Physics 9702 · Paper 2

Kinetic Energy

A force can also speed an object up. The work it does is then stored as energy of motion: .
The exam asks for the formula, for calculations, and, in 2025, for the full derivation.

The formula

E_k = \tfrac{1}{2}mv^2

Symbols

$E_k$ = kinetic energy (J)
$m$ = mass (kg)
$v$ = speed (m s⁻¹)

Speed is squared: double the speed means four times the kinetic energy; 4× the KE means only 2× the speed (9702/12/M/J/24 Q16).
Mass in kg, speed in m s⁻¹, energy in J.

Worked example

Smallest case: one moving ball

A 0.50 kg ball moves at 12 m s⁻¹. Find its kinetic energy.

$E_k = \tfrac{1}{2}(0.50)(12)^2 = 36 \text{ J}$ .

Answer

E_k = 36 \text{ J}

The derivation (a 2025 three-marker)

Worked example

Show that Ek = ½mv²

A constant resultant force F acts on a car of mass m, starting from rest. After displacement s its speed is v. Using the concept of work done, show that its kinetic energy is ½mv². (9702/22/M/J/25 Q1(b)(i))

Work done on the car: $W = Fs$ , and Newton's second law gives $F = ma$ , so $W = mas$ .
From rest, $v^2 = u^2 + 2as$ becomes $v^2 = 2as$ , so $as = \dfrac{v^2}{2}$ .
Substitute: $W = m \times \dfrac{v^2}{2} = \tfrac{1}{2}mv^2$ . All the work became energy of motion, so $E_k = \tfrac{1}{2}mv^2$ .

Answer

E_k = \tfrac{1}{2}mv^2

The three ingredients are $W = Fs$ , $F = ma$ and $v^2 = u^2 + 2as$ . Multiple-choice questions ask exactly which equations the derivation needs (9702/12/O/N/25 Q17).

Changes in kinetic energy

Worked example

One change: a speeding-up car

A truck of mass 9400 kg speeds up from 13 m s⁻¹ to 22 m s⁻¹. Find its gain in kinetic energy. (9702/22/F/M/25 Q3(a)(ii))

$\Delta E_k = \tfrac{1}{2}m(v^2 - u^2) = \tfrac{1}{2}(9400)(22^2 - 13^2)$ .
$\Delta E_k = 4700 \times 315 = 1.5 \times 10^{6} \text{ J}$ .

Answer

\Delta E_k = 1.5 \times 10^{6} \text{ J}

Common mistake

Never compute a KE change as

\tfrac{1}{2}m(\Delta v)^2

. Here that wrong route gives

\tfrac{1}{2}(9400)(9^2) = 3.8 \times 10^{5}

J, four times too small. Square the two speeds first, then subtract.

Kinetic energy from momentum

Because $p = mv$ (lesson 3.06), $E_k = \dfrac{p^2}{2m}$ . Paper 1 uses it to jump from momentum to energy in one line, and to ask for $\Delta E_k = \dfrac{p_2^{\,2} - p_1^{\,2}}{2m}$ when the momentum changes (9702/11/M/J/25 Q19).

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Kinetic Energy

A force can also speed an object up. The work it does is then stored as energy of motion: .
The exam asks for the formula, for calculations, and, in 2025, for the full derivation.

The formula

E_k = \tfrac{1}{2}mv^2

Symbols

$E_k$ = kinetic energy (J)
$m$ = mass (kg)
$v$ = speed (m s⁻¹)

Speed is squared: double the speed means four times the kinetic energy; 4× the KE means only 2× the speed (9702/12/M/J/24 Q16).
Mass in kg, speed in m s⁻¹, energy in J.

Worked example

Smallest case: one moving ball

A 0.50 kg ball moves at 12 m s⁻¹. Find its kinetic energy.

$E_k = \tfrac{1}{2}(0.50)(12)^2 = 36 \text{ J}$ .

Answer

E_k = 36 \text{ J}

The derivation (a 2025 three-marker)

Worked example

Show that Ek = ½mv²

Work done on the car: $W = Fs$ , and Newton's second law gives $F = ma$ , so $W = mas$ .
From rest, $v^2 = u^2 + 2as$ becomes $v^2 = 2as$ , so $as = \dfrac{v^2}{2}$ .
Substitute: $W = m \times \dfrac{v^2}{2} = \tfrac{1}{2}mv^2$ . All the work became energy of motion, so $E_k = \tfrac{1}{2}mv^2$ .

Answer

E_k = \tfrac{1}{2}mv^2

The three ingredients are $W = Fs$ , $F = ma$ and $v^2 = u^2 + 2as$ . Multiple-choice questions ask exactly which equations the derivation needs (9702/12/O/N/25 Q17).

Changes in kinetic energy

Worked example

One change: a speeding-up car

A truck of mass 9400 kg speeds up from 13 m s⁻¹ to 22 m s⁻¹. Find its gain in kinetic energy. (9702/22/F/M/25 Q3(a)(ii))

$\Delta E_k = \tfrac{1}{2}m(v^2 - u^2) = \tfrac{1}{2}(9400)(22^2 - 13^2)$ .
$\Delta E_k = 4700 \times 315 = 1.5 \times 10^{6} \text{ J}$ .

Answer

\Delta E_k = 1.5 \times 10^{6} \text{ J}

Common mistake

Never compute a KE change as

\tfrac{1}{2}m(\Delta v)^2

. Here that wrong route gives

\tfrac{1}{2}(9400)(9^2) = 3.8 \times 10^{5}

J, four times too small. Square the two speeds first, then subtract.

Kinetic energy from momentum