Newton's laws (P2.2)
Gateway A's flagship physics topic. Expect at least one calculation using F = ma, a free-body diagram, and a 6-mark question on safety / stopping distances or momentum.
Newton's three laws
Newton's First Law — inertia
A body continues at rest, or moves at a constant velocity in a straight line, unless acted on by a resultant force.
In other words: if forces are balanced (resultant force = 0), velocity is constant. The body could be stationary or moving at the same speed in the same direction.
Example: a parachutist at terminal velocity. Weight downward = drag upward → resultant = 0 → constant velocity.
Newton's Second Law — F = ma
F (N) = m (kg) × a (m/s²)
The resultant force on a body is proportional to its mass and to its acceleration. Always use resultant force in this equation — not the size of any one force.
Inertial mass: a measure of how difficult it is to change a body's velocity. Higher mass → harder to accelerate.
Newton's Third Law — action–reaction pairs
For every action there is an equal and opposite reaction.
Important: the action and reaction forces act on different objects. So they don't cancel each other out on a single body.
Example: a book on a table. The book pushes down on the table (action); the table pushes up on the book (reaction). The pair are equal in size, opposite in direction, on different bodies.
Resultant force and balanced forces
Forces are vectors — they have direction. To find a resultant on a single line:
- If forces in the same direction: add.
- If forces in opposite directions: subtract.
If the resultant = 0, forces are balanced. The object is either stationary or moving at constant velocity.
If the resultant ≠ 0, forces are unbalanced. The object accelerates in the direction of the resultant.
Free-body diagrams
Draw the body as a dot or rectangle. Draw arrows for every force acting on the body — never forces it exerts on others. Label every force with name and (if known) magnitude.
Common labels:
- Weight (W or mg) — always downward.
- Normal contact force (N or R) — perpendicular to surface.
- Friction (F or f) — opposes motion, parallel to surface.
- Drag (air resistance) — opposes motion through fluid.
- Thrust / push / tension — applied force.
Worked F = ma examples
Example 1. A 1,200 kg car accelerates at 2.5 m/s². Find the resultant force.
- F = m × a = 1,200 × 2.5 = 3,000 N.
Example 2. A 70 kg cyclist experiences a forward force of 200 N from pedalling and a 60 N backward drag/friction.
- Resultant = 200 − 60 = 140 N.
- a = F / m = 140 / 70 = 2 m/s².
Example 3. A 60 kg lift descends at constant velocity. The cable tension is 588 N. What is the cable tension if the lift then accelerates downward at 1.5 m/s²?
- Constant velocity: T = mg = 60 × 9.8 ≈ 588 N. ✓
- Accelerating down at 1.5 m/s²: resultant force on lift is downward, magnitude m × a = 60 × 1.5 = 90 N.
- Resultant = mg − T (downward), so T = mg − ma = 588 − 90 = 498 N.
Stopping distance
stopping distance = thinking distance + braking distance
- Thinking distance is the distance travelled during the driver's reaction time. Increased by tiredness, alcohol, drugs, distractions.
- Braking distance is the distance after the brakes are applied. Increased by greater speed, worn tyres, wet/icy roads, worn brakes.
Doubling the speed roughly quadruples the braking distance (because KE ∝ v²).
Common Gateway-paper mistakes
- Using a single force in F = ma instead of the resultant.
- Forgetting that constant velocity ≠ zero force — it means zero RESULTANT force.
- Mixing up Newton's pair: writing that weight and normal contact force are a Newton-third-law pair (they are NOT — both act on the same body).
- Confusing mass (kg) with weight (N): W = mg.
- Forgetting to convert g to kg in F = ma calculations.
➜Try this— Quick check
- A 0.50 kg ball is thrown with a force of 7.5 N. Acceleration = 7.5 / 0.50 = 15 m/s².
- A 1,500 kg car decelerates from 30 m/s to 0 in 10 s. a = (0 − 30) / 10 = −3 m/s². F = 1,500 × −3 = −4,500 N (i.e. 4,500 N opposing motion).
AI-generated · claude-opus-4-7 · v3-ocr-combined-science