Forces

Forces

:::info Board Coverage AQA Paper 1 & 2 | Edexcel Paper 1 & 2 | OCR A Gateway P2 & P6 | WJEC P2 :::

Explore the simulation above to develop intuition for this topic.

1. Force Diagrams and Vector Quantities

1.1 Scalars and Vectors

Scalar (magnitude only)	Vector (magnitude and direction)
Mass	Weight
Speed	Velocity
Distance	Displacement
Energy	Force
Temperature	Acceleration
Time	Momentum

1.2 Why the Distinction Matters

A scalar tells you “how much.” A vector tells you “how much” and “which way.” When you add two Scalars, you add their magnitudes: 3 kg + 5 kg = 8 kg. When you add two vectors, you must Account for direction: a 3 N force east plus a 5 N force west gives a 2 N force west, not an 8 N Force. Ignoring the vector nature of forces is the single most common source of error in mechanics Problems.

1.3 Force Diagrams

A free body diagram shows all the forces acting on a single object, drawn as arrows from the Centre of the object. The length of each arrow represents the magnitude of the force.

Rules:

• Show only forces acting ON the object
• Forces should be drawn to scale
• Use labelled arrows
• Do not include forces exerted BY the object on other objects

1.4 Resultant Forces

The resultant force is the single force that has the same effect as all the forces acting Together.

Forces in the same direction: add them.

Forces in opposite directions: subtract the smaller from the larger.

Forces at angles: use a vector diagram (triangle or parallelogram of forces).

Worked Example. Two forces of 30 N and 40 N act on an object at right angles. Find the magnitude Of the resultant force.

F = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \mathrm{ N

1.5 Resolving Vectors into Components (Higher Tier)

Any vector can be resolved into two perpendicular components. For a force $F$ at angle $\theta$ to The horizontal:

F_x = F\cos\theta \quad \mathrm{(horizontal component) F_y = F\sin\theta \quad \mathrm{(vertical component)

This technique is essential for inclined plane problems. Instead of working with forces at awkward Angles, you resolve every force into components parallel and perpendicular to the slope, then apply Newton’s second law to each direction independently.

2. Newton’s Laws of Motion

2.1 Newton’s First Law

An object remains at rest or moves with constant velocity unless acted upon by a resultant (net) Force.

This means:

• If the resultant force is zero, the object does not accelerate
• A stationary object stays stationary; a moving object continues at the same speed in the same direction

2.2 The Inertial Frame Caveat

Newton’s first law is only valid in an inertial reference frame — a frame that is not itself Accelerating. If you are in a car that accelerates forward, you feel pushed back into your seat. There is no backward force acting on you; the feeling arises because the car (your reference frame) Is accelerating forward. In the car’s non-inertial frame, Newton’s first law appears to be violated. On Earth’s surface, the frame is approximately inertial because the centripetal acceleration due to Earth’s rotation is very small compared to $g$.

2.3 Newton’s Second Law

The acceleration of an object is proportional to the resultant force and inversely proportional to Its mass:

$F = ma$

Where $F$ is the resultant force (N), $m$ is mass (kg), and $a$ is acceleration (m/s$^2$).

Worked Example. A 1200 kg car accelerates from rest at 2.5 m/s$^2$. Find the driving force (ignore friction).

F = ma = 1200 \times 2.5 = 3000 \mathrm{ N

Worked Example. A resultant force of 45 N acts on a 5 kg object. Find the acceleration.

a = \frac{F}{m} = \frac{45}{5} = 9 \mathrm{ m/s^2

2.4 What $F = ma$ Really Means

The equation $F = ma$ is a vector equation. In component form:

$\sum F_x = ma_x, \qquad \sum F_y = ma_y$

The acceleration is always in the same direction as the resultant force. If the resultant force is Zero, the acceleration is zero (Newton’s first law is a special case of the second). The mass $m$ is A scalar: it tells you how resistant the object is to acceleration. Doubling the mass halves the Acceleration for the same force.

2.5 Newton’s Third Law

When two objects interact, they exert equal and opposite forces on each other.

Key features:

• The forces are equal in magnitude
• The forces are opposite in direction
• The forces act on different objects
• The forces are of the same type (e.g. Both gravitational, both contact)

Example: A book rests on a table. The book exerts a downward force on the table (its weight), And the table exerts an equal upward force on the book (the normal reaction). These are a Newton’s Third law pair? No — the normal reaction and weight are NOT a third law pair because they act on The SAME object. The true pairs are: Earth pulls book down / book pulls Earth up; book pushes table Down / table pushes book up.

:::caution Newton’s third law pairs always act on different objects. The normal reaction and weight Act on the same object, so they are NOT a third law pair.

2.6 Identifying Third Law Pairs: A Systematic Method

To identify the correct third law pair for any force, ask: “What other object is involved in this Interaction?” Then write both forces in the form “Object A exerts force on Object B” and “Object B Exerts force on Object A.” Both forces must be the same type (gravitational, electromagnetic, Contact, etc.) and must be equal and opposite.

• Incorrect: Weight of book (Earth on book) and normal reaction (table on book). Different objects receive the force (book vs. Book), different types (gravitational vs. Contact).
• Correct: Weight of book (Earth on book, gravitational, downward) and gravitational pull of book on Earth (book on Earth, gravitational, upward). Same type, equal magnitude, opposite direction, different objects.

3. Weight, Mass, and Gravity

3.1 Weight and Mass

Mass is a measure of the amount of matter in an object. It is measured in kilograms (kg) and Does not change with location.

Weight is the gravitational force acting on an object. It is measured in newtons (N) and depends On the gravitational field strength.

$W = mg$

Where $W$ is weight (N), $m$ is mass (kg), and $g$ is gravitational field strength (N/kg).

Location	$g$ (N/kg)
Earth’s surface	9.8
Moon’s surface	1.6
Mars’ surface	3.7
Jupiter’s surface	24.8

Worked Example. An astronaut has a mass of 80 kg. Find their weight on Earth and on the Moon.

W_{\mathrm{Earth} = 80 \times 9.8 = 784 \mathrm{ N W_{\mathrm{Moon} = 80 \times 1.6 = 128 \mathrm{ N

3.2 Why Mass and Weight Are Different

Mass is an intrinsic property: it does not depend on where the object is. Weight is a force that Depends on the local gravitational field. An astronaut on the Moon has the same mass as on Earth but Only one-sixth the weight. If you were in deep space, far from any gravitational field, your mass Would be unchanged but your weight would be zero. You would still have inertia (resistance to Acceleration) even though you would be weightless.

3.3 Centre of Mass

The centre of mass is the single point through which the entire weight of an object can be Considered to act.

For a uniform, regular solid, the centre of mass is at its geometric centre.

Finding the centre of mass of an irregular lamina:

1. Suspend the lamina from a point and let it hang freely.
2. Hang a plumb line from the suspension point and draw a vertical line.
3. Repeat from a different suspension point.
4. The centre of mass is where the two lines intersect.

3.4 Centre of Mass and Stability

An object is stable if its centre of mass lies above its base. If the centre of mass moves outside The base, the object topples. This is why:

• A wide-based object is harder to topple than a narrow-based one.
• A low centre of mass makes an object more stable (double-decker buses have heavy engines at the bottom).
• A leaning object topples when the vertical line through its centre of mass falls outside its base.

4. Terminal Velocity

4.1 Drag and Air Resistance

Drag is the resistive force experienced by an object moving through a fluid (liquid or gas). Drag increases with speed.

For many objects at moderate speeds, drag is approximately proportional to the square of the speed:

F_{\mathrm{drag} \propto v^2

This means that doubling the speed quadruples the drag force. This quadratic dependence has enormous Practical consequences: the power required to overcome drag (and therefore the fuel consumption of a Vehicle) increases with the cube of the speed.

4.2 Reaching Terminal Velocity

When an object falls through a fluid:

1. Initially, the only force is weight (downward). Acceleration = $g$.
2. As speed increases, drag increases.
3. The resultant force decreases: F_{\mathrm{net} = W - D.
4. Eventually, drag equals weight: $D = W$. The resultant force is zero, and acceleration is zero.
5. The object continues at a constant speed called terminal velocity.

4.3 Why Terminal Velocity Is Not the Same as Maximum Speed

Terminal velocity is the speed at which the drag force exactly equals the driving force (in free Fall, the weight). It is a dynamic equilibrium: the forces are balanced, so the acceleration is Zero, but the object is still moving. The word “terminal” does not mean the object stops; it means The speed stops increasing.

4.4 Factors Affecting Terminal Velocity

• Mass: Greater mass means greater weight, so higher terminal velocity
• Surface area: Larger surface area means more drag, so lower terminal velocity
• Shape: Streamlined objects experience less drag
• Fluid density: Denser fluids provide more drag

Example: A skydiver with a parachute has a much lower terminal velocity than without one, Because the parachute greatly increases the surface area.

4.5 Terminal Velocity and Motion Graphs

A velocity-time graph for an object falling through a fluid shows:

• An initial linear region (constant acceleration, drag is negligible)
• The curve gradually flattens as drag increases
• The graph becomes horizontal at terminal velocity

The acceleration-time graph shows:

• Initial acceleration = $g$
• Acceleration decreases to zero
• The graph approaches the time axis asymptotically

4.6 Worked Example: Skydiver

A skydiver of mass 80 kg jumps from a plane. At terminal velocity, the drag force equals the weight:

D = mg = 80 \times 9.8 = 784 \mathrm{ N

After opening the parachute, the terminal velocity drops from about 55 m/s to about 5 m/s because The increased surface area produces much more drag at any given speed.

5. Stopping Distances

5.1 Stopping Distance

\mathrm{Stopping distance = \mathrm{thinking distance + \mathrm{braking distance

Thinking distance: The distance travelled during the driver’s reaction time. Depends on:

• Speed (directly proportional)
• Reaction time (affected by tiredness, alcohol, drugs, distractions)

Braking distance: The distance travelled while the brakes are applied. Depends on:

• Speed (proportional to $v^2$)
• Mass of the vehicle
• Condition of brakes and tyres
• Road conditions (wet, icy)
• Gradient of the road

5.2 Why Braking Distance Increases with the Square of Speed

Using the work-energy theorem: the work done by the braking force equals the change in kinetic Energy.

$Fd = \frac{1}{2}mv^2$

For a given braking force $F$The braking distance $d = \frac{mv^2}{2F}$ is proportional to $v^2$. Doubling the speed quadruples the braking distance. Tripling the speed multiplies it by nine. This Is the single most important fact about road safety: small increases in speed produce very large Increases in stopping distance.

5.3 Reaction Time

Typical reaction time: 0.2 — 0.9 seconds.

Worked Example. A car is travelling at 20 m/s. The driver’s reaction time is 0.5 s. The braking Distance is 32 m. Calculate the total stopping distance.

\mathrm{Thinking distance = 20 \times 0.5 = 10 \mathrm{ m \mathrm{Stopping distance = 10 + 32 = 42 \mathrm{ m

5.4 Factors Affecting Reaction Time

• Tiredness: Reaction time increases significantly when fatigued. After 18 hours without sleep, reaction time is comparable to a blood alcohol concentration of 0.05%.
• Alcohol: Even small amounts of alcohol increase reaction time and impair judgement.
• Mobile phones: Using a phone while driving increases reaction time by 30—50%.
• Drugs: Both illegal drugs and certain medications can impair reaction time.
• Age: Reaction time increases with age, particularly after 60.

6. Momentum

6.1 Definition

Momentum is a vector quantity defined as:

$p = mv$

Where $p$ is momentum (kg m/s), $m$ is mass (kg), and $v$ is velocity (m/s).

6.2 Conservation of Momentum

In a closed system (no external forces), the total momentum before a collision equals the total Momentum after the collision.

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

Worked Example. A 2 kg trolley moving at 3 m/s collides with a stationary 1 kg trolley and they Stick together. Find the velocity after the collision.

$2 \times 3 + 1 \times 0 = (2 + 1) \times v$ $6 = 3v$ v = 2 \mathrm{ m/s

6.3 Why Momentum Is Conserved

Newton’s third law tells us that the forces between two colliding objects are equal and opposite. Since these forces act for the same duration, the impulses are equal and opposite, and therefore the Changes in momentum are equal and opposite. The total momentum of the system is unchanged. This is a Direct consequence of Newton’s laws, not a separate assumption.

6.4 Newton’s Second Law and Momentum

$F = \frac{\Delta p}{\Delta t} = \frac{mv - mu}{t}$

This shows that force equals the rate of change of momentum. This form of Newton’s second law is More general than $F = ma$ because it remains valid even when the mass changes (e.g., a rocket Expelling fuel).

6.5 Impact Forces and Safety

For a given change in momentum, increasing the impact time decreases the force:

$F = \frac{\Delta p}{\Delta t}$

Safety features that increase impact time:

• Crumple zones in cars: the front of the car deforms during a crash, extending the stopping time from perhaps 0.01 s to 0.1 s. Since the momentum change is the same, the average force on the passengers is reduced by a factor of 10.
• Seat belts (stretch slightly): distribute the decelerating force over a larger area of the body and extend the stopping time.
• Air bags: provide a cushion that increases the time over which the passenger decelerates.
• Helmets and padding: increase the time over which an impact force acts.
• Crash mats: same principle applied to gymnastics and playgrounds.

6.6 Elastic and Inelastic Collisions (Higher Tier)

Elastic collision: Both momentum and kinetic energy are conserved. Examples: collisions between Hard steel balls, collisions between gas molecules.

Inelastic collision: Momentum is conserved but kinetic energy is not. Some kinetic energy is Transferred to thermal energy, sound, and deformation. Most real-world collisions are inelastic.

Perfectly inelastic collision: The objects stick together after the collision. This maximises The kinetic energy lost.

7. Moments, Levers, and Gears

7.1 Moments

A moment is the turning effect of a force:

$M = Fd$

Where $M$ is moment (Nm), $F$ is force (N), and $d$ is the perpendicular distance from the pivot to The line of action of the force (m).

Principle of moments: For an object in equilibrium, the sum of clockwise moments equals the sum Of anticlockwise moments.

\sum M_{\mathrm{clockwise} = \sum M_{\mathrm{anticlockwise}

Worked Example. A seesaw has a pivot at its centre. A child of mass 30 kg sits 2 m from the Pivot. Where must a child of mass 40 kg sit to balance the seesaw?

$30 \times 9.8 \times 2 = 40 \times 9.8 \times d$ $60 = 40d$ d = 1.5 \mathrm{ m from the pivot

7.2 Why the Distance Must Be Perpendicular

The moment depends on the perpendicular distance from the pivot to the line of action of the force, Not the distance from the pivot to the point where the force is applied. If the force acts at an Angle, the effective moment arm is reduced. This is why pushing a door near the hinge requires much More force than pushing near the handle: the perpendicular distance is much smaller.

7.3 Levers

A lever is a simple machine that uses moments. A small force applied at a large distance from The pivot can balance (or overcome) a large force at a small distance.

Mechanical advantage = load / effort = distance from effort to pivot / distance from load to Pivot.

Classes of lever:

Class	Arrangement	Example
1	Pivot between effort & load	Seesaw, scissors
2	Load between pivot & effort	Wheelbarrow, nutcracker
3	Effort between pivot & load	Tweezers, forearm

7.4 Gears

Gears transfer rotational motion. If a large gear drives a small gear, the small gear rotates Faster but with less turning force (torque).

The ratio of speeds is inversely proportional to the ratio of teeth:

\frac{\mathrm{speed of gear B}{\mathrm{speed of gear A} = \frac{\mathrm{teeth on gear A}{\mathrm{teeth on gear B}

This is a consequence of conservation of energy: if the small gear rotates faster, it must exert Less torque, because $P = \tau \omega$ (power = torque times angular velocity), and the power Transmitted through the gears is (ideally) the same.

8. Hooke’s Law

8.1 Statement

Hooke’s law: The extension of a spring is directly proportional to the force applied, provided The limit of proportionality is not exceeded.

$F = ke$

Where $k$ is the spring constant (N/m) and $e$ is the extension (m).

8.2 Force-Extension Graph

• The graph is a straight line through the origin up to the limit of proportionality
• Beyond this point, the line curves: the spring no longer obeys Hooke’s law
• The gradient of the linear section equals the spring constant $k$
• If the force is removed and the spring does not return to its original length, it has exceeded its elastic limit

8.3 The Spring Constant and Energy Storage

A stiffer spring (higher $k$) stores more energy for the same extension: $E = \frac{1}{2}ke^2$. Two Identical springs in parallel have twice the spring constant (k_{\mathrm{total} = 2k) and store Twice the energy for the same extension. Two identical springs in series have half the spring Constant (k_{\mathrm{total} = \frac{k}{2}).

8.4 Required Practical: Investigating Force and Extension

Method:

1. Measure the natural length of the spring.
2. Add masses one at a time, recording the total mass and the new length.
3. Calculate the extension for each mass.
4. Plot a graph of force (weight = $mg$) against extension.
5. The gradient gives the spring constant.

Sources of error:

• The spring may not return to its original length after large forces (exceeding the elastic limit).
• Parallax error when reading the ruler.
• The ruler may not be perfectly vertical.
• The masses may swing, causing additional stretching.

9. Pressure (Higher Tier)

9.1 Definition

Pressure is the force per unit area:

$P = \frac{F}{A}$

Where $P$ is pressure (pascals, Pa), $F$ is force (N), and $A$ is area (m$^2$).

9.2 Pressure in Fluids

The pressure in a fluid increases with depth:

$P = h\rho g$

Where $h$ is the depth, $\rho$ is the density of the fluid, and $g$ is gravitational field strength.

This explains why dams are thicker at the bottom: the pressure from the water increases linearly With depth, so the base must withstand a much greater force than the top.

9.3 Upthrust and Floating

An object submerged in a fluid experiences an upward force called upthrust equal to the weight Of the fluid displaced. An object floats when its weight equals the upthrust. A steel ship floats Because its hull encloses a large volume of air, making its average density less than that of water.

Common Pitfalls

• Confusing mass and weight. Mass is in kg (scalar); weight is in N (vector). Mass does not change; weight depends on $g$.
• Drawing force arrows on the wrong object in free body diagrams. Show forces on ONE object only.
• Confusing Newton’s third law pairs with balanced forces. Third law pairs act on DIFFERENT objects; balanced forces act on the SAME object.
• Forgetting that momentum is a vector. Assign a positive direction and include signs.
• Using the wrong distance in moment calculations. It must be the PERPENDICULAR distance from the pivot.
• Stating that terminal velocity is when drag equals mass. It is when drag equals WEIGHT.
• Including internal forces on a free body diagram. Only show external forces acting on the object.
• Writing $F = ma$ when the mass is changing (e.g., a burning rocket). Use $F = \frac{dp}{dt}$ instead.
• Assuming friction always opposes motion. Friction opposes relative motion (or the tendency towards it). A block on a conveyor belt may be accelerated by friction.
• Forgetting that the normal force is not always equal to $mg$. On an inclined plane or in a lift, the normal force is different.

Practice Questions

1. A 1500 kg car travelling at 30 m/s decelerates uniformly to rest in 5 seconds. Calculate the decelerating force.

2. Explain, with reference to Newton’s laws, why a passenger in a car is thrown forward when the car stops suddenly.

3. A 60 kg person stands on a bathroom scale in a lift. Calculate the reading on the scale when the lift accelerates upward at 2 m/s$^2$.

4. Two objects collide. Object A (mass 3 kg, velocity 4 m/s to the right) collides with object B (mass 2 kg, velocity 3 m/s to the left). After the collision, A moves at 1 m/s to the left. Find the velocity of B after the collision.

5. A uniform metre rule is pivoted at the 30 cm mark. A 4 N weight hangs from the 10 cm mark. What force must be applied at the 80 cm mark to balance the rule?

6. Explain why a skydiver reaches terminal velocity. Describe what happens when they open their parachute.

7. A spring has a spring constant of 500 N/m. How much energy is stored when it is extended by 0.04 m?

8. Calculate the thinking distance and braking distance for a car travelling at 30 m/s, given a reaction time of 0.6 s and a deceleration of 6 m/s$^2$.

9. A lever has an effort of 50 N applied 0.8 m from the pivot. What load can be lifted at 0.2 m from the pivot?

10. Explain how crumple zones reduce the force on passengers during a collision, using the concept of momentum.

11. A 5 kg block rests on a rough plane inclined at $30^{\circ}$ to the horizontal. The coefficient of friction is 0.3. Determine whether the block slides, and if so, calculate its acceleration.

12. A tennis ball of mass 0.06 kg is hit at 25 m/s. The racket is in contact with the ball for 0.005 s. Calculate the average force exerted on the ball.

13. Two identical springs each have spring constant 200 N/m. Calculate the effective spring constant when they are connected (a) in parallel and (b) in series.

14. A hydraulic press has a small piston of area $0.01$ m$^2$ and a large piston of area $0.5$ m$^2$. If a force of 200 N is applied to the small piston, what force is exerted by the large piston?

15. Explain why a car tyre needs to be inflated to the correct pressure. Include reference to the contact area between the tyre and the road.

10. Derivation: Stopping Distance from the Work-Energy Theorem

The relationship between braking distance and speed can be derived rigorously using the work-energy Theorem. When a vehicle brakes to a stop, the braking force does negative work equal to the change In kinetic energy:

$F d = \frac{1}{2}mv^2$

Solving for $d$:

$d = \frac{mv^2}{2F}$

This derivation reveals two key facts. First, for a given braking force $F$The braking distance is Proportional to $v^2$: doubling the speed quadruples the braking distance. Second, for a given Speed, the braking distance is inversely proportional to the braking force: better brakes (larger $F$) reduce the stopping distance, but the improvement is limited because the maximum braking force Is set by the friction between the tyres and the road.

11. Worked Example: Block on an Inclined Plane with Friction

A 3 \mathrm{ kg block is placed on a rough plane inclined at $35^{\circ}$ to the horizontal. The Coefficient of friction is $0.25$.

Part (a): Does the block slide?

Component of weight down the slope: mg\sin\theta = 3 \times 9.8 \times \sin 35^{\circ} = 16.9 \mathrm{ N.

Normal force: N = mg\cos\theta = 3 \times 9.8 \times \cos 35^{\circ} = 24.1 \mathrm{ N.

Maximum static friction: f_{\max} = \mu_s N = 0.25 \times 24.1 = 6.02 \mathrm{ N.

Since 16.9 \mathrm{ N \gt 6.02 \mathrm{ NThe block slides.

Part (b): Find the acceleration.

$a = g(\sin\theta - \mu\cos\theta) = 9.8(\sin 35^{\circ} - 0.25\cos 35^{\circ})$

= 9.8(0.5736 - 0.25 \times 0.8192) = 9.8(0.5736 - 0.2048) = 9.8 \times 0.3688 = 3.61 \mathrm{ m/s^2

12. Worked Example: Inelastic Collision with Kinetic Energy Loss

Two trolleys collide on a frictionless track. Trolley A has mass 2 \mathrm{ kg and velocity 4 \mathrm{ m/s to the right. Trolley B has mass 3 \mathrm{ kg and velocity 2 \mathrm{ m/s to the Left. They stick together.

Step 1: Choose a sign convention (right = positive).

v_A = +4 \mathrm{ m/s, \qquad v_B = -2 \mathrm{ m/s

Step 2: Apply conservation of momentum.

$m_A v_A + m_B v_B = (m_A + m_B) v_f$

$2(4) + 3(-2) = (2 + 3) v_f$

$8 - 6 = 5 v_f$

v_f = 0.4 \mathrm{ m/s to the right

Step 3: Calculate kinetic energy before and after.

E_{k,i} = \frac{1}{2}(2)(16) + \frac{1}{2}(3)(4) = 16 + 6 = 22 \mathrm{ J

E_{k,f} = \frac{1}{2}(5)(0.16) = 0.4 \mathrm{ J

Kinetic energy lost: 22 - 0.4 = 21.6 \mathrm{ J (transferred to thermal energy and sound).

This illustrates that perfectly inelastic collisions can lose a very large fraction of the initial Kinetic energy.

13. Why Momentum Is Conserved: A Rigorous Argument

Consider two objects colliding. During the collision, object A exerts a force \vec{F}_{A \mathrm{ on B} on object B, and by Newton’s third law, object B exerts an equal and Opposite force \vec{F}_{B \mathrm{ on A} = -\vec{F}_{A \mathrm{ on B} on object A.

By Newton’s second law:

\vec{F}_{A \mathrm{ on B} = \frac{d\vec{p}_B}{dt}, \qquad \vec{F}_{B \mathrm{ on A} = \frac{d\vec{p}_A}{dt}

Adding these:

\frac{d\vec{p}_A}{dt} + \frac{d\vec{p}_B}{dt} = \vec{F}_{B \mathrm{ on A} + \vec{F}_{A \mathrm{ on B} = 0

$\frac{d}{dt}(\vec{p}_A + \vec{p}_B) = 0$

\vec{p}_A + \vec{p}_B = \mathrm{constant

This proof extends to any number of particles. Internal forces always cancel in pairs, so the total Momentum of a closed system is conserved. This is not a separate law — it is a direct consequence Of Newton’s second and third laws.

14. Summary Table: Types of Force

Force	Direction	Formula	Notes
Weight	Vertically downward	$W = mg$	Always present near a gravitational field
Normal reaction	Perpendicular to surface	Adjusts to prevent penetration	Not always equal to $mg$
Friction	Opposes relative motion	$f \leq \mu N$	Static friction adjusts; kinetic is constant
Tension	Along the string, away from object	Equal throughout a massless string	Can only pull, never push
Air resistance (drag)	Opposes motion through fluid	$F_d \propto v^2$ at high speeds	Increases with speed
Upthrust	Upward (in a fluid)	Equals weight of fluid displaced	Explains why objects float

15. Worked Example: Hydraulic Press

A hydraulic press has a small piston of area 0.01 \mathrm{ m^2 and a large piston of area 0.5 \mathrm{ m^2. A force of 200 N is applied to the small piston.

By Pascal’s principle, the pressure is the same throughout the fluid:

P = \frac{F_1}{A_1} = \frac{200}{0.01} = 20000 \mathrm{ Pa

F_2 = P \times A_2 = 20000 \times 0.5 = 10000 \mathrm{ N

The force is multiplied by a factor of $A_2/A_1 = 50$. However, by conservation of energy, the small Piston must move 50 times further than the large piston. The work input equals the work output (assuming no losses).

16. Practice Questions (Additional)

16. A car of mass 1000 kg travelling at 15 \mathrm{ m/s collides with a stationary car of mass 1500 kg. The bumpers lock and they move off together. Calculate their common velocity and the kinetic energy lost in the collision.

17. A 2 \mathrm{ kg block is pushed up a $30^{\circ}$ incline with an initial speed of 8 \mathrm{ m/s. The coefficient of kinetic friction is $0.2$. Calculate how far up the incline the block travels before stopping and whether it slides back down.

18. Explain why a passenger in a car feels thrown forward when the car brakes suddenly. Refer to Newton’s laws in your answer.

19. A uniform metre rule is balanced on a pivot at the 40 cm mark. A 3 N weight hangs from the 10 cm mark and a 5 N weight hangs from the 70 cm mark. Calculate where an additional 2 N weight must be hung to restore balance.

20. A football is kicked from the ground at 20 \mathrm{ m/s at an angle of $35^{\circ}$ to the horizontal. Calculate the maximum height, the time of flight, and the horizontal range. (Assume g = 9.8 \mathrm{ m/s^2 and ignore air resistance.)

Practice Problems

Question 1: Resultant force and acceleration

A 5 \mathrm{ kg object is acted on by two forces: 20 \mathrm{ N to the right and 8 \mathrm{ N to the left. Calculate the resultant force and the acceleration.

Answer

Resultant force = 20 - 8 = 12 \mathrm{ N to the right.

Acceleration = F/m = 12/5 = 2.4 \mathrm{ m/s^2 to the right.

Question 2: Weight and mass on the Moon

An astronaut has a mass of 80 \mathrm{ kg on Earth. The gravitational field strength on the Moon is 1.6 \mathrm{ N/kg. Calculate (a) the astronaut’s weight on Earth and (b) the astronaut’s weight on the Moon.

Answer

(a) W = mg = 80 \times 9.8 = 784 \mathrm{ N.

(b) W = mg = 80 \times 1.6 = 128 \mathrm{ N.

Mass remains the same (80 \mathrm{ kg) on both Earth and the Moon.

Question 3: Stopping distance

A car travelling at 30 \mathrm{ m/s has a thinking distance of 21 \mathrm{ m and a braking distance of 54 \mathrm{ m. Calculate the total stopping distance. Explain why the braking distance increases much more rapidly than the thinking distance when speed doubles.

Answer

Total stopping distance = 21 + 54 = 75 \mathrm{ m.

Thinking distance is proportional to speed ($d_t = vt$), so doubling speed doubles thinking distance. Braking distance is proportional to speed squared ($d_b \propto v^2$Since $KE \propto v^2$ and $W = Fd$), so doubling speed quadruples the braking distance. This is why speeding is so dangerous.

Question 4: Hooke’s Law

A spring has a spring constant of 200 \mathrm{ N/m. Calculate the extension when a 50 \mathrm{ N force is applied. Calculate the energy stored in the spring.

Answer

$F = ke$So e = F/k = 50/200 = 0.25 \mathrm{ m.

Energy stored = \frac{1}{2}ke^2 = 0.5 \times 200 \times 0.25^2 = 0.5 \times 200 \times 0.0625 = 6.25 \mathrm{ J.

Question 5: Terminal velocity

Explain why a skydiver reaches terminal velocity. Describe the forces acting on the skydiver at terminal velocity.

Answer

Initially, the skydiver accelerates downward due to gravity ($F = mg$). As speed increases, air resistance (drag) increases. Drag acts upward, opposing the motion. Eventually, the upward drag force equals the downward gravitational force, so the net force is zero and the skydiver stops accelerating. They continue at a constant speed called terminal velocity.

At terminal velocity: weight $=$ drag. The skydiver is in equilibrium (Newton’s first law).

Worked Examples

Example 1: Newton’s second law

A $2.0\,\text{kg}$ object is pulled across a rough horizontal surface by a horizontal force of $15\,\text{N}$. The frictional force is $5.0\,\text{N}$. Calculate the acceleration.

Solution:

$F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 15 - 5.0 = 10\,\text{N}$

$a = \frac{F_{\text{net}}}{m} = \frac{10}{2.0} = 5.0\,\text{m\,s}^{-2}$

Summary

This topic covers the key concepts of Forces for GCSE Physics. Focus on understanding the fundamental principles, practising with exam-style questions, and applying your knowledge to unfamiliar contexts.

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