NCERT Plus One (Class 11) Physics • Chapter 4

Laws of Motion

Q: Why does a cricket fielder pull his hands backward while catching a fast-moving ball?

Impulse Δp = F_avg · Δt is fixed to stop the ball. Pulling hands back increases impact duration Δt , which reduces average force F_avg = Δp / Δt on the hands and avoids injury.

Q: Action and reaction are equal and opposite, yet they do not cancel each other. Why?

Because action and reaction forces always act on two different bodies . Forces cancel only when they act on the same single body.

Q: State Newton's Second Law of Motion and show how F = ma follows from it for a body of constant mass.

Statement: The rate of change of linear momentum of a body is directly proportional to the applied unbalanced external force and takes place in the direction of force. F ∝ dp dt = d(mv) dt . For constant mass m and SI constant k = 1 : F = m (dv/dt) = ma .

Q: A shell of mass 0.02 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m/s, what is the recoil speed of the gun?

By conservation of momentum (initial p = 0 ): M V + m v = 0 . V = −(m v)/M = −(0.02 × 80)/100 = −1.6 / 100 = −0.016 m/s ( 1.6 cm/s opposite to shell).

Q: Define angle of repose and prove that μ_s = tan θ_r.

Definition: The minimum angle of inclination of a rough plane with horizontal at which a body placed on it just begins to slide down. Proof: At threshold, Normal N = mg cos θ_r and downslope gravity mg sin θ_r balances limiting friction f_max = μ_s N = μ_s mg cos θ_r . Dividing: μ_s = tan θ_r .

Q: State and prove the Law of Conservation of Linear Momentum using Newton's Second and Third Laws.

Statement: When net external force on an isolated system is zero, its total linear momentum remains constant. Proof: Consider two colliding bodies A and B during contact time Δt . By 3rd Law, F_AB = −F_BA . Multiplying by Δt gives equal and opposite impulses: Δp_A = −Δp_B ⇒ Δ(p_A + p_B) = 0 . Hence p_A + p_B = constant ( m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ ).

Q: Find the apparent weight (Normal reaction N) of a 70 kg man standing on a weighing scale in a lift when it moves (a) upward with uniform speed 5 m/s, (b) downward with uniform acceleration 5 m/s², and (c) upward with uniform acceleration 5 m/s². (g = 10 m/s²)

(a) Uniform speed ⇒ a = 0 : N = mg = 70 × 10 = 700 N (70 kg reading). (b) Downward accel 5 m/s² : N = m(g − a) = 70(10 − 5) = 350 N (35 kg reading). (c) Upward accel 5 m/s² : N = m(g + a) = 70(10 + 5) = 1050 N (105 kg reading).

Q: Draw a schematic diagram of forces on a car moving along a circular road banked at angle θ with friction μ_s. Derive the expression for maximum safe speed v_max and hence obtain the optimum speed v₀.

Force Balance on Banked Curve: 1. Vertical ( a_y = 0 ): N cos θ = mg + f sin θ ⇒ N cos θ − μ_s N sin θ = mg . 2. Radial inward (centripetal): N sin θ + f cos θ = mv²/R ⇒ N(sin θ + μ_s cos θ) = mv²/R . Dividing (2) by (1) and dividing top & bottom by cos θ : v_max = √[ R g (μ_s + tan θ) / (1 − μ_s tan θ) ] . Setting μ_s = 0 gives optimum zero-wear speed: v₀ = √(R g tan θ) .

Aristotle's fallacy, inertia, Newton's three laws, impulse, momentum conservation, friction & banked circular roads.

📄 Official NCERT Textbook PDF

Topic 4.2 – 4.4: Aristotle's Fallacy, Inertia & Newton's First Law

Stand inside a KSRTC bus waiting at Palayam stop: the moment the driver hits the accelerator, your upper body tips backward; when brakes slam at a red light, you lurch forward! Why? Your feet ride with the bus floor, while your upper body stubbornly tries to keep doing what it was already doing — that lazy resistance to change is inertia!

ARISTOTLE vs GALILEOAristotle's fallacy: "Force keeps things moving."
Galileo/Newton: Zero net force keeps uniform velocity forever (friction was hiding)!

INERTIA & MASSInertia = resistance to velocity change. Mass is the direct measure of inertia!

NEWTON'S 1ST LAWΣF = 0 ⇔ a = 0 (rest OR uniform straight motion continues unchanged!).

🧠 Memory Trick: More Mass = More Stubbornness (Inertia)! Three kinds: Inertia of Rest (carpet dust), Motion (bus braking), and Direction (mud flying tangent off wheels!).

💡 Simple Kid-Check: For 2,000 years Aristotle got fooled because he saw toy carts stop unless pushed — he missed invisible friction! Galileo imagined a super-smooth icy double ramp: if friction vanishes, a rolling ball keeps gliding across a flat floor forever with ZERO push!

1
Aristotle's Fallacy: Aristotle claimed an external force is needed to maintain uniform motion. Wrong! You only push a bicycle steadily on a road to cancel opposing friction and air drag so net force becomes zero.
2
Galileo's Law of Inertia: On a frictionless double inclined plane, a ball released from height h climbs back to the exact same height h regardless of slope. When the second slope flattens to 0°, the ball travels infinitely at constant velocity!
3
Newton's First Law (Law of Inertia): Every body continues in its state of rest or of uniform motion in a straight line unless compelled by some external unbalanced force to act otherwise.

Equivalence of Rest & Uniform Motion: Net Force F_net = 0 ⇔ Acceleration a = 0

🧒 Everyday Inertia Checklist:
• Rest: Flick a cardboard under a coin — coin drops straight into the glass!
• Motion: Jump off a moving train and you fall forward unless you run a few steps!
• Direction: Car turns sharply right → passengers feel thrown left (trying to go straight)!

Classic 1-mark exam trap: Does a spacecraft gliding through empty deep space with engines switched OFF need a force to keep moving at 10 km/s? NO! With zero friction and zero gravity, zero force maintains constant velocity — force is needed ONLY to change speed or direction! Also: Inertia has NO separate unit — its measure is simply Mass (kg)!

Topic 4.5: Newton's Second Law & Impulse

Would you rather step in front of a bicycle rolling at 5 m/s or a loaded lorry rolling at the exact same 5 m/s? Obviously the bicycle! Why, when both speeds match? Because the lorry packs way more mass times velocity — a giant punch called momentum — and when a fielder catches a fast cricket ball, pulling hands back stretches collision time so the sting drops dramatically!

MOMENTUM & 2ND LAWp = mv | F = dpdt = ma (constant m)

IMPULSE (J)J = F_avg Δt = Δp = m(v − u) | Unit: N·s or kg·m/s

F–t GRAPH RULEArea under Force–Time (F–t) curve = Impulse = Change in Momentum Δp!

🧠 Memory Trick: Impulse = Same Δp, Trade Time for Safety! Catching with soft retreating hands doubles Δt → cuts impact force F = Δp/Δt in HALF!

💡 Simple Kid-Check: Force doesn't just look at velocity — it looks at how fast Momentum (mass × velocity) changes! A tiny pebble at bullet speed OR a slow elephant can pack huge momentum. And Impulse is simply the total momentum kick delivered during a short slap or bounce!

1
Second Law Derivation: Net force F = k dpdt. Choosing SI constant k = 1 defines 1 Newton (N) as the force giving 1 kg an acceleration of 1 m/s²: F = ma!
2
Local Instantaneous Law: Acceleration a at instant t depends ONLY on the net force F acting at that exact instant t (not where the object was earlier!).
3
Impulse–Momentum Theorem: Multiplying force by short collision duration Δt gives J = F_avg Δt = Δp. Increasing stopping time Δt (airbags, cricket gloves, jumping on sand instead of concrete) slashes peak force F!

F = dpdt = ma | 1 N = 10⁵ dyne | Impulse J = Δp = Area under F–t graph

🧒 Super-Easy Rebound Example: A 0.2 kg rubber ball hits a wall perpendicularly at +10 m/s and bounces straight back at −10 m/s in 0.02 s.
• Δp = m(v − u) = 0.2(−10 − 10) = −4 N·s (Magnitude = 2mv = 4 kg·m/s, NOT zero!)
• Average force on ball = |Δp|/Δt = 4 / 0.02 = 200 N!

Watch the ball-bounce sign trap! When a ball of mass m hits a wall at speed v and rebounds at the same speed v, change in momentum is NOT mv − mv = 0! Because velocity reverses direction, Δp = m(−v) − m(+v) = −2mv (magnitude 2mv)! Also, Impulse is NOT a force — its dimension is [MLT⁻¹] (same as momentum), NOT [MLT⁻²]!

Topic 4.6 – 4.7: Newton's Third Law & Conservation of Momentum

Step off an unanchored wooden vallam onto the Alappuzha jetty without tying the rope — as your foot pushes you forward onto the wood, the boat glides backward into the canal! You cannot push anything without getting pushed back equally hard. That partner-push rule also explains why a rifle kicks a soldier's shoulder and why rockets soar in vacuum!

NEWTON'S 3RD LAWF_AB = −F_BA — Action & reaction are equal, opposite, simultaneous, on 2 DIFFERENT bodies!

MOMENTUM CONSERVATIONIf F_ext = 0: p_initial = p_final (m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂)

GUN / BOAT RECOILInitially at rest: M V + m v = 0 ⇒ V = −mM v

🧠 Memory Trick: Action & Reaction NEVER cancel because they sit on DIFFERENT Free Body Diagrams (your shoe pushes Earth; Earth pushes your shoe)!

💡 Simple Kid-Check: Forces ALWAYS come in twin pairs! There is no such thing as a lonely one-sided push in the universe. When a rocket shoots hot gas downward out its tail, that escaping gas pushes the rocket upward — even in airless outer space!

1
Third Law Misconception Busted: Action and reaction occur at the exact same instant (neither causes the other first!). Because F_AB acts on Body A and F_BA acts on Body B, you NEVER add them together when analyzing Body A alone.
2
Deriving Momentum Conservation: For two colliding objects with zero external force, impulses during contact time Δt are F_AB Δt and F_BA Δt = −F_AB Δt. Adding them gives Δp_A + Δp_B = 0 ⇒ Total momentum stays conserved!
3
Explosions & Recoil: A stationary shell explodes or a gun fires a bullet: total initial momentum is 0, so fragments / bullet and gun MUST carry equal and opposite momenta summing to zero!

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (when F_ext = 0) | Recoil: V_gun = −m_bullet · v_bulletM_gun

🧒 Super-Easy Recoil Example: A 4 kg rifle fires a 20 g (0.02 kg) bullet at 400 m/s. Find recoil speed!
V = −(0.02 × 400) / 4 = −8 / 4 = −2 m/s (2 m/s backward!).

Classic 2-mark board question: "If action and reaction are equal and opposite, why don't they cancel each other out so nothing ever moves?" Answer: Cancellation happens ONLY when equal and opposite forces act on the SAME body — action and reaction ALWAYS act on two DIFFERENT bodies! Also, momentum stays conserved in BOTH elastic and inelastic collisions whenever F_ext = 0.

Topic 4.8: Equilibrium of a Particle (Concurrent Force Balance)

Picture a heavy bunch of Onam plantains or a glowing traffic signal hanging quietly over an Ernakulam junction from two angled steel cables! Three invisible tugs meet at that single metal hook — gravity pulling straight down and two slanted cable tensions pulling diagonally up — balancing into complete stillness: equilibrium!

EQUILIBRIUM CONDITIONΣF = 0 ⇒ ΣF_x = 0, ΣF_y = 0, ΣF_z = 0

3 CONCURRENT FORCESForm a closed triangle head-to-tail, OR Lami's Theorem: F₁/sin α = F₂/sin β = F₃/sin γ

LIFT APPARENT WEIGHTUp accel a: N = m(g+a) | Down accel a: N = m(g−a) | Free fall: N = 0

🧠 Memory Trick: Equilibrium does NOT require v = 0! A skydiver falling at steady terminal velocity has ΣF = 0 → Dynamic Equilibrium!

💡 Simple Kid-Check: Equilibrium is a tug-of-war tie! Every pull to the right balances every pull to the left (ΣF_x = 0), and every pull upward balances every pull downward (ΣF_y = 0). Whether the object sits still OR glides steadily at constant velocity, it is in equilibrium!

1
Draw a Clean FBD First! Isolate just ONE body/knot, draw every force acting ON it (weight mg downward, tension T along ropes pulling away, normal N perpendicular to surface).
2
Resolve & Balance Axes: Break tilted tensions into T cos θ and T sin θ, then set Right = Left and Up = Down!
3
Horizontal Push on Hanging Mass: Pull a hanging mass m aside by angle θ with horizontal force F: vertical balance gives T cos θ = mg, horizontal gives T sin θ = F ⇒ Dividing gives F = mg tan θ!

F₁ + F₂ + F₃ = 0 | Lami's Rule: F₁sin α = F₂sin β = F₃sin γ

🧒 Classic NCERT Rope Knot Problem: A 6 kg mass (W = 60 N) hangs from ceiling by a 2 m rope, pulled sideways at its midpoint by horizontal force 50 N. What angle θ does the top half make with vertical?
Top tension balances 60 N vertically and 50 N horizontally ⇒ tan θ = 50 / 60 = 5/6 ⇒ θ ≈ 40°! (Notice the weight of the lower half controls the angle!)

Two classic traps! First: Equilibrium does NOT mean zero velocity — constant velocity (like terminal speed raindrop) is dynamic equilibrium with ΣF = 0! Second: Can you pull a clothesline with a heavy towel hanging in the centre completely straight and horizontal (θ = 0°)? IMPOSSIBLE! Vertical balance 2T sin θ = mg gives T = mg / (2 sin 0°) → ∞ — the wire snaps first!

Topic 4.9: Common Forces & Static, Kinetic & Rolling Friction

Try pushing a heavy wooden almirah across your bedroom floor: at first it refuses to budge even as you push harder and harder (10 N, 30 N, 50 N...) — until suddenly at 60 N it gives a lurch and actually feels slightly easier to keep sliding! Meet nature's smartest self-adjusting grip: friction!

STATIC FRICTION (NO SLIP)Self-adjusting: f_s = F_app up to max f_s(max) = μ_s N (f_s ≤ μ_s N!)

KINETIC & ROLLINGSliding: f_k = μ_k N (constant, μ_k < μ_s).
Rolling: f_r ≪ f_k < f_s(max)!

ANGLE OF REPOSEBlock on incline starts slipping when μ_s = tan θ_r! (Independent of mass & area!)

🧠 Memory Trick: Friction is independent of apparent contact area and depends ONLY on surface roughness (μ) and Normal force (N)!

💡 Simple Kid-Check: Friction does NOT hate motion — it hates sliding/slipping between two surfaces! Without static friction between your slippers and the floor, your feet would spin backward like on wet soap and you could never walk forward!

1
Static Friction (Self-Adjusting): While the object is stationary, static friction matches your push exactly (f_s = F_app) until it hits its ceiling, Limiting Friction f_s(max) = μ_s N.
2
Kinetic & Rolling Friction: Once slipping starts, microscopic bumps don't interlock as deeply, so kinetic friction stays slightly lower (f_k = μ_k N). Wheels roll instead of dragging, cutting rolling friction 100×–1000× smaller!
3
Inclined Plane & Angle of Repose: On a ramp tilted at θ, weight splits into mg cos θ (balanced by Normal N) and mg sin θ down the slope. Slipping begins when mg sin θ = μ_s mg cos θ ⇒ μ_s = tan θ!

f_s ≤ μ_s N | f_k = μ_k N | Ramp Slip Threshold: tan θ_r = μ_s | Spring Force: F = −kx

🧒 Super-Easy Trap-Buster: A 5 kg box sits on a floor with μ_s = 0.4 (N = 50 N ⇒ f_max = 20 N). You push horizontally with only 12 N. What is the friction force?
Since 12 N < 20 N, the box stays still, so friction = 12 N (NOT 20 N!).

THREE super-common exam traps! (1) Static friction is NOT always μ_s N — that is only its MAXIMUM value! While still, f_s = F_applied. (2) Friction depends on Normal force N and surface nature, NOT on the contact surface area (wide vs narrow tyre same mass!). (3) Friction can push objects FORWARD (e.g., walking, car tyres accelerating, or luggage riding inside an accelerating lorry!).

Topic 4.10: Circular Motion on Level & Banked Roads

Drive fast around a flat wet roundabout in Thrissur and your tyres risk skidding outward! Now notice how engineers tilt the tarmac inward like a tilted dish on sharp Munnar and Wayanad mountain curves — that clever road tilt is called banking, letting gravity's Normal reaction help steer you safely around curves even on slippery rain!

LEVEL CIRCULAR ROADFriction alone supplies mv²/R ⇒ v_max = √(μ_s R g)

BANKED ROAD (OPTIMUM)Zero friction needed (no tyre wear!) at v₀ = √(R g tan θ)

BANKED ROAD (MAX SPEED)v_max = √[ R g μ_s + tan θ1 − μ_s tan θ ] (independent of mass m!)

🧠 Memory Trick: Notice μ_s and tan θ ADD on top (μ_s + tan θ) because friction and banking TEAM UP to push the car inward!

💡 Simple Kid-Check: Turning in a circle is NEVER automatic — something real must push your car toward the centre with force mv²/R! On a flat road, only tyre friction does that job. By tilting (banking) the road by angle θ, the big Normal push N tilts inward so N sin θ helps steer you safely!

1
Car on Level Circular Road: Vertical balance gives N = mg; inward static friction supplies centripetal force f_s = mv²/R ≤ μ_s mg ⇒ v_max = √(μ_s R g).
2
Optimum Speed on Banked Road (f = 0): Tilt Normal N at angle θ: N cos θ = mg and N sin θ = mv₀²/R. Dividing both gives tan θ = v₀²/(Rg) ⇒ v₀ = √(R g tan θ) — ZERO sideways friction needed!
3
Maximum Speed on Banked Road (with friction): Both N sin θ and downward-inward friction f cos θ cooperate inward: v_max = √[ Rg(μ_s + tan θ)/(1 − μ_s tan θ) ]. (Below v₀, friction points up the slope to stop inward slipping!)

Level: v_max = √(μ_s R g) | Optimum Banked: v₀ = √(R g tan θ) | Max Banked: v_max = √[ R g μ_s + tan θ1 − μ_s tan θ ]

🧒 Super-Easy Example: A mountain curve of radius 40 m is banked at tan θ = 0.25 (g = 10 m/s²). Find the optimum speed for zero tyre wear!
v₀ = √(R g tan θ) = √(40 × 10 × 0.25) = √100 = 10 m/s (36 km/h) — safe even on zero-friction ice!

Watch THREE high-yield exam traps! (1) Both v_max and v₀ are completely independent of vehicle mass m — a Nano car and a heavy KSRTC bus share the exact same safe speed limit! (2) At optimum speed v₀ = √(Rg tan θ), sideways friction is ZERO (zero tyre wear!). (3) Never mix up the signs in (μ_s + tan θ) / (1 − μ_s tan θ) — plus on top, minus on bottom!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Why does a cricket fielder pull his hands backward while catching a fast-moving ball?

Impulse Δp = F_avg · Δt is fixed to stop the ball. Pulling hands back increases impact duration Δt, which reduces average force F_avg = Δp / Δt on the hands and avoids injury.

Q2 (Kerala HSE Say • 1 Marks): Action and reaction are equal and opposite, yet they do not cancel each other. Why?

Because action and reaction forces always act on two different bodies. Forces cancel only when they act on the same single body.

Q3 (Kerala HSE March • 2 Marks): State Newton's Second Law of Motion and show how F = ma follows from it for a body of constant mass.

Statement: The rate of change of linear momentum of a body is directly proportional to the applied unbalanced external force and takes place in the direction of force.
F ∝ dpdt = d(mv)dt. For constant mass m and SI constant k = 1: F = m (dv/dt) = ma.

Q4 (Kerala HSE Model • 2 Marks): A shell of mass 0.02 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m/s, what is the recoil speed of the gun?

By conservation of momentum (initial p = 0): M V + m v = 0.
V = −(m v)/M = −(0.02 × 80)/100 = −1.6 / 100 = −0.016 m/s (1.6 cm/s opposite to shell).

Q5 (Kerala HSE March • 2 Marks): Define angle of repose and prove that μ_s = tan θ_r.

Definition: The minimum angle of inclination of a rough plane with horizontal at which a body placed on it just begins to slide down.
Proof: At threshold, Normal N = mg cos θ_r and downslope gravity mg sin θ_r balances limiting friction f_max = μ_s N = μ_s mg cos θ_r.
Dividing: μ_s = tan θ_r.

Q6 (Kerala HSE Say • 3 Marks): State and prove the Law of Conservation of Linear Momentum using Newton's Second and Third Laws.

Statement: When net external force on an isolated system is zero, its total linear momentum remains constant.
Proof: Consider two colliding bodies A and B during contact time Δt. By 3rd Law, F_AB = −F_BA.
Multiplying by Δt gives equal and opposite impulses: Δp_A = −Δp_B ⇒ Δ(p_A + p_B) = 0.
Hence p_A + p_B = constant (m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂).

Q7 (NCERT / DHSE Model • 3 Marks): Find the apparent weight (Normal reaction N) of a 70 kg man standing on a weighing scale in a lift when it moves (a) upward with uniform speed 5 m/s, (b) downward with uniform acceleration 5 m/s², and (c) upward with uniform acceleration 5 m/s². (g = 10 m/s²)

(a) Uniform speed ⇒ a = 0: N = mg = 70 × 10 = 700 N (70 kg reading).
(b) Downward accel 5 m/s²: N = m(g − a) = 70(10 − 5) = 350 N (35 kg reading).
(c) Upward accel 5 m/s²: N = m(g + a) = 70(10 + 5) = 1050 N (105 kg reading).

Q8 (Kerala HSE March • 4 Marks): Draw a schematic diagram of forces on a car moving along a circular road banked at angle θ with friction μ_s. Derive the expression for maximum safe speed v_max and hence obtain the optimum speed v₀.

Force Balance on Banked Curve:
1. Vertical (a_y = 0): N cos θ = mg + f sin θ ⇒ N cos θ − μ_s N sin θ = mg.
2. Radial inward (centripetal): N sin θ + f cos θ = mv²/R ⇒ N(sin θ + μ_s cos θ) = mv²/R.
Dividing (2) by (1) and dividing top & bottom by cos θ:
v_max = √[ R g (μ_s + tan θ) / (1 − μ_s tan θ) ].
Setting μ_s = 0 gives optimum zero-wear speed: v₀ = √(R g tan θ).