DSE Physics | Momentum and Impulse Lesson-Prep Guide (for tutors) | sign convention, conservation, four common errors

2026-11-05

Tutor lesson-prep guides · Secondary Physics

Students can recite p = mv yet lose marks the moment a ball bounces, because they never declared a positive direction. This lesson makes that declaration the first line of every momentum calculation.

Objectives — Students can: (1) declare a positive direction first, then write each momentum before and after a collision with its sign; (2) use Ft = change in p to find the average force and explain why a longer contact time gives a smaller force; (3) solve one-dimensional collisions including sticking and rebound, and use the change in kinetic energy to decide whether a collision is elastic.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Vectors	What separates speed from velocity?	Velocity has a direction, speed does not
Newton second law	What is a in F = ma?	Acceleration, the rate of change of velocity
Kinetic energy	What is the formula for kinetic energy?	Half m v squared

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-6 min	Prior check	Three quick questions, especially speed against velocity	Answer orally
6-20 min	Fix the sign, then write momentum	Take right as positive and write the rebound velocity as negative	Students tag three velocities with signs
20-32 min	Impulse and average force	Turn 4.8 N s into 240 N over 0.02 s	Students redo it for 0.2 s and explain
32-44 min	Conservation in collisions	2 kg hits 1 kg and sticks, momenta match before and after	Students set up the equation and solve for v
44-56 min	Elastic against inelastic	Compare 9 J with 6 J and say where the energy went	Students classify two collisions
56-60 min	Exit check	Collect three items, mark at once, tag the act	Work independently

03 Teaching script (read it out)

Scene 1: Act 1: every momentum calculation begins with a positive direction

The marks are lost on direction, not on the formula. Write on the board a ball hitting a wall at 20 m/s and rebounding at 12 m/s, then ask for the change in momentum. Most students answer 0.15 x (20 – 12) = 1.2 because they treat the rebound speed as positive.
The correct opening line is: take rightwards as positive, so before impact u = +20 m/s, and after the bounce v = -12 m/s. Then change in p = m(v – u) = 0.15 x (-12 – 20) = -4.8 N s, where the minus sign means leftwards. Stress that the sign is part of the answer, not a mistake; only when the question asks for magnitude do you write 4.8 N s and state the direction separately.
Put the misconception on the board: momentum is a vector, so an answer without a direction is incomplete.

Students repeat: first line is the positive direction, and the rebound velocity carries a minus sign.

Scene 2: Act 2: impulse is the change in momentum, force is impulse over time

Once the change in momentum is known, that is the impulse, in N s, which equals kg m/s. Continue the same question: contact lasts 0.02 s, so the average force is 4.8 / 0.02 = 240 N. Now change the contact time to 0.2 s, as with a soft pad, and the force becomes 24 N. The same change in momentum over ten times the time gives a tenth of the force.
That single step explains airbags, helmets and bending the knees on landing: the change in momentum is not reduced, the time is stretched, so the peak force drops. Students should state that chain in words rather than recite that an airbag protects you. Add that the area under a force-time graph is the impulse, so a varying force needs its average.

Students repeat: fixed change in momentum, longer time, smaller force.

Scene 3: Act 3: conservation plus kinetic energy decides elasticity

Two situations. First, 2 kg at 3 m/s hits a stationary 1 kg and they stick: momentum before is 6 kg m/s, mass after is 3 kg, so v = 2 m/s. Kinetic energy is 9 J before and 6 J after, so 3 J left as heat and sound, which makes it inelastic.
Second, a 1200 kg car at 15 m/s hits an 800 kg van at 5 m/s from behind and they lock together: v = (18000 + 4000) / 2000 = 11 m/s. Point out that velocities in the same direction share a sign, and opposite directions need one positive and one negative. Close with the rule: momentum is always conserved, kinetic energy only in a perfectly elastic collision, and equal masses in a perfectly elastic collision exchange velocities.

Students repeat: momentum is always conserved, kinetic energy is not.

04 Question types and worked steps

Type	Example	Steps	Answer
Momentum and change in momentum	0.15 kg ball hits a wall at 20 m/s and rebounds at 12 m/s	1 fix the positive direction 2 rebound velocity is negative 3 change in p = m(v – u)	-4.8 N s, leftwards
Impulse and average force	The same impact lasts 0.02 s, find the average force	1 use the change in p as the impulse 2 divide by time 3 state the direction	240 N leftwards
One-dimensional collision	2 kg at 3 m/s hits a stationary 1 kg and sticks	1 total momentum before 2 total mass times v after 3 solve for v	2 m/s
Elasticity test	How much kinetic energy is lost in that collision	1 compute energy before and after 2 subtract 3 a loss above zero means inelastic	9 J becomes 6 J, 3 J lost

Sanity check (teach this): After each answer ask two things: is the direction written, and does the final speed sit between the two original speeds? A 1200 kg car at 15 m/s locking with an 800 kg van at 5 m/s must give something between 5 and 15 m/s; 11 m/s is sensible, while 20 m/s or a negative value means a sign or total-mass error. Kinetic energy after a collision can never exceed the energy before.

05 Independent practice (3 questions)

Question	Answer	What to watch
A 0.15 kg ball hits a wall at 20 m/s and rebounds at 12 m/s in 0.02 s. Find the average force.	240 N away from the wall	Did they make the rebound velocity negative and divide change in p by time
2 kg at 3 m/s hits a stationary 1 kg and they stick. Find the common velocity and the energy lost.	2 m/s; 3 J lost	Did they compute kinetic energy separately before and after
A 1200 kg car at 15 m/s locks with an 800 kg van at 5 m/s. Find the velocity.	11 m/s	Does the answer lie between 5 and 15 m/s

06 Common mistakes and what to say

Mistake	Student writes	Say this
Treating the rebound speed as positive	change in p = 0.15 x (20 – 12) = 1.2 N s	A rebound means it turned round, so the speed is minus twelve.
No declared positive direction	All numbers positive, direction guessed at the end	Write which way is positive in line one, then nothing is a guess.
Wrong mass after sticking	Dividing by one mass only	After sticking the mass is both masses added, three kilograms.
Assuming kinetic energy is always conserved	Using energy conservation to find the speed	Momentum is always conserved, kinetic energy only in a perfectly elastic collision.

07 Homework (paper, section and question numbers)

Redo this chapter — Redo the three items and write the positive direction on the first line
Physics chapter index — Preview Newton laws of motion and note how they connect
Conservation drill — Write one head-on collision of your own and solve it

08 Exit check (last 5 minutes)

Question	Answer	Decision
What is the first line of a momentum calculation	Take one direction as positive	No line means replay Act 1
For a fixed change in momentum, what does a longer time do to the force	It makes the force smaller	Answering larger means replay Act 2
What is always conserved in a collision	Total momentum	Answering kinetic energy means replay Act 3

09 Tutor FAQ

How many equations are needed?

Two: p = mv and Ft = change in p. For collisions use total momentum before equals after.

Can I choose the positive direction?

Yes, either side works, but it must stay the same throughout and the final direction must be stated.

When can kinetic energy be conserved?

Only when the question states a perfectly elastic collision; otherwise use momentum and describe the energy 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.