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 = m g W = mg W = m g Where W W W m m m g g g
Location g g g 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 E a r t h = 80 × 9.8 = 784 N W_{\mathrm{Earth}} = 80 \times 9.8 = 784 \mathrm{ N} W Earth = 80 × 9.8 = 784 N W M o o n = 80 × 1.6 = 128 N W_{\mathrm{Moon}} = 80 \times 1.6 = 128 \mathrm{ N} W Moon = 80 × 1.6 = 128 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:
Suspend the lamina from a point and let it hang freely. Hang a plumb line from the suspension point and draw a vertical line. Repeat from a different suspension point. 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 d r a g ∝ v 2 F_{\mathrm{drag} \propto v^2} F drag ∝ 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:
Initially, the only force is weight (downward). Acceleration = g g g As speed increases, drag increases. The resultant force decreases: F n e t = W − D F_{\mathrm{net}} = W - D F net = W − D Eventually, drag equals weight: D = W D = W D = W 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 velocitySurface area: Larger surface area means more drag, so lower terminal velocityShape: Streamlined objects experience less dragFluid density: Denser fluids provide more dragExample: 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 g 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 = m g = 80 × 9.8 = 784 N D = mg = 80 \times 9.8 = 784 \mathrm{ N} D = m g = 80 × 9.8 = 784 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 S t o p p i n g d i s t a n c e = t h i n k i n g d i s t a n c e + b r a k i n g d i s t a n c e \mathrm{Stopping distance} = \mathrm{thinking distance} + \mathrm{braking distance} Stoppingdistance = thinkingdistance + brakingdistance 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 v^2 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.
F d = 1 2 m v 2 Fd = \frac{1}{2}mv^2 F d = 2 1 m v 2 For a given braking force F F F d = m v 2 2 F d = \frac{mv^2}{2F} d = 2 F m v 2 v 2 v^2 v 2
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.
T h i n k i n g d i s t a n c e = 20 × 0.5 = 10 m \mathrm{Thinking distance} = 20 \times 0.5 = 10 \mathrm{ m} Thinkingdistance = 20 × 0.5 = 10 m S t o p p i n g d i s t a n c e = 10 + 32 = 42 m \mathrm{Stopping distance} = 10 + 32 = 42 \mathrm{ m} Stoppingdistance = 10 + 32 = 42 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 = m v p = mv p = m v Where p p p m m m v v v
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 m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 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 × 3 + 1 × 0 = ( 2 + 1 ) × v 2 \times 3 + 1 \times 0 = (2 + 1) \times v 2 × 3 + 1 × 0 = ( 2 + 1 ) × v 6 = 3 v 6 = 3v 6 = 3 v v = 2 m / s v = 2 \mathrm{ m/s} v = 2 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 = Δ p Δ t = m v − m u t F = \frac{\Delta p}{\Delta t} = \frac{mv - mu}{t} F = Δ t Δ p = t m v − m u This shows that force equals the rate of change of momentum. This form of Newton’s second law is more general than F = m a F = ma F = ma
6.5 Impact Forces and Safety For a given change in momentum, increasing the impact time decreases the force:
F = Δ p Δ t F = \frac{\Delta p}{\Delta t} F = Δ t Δ p 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 = F d M = Fd M = F d Where M M M F F F d d d
Principle of moments: For an object in equilibrium, the sum of clockwise moments equals the sum Of anticlockwise moments.
∑ M c l o c k w i s e = ∑ M a n t i c l o c k w i s e \sum M_{\mathrm{clockwise}} = \sum M_{\mathrm{anticlockwise}} ∑ M clockwise = ∑ M 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 × 9.8 × 2 = 40 × 9.8 × d 30 \times 9.8 \times 2 = 40 \times 9.8 \times d 30 × 9.8 × 2 = 40 × 9.8 × d 60 = 40 d 60 = 40d 60 = 40 d
d = 1.5 m f r o m t h e p i v o t d = 1.5 \mathrm{ m from the pivot} d = 1.5 mfromthepivot 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 = τ ω P = \tau \omega P = τ ω
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 = k e F = ke F = k e Where k k k e e e
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 k 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 k k E = 1 2 k e 2 E = \frac{1}{2}ke^2 E = 2 1 k e 2 k t o t a l = 2 k k_{\mathrm{total}} = 2k k total = 2 k k t o t a l = k 2 k_{\mathrm{total}} = \frac{k}{2} k total = 2 k
8.4 Required Practical: Investigating Force and Extension Method:
Measure the natural length of the spring. Add masses one at a time, recording the total mass and the new length. Calculate the extension for each mass. Plot a graph of force (weight = m g mg m g 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 = F A P = \frac{F}{A} P = A F Where P P P F F F A A A 2 ^2 2
9.2 Pressure in Fluids The pressure in a fluid increases with depth:
P = h ρ g P = h\rho g P = h ρ g Where h h h ρ \rho ρ g g g
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 g 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 = m a F = ma F = ma (e.g., a burning rocket). Use F = d p d t F = \frac{dp}{dt} F = d t d p 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 m g mg m g On an inclined plane or in a lift, the normal force is different.Practice Questions A 1500 kg car travelling at 30 m/s decelerates uniformly to rest in 5 seconds. Calculate the decelerating force.
Explain, with reference to Newton’s laws, why a passenger in a car is thrown forward when the car stops suddenly.
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/s2 ^2 2
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.
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?
Explain why a skydiver reaches terminal velocity. Describe what happens when they open their parachute.
A spring has a spring constant of 500 N/m. How much energy is stored when it is extended by 0.04 m?
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/s2 ^2 2
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?
Explain how crumple zones reduce the force on passengers during a collision, using the concept of momentum.
A 5 kg block rests on a rough plane inclined at 30 ∘ 30^{\circ} 3 0 ∘
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.
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.
A hydraulic press has a small piston of area 0.01 0.01 0.01 2 ^2 2 0.5 0.5 0.5 2 ^2 2
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 = 1 2 m v 2 F d = \frac{1}{2}mv^2 F d = 2 1 m v 2 Solving for d d d
d = m v 2 2 F d = \frac{mv^2}{2F} d = 2 F m v 2 This derivation reveals two key facts. First, for a given braking force F F F v 2 v^2 v 2 F F F
11. Worked Example: Block on an Inclined Plane with Friction A 3 k g 3 \mathrm{ kg} 3 kg 35 ∘ 35^{\circ} 3 5 ∘ 0.25 0.25 0.25
Part (a): Does the block slide?
Component of weight down the slope: m g sin θ = 3 × 9.8 × sin 35 ∘ = 16.9 N mg\sin\theta = 3 \times 9.8 \times \sin 35^{\circ} = 16.9 \mathrm{ N} m g sin θ = 3 × 9.8 × sin 3 5 ∘ = 16.9 N
Normal force: N = m g cos θ = 3 × 9.8 × cos 35 ∘ = 24.1 N N = mg\cos\theta = 3 \times 9.8 \times \cos 35^{\circ} = 24.1 \mathrm{ N} N = m g cos θ = 3 × 9.8 × cos 3 5 ∘ = 24.1 N
Maximum static friction: f max = μ s N = 0.25 × 24.1 = 6.02 N f_{\max} = \mu_s N = 0.25 \times 24.1 = 6.02 \mathrm{ N} f m a x = μ s N = 0.25 × 24.1 = 6.02 N
Since 16.9 N > 6.02 N 16.9 \mathrm{ N \gt 6.02 \mathrm{ N}} 16.9 N > 6.02N
Part (b): Find the acceleration.
a = g ( sin θ − μ cos θ ) = 9.8 ( sin 35 ∘ − 0.25 cos 35 ∘ ) a = g(\sin\theta - \mu\cos\theta) = 9.8(\sin 35^{\circ} - 0.25\cos 35^{\circ}) a = g ( sin θ − μ cos θ ) = 9.8 ( sin 3 5 ∘ − 0.25 cos 3 5 ∘ ) = 9.8 ( 0.5736 − 0.25 × 0.8192 ) = 9.8 ( 0.5736 − 0.2048 ) = 9.8 × 0.3688 = 3.61 m / s 2 = 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} = 9.8 ( 0.5736 − 0.25 × 0.8192 ) = 9.8 ( 0.5736 − 0.2048 ) = 9.8 × 0.3688 = 3.61 m/ s 2 12. Worked Example: Inelastic Collision with Kinetic Energy Loss Two trolleys collide on a frictionless track. Trolley A has mass 2 k g 2 \mathrm{ kg} 2 kg 4 m / s 4 \mathrm{ m/s} 4 m/s 3 k g 3 \mathrm{ kg} 3 kg 2 m / s 2 \mathrm{ m/s} 2 m/s
Step 1: Choose a sign convention (right = positive).
v A = + 4 m / s , v B = − 2 m / s v_A = +4 \mathrm{ m/s, \qquad v_B} = -2 \mathrm{ m/s} v A = + 4 m/s , v B = − 2 m/s Step 2: Apply conservation of momentum.
m A v A + m B v B = ( m A + m B ) v f m_A v_A + m_B v_B = (m_A + m_B) v_f m A v A + m B v B = ( m A + m B ) v f 2 ( 4 ) + 3 ( − 2 ) = ( 2 + 3 ) v f 2(4) + 3(-2) = (2 + 3) v_f 2 ( 4 ) + 3 ( − 2 ) = ( 2 + 3 ) v f 8 − 6 = 5 v f 8 - 6 = 5 v_f 8 − 6 = 5 v f v f = 0.4 m / s t o t h e r i g h t v_f = 0.4 \mathrm{ m/s to the right} v f = 0.4 m/stotheright Step 3: Calculate kinetic energy before and after.
E k , i = 1 2 ( 2 ) ( 16 ) + 1 2 ( 3 ) ( 4 ) = 16 + 6 = 22 J E_{k,i} = \frac{1}{2}(2)(16) + \frac{1}{2}(3)(4) = 16 + 6 = 22 \mathrm{ J} E k , i = 2 1 ( 2 ) ( 16 ) + 2 1 ( 3 ) ( 4 ) = 16 + 6 = 22 J E k , f = 1 2 ( 5 ) ( 0.16 ) = 0.4 J E_{k,f} = \frac{1}{2}(5)(0.16) = 0.4 \mathrm{ J} E k , f = 2 1 ( 5 ) ( 0.16 ) = 0.4 J Kinetic energy lost: 22 − 0.4 = 21.6 J 22 - 0.4 = 21.6 \mathrm{ J} 22 − 0.4 = 21.6 J
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 F ⃗ A o n B \vec{F}_{A \mathrm{ on B}} F A onB F ⃗ B o n A = − F ⃗ A o n B \vec{F}_{B \mathrm{ on A}} = -\vec{F}_{A \mathrm{ on B}} F B onA = − F A onB
By Newton’s second law:
F ⃗ A o n B = d p ⃗ B d t , F ⃗ B o n A = d p ⃗ A d t \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} F A onB = d t d p B , F B onA = d t d p A Adding these:
d p ⃗ A d t + d p ⃗ B d t = F ⃗ B o n A + F ⃗ A o n B = 0 \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 d t d p A + d t d p B = F B onA + F A onB = 0 d d t ( p ⃗ A + p ⃗ B ) = 0 \frac{d}{dt}(\vec{p}_A + \vec{p}_B) = 0 d t d ( p A + p B ) = 0 p ⃗ A + p ⃗ B = c o n s t a n t \vec{p}_A + \vec{p}_B = \mathrm{constant} p A + p B = 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 = m g W = mg W = m g Always present near a gravitational field Normal reaction Perpendicular to surface Adjusts to prevent penetration Not always equal to m g mg m g Friction Opposes relative motion f ≤ μ N f \leq \mu N f ≤ μ 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 ∝ v 2 F_d \propto v^2 F d ∝ v 2 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 m 2 0.01 \mathrm{ m^2} 0.01 m 2 0.5 m 2 0.5 \mathrm{ m^2} 0.5 m 2
By Pascal’s principle, the pressure is the same throughout the fluid:
P = F 1 A 1 = 200 0.01 = 20000 P a P = \frac{F_1}{A_1} = \frac{200}{0.01} = 20000 \mathrm{ Pa} P = A 1 F 1 = 0.01 200 = 20000 Pa F 2 = P × A 2 = 20000 × 0.5 = 10000 N F_2 = P \times A_2 = 20000 \times 0.5 = 10000 \mathrm{ N} F 2 = P × A 2 = 20000 × 0.5 = 10000 N The force is multiplied by a factor of A 2 / A 1 = 50 A_2/A_1 = 50 A 2 / A 1 = 50
16. Practice Questions (Additional) A car of mass 1000 kg travelling at 15 m / s 15 \mathrm{ m/s} 15 m/s
A 2 k g 2 \mathrm{ kg} 2 kg 30 ∘ 30^{\circ} 3 0 ∘ 8 m / s 8 \mathrm{ m/s} 8 m/s 0.2 0.2 0.2
Explain why a passenger in a car feels thrown forward when the car brakes suddenly. Refer to Newton’s laws in your answer.
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.
A football is kicked from the ground at 20 m / s 20 \mathrm{ m/s} 20 m/s 35 ∘ 35^{\circ} 3 5 ∘ g = 9.8 m / s 2 g = 9.8 \mathrm{ m/s^2} g = 9.8 m/ s 2
Practice Problems Question 1: Resultant force and acceleration A 5 k g 5 \mathrm{ kg} 5 kg 20 N 20 \mathrm{ N} 20 N 8 N 8 \mathrm{ N} 8 N
Answer Resultant force = 20 − 8 = 12 N = 20 - 8 = 12 \mathrm{ N} = 20 − 8 = 12 N
Acceleration = F / m = 12 / 5 = 2.4 m / s 2 = F/m = 12/5 = 2.4 \mathrm{ m/s^2} = F / m = 12/5 = 2.4 m/ s 2
Question 2: Weight and mass on the Moon An astronaut has a mass of 80 k g 80 \mathrm{ kg} 80 kg 1.6 N / k g 1.6 \mathrm{ N/kg} 1.6 N/kg
Answer (a) W = m g = 80 × 9.8 = 784 N W = mg = 80 \times 9.8 = 784 \mathrm{ N} W = m g = 80 × 9.8 = 784 N
(b) W = m g = 80 × 1.6 = 128 N W = mg = 80 \times 1.6 = 128 \mathrm{ N} W = m g = 80 × 1.6 = 128 N
Mass remains the same (80 k g 80 \mathrm{ kg} 80 kg
Question 3: Stopping distance A car travelling at 30 m / s 30 \mathrm{ m/s} 30 m/s 21 m 21 \mathrm{ m} 21 m 54 m 54 \mathrm{ m} 54 m
Answer Total stopping distance = 21 + 54 = 75 m = 21 + 54 = 75 \mathrm{ m} = 21 + 54 = 75 m
Thinking distance is proportional to speed (d t = v t d_t = vt d t = v t d b ∝ v 2 d_b \propto v^2 d b ∝ v 2 K E ∝ v 2 KE \propto v^2 K E ∝ v 2 W = F d W = Fd W = F d
Question 4: Hooke’s Law A spring has a spring constant of 200 N / m 200 \mathrm{ N/m} 200 N/m 50 N 50 \mathrm{ N} 50 N
Answer F = k e F = ke F = k e e = F / k = 50 / 200 = 0.25 m e = F/k = 50/200 = 0.25 \mathrm{ m} e = F / k = 50/200 = 0.25 m
Energy stored = 1 2 k e 2 = 0.5 × 200 × 0.25 2 = 0.5 × 200 × 0.0625 = 6.25 J = \frac{1}{2}ke^2 = 0.5 \times 200 \times 0.25^2 = 0.5 \times 200 \times 0.0625 = 6.25 \mathrm{ J} = 2 1 k e 2 = 0.5 × 200 × 0.2 5 2 = 0.5 × 200 × 0.0625 = 6.25 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 = m g F = mg F = m g
At terminal velocity: weight = = =
Worked Examples Example 1: Newton’s second law
A 2.0 kg 2.0\,\text{kg} 2.0 kg 15 N 15\,\text{N} 15 N 5.0 N 5.0\,\text{N} 5.0 N
Solution:
F net = F applied − F friction = 15 − 5.0 = 10 N F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 15 - 5.0 = 10\,\text{N} F net = F applied − F friction = 15 − 5.0 = 10 N a = F net m = 10 2.0 = 5.0 m s − 2 a = \frac{F_{\text{net}}}{m} = \frac{10}{2.0} = 5.0\,\text{m\,s}^{-2} a = m F net = 2.0 10 = 5.0 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.
Cross-References Energy — Work done by forces transfers energy; kinetic energy and potential energy connect forces to energy conservation.Electricity — Electric motors use forces to convert electrical energy into mechanical motion.Magnetism — Magnetic forces on current-carrying conductors are a key application of force concepts.Algebra — Force calculations require solving equations and rearranging formulas.