Chapter 4: Laws of Motion Long Questions | physics Class 11 CBSE

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Chapter 4: Laws of Motion Long Questions | physics Class 11 CBSE | ByteNotes.in

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Question 1

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Describe Galileo's contribution to the law of inertia, contrast it with Aristotle's view, and explain how Newton extended Galileo's ideas into his First Law of Motion.

Aristotle incorrectly held that a continuous external force is necessary to keep a body in motion; in the absence of force, bodies come to rest, which matched common experience but was fundamentally wrong.
The flaw in Aristotle's view is that he ignored friction—bodies stop because of friction, not because of the absence of a driving force; if friction were absent, uniform motion would persist indefinitely.
Galileo studied motion on inclined planes and used the double inclined plane experiment to show that a ball always rises to its original height; on a horizontal frictionless surface it would move forever.
Galileo concluded that rest and uniform linear motion are equivalent states, both characterised by zero net force and zero acceleration; changing this state requires an external force.
Newton formalised Galileo's insight as his First Law: every body continues in its state of rest or uniform motion in a straight line unless compelled by an external force; this defines both force and inertia.
Inertia, meaning resistance to change, is the key property: bodies resist changes to their state of motion; the more massive a body, the greater its inertia and the greater the force needed to change its motion.

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Question 2

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Derive Newton's Second Law of Motion starting from the concept of momentum, and discuss four important features of this law.

Momentum p = mv is the product of mass and velocity; qualitative observations show that the greater the rate of change of momentum, the greater the force needed—establishing the foundation of the Second Law.
Mathematically, F ∝ Δp/Δt; taking the limit as Δt → 0, F = k(dp/dt); choosing k = 1 in SI units gives F = dp/dt = ma for a body of fixed mass.
The SI unit of force is the newton: 1 N is the force that causes an acceleration of 1 m s⁻² to a mass of 1 kg; thus 1 N = 1 kg m s⁻².
Feature 1: The Second Law is consistent with the First Law—when F = 0, a = 0, confirming that zero force implies zero acceleration (uniform motion or rest).
Feature 2: It is a vector law equivalent to three component equations (Fx = max, Fy = may, Fz = maz); a force changes only the velocity component along its own direction, leaving perpendicular components unchanged.
Feature 3: It is a local law—the acceleration at any instant depends only on the net force at that instant, not on the history of motion; Feature 4: it applies to a system of particles if F is total external force and a is acceleration of the centre of mass.

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Question 3

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Analyse Newton's Third Law of Motion in detail, clarifying common misconceptions about action-reaction pairs and internal forces, and illustrate with the example of a gun firing a bullet.

Newton's Third Law states: forces always occur in pairs; the force on body A by B (FAB) is equal in magnitude but opposite in direction to the force on body B by A (FBA): FAB = −FBA.
Misconception 1: 'action causes reaction' — this is wrong; the two forces act simultaneously, there is no cause-effect relation, and either force may be labelled the action.
Misconception 2: 'action and reaction cancel each other' — this is wrong; they act on different bodies, so they cannot be added to get zero net force on either body individually.
When a gun fires a bullet, the force F on the bullet by the gun and the recoil force −F on the gun by the bullet act for the same time interval Δt; each equals the change in momentum of the respective body.
Since both bullet and gun are initially at rest, the bullet gains forward momentum pb and the gun gains recoil momentum pg = −pb; their sum is zero, illustrating conservation of momentum from the Third Law.
Internal forces between parts of the same system cancel in pairs by the Third Law, making only external forces relevant for the motion of the system as a whole—this enables the Second Law to apply to multi-body systems.

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Question 4

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Compare static friction and kinetic friction comprehensively, discuss the laws governing each, and explain the role of the coefficient of friction in practical situations.

Static friction (fs) acts between surfaces at rest relative to each other; it opposes impending relative motion and is self-adjusting—it equals the applied force until it reaches its maximum value (fs)max = μsN.
Kinetic (sliding) friction (fk) acts between surfaces in relative motion; it opposes actual relative motion and has a fixed value fk = μkN for given surfaces; it is nearly independent of velocity.
Both types are independent of the area of contact; both are proportional to the normal force N; the coefficients μs and μk depend only on the nature of the pair of surfaces in contact.
Experimentally μk < μs: it is harder to start motion than to maintain it; once sliding begins, the required force decreases, which is why a body accelerates after the applied force just exceeds (fs)max.
In a box accelerating with a train, static friction provides the required centripetal/accelerating force; for a car braking, kinetic friction acts between tyres and road to decelerate the vehicle.
Friction laws are empirical (not fundamental laws); they hold approximately and are useful for macroscopic calculations, though the microscopic origin lies in electrical forces between charged surface constituents.

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Question 5

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Explain the law of conservation of momentum, derive it from Newton's Laws for a two-body collision, and discuss whether it applies to both elastic and inelastic collisions.

The law of conservation of momentum states: the total momentum of an isolated system of particles remains constant when no net external force acts on the system.
For two bodies A and B colliding with initial momenta pA and pB, the Second Law gives FAB Δt = p'A − pA and FBA Δt = p'B − pB, where primes denote final momenta.
By Newton's Third Law, FAB = −FBA; substituting gives p'A − pA = −(p'B − pB), which simplifies to p'A + p'B = pA + pB, proving total momentum is conserved.
This derivation requires no assumption about the type of collision; conservation of momentum holds for both elastic and inelastic collisions alike.
In elastic collisions, an additional condition applies: total kinetic energy before equals total kinetic energy after; in inelastic collisions, kinetic energy is not conserved (converted to heat or deformation), but momentum always is.
Practical applications include gun recoil, rocket propulsion, and collision analysis; in all these isolated systems, the vector sum of momenta before equals the vector sum after the event.

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Question 6

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Analyse the motion of a car taking a circular turn on a level road and on a banked road, deriving the maximum safe speed in each case and comparing the two situations.

On a level road, three forces act on the car: weight mg downward, normal reaction N upward, and static friction f along the surface; since there is no vertical acceleration, N = mg.
The centripetal force required is mv²/R directed inward; on a level road this is provided entirely by static friction: f = mv²/R ≤ μsN = μsmg.
Thus v² ≤ μsRg, giving maximum safe speed on a level road: vmax = √(μsRg), which is independent of the car's mass.
On a banked road at angle θ, both the horizontal component of normal reaction (N sinθ) and the horizontal component of friction (f cosθ) contribute to the centripetal force, reducing dependence on friction.
At optimum speed v₀ = √(Rg tanθ), no friction is needed at all; the normal reaction alone provides the centripetal force, minimising tyre wear.
The maximum speed on a banked road is vmax = √[Rg(μs + tanθ)/(1 − μs tanθ)], which is greater than √(μsRg) on a flat road, showing that banking always allows higher safe speeds for the same radius and friction coefficient.

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Question 7

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Discuss the concept of equilibrium of a particle under three concurrent forces, explain the free-body diagram technique, and illustrate its application to a suspended mass with a horizontal force applied at the midpoint of the rope.

Equilibrium requires the net external force to be zero; for three concurrent forces F₁, F₂, F₃, the condition is F₁ + F₂ + F₃ = 0, meaning the resultant of any two must be equal and opposite to the third.
Graphically, three equilibrium forces are represented by sides of a closed triangle with arrows in the same sense; this extends to n forces forming a closed polygon.
A free-body diagram (FBD) isolates a chosen system and shows all external forces acting on it—including gravity, normal reactions, tension, applied forces—but excludes forces the system exerts on others.
For a mass of 6 kg suspended by a rope with a 50 N horizontal force at the midpoint P: the lower half of rope has tension T₂ = mg = 60 N (from the FBD of the mass).
For point P under three forces (T₁ upward-left at angle θ, T₂ = 60 N downward, 50 N horizontal): horizontal equilibrium gives T₁ sinθ = 50 N; vertical equilibrium gives T₁ cosθ = 60 N.
Dividing gives tanθ = 50/60 = 5/6, so θ = tan⁻¹(5/6) ≈ 40°; the answer is independent of rope length and the position of the applied force, demonstrating the power of FBD analysis.

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Question 8

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Describe the systematic procedure for solving problems in mechanics using free-body diagrams, and apply it to a block-trolley system connected by a string over a pulley with kinetic friction acting on the trolley.

Step 1: Draw a schematic diagram of the entire assembly, showing all bodies, links, supports, and connections; Step 2: choose a convenient sub-system to analyse.
Step 3: Draw the FBD of the chosen sub-system, showing all forces from the environment on it (gravity, normal reaction, tension, friction); do not include forces the sub-system exerts on others.
Step 4: Apply Newton's Second Law (ΣF = ma) for each chosen sub-system; use Newton's Third Law when connecting equations for different sub-systems—if A exerts force F on B, B exerts −F on A.
For a 3 kg block hanging and a 20 kg trolley on a surface with μk = 0.04: FBD of the block gives 30 − T = 3a; FBD of the trolley gives T − fk = 20a where fk = μkN = 0.04 × 200 = 8 N.
Solving: 30 − T = 3a and T − 8 = 20a; adding gives 22 = 23a, so a = 22/23 ≈ 0.96 m s⁻²; substituting back gives T ≈ 27.1 N.
The key insight is that the string being inextensible means both the block and trolley have the same magnitude of acceleration; internal forces (like tension) must be consistently accounted for in each FBD.

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Question 9

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Analyse the common forces in mechanics—gravitational, normal reaction, tension, spring force, friction—and explain how all contact forces fundamentally originate from electrical forces.

Gravitational force is the only non-contact force in everyday mechanics; it acts at a distance through space without any medium and is the force that gives every object its weight mg near the earth's surface.
Normal reaction is the contact force component perpendicular to the surfaces in contact; it is a self-adjusting force that prevents surfaces from penetrating each other, balancing the perpendicular component of applied force.
Tension is the restoring force in a string when it is taut; for an inextensible, massless string, tension T is uniform throughout; spring force follows Hooke's Law: F = −kx, where k is the spring constant and x is displacement.
Friction is the contact force component parallel to surfaces in contact; static friction opposes impending motion (fs ≤ μsN) while kinetic friction opposes actual sliding (fk = μkN).
At the microscopic level, all contact forces arise from electromagnetic (electrical) forces between the nuclei and electrons of atoms in the two bodies; elasticity, molecular collisions, and surface interactions all trace back to inter-atomic electrical forces.
Despite this common origin, contact forces are treated as distinct empirical types in macroscopic mechanics because their microscopic derivation is too complex to be useful in solving practical problems at the macroscopic scale.

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Question 10

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Explain why friction has a dual role in practical life—being both harmful and beneficial—and discuss rolling friction and methods to reduce friction in machines.

Friction is harmful in machines with moving parts: it opposes relative motion between parts, dissipates mechanical energy as heat, causes wear, and reduces efficiency; lubricants and smooth finishes reduce kinetic friction.
Ball bearings placed between moving parts replace sliding friction with much smaller rolling friction, significantly reducing power dissipation; a thin air cushion between surfaces can virtually eliminate contact friction.
Friction is essential for walking: the frictional force between feet and ground is the only external horizontal force that propels us forward; on a frictionless surface like ice, walking is impossible.
Static friction between car tyres and the road provides the centripetal force for turning and the driving force for acceleration; brakes rely on kinetic friction to decelerate vehicles quickly.
Rolling friction is much smaller than sliding friction (by 2–3 orders of magnitude) because in ideal rolling only a single point is in contact with the surface, and that point has zero relative motion at every instant.
In practice, rolling friction arises from momentary deformation of the contact region, which creates a finite contact area and a net resistive component of the contact force; discovery of the wheel was a major advance because it replaced sliding friction with rolling friction.

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Chapter 4: Laws of Motion | Long Questions

Describe Galileo's contribution to the law of inertia, contrast it with Aristotle's view, and explain how Newton extended Galileo's ideas into his First Law of Motion.

Derive Newton's Second Law of Motion starting from the concept of momentum, and discuss four important features of this law.

Analyse Newton's Third Law of Motion in detail, clarifying common misconceptions about action-reaction pairs and internal forces, and illustrate with the example of a gun firing a bullet.

Compare static friction and kinetic friction comprehensively, discuss the laws governing each, and explain the role of the coefficient of friction in practical situations.

Explain the law of conservation of momentum, derive it from Newton's Laws for a two-body collision, and discuss whether it applies to both elastic and inelastic collisions.

Analyse the motion of a car taking a circular turn on a level road and on a banked road, deriving the maximum safe speed in each case and comparing the two situations.

Discuss the concept of equilibrium of a particle under three concurrent forces, explain the free-body diagram technique, and illustrate its application to a suspended mass with a horizontal force applied at the midpoint of the rope.

Describe the systematic procedure for solving problems in mechanics using free-body diagrams, and apply it to a block-trolley system connected by a string over a pulley with kinetic friction acting on the trolley.

Analyse the common forces in mechanics—gravitational, normal reaction, tension, spring force, friction—and explain how all contact forces fundamentally originate from electrical forces.

Explain why friction has a dual role in practical life—being both harmful and beneficial—and discuss rolling friction and methods to reduce friction in machines.

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