Cambridge AS & A Level9702

Dynamics

Physics 9702 Chapter Notes

What this chapter covers

Dynamics - Momentum and Newton’s laws of motionDynamics - Non-uniform motionDynamics - Linear momentum and its conservation
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In this chapter

  1. 1Newton's First Law and Inertia
  2. 2Newton's Second Law: Force, Mass, and Acceleration
  3. 3Weight, Mass, and Gravity
  4. 4Free-Body Diagrams and Common Forces
  5. 5Newton's Third Law of Motion
  6. 6SI Units and Equation Homogeneity

1. Newton's First Law and Inertia

Newton's First Law of Motion describes what happens to an object when the forces acting on it are balanced. It states that an object will remain at rest, or continue to move at a constant velocity, unless acted upon by a resultant force. This tendency of an object to resist changes in its state of motion is called inertia. The more mass an object has, the more inertia it has, meaning it is harder to start it moving, stop it, or change its direction. For an object moving at a constant velocity (which includes being stationary, where velocity is zero), the resultant force is zero. This doesn't mean no forces are acting on it; it means all the forces are balanced and cancel each other out.

Key term

Inertia: The property of an object that resists any change in its state of rest or uniform motion in a straight line.

Examiner insight

Examiners often test the First Law by asking for an explanation of motion at a constant velocity, where students must state that the resultant force is zero.

Common pitfall

Stating that 'no forces' are acting on a stationary object, instead of correctly stating that the 'resultant force is zero' or that the forces are balanced.

Worked example 13 marks

A car is travelling along a straight, level road at a constant speed of 25 m/s. The engine provides a forward thrust of 1200 N.(a) What is the resultant force on the car?(b) What is the magnitude of the total resistive force (air resistance and friction) acting on the car?

  1. 1

    Step 1: Identify the state of motion. The car is moving at a constant speed (and constant velocity as the road is straight).

  2. 2

    Step 2: Apply Newton's First Law. Since the velocity is constant, the acceleration is 0 m/s². Therefore, the resultant force on the car must be zero.

  3. 3

    Answer (a): The resultant force on the car is 0 N.

  4. 4

    Step 3: Consider the forces acting horizontally. There is a forward thrust of 1200 N. For the resultant force to be zero, the backward resistive forces must be equal in magnitude and opposite in direction to the forward thrust.

  5. 5

    Step 4: Equate the forces. Forward Thrust = Total Resistive Force. Therefore, the total resistive force is 1200 N.

  6. 6

    Answer (b): The magnitude of the total resistive force is 1200 N.

Recap

  • An object at rest stays at rest if the resultant force is zero.
  • An object moving at a constant velocity continues at that velocity if the resultant force is zero.
  • Inertia is the resistance of an object to a change in its motion.
  • Mass is the measure of an object's inertia.
  • Constant velocity implies zero acceleration and therefore zero resultant force.

Quick check

  1. State Newton's First Law of Motion.1 mark
  2. A hockey puck slides on frictionless ice at a constant velocity. What is the resultant force on the puck?1 mark

2. Newton's Second Law: Force, Mass, and Acceleration

Newton's Second Law explains what happens when there is a resultant (or unbalanced) force acting on an object. It states that the acceleration of an object is directly proportional to the resultant force acting on it, and inversely proportional to its mass. The direction of the acceleration is always the same as the direction of the resultant force. This relationship is summed up in the most important equation in dynamics: F = ma. Here, 'F' is the resultant force in newtons (N), 'm' is the mass in kilograms (kg), and 'a' is the acceleration in metres per second squared (m/s²). To use this equation correctly, you must first find the resultant force by adding up all the force vectors acting on the object.

F = ma

Key term

Resultant Force: The single force that has the same effect as all the individual forces acting on an object combined.

Examiner insight

Candidates who explicitly calculate the resultant force as a separate step before applying F=ma are less likely to make mistakes.

Common pitfall

Using only one force (e.g., the engine thrust) in the F=ma equation, while ignoring other forces like friction or weight.

Fun fact

The 'g-force' experienced by fighter pilots and astronauts is a measure of acceleration. 1g is the acceleration due to gravity on Earth (9.81 m/s²). A pilot pulling 9g is accelerating at 9 times that rate!

Worked example 13 marks

A box of mass 5.0 kg is pushed along a rough horizontal floor by a force of 30 N. The frictional force opposing the motion is 10 N. Calculate the acceleration of the box.

  1. 1

    Step 1: Identify all horizontal forces acting on the box. There is a pushing force of 30 N and a frictional force of 10 N in the opposite direction.

  2. 2

    Step 2: Calculate the resultant force (F). The forces are in opposite directions, so we subtract them. F = Pushing Force - Frictional Force.

  3. 3

    F = 30 N - 10 N = 20 N. The direction of the resultant force is the direction of the push.

  4. 4

    Step 3: State Newton's Second Law: F = ma.

  5. 5

    Step 4: Rearrange the formula to make acceleration 'a' the subject: a = F/m.

  6. 6

    Step 5: Substitute the known values into the equation. a = 20 N / 5.0 kg.

  7. 7

    Step 6: Calculate the result. a = 4.0 m/s².

  8. 8

    Answer: The acceleration of the box is 4.0 m/s².

Worked example 24 marks

A rocket of mass 8.0 x 10⁴ kg has an engine that provides an upward thrust of 1.2 x 10⁶ N. Calculate its initial upward acceleration on take-off. (Use g = 9.81 m/s²).

  1. 1

    Step 1: Identify the vertical forces. There is an upward thrust (T) and a downward weight (W).

  2. 2

    Step 2: Calculate the weight of the rocket. W = mg = (8.0 x 10⁴ kg) × (9.81 m/s²) = 7.848 x 10⁵ N.

  3. 3

    Step 3: Calculate the resultant upward force (F). F = Thrust - Weight.

  4. 4

    F = (1.2 x 10⁶ N) - (7.848 x 10⁵ N) = 4.152 x 10⁵ N.

  5. 5

    Step 4: Apply Newton's Second Law, F = ma, to find the acceleration.

  6. 6

    a = F / m = (4.152 x 10⁵ N) / (8.0 x 10⁴ kg) = 5.19 m/s².

  7. 7

    Answer: The initial upward acceleration is 5.2 m/s² (to 2 significant figures).

Recap

  • A resultant force causes an object to accelerate.
  • Acceleration is directly proportional to resultant force (a ∝ F).
  • Acceleration is inversely proportional to mass (a ∝ 1/m).
  • The formula F = ma links resultant force, mass, and acceleration.
  • Resultant force and acceleration are always in the same direction.

Quick check

  1. A resultant force of 50 N acts on a 10 kg object. What is its acceleration?1 mark
  2. If the resultant force on an object is doubled, what happens to its acceleration?1 mark

3. Weight, Mass, and Gravity

It is crucial to distinguish between mass and weight. Mass is a measure of the amount of 'stuff' or matter in an object and is a scalar quantity measured in kilograms (kg). An object's mass is the same everywhere in the universe. Weight, on the other hand, is the force of gravity acting on an object's mass. It is a vector quantity, measured in newtons (N). An object's weight depends on its location. The relationship between weight and mass is given by the formula W = mg, which is a direct application of F=ma. Here, 'W' is the weight, 'm' is the mass, and 'g' is the acceleration of free fall (also known as the gravitational field strength). On Earth, g is approximately 9.81 m/s² or 9.81 N/kg.

W = mg

Key term

Weight: The gravitational force exerted on an object due to its mass.

Examiner insight

Clear distinction between the concepts of mass (a scalar property of a body) and weight (a vector force) is essential for high marks.

Common pitfall

Confusing mass and weight. They are different quantities with different units and should not be used interchangeably.

Worked example 14 marks

An astronaut has a mass of 75 kg.(a) Calculate her weight on Earth, where g = 9.81 m/s².(b) The astronaut travels to the Moon, where the gravitational field strength is 1.6 N/kg. What is her mass and weight on the Moon?

  1. 1

    Step 1 (a): State the formula for weight: W = mg.

  2. 2

    Step 2 (a): Substitute the values for mass and g on Earth. W_Earth = 75 kg × 9.81 m/s².

  3. 3

    Step 3 (a): Calculate the weight on Earth. W_Earth = 735.75 N. Rounding to 3 s.f. gives 736 N.

  4. 4

    Answer (a): The astronaut's weight on Earth is 736 N.

  5. 5

    Step 1 (b): State the astronaut's mass on the Moon. Mass is an intrinsic property and does not change with location. Mass on Moon = 75 kg.

  6. 6

    Step 2 (b): Calculate the weight on the Moon using W = mg, with the Moon's value for g.

  7. 7

    Step 3 (b): W_Moon = 75 kg × 1.6 N/kg = 120 N.

  8. 8

    Answer (b): Her mass on the Moon is 75 kg and her weight is 120 N.

Recap

  • Mass is the amount of matter in an object, measured in kg.
  • Weight is the force of gravity on an object, measured in N.
  • Mass is constant everywhere, but weight depends on the local gravitational field strength (g).
  • The formula linking weight and mass is W = mg.
  • On Earth, g is approximately 9.81 m/s² (or 9.81 N/kg).

Quick check

  1. What is the weight of a 2.0 kg bag of flour on Earth? (Use g = 9.81 N/kg).1 mark
  2. An object has a weight of 49 N on Earth. What is its mass? (Use g = 9.8 N/kg).1 mark

4. Free-Body Diagrams and Common Forces

To solve dynamics problems, you must first identify all the forces acting on the object of interest. A free-body diagram is an essential tool for this. It is a simplified diagram that represents the object as a dot or a box, with arrows drawn from the centre to represent all the forces acting *on* it. Each arrow should be labelled with the name or symbol of the force, and its length can be used to represent the relative magnitude of the force. Common forces you will encounter include:

  • Weight (W): The force of gravity, always acting vertically downwards.
  • Normal Reaction (R or N): A contact force from a surface, always acting perpendicular to the surface.
  • Tension (T): The pulling force transmitted through a string, rope, or cable.
  • Friction (f): A force that opposes motion or attempted motion between surfaces in contact.
  • Air Resistance (or Drag): A type of friction that opposes the motion of an object through a fluid (like air or water).

Key term

Free-Body Diagram: A diagram showing an object isolated from its surroundings, with all the forces that act on the object represented by arrows.

Examiner insight

A large, clear, and correctly labelled free-body diagram is often the key to solving a complex dynamics problem and can earn marks on its own.

Common pitfall

Including forces that the object exerts on other things in its free-body diagram. For example, in a diagram for a book on a table, do not include the force of the book on the table.

Worked example 13 marks

A wooden block of mass 2.0 kg is at rest on a rough slope inclined at 30° to the horizontal. Draw a labelled free-body diagram showing the three forces acting on the block.

  1. 1

    Step 1: Represent the block as a box or dot on a slope.

  2. 2

    Step 2: Draw the weight (W). This force always acts vertically downwards, from the centre of the block towards the centre of the Earth.

  3. 3

    Step 3: Draw the normal reaction force (R). This force is exerted by the slope on the block. It acts perpendicular (at 90°) to the surface of the slope, pointing away from it.

  4. 4

    Step 4: Draw the frictional force (f). Since the block is at rest and would slide down without friction, the frictional force must be acting up the slope, opposing the tendency to move.

  5. 5

    Step 5: Label all three forces clearly with their names or standard symbols (W, R, f). The diagram should show only the block and the three force arrows originating from it.

Recap

  • A free-body diagram shows all the forces acting on a single object.
  • Weight (W) always acts vertically downwards.
  • Normal Reaction (R) acts perpendicular to a surface.
  • Tension (T) is a pulling force in a string or rope.
  • Friction (f) and air resistance oppose motion.

Quick check

  1. A child is on a swing at the lowest point of its motion. Name the two forces acting on the child.2 marks
  2. In which direction does the normal reaction force always act?1 mark

5. Newton's Third Law of Motion

Newton's Third Law deals with interacting objects. It is often stated as: 'For every action, there is an equal and opposite reaction.' This means that if object A exerts a force on object B, then object B simultaneously exerts a force on object A that is equal in magnitude and opposite in direction. These two forces are known as an action-reaction pair. It is critical to remember four things about these pairs: they are equal in magnitude, opposite in direction, act on two different objects, and are of the same type (e.g., both are gravitational or both are contact forces). Because they act on different objects, they never cancel each other out.

Key term

Action-Reaction Pair: A pair of forces that are equal in magnitude, opposite in direction, of the same type, and act on two different interacting objects.

Examiner insight

To get full marks when stating Newton's third law pairs, you must explicitly mention both objects and the type of force (e.g., 'The gravitational force of the Earth on the Moon' and 'The gravitational force of the Moon on the Earth').

Common pitfall

Confusing a Newton's third law pair with a balanced pair of forces. The weight of a book and the normal reaction from the table are a balanced pair acting on one object (the book), not an action-reaction pair.

Fun fact

When you walk, you push the Earth backwards with your feet. Due to Newton's Third Law, the Earth pushes you forwards, causing you to move. Because the Earth's mass is so enormous, its resulting acceleration is immeasurably small.

Worked example 14 marks

A book with a weight of 5 N rests on a horizontal table.(a) Identify the force that is the 'reaction' part of the action-reaction pair with the book's weight.(b) Identify the action-reaction pair for the contact force the table exerts on the book.

  1. 1

    Step 1 (a): The 'action' force is the weight of the book. This is the gravitational force exerted *by the Earth on the book*.

  2. 2

    Step 2 (a): According to Newton's Third Law, the 'reaction' force must be exerted *by the book on the Earth*. It must be of the same type (gravitational), equal in magnitude (5 N), and opposite in direction (upwards).

  3. 3

    Answer (a): The reaction force is the gravitational force of 5 N exerted by the book on the Earth.

  4. 4

    Step 1 (b): The contact force the table exerts on the book is the normal reaction force, acting upwards on the book.

  5. 5

    Step 2 (b): The action-reaction pair to this force must be exerted *by the book on the table*. It is a contact force of equal magnitude, acting downwards on the table.

  6. 6

    Answer (b): The pair consists of the upward contact force from the table on the book, and the equal and opposite downward contact force from the book on the table.

Recap

  • If object A pushes object B, object B pushes object A with an equal and opposite force.
  • Action-reaction pairs are always equal in magnitude and opposite in direction.
  • Action-reaction pairs always act on two different objects.
  • Action-reaction pairs are always of the same type of force (e.g., both gravitational).
  • Because they act on different bodies, action-reaction forces never cancel out.

Quick check

  1. A cannon fires a cannonball. The force on the cannonball is F. What is the magnitude and direction of the force on the cannon?2 marks

6. SI Units and Equation Homogeneity

In physics, all quantities can be expressed in terms of a few fundamental or 'base' quantities. The internationally agreed system (SI) has seven base units, including the kilogram (kg) for mass, metre(m) for length, and second(s) for time. All other units are 'derived units', formed by combinations of base units. For example, the unit of force, the newton (N), is a derived unit. Using F=ma, we can see that 1 N = 1 kg × 1 m/s², so the base units for the newton are kg m s⁻². The principle of homogeneity states that for any physical equation to be valid, the base units on both sides of the equation must be identical. We can use this principle to check the correctness of formulas. Prefixes are used with units to denote multiples or submultiples of 10, such as kilo- (k, 10³), centi- (c, 10⁻²), and milli- (m, 10⁻³).

Key term

Homogeneity: The principle that for a physical equation to be valid, the base units or dimensions must be the same on both sides.

Examiner insight

Questions on homogeneity are a test of careful algebraic manipulation. Write out each step clearly to avoid losing track of powers and to secure method marks.

Common pitfall

Forgetting to square or cube units when they are part of a squared or cubed term in an equation, for example, using m/s instead of m²/s² for a v² term.

Worked example 13 marks

The formula for kinetic energy is E_k = ½mv². Show that this equation is homogeneous.

  1. 1

    Step 1: Identify the units of the left-hand side (LHS). Energy (E_k) is measured in joules (J). A joule is the work done when a force of 1 N moves an object 1 m. So, J = N m.

  2. 2

    Step 2: Express the LHS in SI base units. We know N = kg m s⁻². Therefore, J = (kg m s⁻²) × m = kg m² s⁻².

  3. 3

    Step 3: Identify the units of the right-hand side (RHS). The term is ½mv². The constant ½ has no units.

  4. 4

    Step 4: Express the RHS in SI base units. The units are those of mass × (velocity)². Units = kg × (m s⁻¹)² = kg × (m² s⁻²).

  5. 5

    Step 5: Compare the LHS and RHS. LHS units = kg m² s⁻². RHS units = kg m² s⁻². The base units are identical.

  6. 6

    Answer: Since the base units on both sides are the same (kg m² s⁻²), the equation is homogeneous.

Worked example 23 marks

A student suggests that the period T of a simple pendulum is given by T = 2π√(g/L), where L is its length and g is the acceleration of free fall. Use a homogeneity check to determine if this equation can be correct.

  1. 1

    Step 1: Analyse the LHS. The unit of period (T) is seconds (s).

  2. 2

    Step 2: Analyse the RHS. The term 2π is a dimensionless constant. We need to find the units of √(g/L).

  3. 3

    Step 3: Find the units of g/L. Unit of g is m s⁻². Unit of L is m. So, units of g/L are (m s⁻²) / m = s⁻².

  4. 4

    Step 4: Find the units of √(g/L). This is the square root of the units from Step 3. √(s⁻²) = s⁻¹.

  5. 5

    Step 5: Compare LHS and RHS. LHS unit is s. RHS unit is s⁻¹. These are not the same.

  6. 6

    Answer: The equation is not homogeneous because the units do not match (s ≠ s⁻¹). Therefore, the equation cannot be correct.

Recap

  • All physical quantities have units that can be broken down into SI base units (kg, m, s, etc.).
  • The newton (N) has base units of kg m s⁻².
  • An equation is homogeneous if the base units on both sides are identical.
  • Homogeneity analysis can be used to check the validity of an equation.
  • Remember to handle powers correctly when analysing units (e.g., for v², the units m/s become m²/s²).

Quick check

  1. Express the unit of pressure, the pascal (Pa = N/m²), in SI base units.2 marks

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A 1200 kg car accelerates uniformly from rest to a speed of 18 m/s in 10 s. Calculate (a) the acceleration of the car, and (b) the resultant force required to produce this acceleration.4 marks
  2. Explain the difference between the mass of an object and its weight. Your answer should refer to their definitions, units, and how they vary with location.3 marks
  3. A skydiver of mass 75 kg falls at a constant terminal velocity. What is the magnitude of the air resistance acting on her? Explain your reasoning. (Use g = 9.81 m/s²).3 marks
  4. State Newton's Third Law of Motion. A footballer kicks a ball. Identify the action-reaction force pair involved in the interaction between the foot and the ball.3 marks
  5. A lift and its occupants have a total mass of 800 kg. Calculate the tension in the supporting cable when the lift is (a) stationary, (b) accelerating upwards at 1.5 m/s², and (c) accelerating downwards at 1.5 m/s². (Use g = 9.81 m/s²).6 marks
  6. A boat of mass 450 kg is moving at a constant velocity. The engine provides a thrust of 800 N. A sudden gust of wind provides an extra forward force of 250 N. If the water resistance remains constant, what is the initial acceleration of the boat?4 marks
  7. The drag force F acting on a sphere moving through a fluid can be described by the equation F = 6πηrv, where r is the radius of the sphere, v is its speed, and η is the coefficient of viscosity of the fluid. Determine the SI base units of viscosity, η.3 marks
  8. A block of mass 5.0 kg is pulled up a smooth (frictionless) slope inclined at 20° to the horizontal by a rope parallel to the slope. If the block accelerates at 0.80 m/s², calculate the tension in the rope. (Use g = 9.81 m/s²).5 marks
  9. A helicopter of mass 1500 kg is hovering at a constant height. (a) Draw a free-body diagram showing the forces acting on the helicopter. (b) Calculate the magnitude of the upward lift force from the rotors. (c) The pilot increases the lift force to 18 000 N. Calculate the initial upward acceleration.5 marks
  10. An object is said to be in equilibrium. What does this imply about its motion and the forces acting on it? Give an example of an object in equilibrium.2 marks

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