Chapter 5: Laws of Motion (Class 11 Physics) 

1. What is inertia?

Property of a body to resist changes in its state of rest or uniform motion.

2. State Newton’s first law of motion.

Body remains in state of rest or uniform motion unless an external unbalanced force acts on it.

3. Which law is also called the law of inertia?

Newton’s first law of motion.

4. Define linear momentum.

Product of mass and velocity (p = mv). Vector quantity.

5. State Newton’s second law of motion (in terms of momentum).

Rate of change of momentum is directly proportional to applied external force (F = dp/dt).

6. Write the common form of Newton’s second law for constant mass.

F = ma (Force = mass × acceleration).

7. What is impulse?

Product of average force and the time it acts (F_avg × Δt). Also equal to change in momentum.

8. State Newton’s third law of motion.

To every action, there is an equal and opposite reaction.

9. Do action and reaction forces act on the same body?

No, they act on different bodies.

10. State the law of conservation of linear momentum.

If net external force on a system is zero, its total linear momentum remains constant.

11. What are concurrent forces?

Forces whose lines of action pass through a common point.

12. What is the condition for equilibrium of concurrent forces?

The vector sum of all forces is zero (F_net = 0).

13. Name the friction that opposes the start of motion.

Static friction (specifically, limiting friction).

14. Name the friction that acts during motion.

Kinetic (or sliding) friction.

15. Which is generally greater: static friction or kinetic friction?

Maximum static (limiting) friction.

16. What is rolling friction?

Resistance when a body rolls over a surface.

17. Give an example of a lubricant.

Oil, Grease, Graphite.

18. What provides the centripetal force for a car turning on a level road?

Static friction between tyres and road.

19. What is banking of roads?

Raising the outer edge of a curved road relative to the inner edge.

20. Define Centripetal force.

The net force directed towards the center, required for circular motion (mv²/r).

1. Explain inertia of rest and inertia of motion with one example each.

• Inertia of Rest: Tendency to resist change from state of rest. Ex: Passengers fall backward when bus starts suddenly.
• Inertia of Motion: Tendency to resist change from state of uniform motion. Ex: Passengers fall forward when bus stops suddenly.

2. Why is Newton’s second law considered the ‘real’ law of motion?

• Because the first law (inertia) and third law (action-reaction) can be derived from the second law (F=dp/dt or F=ma).
• First Law: If F=0, then dp/dt=0, meaning momentum p (mv) is constant. If m is constant, v is constant (definition of first law).
• Third Law: Can be derived by considering the change in momentum of an isolated system of two interacting bodies.

3. Explain why action and reaction forces do not cancel each other.

• Because they act on **different bodies**.
• Example: When a book rests on a table, Action (Book’s weight ON Table) acts on the Table. Reaction (Table’s upward force ON Book) acts on the Book.
• Since they act on different objects, they cannot cancel out. Cancellation requires forces acting on the same object.

4. Give two applications of the law of conservation of momentum.

• Recoil of a Gun: Gun moves backward when bullet moves forward to conserve zero initial momentum.
• Rocket Propulsion: Expelled gas momentum downwards equals rocket momentum upwards.
• Collisions between objects.

5. Why is it difficult to walk on a sandy surface?

• Walking requires pushing the ground backward (action) to get a forward push (reaction) from the ground due to friction.
• Sand particles shift easily and offer very little backward static frictional force (reaction force) against the push of the foot.
• Without sufficient reaction force, it’s hard to move forward.

6. Explain static friction, limiting friction, and kinetic friction.

• Static (f_s): Opposes tendency of motion when at rest. Self-adjusting (0 ≤ f_s ≤ f_s(max)).
• Limiting (f_s(max)): Maximum value of static friction just before sliding starts (f_s(max) = μ_sN).
• Kinetic (f_k): Acts when object is sliding. Roughly constant (f_k = μ_kN). Generally f_k < f_s(max).

7. State the laws of limiting friction.

• Limiting friction depends on nature of surfaces.
• Limiting friction is directly proportional to normal reaction (f_s(max) ∝ N).
• Limiting friction is independent of the apparent area of contact.
• (Kinetic friction is largely independent of relative velocity).

8. How does lubrication reduce friction?

• Lubricants (oil/grease) form a thin layer between surfaces.
• Prevents direct contact of surface irregularities (reduces interlocking).
• Allows surfaces to slide over the lubricant layer, which offers much less resistance than solid-solid friction.

9. Why are wheels advantageous?

• Wheels replace sliding friction with much smaller rolling friction.
• Rolling friction requires significantly less force to overcome.
• Makes movement of heavy objects much easier and energy efficient.

10. Define centripetal force and write its formula.

• Net force directed towards the center required for uniform circular motion.
• Formula: F_c = mv²/r = mω²r.
• It’s not a new force type, but the resultant providing the inward acceleration.

11. Why is a road banked?

• To allow vehicles to turn safely at higher speeds without solely relying on friction.
• The horizontal component of the normal reaction (from the banked surface) helps provide the necessary centripetal force.

12. A force of 50 N is applied to a 10 kg block resting on a surface (μk = 0.4). Find the acceleration. (g=10 m/s²)

• Normal Reaction N = mg = 10×10 = 100 N.
• Kinetic Friction fk = μk N = 0.4 × 100 = 40 N.
• Net Force Fnet = Applied Force – Kinetic Friction = 50 N – 40 N = 10 N.
• Acceleration a = Fnet / m = 10 N / 10 kg = 1 m/s².

13. Calculate the impulse required to stop a 1500 kg car moving at 25 m/s.

• Impulse = Change in Momentum = p_f – p_i.
• p_i = mu = 1500 × 25 = 37500 kg m/s.
• p_f = mv = 1500 × 0 = 0 (stops).
• Impulse J = 0 – 37500 = -37500 kg m/s (or N s).
• Magnitude is 37500 N s.

14. Why does a cricketer move his hands backwards while catching a ball?

• To increase the time (Δt) over which the ball’s momentum is brought to zero.
• From Impulse-Momentum theorem, F × Δt = Δp (change in momentum).
• Since Δp is fixed (ball stops), increasing Δt reduces the average force (F = Δp / Δt) exerted by the ball on the hands.
• This reduces the impact and prevents injury.

15. State Lami’s theorem.

• If three concurrent forces acting on a point are in equilibrium,
• Then each force is proportional to the sine of the angle between the other two forces.
• F₁sin α = F₂sin β = F₃sin γ (α, β, γ are angles opposite F₁, F₂, F₃ resp.)

16. Can kinetic friction be greater than limiting static friction?

• No, generally kinetic friction is always less than limiting static friction.
• It takes more force to start an object moving (overcome max static friction) than to keep it moving (overcome kinetic friction).

1. State and explain Newton’s Three Laws of Motion with examples.

1. First Law (Inertia): Body stays at rest or in uniform motion unless acted upon by external unbalanced force. Ex: Ball at rest stays at rest until kicked. (Viramavastha ya ek saman gati tab tak jab tak external force na lage).
2. Second Law (F=dp/dt or F=ma): Rate of change of momentum is proportional to applied force. Gives measure of force. Ex: Harder kick = more acceleration. (Momentum change ka rate applied force ke proportional. Force ka measure deta hai).
3. Third Law (Action-Reaction): Every action has equal & opposite reaction acting on different bodies. Ex: Gun recoil, Walking. (Har kriya ki saman aur vipreet pratikriya alag bodies par).

2. State and prove the Law of Conservation of Linear Momentum using Newton’s laws.

Statement: If net external force on a system is zero, its total linear momentum remains constant.

Proof using Second Law:

• Newton’s 2nd Law: Fext = dpdt, where p is total momentum.
• If F_ext = 0, then dpdt = 0.
• The derivative of a quantity is zero only if that quantity is constant.
• Therefore, p = constant. (Total linear momentum is conserved).

Proof using Third Law (for 2 bodies):

• Consider isolated system of bodies A and B colliding.
• Force on A by B (F_AB) and Force on B by A (F_BA).
• By 3rd Law: F_AB = – F_BA.
• By 2nd Law: F_AB = dp_A/dt and F_BA = dp_B/dt.
• So, dp_A/dt = – dp_B/dt => dp_A/dt + dp_B/dt = 0 => d/dt (p_A + p_B) = 0.
• This means (p_A + p_B) = constant. (Total momentum conserved).

3. What is Friction? Discuss Static, Limiting, Kinetic, and Rolling friction. State laws of friction.

• Friction: Force opposing relative motion between surfaces in contact.
• Static Friction (f_s): Opposes tendency of motion when at rest. Self-adjusting (0 ≤ f_s ≤ f_s(max)).
• Limiting Friction (f_s(max)): Maximum static friction just before sliding (f_s(max) = μ_sN).
• Kinetic Friction (f_k): Acts during sliding. Roughly constant (f_k = μ_kN). Generally f_k < f_s(max).
• Rolling Friction (f_r): Resistance when rolling. Very small compared to sliding (f_r << f_k). Caused by deformation.
• Laws of Limiting Friction: Opposes motion tendency; Proportional to Normal Reaction (N); Depends on surface nature; Independent of contact area.

4. Explain Centripetal Force. Derive the expression F_c = mv²/r.

Centripetal Force: The net force towards the center needed to keep an object in uniform circular motion (UCM).

Derivation:

• In UCM, velocity vector v changes direction, so there’s acceleration a_c.
• This acceleration a_c is directed towards the center (centripetal) and has magnitude v²/r.
• By Newton’s Second Law (F = ma), the net force causing this acceleration must also be towards the center.
• This net inward force is called Centripetal Force (F_c).
• F_c = mass × centripetal acceleration = m × a_c.
• F_c = mv²/r (or mω²r).

5. Analyze the motion of a vehicle on a banked road (deriving max safe speed with friction).

On a banked road (angle θ), forces are: Weight (mg), Normal Reaction (N perpendicular to road), Friction (f, up or down slope depending on speed).

• Resolve N and f into horizontal and vertical components.
• Vertical Equilibrium: N cosθ = mg + f sinθ (Assuming f acts downwards for max speed).
• Horizontal Motion: Net inward force = Centripetal Force. N sinθ + f cosθ = mv²/r.
• Friction f ≤ μs N (using static friction for maximum speed without slipping). Let f = μs N.
• From vertical: N(cosθ – μs sinθ) = mg => N = mg / (cosθ – μs sinθ).
• Substitute f=μs N and N into horizontal eqn: (N sinθ + μs N cosθ) = mv²/r => N(sinθ + μs cosθ) = mv²/r.
• [mg / (cosθ - μs sinθ)] × (sinθ + μs cosθ) = mv²/r
• Cancel ‘m’: v² = [ rg (sinθ + μs cosθ) / (cosθ - μs sinθ) ]
• Divide numerator and denominator by cosθ:
• v² = [ rg (tanθ + μs) / (1 - μs tanθ) ]

Max Safe Speed: v_max = √[ rg (μs + tanθ) / (1 – μs tanθ) ]

6. Explain the concept of Impulse and Impulse-Momentum Theorem. How is it applied in daily life?

• Impulse (J): Measure of total effect of a force over time. J = F_avg × Δt.
• Impulse-Momentum Theorem: States that Impulse acting on an object equals the change in its linear momentum (J = Δp = p_f – p_i). Derived from F=Δp/Δt.
• Application Idea: For a given change in momentum (Δp), if the time of impact (Δt) is increased, the average force (F=Δp/Δt) experienced decreases.
• Examples:
  • Cricketer pulling hands back: Increases Δt, reduces F on hands.
  • Shock absorbers in vehicles: Increase Δt of impact, reduce F on passengers.
  • Packing fragile items with straw/foam: Increases Δt during impact, reduces F on items.
  • Jumping on soft ground: Increases Δt of stopping, reduces F on legs.

7. Derive the expression for the recoil velocity of a gun.

• Consider a gun (mass M) firing a bullet (mass m).
• Before firing, gun and bullet are at rest. Total initial momentum P_i = 0.
• After firing, let bullet velocity be v and gun recoil velocity be V. Total final momentum P_f = mv + MV.
• Assuming the system (gun + bullet) is isolated during the short firing process, net external force is zero.
• By Law of Conservation of Linear Momentum: P_i = P_f.
• 0 = mv + MV.
• MV = -mv.
• Recoil Velocity V = – (m/M) v.
• Negative sign indicates gun recoils in the opposite direction to the bullet. Magnitude V = (m/M)v.

8. Discuss static and kinetic friction using a graph showing the variation of frictional force with applied force.

Graph Concept:

• Plot Applied Force (F_app) on x-axis and Frictional Force (f) on y-axis.
• Region 1 (Static Friction): As F_app increases from 0, static friction (f_s) also increases equally to match it (f_s = F_app). The object remains at rest. Graph is a straight line passing through origin with slope 1.
• Point 2 (Limiting Friction): f_s reaches its maximum value, f_s(max) = μ_sN. This is the point just before sliding begins. Graph peaks here.
• Region 3 (Kinetic Friction): Once F_app exceeds f_s(max), the object starts sliding. Friction force slightly drops to a constant value called kinetic friction (f_k = μ_kN). Graph drops slightly and becomes a horizontal line at f_k level (as kinetic friction is roughly independent of applied force or velocity).

9. What is equilibrium of a particle? Explain static and dynamic equilibrium.

A particle is in **equilibrium** if the **net external force** acting on it is zero (F_net = ΣF = 0).

• If F_net = 0, then by Newton’s 2nd Law, acceleration a = 0.
• This means the velocity v of the particle is constant (could be zero or non-zero).

Types of Equilibrium:

• Static Equilibrium: If the particle is in equilibrium AND is **at rest** (v=0), it is in static equilibrium. Ex: A book resting on a table.
• Dynamic Equilibrium: If the particle is in equilibrium AND is moving with a **constant non-zero velocity** (a=0 but v≠0, constant), it is in dynamic equilibrium. Ex: A raindrop falling with constant terminal velocity.

10. Explain why lubrication helps, mentioning fluid friction.

• Friction between solid surfaces arises due to interlocking of irregularities.
• Lubricants (liquids like oil, or sometimes gases) form a thin film between the solid surfaces.
• This prevents direct solid-to-solid contact.
• Motion now involves the sliding of solid surface over the lubricant layer, or layers of lubricant sliding over each other.
• The friction between solid and fluid layers, or within fluid layers (called viscous force or internal fluid friction), is generally much lower than solid-solid friction.
• Thus, the overall frictional force is significantly reduced, allowing easier movement, less wear, and less heat generation.

11. Can friction be zero? Explain circumstances.

• Yes, theoretically friction can be considered zero in ideal situations.
• 1. Perfectly Smooth Surfaces: If two surfaces in contact were perfectly smooth with no irregularities at the atomic level, there would be no interlocking, and friction (in the traditional sense) would be zero. This is practically unattainable.
• 2. No Contact: Friction is a contact force. If there is no contact between surfaces (e.g., an object moving in perfect vacuum far from other objects), there is no contact friction.
• 3. Rolling Friction (Approx Zero): While not truly zero, rolling friction is often so small compared to static/sliding friction that it is sometimes approximated as zero in simple physics problems, especially for ideal rolling without deformation.
• 4. Zero Normal Force: Since friction f ≤ μN, if the normal force N pressing surfaces together is zero, the frictional force will also be zero. (e.g., object in free space).

12. Give reasons: (a) It is dangerous to jump out of a moving bus. (b) Dust comes out of a carpet when beaten with a stick.

• (a) Jumping from Moving Bus:
  • When inside the bus, your body shares the bus’s forward velocity (Inertia of Motion).
  • When you jump out and your feet touch the ground, your feet are suddenly brought to rest by friction with the ground.
  • However, the upper part of your body continues to move forward due to inertia of motion.
  • This difference in motion causes you to fall forward, potentially causing injury.
• (b) Beating a Carpet:
  • Initially, both the carpet and the dust particles within it are at rest.
  • When the carpet is beaten with a stick, the carpet is suddenly set into motion.
  • Due to Inertia of Rest, the dust particles tend to remain in their state of rest.
  • As the carpet moves away, the dust particles fall down due to gravity, thus getting separated from the carpet.

13. Derive the expression for maximum safe speed on a level circular road.

• For a car (mass m) on a level circular road (radius r) moving at speed v:
• The required inward force to turn (Centripetal Force) = mv²/r.
• This force must be provided by the friction between the tyres and the road.
• The maximum available frictional force is the limiting static friction, f_s(max) = μ_sN.
• On a level road, Normal Reaction N = Weight mg.
• So, f_s(max) = μ_smg.
• For safe turning without skidding, the required centripetal force must be less than or equal to the maximum available static friction:
• mv² / r ≤ μsmg
• Cancel ‘m’ from both sides: v² / r ≤ μsg
• v² ≤ μsrg
• Therefore, the maximum safe speed is when equality holds: vmax = √(μsrg).

14. State the difference between Mass and Weight.

• Mass (m): Amount of matter in a body. It’s a measure of inertia. It’s a scalar quantity. Unit: kg. It remains constant everywhere.
• Weight (W): Gravitational force exerted on a body by the Earth (or another celestial body). W = mg (where g is acceleration due to gravity). It’s a vector quantity (directed towards Earth’s center). Unit: Newton (N). It varies with the value of ‘g’ (changes with altitude, latitude, planet).

15. A man of mass 70 kg stands on a weighing scale in a lift which is moving (a) upwards with uniform speed of 10 m/s (b) downwards with uniform acceleration of 5 m/s² (c) upwards with uniform acceleration of 5 m/s². What would be the reading of the scale in each case? (g=9.8 m/s²)

Reading of weighing scale = Normal Reaction (R) exerted by the scale on the man.

Net force on man = ma (Newton’s 2nd Law). Forces are Weight (mg downwards) and Normal Reaction (R upwards).

• (a) Upwards with uniform speed: Uniform speed => a = 0. Net Force = 0. Upward force = Downward force. R – mg = 0 => R = mg. R = 70 × 9.8 = 686 N. (Reads true weight).
• (b) Downwards with uniform acceleration a=5 m/s²: Net force is downwards. mg – R = ma. R = mg – ma = m(g – a). R = 70 (9.8 – 5) = 70 × 4.8 = 336 N. (Reads less than true weight – apparent weight decrease).
• (c) Upwards with uniform acceleration a=5 m/s²: Net force is upwards. R – mg = ma. R = mg + ma = m(g + a). R = 70 (9.8 + 5) = 70 × 14.8 = 1036 N. (Reads more than true weight – apparent weight increase).