CIE AS/A Level Maths-4.4 Newton’s laws of motion - Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-4.4 Newton’s laws of motion- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-4.4 Newton’s laws of motion- Study Notes
Key Concepts:
- Newton’s Laws of Motion with Constant Forces
- Mass and Weight
- Motion of a Particle on Vertical and Inclined Planes
- Motion of Connected Particles
Newton’s Laws of Motion with Constant Forces
Newton’s Laws of Motion with Constant Forces
Newton’s Laws Recap
- First Law: A particle remains at rest or in uniform motion in a straight line unless acted upon by a resultant force.
- Second Law: The resultant force on a particle is proportional to the rate of change of momentum. For constant mass, this gives \( F = ma \).
- Third Law: If body A exerts a force on body B, then body B exerts an equal and opposite force on body A.
Forces that may act on a particle:
- Weight: \( W = mg \), acting vertically downwards.
- Normal Reaction (R): The contact force perpendicular to a surface.
- Tension (T): A pulling force through a string, rope, or chain.
- Friction (F): A force opposing relative motion, \( F \leq \mu R \).
- Thrust: A pushing force transmitted through a rigid rod.
Method for solving problems:
- Draw a free-body diagram for each particle.
- Choose a direction as positive (usually along the line of motion).
- Resolve forces horizontally and vertically as needed.
- Apply Newton’s 2nd Law: \( F = ma \) in each direction.
- If there are connected particles, set up simultaneous equations.
Example:
A particle of mass \(5 \ \text{kg}\) is pulled along a rough horizontal surface by a horizontal force of \(40 \ \text{N}\). The coefficient of friction is \(0.2\). Find the acceleration of the particle.
▶️ Answer/Explanation
Step 1: Forces acting:
Weight = \( W = 5g = 49 \ \text{N}\).
Normal reaction = \( R = 49 \ \text{N}\).
Friction = \( F = \mu R = 0.2 \times 49 = 9.8 \ \text{N}\).
Step 2: Apply Newton’s 2nd Law horizontally:
Resultant force = \( 40 – 9.8 = 30.2 \ \text{N}\).
\( ma = 30.2 \implies a = \dfrac{30.2}{5} = 6.04 \ \text{m/s}^2 \).
Final Answer: The acceleration is \( \boxed{6.04 \ \text{m/s}^2} \).
Example:
Two particles of masses \(4 \ \text{kg}\) and \(6 \ \text{kg}\) are connected by a light inextensible string passing over a smooth pulley. Find the acceleration of the system and the tension in the string.
▶️ Answer/Explanation
Step 1: The heavier mass (6 kg) will move downward, lighter (4 kg) moves upward.
Step 2: Let acceleration = \(a\), tension = \(T\).
For \(6 \ \text{kg}\): \( 6g – T = 6a \).
For \(4 \ \text{kg}\): \( T – 4g = 4a \).
Step 3: Add equations:
\( (6g – T) + (T – 4g) = 6a + 4a \).
\( 2g = 10a \implies a = \dfrac{2g}{10} = 1.96 \ \text{m/s}^2 \).
Step 4: Substitute to find \(T\):
\( 6g – T = 6a \implies 58.8 – T = 11.76 \implies T = 47.04 \ \text{N}\).
Final Answer:
Acceleration = \( \boxed{1.96 \ \text{m/s}^2} \)
Tension = \( \boxed{47.0 \ \text{N}} \)
Mass and Weight
Mass and Weight
Mass (\(m\)) is a scalar quantity that measures the amount of matter in an object. It is independent of location and is measured in kilograms (kg).
Weight (\(W\)) is the force due to gravity acting on a mass. It depends on both the mass and the local gravitational field strength.
The relationship is given by:
\( W = mg \)
where:
- \(W\) = weight (in newtons, N)
- \(m\) = mass (in kilograms, kg)
- \(g\) = gravitational field strength (in \(\text{m/s}^2\))
For A Level mechanics, unless otherwise stated, take \( g \approx 10 \ \text{m/s}^2 \).
- Weight is a vector quantity, always directed vertically downwards (towards the centre of the Earth).
Example:
A box of mass \(8 \ \text{kg}\) is resting on the ground. Find its weight.
▶️ Answer/Explanation
We use the formula \( W = mg \).
\( W = 8 \times 10 = 80 \ \text{N} \)
The weight acts vertically downwards.
\(\boxed{80 \ \text{N downward}}\)
Example:
A rocket has a mass of \(1200 \ \text{kg}\). Find the force of gravity acting on it when it is near the Earth’s surface.
▶️ Answer/Explanation
Using \( W = mg \):
\( W = 1200 \times 10 = 12,000 \ \text{N} \)
Thus, the rocket’s weight is:
\(\boxed{12,000 \ \text{N downward}}\)
Motion of a Particle on Vertical and Inclined Planes
Motion of a Particle on Vertical and Inclined Planes
When a particle moves under constant acceleration, the kinematics equations can be used:
\( v = u + at \)
\( s = ut + \dfrac{1}{2}at^2 \)
\( v^2 = u^2 + 2as \)
Vertical motion:
- The only force acting (if air resistance is neglected) is weight \( W = mg \) downward.
- If upwards is taken as positive, then acceleration \( a = -g \).
Inclined plane motion (angle \(\theta\)):
- Component of weight parallel to plane = \( mg\sin\theta \).
- Component of weight perpendicular to plane = \( mg\cos\theta \) (balanced by normal force).
- If there is friction (\( F = \mu R \)), it always opposes motion.
- Net force along the plane gives acceleration using Newton’s 2nd Law:
\( F_{\text{net}} = ma \)
Key idea with rough planes:
The acceleration is different when moving up and when moving down the plane, because the frictional force always acts opposite to the direction of motion.
Example:
A ball is thrown vertically upward with a speed of \(20 \ \text{m/s}\). Find the maximum height reached (take \( g = 10 \ \text{m/s}^2\)).
▶️ Answer/Explanation
At maximum height, velocity \( v = 0 \).
Using equation: \( v^2 = u^2 + 2as \)
\( 0 = (20)^2 + 2(-10)h \)
\( 0 = 400 – 20h \)
\( h = 20 \ \text{m} \)
\(\boxed{20 \ \text{m}}\)
Example:
A block of mass \(5 \ \text{kg}\) is projected up a rough incline at \(30^\circ\) to the horizontal with initial velocity \(8 \ \text{m/s}\). The coefficient of friction is \(\mu = 0.2\). Find:
- (a) The distance it moves up the plane before coming to rest.
- (b) The acceleration as it comes back down the plane.
▶️ Answer/Explanation
Step 1: Forces up and down the incline
When moving up the plane, forces opposing motion = \( mg\sin\theta + \mu mg\cos\theta \).
\( F = 5 \times 10 (\sin 30^\circ + 0.2\cos 30^\circ) \)
\( F = 50 (0.5 + 0.2 \times 0.866) \)
\( F = 50 (0.5 + 0.173) = 50 \times 0.673 = 33.65 \ \text{N} \)
So, acceleration up the plane:
\( a = \dfrac{F}{m} = \dfrac{33.65}{5} \approx 6.73 \ \text{m/s}^2 \) (down the plane)
Step 2: Distance before coming to rest
Using \( v^2 = u^2 + 2as \), with \( v=0, u=8, a=-6.73 \):
\( 0 = 64 – 2(6.73)s \)
\( s = \dfrac{64}{13.46} \approx 4.75 \ \text{m} \)
Step 3: Acceleration down the plane
Now forces down the plane = \( mg\sin\theta – \mu mg\cos\theta \).
\( F = 50(0.5 – 0.173) = 50 \times 0.327 = 16.35 \ \text{N} \)
\( a = \dfrac{16.35}{5} = 3.27 \ \text{m/s}^2 \) (down the plane)
Final Answer:
Distance up plane: \(\boxed{4.75 \ \text{m}}\)
Acceleration down plane: \(\boxed{3.27 \ \text{m/s}^2}\)
Motion of Connected Particles
Motion of Connected Particles
Connected particle systems usually involve two or more particles joined by a light inextensible string or a rigid tow-bar.
- The string is assumed to be light (massless) and inextensible (does not stretch), and the pulley is assumed to be smooth (no friction).
- This means the tension is the same throughout the string, and the connected particles have the same acceleration.
- Newton’s Second Law (\(F = ma\)) is applied separately to each particle, then equations are combined to solve for unknowns such as tension and acceleration.
Typical models include:
- Two particles on either side of a smooth pulley (Atwood’s machine type problems).
- A particle on a rough plane connected to a hanging particle.
- A car and trailer connected by a tow-bar or rope.
Example:
Two particles \(A\) of mass \(4 \ \text{kg}\) and \(B\) of mass \(6 \ \text{kg}\) are connected by a light inextensible string passing over a smooth pulley.
Find the acceleration of the system and the tension in the string.
▶️ Answer/Explanation
Let acceleration of the system be \(a\).
Since \(B\) is heavier, it moves downward and \(A\) moves upward.
For \(A\): \(T – 4g = 4a\)
For \(B\): \(6g – T = 6a\)
Adding equations:
\((6g – T) + (T – 4g) = 6a + 4a\)
\(2g = 10a\)
\(a = 0.2g = 2 \ \text{m/s}^2\) (taking \(g = 10 \ \text{m/s}^2\))
Now substitute into equation for \(A\):
\(T – 40 = 8 \Rightarrow T = 48 \ \text{N}\)
Final Answer: \(\boxed{a = 2 \ \text{m/s}^2}, \ \boxed{T = 48 \ \text{N}}\)
Example:
A car of mass \(800 \ \text{kg}\) pulls a trailer of mass \(200 \ \text{kg}\) along a horizontal road. The driving force of the engine is \(2000 \ \text{N}\). Find the acceleration of the system and the tension in the tow-bar (assume no resistance).
▶️ Answer/Explanation
Total mass of system = \(800 + 200 = 1000 \ \text{kg}\).
Total force = \(2000 \ \text{N}\)
Acceleration = \(F/m = 2000 / 1000 = 2 \ \text{m/s}^2\)
Now consider trailer alone:
Tension = \(m a = 200 \times 2 = 400 \ \text{N}\)
Final Answer: \(\boxed{a = 2 \ \text{m/s}^2}, \ \boxed{T = 400 \ \text{N}}\)