Motion & Forces: kinematics and Newtonian dynamics
Key idea: Connect motion representations to the force and momentum models that explain changes in motion.
Before you start: Kinematics GraphsNewton’s Laws, Vectors and Problem Solving
By the end, you can
- Define and distinguish position, distance, displacement, speed, velocity and acceleration.
- Interpret position–time and velocity–time graphs using gradients and areas, including non-uniform acceleration.
- Derive and apply the equations of uniformly accelerated motion, including vertical motion without air resistance.
- Use inertia, momentum and Newton’s laws to connect resultant force to change in momentum.
Starting-point self-check
1. Check your starting point
Attempt both alignment groups without notes and mark the first representation or law you could not justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Dynamics outcomes 3(f)–(i)
Question 1
A 1200 kg car changes velocity from 10 m s⁻¹ east to 16 m s⁻¹ east in 3.0 s. Find its momentum change and average resultant force.
Check the model response
Δp = m(v − u) = 1200(6) = 7200 kg m s⁻¹ east. Average resultant force = Δp/Δt = 2400 N east.
Question 2
A book rests on a table. Identify the Newton’s-third-law partner of the upward normal force exerted by the table on the book.
Check the model response
It is the downward normal force exerted by the book on the table. The book’s weight is not its partner because both weight and normal force act on the book.
repair
2. Repair the five common breaks
Use only the repair matching an error, then redraw the graph or force model.
Dynamics outcomes 3(f)–(i)
Check this idea
Misconception: An object needs a forward resultant force to keep moving at constant velocity.
Repair: Inertia means constant velocity persists when resultant force is zero. A resultant force produces acceleration or, more generally, a rate of change of momentum.
Check this idea
Misconception: Newton’s-third-law forces cancel because they act on the same object.
Repair: An interaction pair is equal and opposite but acts on two different bodies. Only forces on the same chosen body are combined to find its resultant.
worked example
3. Follow four worked models
Follow how each solution selects a representation, sign convention and physical law.
Dynamics outcomes 3(f)–(i)
Model 1
A 0.20 kg ball moving at 15 m s⁻¹ east rebounds at 10 m s⁻¹ west in 0.050 s. Find the average force on the ball.
Check the model response
Take east positive. Δp = 0.20(−10 − 15) = −5.0 kg m s⁻¹. F = Δp/Δt = −5.0/0.050 = −100 N, so the average force is 100 N west.
Model 2
A 5.0 kg crate is pulled right by 22 N while resistance is 7.0 N left. Explain its motion using Newton’s laws and find its acceleration.
Check the model response
The resultant force is 22 − 7 = 15 N right. Newton’s second law gives a = F/m = 15/5.0 = 3.0 m s⁻² right. The pull and resistance are not a third-law pair because both act on the crate.
guided practice
4. Guided practice
Use each hint only to set up axes, graph quantities or the chosen system.
Dynamics outcomes 3(f)–(i)
Question 1
A 900 kg vehicle experiences a 2700 N driving force and 900 N resistance. Find its acceleration.
Hint: Add forces with signs before applying F = ma.
Check the model response
Resultant force = 2700 − 900 = 1800 N forward. a = F/m = 1800/900 = 2.0 m s⁻² forward.
Question 2
A swimmer pushes water backward. State the interaction pair and explain why the swimmer accelerates forward.
Hint: Put each force on its own receiving body.
Check the model response
The swimmer exerts a backward force on the water; the water exerts an equal forward force on the swimmer. The swimmer accelerates according to the resultant force on the swimmer, not by adding the force acting on the water.
independent practice
5. Independent practice
Solve without repair notes and show axes, graph interpretation and force-system boundaries.
Dynamics outcomes 3(f)–(i)
Question 1
Define inertia and momentum. Compare the momenta of a 1500 kg car at 12 m s⁻¹ and a 3000 kg truck at 6.0 m s⁻¹.
Check the model response
Inertia is resistance to change of velocity and is measured by mass. Momentum p = mv. Each vehicle has momentum 18 000 kg m s⁻¹ in its direction of motion.
Question 2
A resultant force varies so that an object’s momentum increases from 6 to 30 kg m s⁻¹ in 4.0 s. State the governing law and find the average force. If its constant mass is 3.0 kg, find the average acceleration.
Check the model response
Newton’s second law is F = dp/dt. Average force = (30 − 6)/4.0 = 6.0 N. For constant mass, F = ma, so average acceleration = 6.0/3.0 = 2.0 m s⁻².
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Dynamics outcomes 3(f)–(i)
Question 1
State all three Newton’s laws. For the second law, give both the momentum form and its constant-mass form.
Check the model response
First: a body remains at rest or in uniform straight-line motion unless acted on by a resultant force. Second: resultant force equals rate of change of momentum, F = dp/dt; for constant mass, F = ma. Third: interacting bodies exert equal and opposite forces on each other.
Question 2
A 0.40 kg trolley changes velocity from 3.0 m s⁻¹ right to 2.0 m s⁻¹ left in 0.20 s. Calculate the average resultant force and identify the force the trolley exerts on the agent causing this change.
Check the model response
Take right positive. Δp = 0.40(−2.0 − 3.0) = −2.0 kg m s⁻¹, so F = −2.0/0.20 = −10 N: 10 N left on the trolley. The trolley exerts an equal 10 N force right on the interacting agent.
Re-test practice
7. Delayed re-test practice
Return after three days and solve these fresh prompts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Dynamics outcomes 3(f)–(i)
Question 1
A 2.5 kg body’s momentum changes by 15 kg m s⁻¹ north in 0.50 s. Find its average resultant force and, for constant mass, its velocity change.
Check the model response
Average force = Δp/Δt = 15/0.50 = 30 N north. Δv = Δp/m = 15/2.5 = 6.0 m s⁻¹ north.
Question 2
A bus moves at constant velocity while its engine provides 5.0 kN forward. Explain the resultant force and identify the Newton’s-third-law partner of the road’s forward force on the tyres.
Check the model response
Constant velocity means zero resultant force, so total resistance is 5.0 kN backward. The third-law partner is the tyres’ backward force on the road, not a resistance force on the bus.