Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 4: Mechanics · official syllabus
NEWTON'S LAWS LAB
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Four short missions on resultant force, vertical motion, a rough incline and connected particles. Draw forces on each body, choose a direction, then use the resultant force = mass x acceleration.
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Mission 1 of 4
Use the resultant
A 3 kg particle moves right on a horizontal surface. A 21 N force pulls right and a constant 3 N resistance acts left. Take g = 10 m/s^2; there is no vertical acceleration. Set its rightward acceleration. The teal acceleration arrow is not an extra force.
Which force belongs on the force side of F = ma?
Fixed force diagram: weight/reaction 30 N, pull 21 N, resistance 3 N. The separate teal arrow is acceleration on a different scale, not a force.
Your acceleration is 3.0 m/s^2. Subtract the 3 N resistance before dividing by 3 kg.
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
Use the resultant force, not just the driving force 4.4
Newton's first law: a particle stays at rest or moves with constant velocity when the resultant force is zero. For constant mass, Newton's second law gives resultant force = mass x acceleration. Resolve forces along a chosen positive direction and write the signed sum as ma.
Draw forces acting on the chosen particle, not forces it exerts on other bodies. Weight acts downward with W=mg. Mass is measured in kg; weight and other forces are measured in N. Use g=10 m/s^2 in these examples, and the stated value in an exam question.
Worked example: a 3 kg particle moves right under a 21 N rightward pull and a 3 N leftward resistance. The resultant is 18 N, so a=18/3=6 m/s^2 to the right. Dividing the driving force 21 N alone by the mass would ignore resistance.
Vertical motion and tension 4.4
Choose upward as positive if it simplifies the problem. For a particle lifted by a string, T-mg=ma. A downward acceleration then has a negative value in that equation. Changing the positive direction changes signs, not the physical answer.
Worked example: a 4 kg particle accelerates upward at 2 m/s^2 with g=10. T-40=4(2), so T=48 N. For downward acceleration of 2 m/s^2 with the same upward-positive convention, T-40=4(-2), so T=32 N.
A light taut string transmits tension and pulls along itself. It cannot push. A zero or negative calculated tension is a warning to check whether the assumed string can remain taut and whether the motion model is valid.
Inclined planes: resolve parallel and perpendicular 4.4
On a plane inclined at theta to the horizontal, weight has component mg sin theta down the plane and mg cos theta into it. If the particle stays on the plane with no other perpendicular forces, perpendicular acceleration is zero and R=mg cos theta. Choose the positive direction along the plane for the motion equation.
For a smooth plane with no other along-plane force, down-plane acceleration is g sin theta. On a rough plane, include friction in the direction opposing motion. When using the sliding-friction model F=muR, state it and do not confuse it with the general static bound F<=muR.
Worked example: a 5 kg particle slides down a rough plane with sin theta=3/5, cos theta=4/5, mu=1/4 and g=10. R=50(4/5)=40 N and friction is 10 N up the plane. Thus 30-10=5a, so a=4 m/s^2 down the plane.
Friction changes direction when the motion reverses 4.4
A particle sliding up a rough plane has both the downhill weight component and friction opposing its uphill velocity. If uphill is positive, ma=-mg sin theta-mu mg cos theta under the sliding-friction model.
For the same plane with sin theta=3/5, cos theta=4/5, mu=1/4 and g=10, the acceleration while sliding up is -8 m/s^2 in an uphill-positive convention. While sliding down, friction reverses and the down-plane acceleration is +4 m/s^2 in a downhill-positive convention.
At the instant the particle stops, do not automatically continue either sliding equation. Check static friction and any applied forces to decide whether it remains at rest or starts moving again. In this example the downhill weight component 30 N exceeds limiting friction 10 N, so it cannot remain at rest under those forces alone.
Connected particles: equations for each body 4.4
For a light inextensible taut string over a fixed smooth pulley, use a common acceleration magnitude along the string and equal tension in its two portions. Write a force equation for each particle in its own direction of motion, then solve simultaneously. State the string and pulley assumptions; they are what justify these shared quantities.
Worked example: A (3 kg) is on a smooth horizontal table and connects over the edge pulley to hanging B (2 kg). With g=10 and the string taut, A accelerates towards the pulley and B downward. For A, T=3a. For B, 20-T=2a. Adding gives a=4 m/s^2, then T=12 N.
Using both particles as one system removes the internal tension and gives the acceleration quickly, but it does not directly give the tension. Return to an individual particle's equation. If one particle reaches the ground or the string goes slack, the original connected-motion equations no longer describe the next stage.
Towing, tension, thrust and model limits 4.4
A light rope can pull by tension but cannot push. A light rigid rod or tow-bar can transmit a push (thrust) or a pull. Draw its force in the assumed direction, solve, and interpret any sign change against what the connector can physically do.
Worked example: a 600 kg car tows a 200 kg trailer on level ground with acceleration 1 m/s^2. The trailer's resistance is 100 N and the car's is 300 N. Trailer equation: T-100=200(1), so T=300 N. Whole-system equation: D-300-100=800(1), so driving force D=1200 N.
Include any resistance specified in the question; do not invent an air-resistance term. With constant forces and constant mass the acceleration is constant, so constant-acceleration kinematics can then describe the motion while those conditions hold. A force model and a kinematics equation answer different parts of the problem.
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