.4)^2 = 1.4 \text{ J}$ .
Final KE: $1.4 + 4.3 = 5.7 \text{ J}$ .

Answer: 4.3 J lost; 5.7 J final KE.

Two traps that repeat: an object falling at constant speed gains no KE, so its lost GPE goes to thermal energy of the air, not to motion (9702/11/M/J/25 Q20); a projectile at the top of its arc keeps its horizontal motion, so a 45° launch keeps half its kinetic energy at the peak, not zero (9702/12/F/M/25 Q18).

List where the energy starts, where it usefully goes, and what becomes heat. Every term is a number of joules, and the two totals must be equal.