DSE Physics | Newton laws and free-body analysis: a tutor lesson guide

DSE Physics | Newton laws and free-body analysis: a tutor lesson guide | From F = ma to exam questions

2026-11-04

Tutor lesson-prep guides · Secondary Physics

Students can recite F equals ma, then freeze when a slope and friction appear together. The gap is not the formula but counting forces and naming the body each force acts on. This lesson uses one 2 kg trolley: draw first, write the equation second, explain collisions last.

Objectives — Students can: (1) draw a complete free-body diagram for a named object and state what applies each force; (2) use F = ma on horizontal motion with friction to find the net force and acceleration; (3) use impulse to explain why a short contact time produces a large force, and separate action from reaction.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Weight and mass	What is the weight of a 30 kg student?	294.3 N, multiply by 9.81
Resolving vectors	A 10 N force at 30 degrees: horizontal part?	8.66 N
Acceleration	What is the unit of acceleration?	m/s squared, the rate of change of velocity

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Three-question check	Weight, components, meaning of acceleration	Answer orally
8-24 min	The three laws	Inertia, F = ma, action and reaction	Give one example each
24-38 min	Free-body diagram	2 kg trolley, 6 N pull, friction 0.15	Draw and write the equation
38-48 min	Impulse	0.5 kg ball hits a wall at 8 and leaves at 6	Find change in momentum and force
48-56 min	Timed practice	Two calculations and one explanation	Work alone
56-60 min	Exit check	Three checkpoints, one question each	Self-assess and mark gaps

03 Teaching script (read it out)

Scene 1: Three laws: always name the body first

The first law says no net force means no change in velocity, not no motion; compare a spacecraft with a braking train. The second law is one equation: net force equals mass times acceleration, so a 2 kg trolley under 6 N has 3 m/s squared, but real questions subtract friction first. The third law is where marks are lost: action and reaction act on two different bodies, so they never cancel. Students must say which body each force acts on and who applies it.

For each law, students give one example and name both bodies.

Scene 2: Free-body diagram: four steps, none optional

Circle the object, draw weight 19.6 N downwards, draw the normal force upwards, then add the applied force and friction. With a coefficient of 0.15 on 2 kg, friction is 2.94 N, so the net force is 6 minus 2.94, which is 3.06 N, giving 1.53 m/s squared. Stress one point: friction opposes relative motion, not the applied force; if the pull is below the maximum static friction, the trolley does not move and the net force is zero.

Students draw their own diagram; the tutor only checks for missing forces and wrong directions.

Scene 3: Impulse: why short contacts hit hard

A 0.5 kg ball hits a wall at 8 m/s and leaves at 6 m/s, so the change in momentum is 0.5 times (8 plus 6), which is 7.0 kg m/s, pointing away from the wall. Over 0.02 s the average force is 350 N. Explain why the speeds add: the velocities point opposite ways, so the two momentum values have opposite signs. Stretch the same change over 0.2 s and the force drops to 35 N, which is the whole argument for airbags and padding.

Students change the contact time with the same numbers and watch the force move.

04 Question types and worked steps

Type	Example	Steps	Answer
Calculation with friction	2 kg trolley, 6 N pull, coefficient 0.15	Draw, find weight, find friction, net force, divide by mass	1.53 m/s squared
Components on a slope	10 N push at 30 degrees to the horizontal	Resolve to 8.66 N and 5.00 N, use the horizontal part	Answer by component, never divide 10 N
Explanation of impact force	Why is padding safer at the same momentum change?	State fixed change, longer time, smaller force	Must state the fixed change first
Diagram question	Block on an inclined plane	Weight, normal force, friction, directions labelled	Never draw a made-up sliding force

Sanity check (teach this): The numbers must agree: the net force 3.06 N is smaller than the 6 N pull, and 1.53 is smaller than the frictionless 3. A 30 kg object weighs 294.3 N, about three times a student body weight, so an answer of 30 N confuses mass with weight. An impact force of 350 N dwarfs the ball weight of 4.9 N because the contact is brief, not because the arithmetic is wrong. If a student gets more acceleration than the frictionless case, friction was added instead of subtracted.

05 Independent practice (3 questions)

Question	Answer	What to watch
3 kg trolley, 9 N pull, 3 N friction: acceleration	2 m/s squared	Subtract friction before dividing
0.2 kg ball stops from 5 m/s in 0.01 s: average force	100 N	Only the incoming momentum here
Explain why a wardrobe still has friction when it will not move	Static friction equals the push, net force is zero	Say equal and opposite

06 Common mistakes and what to say

Mistake	Student writes	Say this
Treating action and reaction as a balanced pair	Concludes the object cannot move	Ask which body each force acts on; different bodies cannot cancel.
Leaving weight out of the diagram	Draws only the pull and friction	Use the fixed order: weight, normal, applied, friction.
Dividing the pull by mass and stopping	Answers 3 m/s squared for 6 N on 2 kg	Ask where friction went; the answer is 1.53.
Subtracting momenta on a rebound	Answers 1.0 kg m/s	Make them mark velocity arrows; opposite directions means adding.

07 Homework (paper, section and question numbers)

Physics chapter index — All chapters
Physics practice papers — Forces: calculation questions 1 and 3, explanation question 2

08 Exit check (last 5 minutes)

Question	Answer	Decision
Diagram	Draw the three forces on a block on a slope	Missing weight: return to the four steps
Friction	2 kg with coefficient 0.15: friction value?	Says 6 N: must multiply by weight
Impulse	0.5 kg from 8 to 6: change in momentum?	Says 4: redo with velocity arrows

09 Tutor FAQ

Must students memorise the wording of the laws?

No, but they must state the condition and the bodies involved; explanation marks sit there.

How should signs be handled?

Fix a positive direction first and let it into every term; adding signs at the end loses marks.

Can momentum and energy be used together?

Only in elastic collisions; otherwise momentum alone, and say energy is lost.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.