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Energy · 4.1

Kinetic Energy

A moving object can transfer energy by doing work. Mass matters, but speed matters twice.

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Simple definition
Kinetic energy is the energy an object has because it is moving.
Example
A moving bicycle has kinetic energy; braking transfers some of it into thermal energy.
01

Build the model

Start with the energy story.

Kinetic energy is a scalar store carried by motion. Direction does not appear in the formula, so two objects moving in opposite directions can have the same kinetic energy.

Motion storeK = 1/2 mv²

Energy of motion: it grows with mass, and with the square of speed — twice as fast means four times the energy.

m in kilograms, v in metres per second

Work-energy theoremWnet = ΔK

The total work done on an object tells you exactly how much its energy of motion changed.

Net work changes the kinetic energy

Energy unit1 J = 1 kg·m²/s²

The joule, written out in basic units — every energy formula on this page produces this same unit.

A joule is a unit of transferred or stored energy

01

Mass is linear

At one fixed speed, doubling the mass doubles K.

02

Speed is squared

Doubling the speed multiplies K by four; tripling it multiplies K by nine.

03

Net work changes K

Positive net work speeds an object up. Negative net work removes kinetic energy.

04

One store among several

Kinetic energy is one entry on a list you should be able to give in full: kinetic, in anything moving; gravitational potential, in anything raised in a gravitational field; chemical, in fuel, food or a battery; elastic strain, in anything stretched, squashed or twisted; nuclear, the store emptied by fission and fusion; electrostatic, in charges held apart; and internal or thermal, the random energy of the particles inside an object.

05

Four ways energy moves

Energy does not stay put. It is transferred between stores along four pathways: mechanical work done by a force, electrical work done by a current, heating, and waves such as light and sound. A full answer names three things — the store that emptied, the pathway it travelled by, and the store that filled.

02

Change one variable at a time

Test the model live.

Kinetic energy5,760 J

If the cart stopped, up to this much energy could be transferred by net work.

03

Catch the common trap

Explain before calculating.

A cart keeps the same mass but its speed doubles. What happens to its kinetic energy?

Choose one answer, then use the feedback to check your model.

04

Worked examples

Model, substitute, check.

EasyFind the kinetic energy of a 0.40 kg ball moving at 5.0 m/s.
  1. K = ½mv² = ½ × 0.40 × 25.
  2. K = 5.0 J.

AnswerK = 5.0 J

MediumA 1,200 kg car speeds up from 10 m/s to 20 m/s. How much net work is done?
  1. Kinitial = 1/2(1,200)(10²) = 60,000 J.
  2. Kfinal = 1/2(1,200)(20²) = 240,000 J.
  3. Wnet = ΔK = 240,000 - 60,000 J.

AnswerWnet = +180,000 J = 180 kJ

HardTrolley A (2.0 kg at 6.0 m/s) and trolley B (8.0 kg at 3.0 m/s) are compared. Which has more kinetic energy, and which has more momentum?
  1. K(A) = ½ × 2.0 × 36 = 36 J; K(B) = ½ × 8.0 × 9.0 = 36 J — equal kinetic energies.
  2. p(A) = 2.0 × 6.0 = 12 kg·m/s; p(B) = 8.0 × 3.0 = 24 kg·m/s.
  3. Same energy, but B carries double the momentum — the two quantities scale differently with v.

AnswerEqual K (36 J each); B has twice the momentum

ChallengingA car's braking distance from 30 m/s is 45 m. Using energy reasoning, predict its braking distance from 60 m/s with the same braking force.
  1. Braking: Fd = ½mv², so d ∝ v² for fixed force.
  2. Doubling v quadruples the kinetic energy to remove.
  3. d = 4 × 45 = 180 m.

Answer180 m — double the speed, four times the distance

Exam diagrams for this topic2 figures to inspect and practiseQuestions, hints and marking points in one compact subsection.

See it. Read it. Work it.

These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.

  1. 01

    InspectRead the figure comment.

  2. 02

    TraceFollow labels, arrows and axes.

  3. 03

    AnswerWork one part at a time.

  4. 04

    CheckReveal hints and marking points.

IGCSE

01Fig. 5.1EnergyIGCSE
Ball released from rest above the ground and the height it rebounds to1.8 mreleased from restball, mass 0.50 kg1.2 mhighest point after the bounce

Figure comment

Fig. 5.1A ball of mass 0.50 kg is drawn twice above level ground. On the left it is at the point of release, with a dimension line marking 1.8 m from the ground up to the ball. On the right it is at the highest point reached after the bounce, with 1.2 m marked in the same way. Dashed vertical lines show the path down and the path back up.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

1.7.1 EnergyOpen IGCSE question · Motion, forces and energy Q5
Guided questions 5 parts

Reading cue. Both 1.8 m and 1.2 m run from the ground up to the ball, not from one to the other, and 1.2 m is the top of the rebound, where the ball is momentarily at rest.

  1. aIdentify Identify the point on the path drawn in Fig. 5.1 at which the ball has its greatest kinetic energy, and state the height of the ball above the ground there.

    recall2 marks

    Check answer 2 marks
    1. just before it reaches the ground, at the bottom of the fall (1)
    2. height = 0 m, at ground level (1)

  2. bDetermine Determine the speed of the ball just before it strikes the ground. Take the gravitational field strength as 9.8 N/kg and ignore air resistance.

    routine3 marks

    Check answer 3 marks
    1. kinetic energy gained equals gravitational potential energy lost: (1/2)mv² = mgh (1)
    2. v² = 2 × 9.8 × 1.8 = 35.28 (m/s)² (1)
    3. v = 5.9 m/s (1)

  3. cCalculate Calculate the percentage of its kinetic energy at the ground that the ball still has immediately after the bounce, and show that this percentage does not depend on the mass of the ball.

    demanding3 marks

    Check answer 3 marks
    1. energy immediately after the bounce is mgh with h = 1.2 m; energy at the ground before the bounce is mgh with h = 1.8 m (1)
    2. ratio = 1.2/1.8 = 0.67, so 67% (1)
    3. m and g appear in both expressions and cancel, so only the two marked heights decide the percentage (1)

  4. dEstimate The ball is allowed to bounce a second time. Assuming the same percentage of energy is lost at every bounce, estimate the height of the ball at the top of its second rebound, and suggest one reason why the true height would be lower than your estimate.

    top of the paper3 marks

    Check answer 3 marks
    1. height = 1.2 x (1.2/1.8) (1)
    2. = 0.80 m (1)
    3. reason, any one: air resistance also removes energy during the flight; the fraction lost at each bounce is not really constant, since the ball deforms differently at different speeds (1)

Transfer challenge

A pendulum bob is pulled to one side until it is 0.25 m above its lowest point and then released. On the far side it rises to 0.20 m above its lowest point. Determine the speed of the bob as it passes through its lowest point, and calculate the percentage of its energy lost in that swing. Take the gravitational field strength as 9.8 N/kg.

Check answer 4 marks
  1. (1/2)mv² = mgh, so v² = 2 × 9.8 × 0.25 = 4.9 (m/s)² (1)
  2. v = 2.2 m/s (1)
  3. energy lost is in the ratio (0.25 - 0.20)/0.25, because energy is proportional to height (1)
  4. = 20% (1)
02Fig. 8.1Momentum · EnergyIGCSE
Two trolleys on a level track, before the collision and after itbefore the collisionAB2.5 m/sat rest0.80 kg1.2 kgafter the collisionABvjoined together

Figure comment

Fig. 8.1Two panels, one above the other. In the upper panel, labelled before the collision, trolley A of mass 0.80 kg stands on a level track with an arrow showing it moving to the right at 2.5 m/s towards trolley B of mass 1.2 kg, which is at rest. In the lower panel, labelled after the collision, the two trolleys are drawn joined together and moving to the right with a speed marked v.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

1.6 Momentum1.7.1 EnergyOpen IGCSE question · Motion, forces and energy Q8
Guided questions 5 parts

Reading cue. In the upper panel the 2.5 m/s belongs to A alone, B at rest carrying none; in the lower panel v is the speed of the joined 2.0 kg, not of either trolley on its own.

  1. aState State the momentum of trolley B before the collision, and give the reason from the upper panel of Fig. 8.1.

    recall2 marks

    Check answer 2 marks
    1. zero (0 kg m/s) (1)
    2. B is at rest, so its velocity is zero, and momentum is mass x velocity (1)

  2. bDetermine After the collision the joined trolleys move to the right at 1.0 m/s. Determine the change in momentum of trolley A, and give its direction.

    routine3 marks

    Check answer 3 marks
    1. momentum of A before = 0.80 × 2.5 = 2.0 kg m/s to the right (1)
    2. momentum of A after = 0.80 × 1.0 = 0.80 kg m/s to the right (1)
    3. change = 1.2 kg m/s, directed to the left, that is opposite to A's motion (1)

  3. cCalculate The two trolleys are in contact for 0.15 s. Calculate the average force that trolley B exerts on trolley A during the collision.

    demanding3 marks

    Check answer 3 marks
    1. force = change in momentum / time taken (1)
    2. = 1.2 / 0.15 (1)
    3. = 8.0 N, acting to the left on A (1)

  4. dExplain Using both panels of Fig. 8.1, explain why the momentum gained by trolley B is exactly equal to the momentum lost by trolley A, even though B is the heavier trolley and its velocity changes by less.

    top of the paper4 marks

    Check answer 4 marks
    1. the force B exerts on A is equal in size and opposite in direction to the force A exerts on B (1)
    2. the two trolleys are in contact for the same length of time, so force x time is the same for both and the changes in momentum are equal and opposite (1)
    3. A's velocity changes by 1.5 m/s, giving 0.80 × 1.5 = 1.2 kg m/s; B's changes by 1.0 m/s, giving 1.2 × 1.0 = 1.2 kg m/s (1)
    4. B's larger mass is offset exactly by its smaller change in velocity, so the total momentum of the two trolleys is unchanged (1)

Transfer challenge

A skater of mass 50 kg stands at rest on ice holding a ball of mass 2.0 kg. She throws the ball horizontally away from her at 6.0 m/s. Determine the speed at which she moves backwards, and state the total momentum of the skater and ball after the throw.

Check answer 4 marks
  1. momentum of ball after the throw = 2.0 × 6.0 = 12 kg m/s (1)
  2. the skater must carry 12 kg m/s in the opposite direction, since the total was zero before (1)
  3. speed = 12/50 = 0.24 m/s (1)
  4. total momentum after the throw = zero, the same as before (1)

Cambridge 9702 · focused teaching and practice

Complete the objective

Open a topic for its explanation, worked example and original practice. The AS and additional A-Level requirements remain separate. Practical preparation does not replace supervised laboratory work.

AS · 5.2.3

Derive kinetic energy from motion equations

Teaching, example & practice

General browsing: select Cambridge AS or A2 and your examination year for a recommended route.

For a body of constant mass accelerated in a straight line by a constant resultant force, F=ma and net work is W=Fs. The constant-acceleration relation v²−u²=2as gives as=(v²−u²)/2.

Substitution gives W=mas=½m(v²−u²)=½mv²−½mu². The work changes the energy associated with motion, so E_K=½mv² when its value at rest is set to zero. This is an algebraic derivation: no calculus is required.

The calculation uses net force and net work. In a system with friction or other transfers, the work supplied by one agent need not equal the kinetic-energy increase. Speed is squared, so opposite velocity directions can have equal kinetic energy.

W_net=Fs=mas

v²−u²=2as

W_net=½mv²−½mu²=ΔE_K

Worked example

A 2.0 kg trolley accelerates from 3.0 to 5.0 m s⁻¹. Determine the net work and show why ½m(v−u)² is wrong.

  1. ΔE_K=½ × 2.0 × (25−9)=16 J.
  2. The incorrect expression gives ½ × 2 × (5−3)²=4 J; a difference of squares is not the square of a difference.

Net work=16 J.

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