Forces

Question 1

Average Speed

Answer 1

Distance over time for the entire region of interest.

Question 2

Free-Fall

Answer 2

the only force acting on the object is the force of gravity.

Question 3

Projectile Motion

Answer 3

Projectile Motion:

• Motion of an object fired from a point with only gravity acting on it.
• Key Concepts:
  • Time of flight: How long the projectile is in the air.
  • Maximum height: Height where the projectile is momentarily at rest.
  • Range: Horizontal distance travelled.
• Horizontal Motion:
  
  • Constant velocity (no acceleration).
• Vertical Motion:
  
  • Constant acceleration due to gravity.

Types of Projection:

1. Vertical projection: Straight up or down.
2. Horizontal projection: Fired horizontally.
3. Projection at an angle: Fired at an angle to the horizontal.

Problem-Solving Tips:

• Split motion into horizontal and vertical components. (Suvat both)
• Analyse each component separately.

Effect of Air Resistance on Projectiles:

• Air resistance decreases the horizontal component of a projectile’s velocity.
• This reduces:
  
  • The range (horizontal distance travelled).
  • The maximum height reached.
• Compared to a scenario with no air resistance, both range and height are reduced.

Question 4

Instantaneous Speed

Answer 4

The exact speed of an object at a specific given point.
(Draw tangent)

Question 5

Reaction Time

Answer 5

The time taken to process a stimulus and trigger a response to it.
It is affected by alcohol, drugs and tiredness.

Question 6

Archimedes’ Principle

Answer 6

• The upwards force acting on an object submerged in a fluid, is equal to the weight of the fluid it displaces.
• (Upthrust = 𝜌_liquid 𝑉_object 𝑔)
• (Upthrust = (h₂ - h₁)𝜌gA)

Question 7

Centre of Mass

Answer 7

Centre of Gravity (Mass):

• The single point where an object’s mass can be considered to act.
• An object will topple if its centre of mass moves past its pivot (direction of moment changes).

Stability:

• Stability depends on the position of the centre of mass:
  
  • An object is stable if its centre of mass lies above its base.
  • An object topples if its centre of mass moves outside its base.
• Wider base and lower centre of mass increase stability.
• Narrower base and higher centre of mass make an object more likely to topple.

Centre of Mass Properties:

• Does not depend on the gravitational field.
• Can lie inside or outside the body.
• Can shift depending on the shape of the body.

Question 8

Couple

Answer 8

• A couple consists of a pair of equal and opposite coplanar forces that act to produce rotation only.
• A couple has the following characteristics:
  
  • Equal in magnitude
  • Opposite in direction
  • Perpendicular to the distance between them
  • Zero acceleration (resultant force)

Question 9

Drag

Answer 9

Drag Forces:
- Forces that oppose motion of an object moving through a fluid (gas or liquid).

• Characteristics:
  
  • Act in the opposite direction to motion.
  • Slow down objects or keep them moving at a constant speed.
  • Convert kinetic energy into heat and sound.

Factors of drag
- Cross-sectional area in contact with fluid
- Density of the fluid.
- Speed of the object (d∝v²)

(Factors for air resistance)
- Altitude.
- Temperature.
- Humidity.

Question 10

Equilibrium

Answer 10

Equilibrium:

• For an object to be in equilibrium:
  
  • The resultant force must be zero.
  • The resultant moment must be zero.
• An object in equilibrium will:
  
  • Remain at rest or move at a constant velocity.
  • Not rotate.

Coplanar Forces in Equilibrium:

• Coplanar forces can be represented by closed vector triangles.
• When vectors are joined, they form a closed path.
• In exam questions, diagrams are often drawn to scale – use a ruler for accuracy.

Question 11

Free-Body Diagram

Answer 11

Free Body Diagrams:
- Used to model forces acting on an object.
- Each force is represented as a vector arrow:
- Scaled to the magnitude of the force.
- Points in the direction the force acts.
- Labelled with the force’s name.

Uses of Free Body Diagrams:
- Identify which forces act in which plane.
- Resolve the net force in a specific direction.

Calculating Net Force:
- Use labelled angles and magnitudes.
- Resolve each force into horizontal and vertical components.

Question 12

Moment of Force

Answer 12

Moments:
- A moment is the turning effect of a force, causing objects to rotate about a pivot.

• Formula:
  
  • Moment (N m) = Force (N) × Perpendicular distance from the pivot (m).
• SI Unit: Newton metres (N m) to the pivot.**

Key Points:
- The pivot is the point about which an object rotates.
- Anything can act as a pivot (and create simultaneous equations)
- Perpendicular distance is crucial: only the component of force perpendicular to the pivot creates a moment.
- Drawing forces on a diagram helps identify which forces contribute to the turning effect.
- Choosing a pivot can simplify calculations by eliminating the reaction force at that point.

Question 13

Newton’s Second Law

Answer 13

• The sum of the forces acting on an object is equal to the rate of change of momentum of the object.
• This is also expressed as the net force acting on an object equaling the product of the object’s mass and acceleration.

Question 14

Principle of Moments

Answer 14

For an object to be in equilibrium, the sum of the clockwise moments acting about a point must be equal to the sum of the anticlockwise moments acting about the same point.
(It can be spinning)

Question 15

Terminal Velocity

Answer 15

Terminal Velocity:

• Occurs when the resistive force (drag) equals the driving force (weight).
• Initially: Only weight (W = mg) acts, causing acceleration.
• As velocity increases, drag force increases, reducing resultant force and acceleration.
• When drag force = weight,
  resultant force = 0,
  and acceleration stops
  – object reaches terminal velocity.

• Velocity-time graph:
  • Acceleration (gradient) decreases until it reaches zero at terminal velocity.
• Parachute deployment:
  
  • Causes deceleration to a lower terminal velocity, reducing landing impact.
• Misconception: Skydivers do not move upwards when parachutes deploy – they simply decelerate.

Question 16

Conservation of Energy

Answer 16

Principle of Conservation of Energy:

• Energy cannot be created or destroyed, only transferred between forms.
• The total energy in a closed system remains constant.

Energy Dissipation:

• Wasted energy is lost to the surroundings, often as heat, light, or sound.
• Energy not transferred to useful stores is considered wasted.

Energy Transfers and Stores:

• Examples:
  
  • Gravitational potential energy → kinetic energy (falling object).
  • Chemical energy → electrical and light energy (battery).
  • Elastic potential energy → kinetic energy (spring).
• Work done against resistive forces (e.g., friction) also dissipates energy.

Question 17

Elastic Deformation

Answer 17

• If a material deforms with elastic behaviour, it will return to its original shape when the deforming forces are removed.
• The object will not be permanently deformed.

Question 18

Elastic Potential Energy

Answer 18

• The energy stored in an object when it is stretched is equal to the work done to stretch the object.
• This energy can be determined from the area under a force-extension graph.

Question 19

Hooke’s Law

Answer 19

Hooke’s Law:

• The extension of an elastic object is directly proportional to the force applied, up to the limit of proportionality.

Key Points:

• Force constant (k): Measures stiffness; larger k = stiffer material.
• Applies to both extensions (increase in length) and compressions (decrease in length).
• (1/K _total for springs in series)
• Force-extension graph:
  
  • Gradient = force constant (k) if force is on the y-axis and extension on the x-axis.
  • If axes are swapped, gradient = 1/k.

Question 20

Plastic Deformation

Answer 20

• It will not return to its original shape when the deforming forces are removed.
• The object will be permanently deformed.

Question 21

Strain

Answer 21

The ratio of an object’s extension to its original length. It is a ratio of two
lengths and so has no unit.

Question 22

Types of deformation

Answer 22

Deformation:

• Forces can change the motion, size, or shape of a body.
• Tensile forces: Stretch a body (e.g., pulling a spring).
• Compressive forces: Compress a body (e.g., pushing a spring).

Example:

• A spring extends under tensile force and compresses under compressive force.

Question 23

Ultimate Tensile Strength

Answer 23

The maximum stress than an object can withstand before fracture occurs

Question 24

Young Modulus

Answer 24

• The ratio of stress to strain for a given material.
• Its unit is the Pascal (Pa).

Question 25

Conservation of Momentum

Answer 25

The total momentum of a system before an event must be equal to the total momentum of the system after the event, assuming no
external forces act.
(kgms-¹) Or (Ns)

Question 26

Types of Collisions

Answer 26

• An elastic collision is one in which the total kinetic energy of the system before the collision is equal to the total kinetic energy of the system after the collision.
• A collision is inelastic if kinetic energy is not conserved.
• Prove by working out KE before and after and compare values

Question 27

Impulse

Answer 27

• The change of momentum of an object when a force acts on it is called impulse.
• It is equal to the product of the force acting on the object and the length of time over which it acts.
  (Area under a force-time graph)

Question 28

Newton’s First Law

Answer 28

• Object will remain in its current state of motion unless acted on by a resultant force.
• For an object to accelerate, it requires a resultant force.

Newton

• The force that will give a mass of 1 kg an acceleration of 1 m s–2
• The SI unit for force is kg m s–2

Question 29

Newton’s Third Law

Answer 29

Every action has an equal and opposite reaction. If an object exerts a force on another object, then the other object must exert a force back, that is opposite in direction and equal in magnitude.

Question 30

Car safety

Answer 30

• In a car collision, airbags and seatbelts increase the time (𝑡) for the driver’s deceleration, reducing the force (𝐹) exerted on the body (𝐹 = (𝑚𝑣 - 𝑚𝑢) / 𝑡).
• Crumple zones absorb energy by deforming, further reducing the force on the driver.

Question 31

Pressure in a liquid

Answer 31

Pressure in a Fluid Column:

• An object in a fluid feels pressure from the weight of the fluid above it.
• Pressure at the base of the fluid is the same in all directions.

Equations:

1. Weight of the column:
   
   • W = m × g = ρ × A × h × g.
2. Pressure at the base:
   
   • P = W / A = ρ × h × g.
3. Hydrostatic pressure change:
   
   • ΔP = ρ × g × Δh.

Key Points:

• Pressure increases with depth because of the weight of the fluid above.
• Atmospheric pressure might need to be added in some cases.
• Volume divided by cross-sectional area equals height (h).

Question 32

Time graph

Answer 32

Displacement-Time Graph:

• Gradient (slope) = Velocity
• Y-intercept = Initial Displacement
• Straight Diagonal Line = Constant Velocity
• Curved Line = Acceleration
• Horizontal Line (Zero Slope) = Object at Rest

Velocity-Time Graph:

• Slope = Acceleration
• Y-intercept = Initial Velocity
• Straight Line = Uniform Acceleration
• Curved Line = Non-Uniform Acceleration
• Horizontal Line (Zero Slope) = Constant Velocity
• Area Under the Curve = Displacement or Distance Travelled

Acceleration-Time Graph:

• Y-intercept = Initial Acceleration
• Horizontal Line (Zero Slope) = Constant Acceleration
• Area Under the Curve = Change in Velocity

Question 33

Change in Momentum if it rebounds (hits a wall)

Answer 33

• Change in momentum is only due to horizontal velocities
• ∆p = m(v_f - v_i)

Question 34

Torque

Answer 34

• The size of a turning effect
• τ = Fd

Where:

τ= torque (N m)

F= one of the forces (N)

d= perpendicular distance between the forces (m)

Question 35

Mass on a slope

Answer 35

Forces on an Object on a Slope

• Weight (mg): Acts vertically downward.
• Reaction Force (Fr): Acts perpendicular to the slope, preventing the object from sinking into the surface.
• Parallel Component (mgsinθ): The force pulling the object down the slope.
• Perpendicular Component (mgcosθ): Contributes to the normal force and affects friction.

Object’s Motion

• Stationary/Constant Speed:
  
  • An equal and opposite force (e.g., friction) balances the force pulling it down the slope.
• Moving Down the Slope:
  
  • No Friction: Gravitational Potential Energy (GPE) converts entirely into Kinetic Energy (KE).
  • With Friction: Some GPE is converted into work against friction, reducing the KE at the bottom.

Question 36

Time graph of a bouncing ball

Answer 36

1. Upwards Motion:
   
   • Positive velocity decreases (decelerates) until reaching the highest point.
2. At Point A (Highest Point):
   
   • Maximum displacement.
   • Velocity = 0 (momentarily).
   • Velocity changes from positive to negative (direction change).
   • Acceleration (g) remains constant and downward.
3. At Point B (Lowest Point):
   
   • Minimum displacement (ball on the ground).
   • Velocity changes from negative to positive (instantaneous direction change).
   • Speed (magnitude) remains the same.
   • Momentary acceleration due to change in velocity.

Question 37

Stopping distance

Answer 37

Stopping Distance:
- Stopping distance = Thinking distance + Braking distance.
- Increases considerably with speed:
- Thinking distance increases proportionally with speed.
- Braking distance increases proportionally to the square of speed (u²).

Thinking Distance:
- Distance travelled before brakes are applied.
- Formula:
- Thinking distance = Initial speed (u) × Reaction time.
- Factors affecting thinking distance:
- Initial speed.
- Intoxication (alcohol/drugs).
- Distractions (e.g., mobile phones).
- Tiredness (slower reaction times).

Braking Distance:
- Distance travelled after brakes are applied.
- Formula:
- Work done by brakes = Braking force × Braking distance = ½ mv².
- Factors affecting braking distance:
- Initial speed.
- Mass of the vehicle.
- Poor road conditions (e.g., icy, wet).
- Car conditions (e.g., worn brakes).

Question 38

Tension, Upthrust, Friction, Normal Contact Force, Lift

Answer 38

Tension (T):
- The force experienced by a cable, rope, or string when:
- Pulled, hung, rotated, or supported.

Normal Contact Force (N or R):
- The force arising when an object rests against another object.
- Acts perpendicular (90°) to the plane of contact.
- Also called the reaction force (from Newton’s Third Law).

Upthrust:
- The upward buoyancy force acting on an object in a fluid (liquid or gas).
- Always acts upwards.

Friction (F or Fr):
- The force arising when two surfaces are in contact.
- Always opposes motion.
- Acts at the point of contact and in the opposite direction to motion.

Lift
- An upwards force on an object moving through a fluid.
- Acts perpendicular to fluid flow.
- Example: Aeroplane wings push air down, creating an equal and opposite reaction (lift) due to Newton’s Third Law.

Question 39

Work Done

Answer 39

Work Done:

• Energy transferred when a force moves an object over a distance.
• If the force acts in the direction of motion, the object gains energy.
• If the force acts opposite to the motion, the object loses energy.

The Joule (J):

• Unit of energy or work.
• SI unit: kg m² s⁻².
• Definition: Energy transferred when a 1 N force moves an object 1 m parallel to its motion.

Calculating Work Done:

• General formula: W = Fx cos θ
  
  • θ = angle between force and displacement.
• For vertical motion, use sin θ
• Always use the component of force parallel to the displacement.

Common Exam Mistake:

• Choosing the incorrect force (not parallel to motion).
• Resolve the force vector to find the parallel component.

Question 40

Derivation of Kinetic Energy

Answer 40

1. Mass at rest accelerates to speed v over distance d.
2. Work done: W = F × d.
3. Force: F = ma (Newton’s Second Law).
4. SUVAT equation: v² = u² + 2as.
   
   • Initial speed u = 0, distance s = d.
   • Equation simplifies to: v² = 2ad.
5. Rearrange for acceleration: a = v² / 2d.
6. Substitute a into F = ma: F = mv² / 2d.
7. Substitute F into work done: W = (mv² / 2d) × d = ½mv².
8. Kinetic energy due to speed: KE = ½mv².

Question 41

Derivation of P = F × v

Answer 41

1. Power is the rate of change of work:
   P = W / t.
2. Work done: W = F × d.
3. At constant velocity, distance d = v × t.
   
   • Therefore, W = F × v × t.
4. Substitute W into the power equation:
   P = (F × v × t) / t.
5. Cancel t:
   P = F × v.

Question 42

Force-Extension Graph

Answer 42

Force-Extension Graph:

• Hooke’s Law is obeyed up to the limit of proportionality, shown by a straight line through the origin.
• Beyond this point, the graph curves, and Hooke’s Law no longer applies.

Key Features:

1. Limit of Proportionality:
   
   • Point where extension is no longer proportional to the applied force.
   • Identified where the graph starts to curve (flatten out).
2. Force Constant (k):
   
   • Gradient of the straight part of the graph (linear region).
   • Represents stiffness of the material.
3. Elastic Limit:
   
   • Maximum stretch where the material can still return to its original length.
   • Always after the limit of proportionality.

Exam Tip:
- k is the gradient only in the linear region and must pass through zero where Hooke’s Law is obeyed.

Question 43

Type of materials

Answer 43

Metal Wire:

• Obeys Hooke’s Law and exhibits elastic deformation until its elastic limit.
• Loading and unloading curves are identical up to the elastic limit.
• Beyond this limit, it experiences plastic deformation, leading to permanent extension.
• Unloading curve has the same gradient as the loading curve.
• Brittle objects show very little strain before breaking and has a high YM
  
  • Large Plastic Region → Can be drawn into a wire (ductile)

Rubber:

• Does not obey Hooke’s Law and does not experience plastic deformation.
• Hysteresis Loop: The area between the loading and unloading curves.
  
  • Represents work done in stretching the material.
  • Energy is transferred to thermal energy when the force is removed.
  • Unloading curve is always below the stretching curve.
  • Area between loading and unloading = Net work done (thermal energy dissipated).
  • Area under loading to the x axis = Minimum energy required to stretch the material to extension e.

Polymeric Materials (Polyethene):

• Does not obey Hooke’s Law and undergoes plastic deformation immediately.
• Easily stretched into new shapes, but does not return to its original form.
• Used in applications where permanent deformation is desirable, e.g., plastic bags.