Topic 1
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Topic 1 — Key Facts for DU Admission (Bangladesh) Core concept: Newton’s Laws of Motion — the bedrock of classical mechanics that explains how forces change the state of motion of material bodies High-yield point: F = ma is the most frequently tested formula; action-reaction pairs act on different bodies ⚡ Exam tip: Draw free body diagrams for every numerical problem; at least one question from Newton’s laws appears annually in DU admission tests
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Topic 1 — DU Admission (Bangladesh) Study Guide Overview: Newton’s laws of motion form the dynamical foundation of physics, connecting forces acting on bodies to their acceleration, momentum, and equilibrium states Core principles: Three laws of motion, conservation of linear momentum, impulse-momentum relationship Key points: Inertia and mass relationship, free body diagram construction, pulley systems, friction behavior, centripetal force as causative agent Study strategy: Master free body diagrams before attempting numerical problems; always identify all forces and their lines of action first
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Newton’s Laws of Motion — Complete Study Notes
Foundational Concepts in Dynamics
Classical mechanics splits naturally into two intellectual domains. Kinematics concerns itself with the mathematical description of motion — position, velocity, acceleration — without inquiry into what produces or changes that motion. Dynamics, by contrast, asks why motion changes, identifying forces as the causal agents. Newton’s three laws, published in Philosophiæ Naturalis Principia Mathematica in 1687, codify the relationship between force and motion and remain the cornerstone of engineering, astronomy, and physics education worldwide. For the DU admission examination, a thorough grasp of these laws and their applications determines success in a significant portion of the physics paper.
Newton’s First Law: The Law of Inertia
“Every body continues in its state of rest or of uniform motion in a straight line unless and until compelled by external forces to change that state.”
This law shatters the ancient intuition that a force is required to sustain motion. In the Aristotelian view, constant motion demanded a constant force; Newton showed that uniform motion requires no force at all — only a change in motion (acceleration) demands a force. The key concept is inertia: the innate resistance of any material body to changes in its state of motion. Mass serves as the quantitative measure of inertia. A body with greater mass possesses greater inertia and resists acceleration or deceleration more stubbornly.
The law implicitly defines the inertial reference frame — a coordinate system in which Newton’s first law holds true. Any reference frame moving at constant velocity relative to an inertial frame is itself inertial. Earth approximates an inertial frame closely enough for most laboratory problems, though it rotates and therefore constitutes a non-inertial frame for very precise measurements.
Practical manifestations of inertia: When a bus accelerates forward, passengers feel a backward lurch — their bodies tend to remain at rest while the bus moves beneath them. A coin placed on a card flicked horizontally falls into the cup below because the coin retains its initial zero horizontal velocity. Dust is dislodged from a carpet because the carpet moves (under the beating action) while dust particles resist that motion. Astronauts in orbit experience weightlessness because spacecraft and contents accelerate identically under gravity, with no contact forces to differentiate their states.
Newton’s Second Law: The Law of Acceleration
“The rate of change of momentum of a body is directly proportional to the net force applied and takes place in the direction of that force.”
The second law delivers the quantitative heart of classical dynamics. When multiple forces act simultaneously on a body, each contributes to the net force vector; what matters is the vector sum of all external forces, not any individual force in isolation.
The fundamental equation:
F = ma
Where F represents the net external force measured in Newtons (N), m is the inertial mass in kilograms (kg), and a is the resulting acceleration in meters per second squared (m/s²). One Newton is precisely the force that accelerates a one-kilogram mass at one meter per second squared: 1 N = 1 kg·m/s².
This formulation — F = ma — is actually a simplified special case of the more general momentum-based definition. Since momentum p = mv, the second law in its most fundamental form reads:
F = dp/dt = d(mv)/dt
When mass remains constant (as it does for most problems involving speeds far below relativistic velocities), this reduces to F = ma. For rocket motion, where fuel ejection causes mass to change during flight, the full form must be used.
Key exam points:
- F in F = ma refers to the net (resultant) force, not any single applied force
- Acceleration direction always matches the direction of the net force
- If ΣF = 0, then a = 0 and the body is in translational equilibrium
- For each spatial direction (x, y, z), write the separate equation ΣF = ma
Newton’s Third Law: The Law of Action and Reaction
“To every action there is always opposed an equal reaction: or, the mutual actions of any two bodies are always equal and directed to opposite parts.”
This law establishes that forces never appear in isolation. Whenever body A exerts a force on body B, body B simultaneously exerts an equal and opposite force on body A. Three properties characterize these action-reaction pairs: they are equal in magnitude, opposite in direction, and always act on different bodies. This last property is the most frequently misunderstood aspect.
Common errors to avoid: A frequent mistake is supposing that equal and opposite forces acting on the same body cancel, producing zero net force. They cannot act on the same body — by definition, action and reaction act on mutually different bodies. The weight of a book resting on a table (action: book pushes down on table) is balanced not by a reaction force on the book, but by the table pushing up on the book (reaction: table pushes up on book). These forces appear equal and opposite because they form an action-reaction pair, but they act on different bodies and therefore do not cancel.
Practical examples: Walking involves the foot pushing backward against the ground (action); the ground responds by pushing the foot forward (reaction), propelling the walker. A rocket ejects hot gas particles downward (action); the rocket is driven upward by the reaction force. A swimmer’s hands push water backward (action); the water pushes the swimmer forward (reaction).
Linear Momentum and Impulse
Momentum, the quantity of motion a body possesses, is defined as:
p = mv
Momentum is a vector quantity with direction matching velocity. The SI unit is kilogram-meter per second (kg·m/s). For a system of particles, total momentum equals the vector sum of individual momenta. The law of conservation of linear momentum states that in the absence of external forces, the total momentum of a closed system remains constant. This principle governs collision problems throughout the syllabus.
Impulse connects force and momentum change:
J = FΔt = Δp = mv_f − mv_i
The impulse-momentum theorem confirms that a force applied over a time interval changes momentum by exactly the impulse delivered. The area under a force-versus-time graph equals the impulse. For variable forces, J = ∫F dt.
Free Body Diagrams: The Problem-Solving Framework
Every mechanics problem in the DU admission exam should begin with a meticulously constructed free body diagram (FBD). This diagram isolates the body of interest and represents every external force acting upon it as a vector with correct direction and (qualitatively) correct magnitude.
Systematic procedure:
- Sketch the body as a simple dot or box, noting its mass
- Identify every object in the environment that contacts or interacts with the body — each constitutes a potential force source
- For each interaction, draw the appropriate force vector: gravitational weight mg pointing downward; normal force N perpendicular to and away from any contacting surface; tension T along any rope or cable pulling on the body; applied force F in the specified direction; friction f parallel to the surface opposing relative motion
- Label each force clearly with its symbol
- Establish a convenient coordinate system (typically aligned with inclined planes or principal force directions)
- Decompose any angled force into components along the chosen axes
- Apply ΣF_x = ma_x and ΣF_y = ma_y for each axis separately
Essential reminders:
- The normal reaction N is perpendicular to the contact surface — never vertical unless the surface is horizontal
- Weight mg always acts vertically downward, never perpendicular to an inclined plane
- Friction always opposes the relative motion or impending motion, not the applied force
- Tension pulls away from the body along the rope
- For action-reaction pairs: never draw both forces on the same FBD
Special Problem Categories
Objects on Horizontal Surfaces
Weight acts downward (mg), the normal force acts upward (N = mg on a flat horizontal surface in equilibrium), and an applied force F may act horizontally or at an angle. Kinetic friction magnitude: f_k = μ_k N, always opposing the direction of motion.
Inclined Plane Analysis
The weight vector mg must be resolved into two components relative to the inclined plane surface:
- Parallel to plane: F_∥ = mg sinθ — acts to pull the body down the slope
- Perpendicular to plane: F_⊥ = mg cosθ — presses the body against the surface
The normal force equals N = mg cosθ (for no other vertical forces). Acceleration down an incline with kinetic friction: a = g(sinθ − μ_k cosθ). Without friction, a = g sinθ.
Pulley Systems
For an ideal massless frictionless pulley, tension transmits unchanged through the rope: T₁ = T₂ = T. For a fixed pulley, only the direction of tension changes. For a movable pulley where the rope supports the load on both sides, mechanical advantage doubles the effective force: 2T = mg.
Atwood’s machine — two masses m₁ and m₂ connected by a string over a frictionless pulley:
- Acceleration: a = (|m₁ − m₂| × g) / (m₁ + m₂)
- Tension: T = (2 × m₁ × m₂ × g) / (m₁ + m₂)
The larger mass descends; acceleration direction follows whichever mass is heavier.
Connected Bodies
For two blocks A and B in contact on a frictionless surface pulled by a force F:
- System acceleration: a = F / (m_A + m_B)
- Contact force between blocks: F_AB = m_B × a (force that A exerts on B)
- By Newton’s third law, B exerts an equal opposite force on A
Friction Phenomena
Friction originates from microscopic asperities (tiny hills and valleys) on surfaces interlocking and resisting relative motion. Two categories matter:
Static friction (f_s): Prevents the initiation of relative motion. It adjusts itself exactly to match the applied force up to its maximum value:
- 0 ≤ f_s ≤ f_s(max) = μ_s N
- At the verge of motion, f_s = f_s(max)
Kinetic friction (f_k): Opposes ongoing motion. It remains approximately constant:
- f_k = μ_k N
- Kinetic coefficient is always less than static coefficient: μ_k < μ_s
Rolling friction is much smaller than sliding friction — the principle underlying wheel and bearing design.
Circular Motion and Newton’s Second Law
When a particle moves in a circular path with constant speed, its velocity vector continuously changes direction. This change requires a centripetal (center-seeking) force directed toward the circle’s center:
F_c = mv²/r = mrω²
This is not a new category of force — it is whatever physical force actually provides the radial pull: tension in a string for a whirling mass, gravitational pull for orbiting satellites, friction for a car rounding a curve, or normal force for a car on a banked curve. Without adequate centripetal force, the body travels in a straight line tangent to the circular path (Newton’s first law).
DU Admission Examination Patterns
DU question setters favor problems combining Newton’s second law with free body diagrams in these recurring configurations:
- Single block on horizontal or inclined surface — find acceleration and normal force
- Two masses connected via pulley — determine acceleration, tension, and system behavior
- Stacked blocks with applied force — find common acceleration and interfacial forces
- Conservation of momentum in collision — classify as elastic or inelastic
- Impulse-momentum applications — impact time questions and force reduction
Must-Remember Formulas
| Formula | Physical Meaning |
|---|---|
| F = ma | Net force equals mass times acceleration |
| W = mg | Weight (gravitational force on mass m) |
| p = mv | Linear momentum |
| J = FΔt | Impulse equals force times time interval |
| f_k = μ_k N | Kinetic friction magnitude |
| f_s(max) = μ_s N | Maximum static friction |
| a = g(sinθ − μcosθ) | Acceleration down incline with friction |
| a = (m₁−m₂)g/(m₁+m₂) | Atwood’s machine acceleration |
| F_c = mv²/r | Centripetal force requirement |
| v_f = (m₁v₁ + m₂v₂)/(m₁+m₂) | Perfectly inelastic collision final velocity |
Examination Strategy
- Read the problem completely before drawing any diagram — identify what is given and what is requested
- Draw the free body diagram first — this single step prevents more errors than any other
- Choose axes wisely — align one axis with the direction of acceleration for simpler equations
- Check units at every step — N = kg·m/s²; confirm all forces are in Newtons
- Identify action-reaction pairs correctly — never place both members of a pair on the same diagram
- Substitute numerical values last — keep expressions symbolic through algebraic manipulation to minimize arithmetic errors
- Ask whether the answer is physically reasonable — negative mass or acceleration in an unexpected direction signals error
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