A.3.9—Kinetic energy

Kinetic-energy forms

Translational kinetic energy is

$E_k=\frac12mv^2=\frac{p^2}{2m}$

Choose the known quantity

Use $\frac12mv^2$ when mass and speed are given, or $p^2/(2m)$ when momentum is given. Kinetic energy is scalar and cannot be negative.

Worked example from local Question Bank row 35674

For $m=0.14\,\mathrm{g}=1.4\times10^{-4}\,\mathrm{kg}$ and $v=3.1\,\mathrm{m\,s^{-1}}$ ,

$E_k=\frac12(1.4\times10^{-4})(3.1)^2=6.7\times10^{-4}\,\mathrm{J}=0.67\,\mathrm{mJ}$

Converting grams to kilograms before substitution keeps the energy unit in joules.

Common trap

Doubling speed quadruples kinetic energy; do not scale it linearly with speed.

How it is assessed

The evidence asks for final speed after power/resistance information and asks for energy transferred by a constant resultant force.

Command terms

Calculate / Identify

What earns marks

Select the kinetic-energy form matching the given quantities and keep speed in m s⁻¹, mass in kg and momentum in kg m s⁻¹. If a force accelerates an object from rest, use the work–energy link to identify the transferred energy.

Watch for

Using momentum directly as energy or forgetting the square on speed.

Representative question

[Maximum number: 2]

Calculate the final speed of the car.

A different car travels on a horizontal road at a constant speed of $45 \mathrm{~m} \mathrm{~s}^{-1}$ . The engine of the car develops a power of 140 kW . The resistive force $F_{\mathrm{d}}$ acting on the car is given by

Area $=2.4 \times 10^{5} \mathrm{~J}$ $\ll 2.4 \times 10^{5}=\frac{1}{2} \times 1.6 \times 10^{3} \times v^{2} \Rightarrow \gg v=17 \mathrm{~m} \mathrm{~s}^{-1}$

A.3.9—Kinetic energy

Calculate Translational Kinetic Energy

A.3.9 Exam Analysis

Assessment in practice

Question 1