Answer: Inertia is the property of matter by which a body resists any change to its state of motion — whether at rest or moving with uniform velocity. Mass is the quantitative measure of inertia: a larger mass implies greater inertia and therefore greater resistance to change. For example, it is harder to start pushing a heavy car (large mass) than a bicycle (small mass); likewise, when a bus stops suddenly passengers lurch forward — their bodies tend to keep moving due to inertia. Another example: a tablecloth can be quickly pulled from under dishes without disturbing them significantly because the dishes’ inertia resists the rapid horizontal change.

Answer: Newton's first law (law of inertia) states that a body remains at rest or moves with uniform velocity in a straight line unless acted upon by a net external force. This law implicitly defines inertial reference frames: frames in which Newton’s first law holds are called inertial frames. In an inertial frame, free bodies (no net force) move uniformly. Non-inertial frames (accelerating frames) show apparent forces (e.g., centrifugal force) and Newton’s first law does not hold without introducing fictitious forces. Thus, the first law helps distinguish frames where Newtonian mechanics applies directly.

Answer: Mass is the intrinsic amount of matter or the measure of inertia of a body; its SI unit is kilogram (kg) and it does not depend on location. Weight is the gravitational force acting on the mass: W = m·g, measured in newtons (N). Weight depends on the local acceleration due to gravity g, which varies slightly with altitude and planetary body. For example, a person of mass 60 kg has the same mass on Earth and Moon, but the weight on the Moon is about one-sixth of that on Earth because lunar gravity is smaller.

Answer: Momentum p of a body is defined as the product of its mass and velocity: p = m·v. It is a vector quantity pointing in the direction of velocity. Impulse J is the product of force and the time over which it acts: J = F·Δt. The impulse–momentum theorem states that impulse equals the change in momentum: J = Δp. Practically, this explains why longer impact time reduces force for the same change in momentum — helmets and airbags increase impact time to reduce injury, and crumple zones in cars absorb impact increasing Δt so peak forces are smaller.

Answer: Static equilibrium refers to the state where an object is at rest and remains so because the net force and net torque acting on it are zero. Newton’s first and second laws imply that if the vector sum of all forces acting on an object is zero, acceleration is zero. To test equilibrium, draw a free-body diagram, resolve all forces into components, and verify ΣF_x = 0 and ΣF_y = 0 (and Στ = 0 if rotation is considered). If these conditions hold, the object is in equilibrium.

Answer: Newton's first law states that bodies maintain their state of motion unless a net external force acts. When a vehicle stops suddenly, passengers tend to continue moving forward with their previous velocity due to inertia. Seat belts supply the external force needed to decelerate the passenger safely and bring them to rest relative to the vehicle over a controlled time interval, reducing the risk of injury. Without seat belts, passengers may hit the windshield or be ejected from the vehicle.

Answer: Newton's second law states that the net force acting on a body is equal to the rate of change of its linear momentum: F_net = dp/dt. For constant mass m, p = m·v so dp/dt = m·dv/dt = m·a. Therefore F_net = m·a. This is a vector equation: the direction of acceleration is the direction of the net force. If multiple forces act, vector sum gives the net force. The law applies in inertial frames and provides quantitative link between forces and motion.

Answer: Take right as positive. Net force = (10 + 5) − 15 = 0 N. Since F_net = 0, acceleration a = F_net/m = 0/5 = 0 m/s². The body moves with constant velocity (or remains at rest). This demonstrates that equal and opposite resultant forces cancel resulting in zero acceleration.

Answer: From F = m·a, for a fixed net force F, acceleration a = F/m. Thus heavier objects (larger m) experience smaller acceleration for the same force. Example: a 10 N force on a 2 kg mass gives a = 10/2 = 5 m/s²; on a 5 kg mass the same 10 N gives a = 10/5 = 2 m/s² — the heavier mass accelerates less.

Answer: F = 12 N, m = 4 kg → a = F/m = 12/4 = 3 m/s². Initial velocity u = 0; after t = 5 s, v = u + at = 0 + 3×5 = 15 m/s. Thus acceleration 3 m/s² and velocity after 5 s is 15 m/s.

Answer: F = m·a (with m constant) is valid for non-relativistic speeds and in inertial frames. It doesn't directly apply when mass varies with time (e.g., rocket losing mass) without using the full momentum form F_net = dp/dt. Also at velocities approaching speed of light relativistic dynamics modify momentum relation, and in non-inertial (accelerating) frames fictitious forces must be included. Further, at atomic scales quantum mechanics supersedes classical F = m·a.

Answer: General second law: F_net = dp/dt where p = m·v. If m varies with time, dp/dt = m·dv/dt + v·dm/dt. For a rocket expelling mass, thrust arises from mass ejection; one must account for momentum carried away by expelled fuel. Applying the law requires accounting for system boundaries and the momentum flux across them; simply using F = m·a fails if mass changes without this correction.

Answer: Procedure: (1) Draw free-body diagram for the object. (2) Choose coordinate axis (usually along motion). (3) Resolve forces and compute net force ΣF. (4) Apply F_net = m·a and solve for a or other unknowns. Example: A 10 kg block is pulled by 50 N horizontally on a surface with kinetic friction μ_k = 0.2. Normal N = mg = 98 N. Friction f_k = μ_k N = 0.2×98 = 19.6 N. Net force = 50 − 19.6 = 30.4 N. Acceleration a = F_net/m = 30.4 / 10 = 3.04 m/s².

Answer: Newton’s third law: For every action there is an equal and opposite reaction. Action and reaction act on two different bodies, hence they cannot cancel because cancellation would require them to act on the same body. Example: when you push a wall, the wall pushes back on you equally; the forces act on different bodies (wall and you). In rocket propulsion, expelled gas exerts downward action on gas, reaction pushes rocket upward.

Answer: Let mass = m, acceleration of elevator upward = a, g downward. Forces on person: weight mg downward, normal reaction R upward from scale. Applying Newton’s second law upward as positive: R − mg = m·a → R = m(g + a). Thus the scale reads a value proportional to normal reaction; when elevator accelerates upward reading increases; when it accelerates downward reading decreases. The scale measures normal force, not mass directly.

Answer: During walking the foot pushes the ground backward and slightly downward (action). According to Newton III, the ground exerts an equal and opposite reaction force on the foot, forward and slightly upward. The horizontal component of reaction accelerates the body forward. Friction between shoe and ground prevents slipping and allows the horizontal action–reaction force to be effective. Thus, propulsion in walking arises from action–reaction pairs and static friction.

Answer: Rockets carry propellant which is expelled at high speed backwards relative to the rocket. Action: rocket pushes exhaust gases backward. Reaction: gases exert equal and opposite force on rocket, pushing it forward. This does not require atmosphere — momentum conservation suffices. Diagram: rocket body with backward exhaust velocity; show momentum of expelled gas and forward momentum of rocket. Thrust = mass flow rate × exhaust velocity.

Answer: Before firing, total momentum is zero. After firing, bullet of mass m_b attains velocity v_b forward; gun of mass m_g recoils with velocity v_g backward. Conservation: m_b v_b + m_g v_g = 0 → v_g = −(m_b/m_g) v_b. Newton III also explains forces during explosion: gas exerts forward force on bullet and equal opposite force on gun. Example: bullet 0.01 kg, v_b = 400 m/s, gun mass 2 kg → v_g = −(0.01/2)*400 = −2 m/s (recoil backward).

Answer: Static friction acts when two surfaces are in contact without relative motion; it adjusts up to a maximum value f_s(max) = μ_s N to prevent motion. Kinetic (sliding) friction acts during relative motion and is typically f_k = μ_k N. μ_s is usually greater than μ_k because starting motion requires breaking microscopic contacts (asperities) formed during rest, demanding larger force; once sliding, fewer adhesion points remain and average resistance falls.

Answer: Normal N = mg = 20×9.8 = 196 N. Maximum static friction f_s(max) = μ_s N = 0.3 × 196 = 58.8 N. The applied horizontal force must exceed 58.8 N to start motion. So minimum ≈ 58.8 N.

Answer: Lubrication places a thin fluid layer between surfaces, reducing direct asperity contact and lowering both static and kinetic friction by allowing surfaces to slide over the fluid. It also carries heat and can reduce wear. Lubrication may be insufficient under extreme loads, high temperatures, or contaminated surfaces where film breaks down, or in boundary lubrication regimes where asperity contact still occurs — mechanical design and material choice must complement lubrication.

Answer: Friction converts useful mechanical energy into thermal energy, causing energy loss (dissipation) and wear. Engineers mitigate unwanted friction by using lubricants, low-friction materials (e.g., Teflon), bearings (ball/roller), polished surfaces, precise alignment, and cooling systems. In some cases, controlled friction is used (brakes) where dissipation is desired.

Answer: Forces along plane: component of gravity mg sinθ downwards, kinetic friction f_k = μ_k N = μ_k mg cosθ opposing downwards. Net force = mg sinθ − μ_k mg cosθ = mg(sinθ − μ_k cosθ). So a = F_net/m = g(sinθ − μ_k cosθ). Limiting cases: if μ_k = 0 (frictionless), a = g sinθ. If angle small and μ_k large so sinθ < μ_k cosθ, expression gives negative acceleration (block would not slide; in reality static friction prevents motion).

Answer: In elastic collisions, both momentum and kinetic energy of the system are conserved (no kinetic energy lost to deformation or heat). Example: collisions between gas molecules (idealized). In inelastic collisions, momentum is conserved but kinetic energy is not — some is converted into internal energy, deformation, sound. A perfectly inelastic collision maximizes kinetic energy loss and colliding bodies stick together. Example: car crash (mostly inelastic).

Answer: For perfectly inelastic collision where they stick, total momentum before = m1 u1 + m2 (−u2 if opposite direction chosen); careful sign convention needed. If u1 and u2 are with opposite senses and we take right positive and initial velocities accordingly, final common velocity v = (m1 u1 + m2 u2) / (m1 + m2) when both velocities taken in chosen sign sense. This follows from conservation of momentum (no external force in the collision timescale).

Answer: Initial momentum = 0.2×4 = 0.8 kg·m/s. Total mass = 1.0 kg. Final velocity v = 0.8 / 1.0 = 0.8 m/s. Initial KE = 0.5×0.2×4² = 0.5×0.2×16 = 1.6 J. Final KE = 0.5×1.0×0.8² = 0.5×0.64 = 0.32 J. Loss = 1.6 − 0.32 = 1.28 J lost as heat/deformation.

Answer: Conservation of momentum states total momentum of isolated system is constant. For a firing weapon, bullet and gun form closed system: forward momentum of bullet equals backward momentum of gun (recoil). For spacecraft stage separation, propellant expelled backward imparts forward momentum to remaining craft. Rocket staging design exploits momentum exchange and conservation to change velocity of vehicle stages efficiently.

Answer: Strategy: (1) Read carefully and list knowns/unknowns. (2) Draw clear diagrams and free-body diagrams for each body. (3) Choose coordinate axes and sign conventions for each mass. (4) Write Newton’s second law for each body (ΣF = m·a), ensuring tension and acceleration sign consistency. (5) Solve simultaneous equations and check units/limits. Example — Atwood machine: masses m1 and m2 with m1 > m2. For m1 down: m1 g − T = m1 a; for m2 up: T − m2 g = m2 a. Add equations: (m1 − m2) g = (m1 + m2) a → a = (m1 − m2) g / (m1 + m2). Tension T = m1(g − a).

Answer: Let m1 = 2 kg (up), m2 = 3 kg (down). Acceleration a = (m2 − m1) g / (m1 + m2) = (3 − 2) g / 5 = g/5 ≈ 1.96 m/s² (with g = 9.8). Tension T from m1’s equation: T − m1 g = m1 a → T = m1(g + a) = 2(9.8 + 1.96) = 2×11.76 = 23.52 N.

Answer: Common mistakes: (1) Confusing mass and weight — always track units. (2) Wrong sign conventions — be consistent in chosen positive direction. (3) Forgetting that action–reaction act on different bodies. (4) Neglecting friction when problem states surfaces are rough, or incorrectly applying friction formulas. (5) Misapplying F = m·a when mass varies. Advice: always draw free-body diagrams, list forces with directions, convert units to SI, write ΣF = m·a component-wise, and show intermediate steps — examiners award method marks. Check limiting cases for plausibility (e.g., zero friction, infinite mass).