Work, Energy and Power - Vardaan Master Notes

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References

Compiled from: NCERT Physics Part-1 (Work, Energy and Power) • H.C. Verma — Concepts of Physics Vol-1 (Ch. 8: Work and Energy) • D.C. Pandey — Understanding Physics: Mechanics Part-1 (Work, Energy & Power, Circular Motion) • Pradeep's Fundamental Physics XI • S.L. Arora • Cengage (B.M. Sharma) Mechanics-I • Errorless Physics. Target exams: Boards • JEE Main • JEE Advanced • NEET • NDA.

1. Work Done by a Constant Force

Definition of Work $$W = \vec{F}\cdot\vec{s} = Fs\cos\theta$$

θ = angle between force and displacement. Work is a scalar (dot product!). SI unit: joule (J) = 1 N·m; dimension $[ML^2T^{-2}]$. Other units: 1 erg = 10⁻⁷ J; 1 eV = 1.6 × 10⁻¹⁹ J; 1 kWh = 3.6 × 10⁶ J; 1 calorie = 4.186 J.

1.1 Positive, Negative and Zero Work (Complete Case List)

Sign of W	Condition	Classic Examples (Boards love these)
Positive (0° ≤ θ < 90°)	Force has a component along displacement	Gravity on a falling body; engine force on an accelerating car; stretching force on a spring (by the agent)
Negative (90° < θ ≤ 180°)	Force opposes the displacement	Friction on a sliding block; gravity on a rising ball; braking force; work by a spring on the stretching agent
Zero	θ = 90°, or s = 0, or F = 0	Gravity on a body moved horizontally; centripetal force in circular motion; tension in a whirling string; a coolie walking with a load on his head (idealised); pushing a rigid wall (s = 0)

⭐ EXAM TRAP — WORK CAN BE ZERO EVEN WITH BIG FORCES

Centripetal force does zero work because F ⊥ v at every instant — this is why speed stays constant in uniform circular motion. A man holding a heavy suitcase stationary does no work in physics (s = 0), though he tires physiologically. Both are repeat 1-markers in Boards/NEET.

1.2 Work in Component Form

$W = \vec{F}\cdot\vec{s} = F_xs_x + F_ys_y + F_zs_z$. Work depends on the frame of reference (displacement does!) — work by the same force can differ for observers in relative motion (an HCV subtlety asked in JEE Advanced).

Image Placeholder — Fig. 1

W = Fs cos θ with the three sign cases (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a physics textbook diagram with one main panel and three small panels below it, LANDSCAPE orientation (16:9), pure white background (#ffffff). MAIN PANEL (top): a grey block on a horizontal black floor line displaced to the right; a long green (#43a047) horizontal arrow under the floor labelled "displacement s"; from the block's centre a red (#d32f2f) force arrow F inclined at angle θ ≈ 35° above the horizontal, with the angle arc between F and s labelled "θ"; dashed grey components of F: horizontal "F cos θ (does the work)" and vertical "F sin θ (does no work)"; boxed formula beside it: "W = F s cos θ = F⃗ · s⃗". BOTTOM ROW of three mini panels, each a block with a displacement arrow s (green, rightward) and a force arrow (red): PANEL A titled "Positive work (θ < 90°)" — force arrow pointing right along s, caption "e.g. gravity on a falling body"; PANEL B titled "Negative work (θ > 90°)" — force arrow pointing LEFT against s, caption "e.g. friction on a sliding block"; PANEL C titled "Zero work (θ = 90°)" — force arrow pointing straight UP perpendicular to s, caption "e.g. centripetal force, or gravity in horizontal motion". Style: flat minimalist vector textbook figure, thin black outlines, green displacement and red force arrows exactly as specified, dashed grey component lines, clean sans-serif labels, no gradients, no shadows, pure white background.

One figure = the full sign analysis of Section 1.1.

✍ IN-TEXT PRACTICE 1.1 — Boards / NDA

Q. A body is displaced by $\vec{s} = (3\hat{i} + 4\hat{j})$ m under a force $\vec{F} = (4\hat{i} + \hat{j})$ N. Find the work done and the angle case (positive/negative/zero).

Dot product: $W = (4)(3) + (1)(4) = \mathbf{16 \text{ J}}$

Sign: positive → the angle between F and s is acute.

2. Work Done by a Variable Force

Variable Force $$W = \int_{x_1}^{x_2} F(x)\,dx = \text{area under the } F\text{–}x \text{ graph}$$

Area below the x-axis counts as negative work. This single idea handles springs, gravity over large heights, and every graph question of this chapter.

✍ IN-TEXT PRACTICE 2.1 — JEE Main (Integration)

Q. A force $F = (3x^2 - 2x)$ N acts on a particle along x. Find the work done as it moves from x = 0 to x = 2 m.

Integrate: $W = \displaystyle\int_0^2 (3x^2 - 2x)\,dx = [x^3 - x^2]_0^2 = 8 - 4 = \mathbf{4 \text{ J}}$

3. Kinetic Energy & the Work–Energy Theorem

Kinetic Energy $$K = \tfrac{1}{2}mv^{2} = \frac{p^{2}}{2m}, \qquad p = \sqrt{2mK}$$

Always positive, frame-dependent, scalar. Percentage tricks (NEET/NDA): if p increases by 100% (doubles), K becomes 4× (300% increase); if K increases by 300%, p doubles (100% increase); for small changes, %ΔK ≈ 2 × %Δp.

3.1 Work–Energy Theorem (Both Derivations)

Statement: The net work done by all forces on a body equals the change in its kinetic energy: $W_{net} = K_f - K_i$.

Derivation (constant force): $v^2 = u^2 + 2as \Rightarrow \tfrac12mv^2 - \tfrac12mu^2 = (ma)s = Fs = W$ ✓

Derivation (variable force): $$W = \int F\,dx = \int m\frac{dv}{dt}\,dx = \int mv\,dv = \tfrac12mv^2 - \tfrac12mu^2$$ The theorem holds for all forces — conservative or not (friction included) and in any inertial frame. It is the integral form of Newton's second law.

⭐ HOW TO USE IT (DC PANDEY'S RULE)

Whenever a question gives forces/distances and asks for speed (or vice-versa) without asking for time, use the work–energy theorem instead of F = ma + kinematics — one line instead of three. Example: penetration of bullets into planks, blocks sliding on rough patches, pendulum speeds.

✍ IN-TEXT PRACTICE 3.1 — NEET (Bullet–planks classic)

Q. A bullet loses 1/20 of its velocity passing through one plank. How many such planks will just stop it?

Energy lost per plank (same resistive work W each): after one plank $v = \tfrac{19}{20}u$, so $W = \tfrac12mu^2\left[1 - \left(\tfrac{19}{20}\right)^2\right] = \tfrac12mu^2\cdot\tfrac{39}{400}$

Number of planks: $n = \dfrac{\tfrac12mu^2}{W} = \dfrac{400}{39} \approx 10.3 \Rightarrow \mathbf{11 \text{ planks}}$ (round UP to stop it fully).

✍ IN-TEXT PRACTICE 3.2 — Boards (Rough patch)

Q. A 2 kg block moving at 10 m/s slides onto a rough patch with μ = 0.5 (g = 10 m/s²). How far does it travel before stopping?

Work–energy: $-\mu mg\,s = 0 - \tfrac12mv^2$

$s = \dfrac{v^2}{2\mu g} = \dfrac{100}{2\times0.5\times10} = \mathbf{10 \text{ m}}$

4. Potential Energy & Conservative Forces

4.1 Conservative vs Non-Conservative Forces

Property	Conservative (gravity, spring, electrostatic)	Non-conservative (friction, viscous drag)
Work over a path	Depends only on end points	Depends on the path (longer path → more loss)
Work over a closed loop	Zero	Non-zero (negative)
Potential energy	Can be defined: $\Delta U = -W_{cons}$	Cannot be defined
Mechanical energy	Conserved	Dissipated as heat/sound

4.2 Gravitational PE & the Relation F = −dU/dx

Potential Energy $$U = mgh \ (\text{near Earth's surface, reference at } h = 0)$$ $$F = -\frac{dU}{dx} \qquad\Longleftrightarrow\qquad \Delta U = -\int \vec{F}_{cons}\cdot d\vec{s}$$

Only differences in U are physical — the zero level is our choice. Force points from higher U toward lower U ("downhill" on the U–x curve).

4.3 Equilibrium from the U–x Curve (JEE Favourite)

Type	Condition	U–x shape	Behaviour on displacement
Stable	$\frac{dU}{dx}=0$, $\frac{d^2U}{dx^2} > 0$	Minimum (valley)	Restoring force pushes it back (oscillates)
Unstable	$\frac{dU}{dx}=0$, $\frac{d^2U}{dx^2} < 0$	Maximum (hilltop)	Force pushes it further away
Neutral	$\frac{dU}{dx}=0$, $\frac{d^2U}{dx^2} = 0$	Flat region	Stays wherever displaced

Image Placeholder — Fig. 2

Potential energy curve with all three equilibria and allowed regions (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a single large physics graph, LANDSCAPE orientation (16:9), pure white background (#ffffff). Axes: horizontal "x (position)" with arrowhead, vertical "U (potential energy)" with arrowhead, origin "O". Draw a smooth blue (#1e88e5) curve that: starts high on the left, descends into a clear VALLEY (minimum) at point A, rises to a HILLTOP (maximum) at point B, then descends and flattens into a horizontal PLATEAU at point C on the right. Label the three marked points with dots and callouts: "A — stable equilibrium (dU/dx = 0, d²U/dx² > 0): particle oscillates in the valley"; "B — unstable equilibrium (d²U/dx² < 0): slightest push moves it away"; "C — neutral equilibrium (U constant)". Draw a dashed green (#43a047) horizontal line across the graph labelled "total mechanical energy E"; shade lightly in green (#e8f5e9) the region(s) of the x-axis where the blue U curve lies BELOW the E line, labelled "allowed region: K = E − U ≥ 0"; mark the two intersections of E with the curve with small dots labelled "turning points (K = 0, v = 0)". Add two short red (#d32f2f) force arrows on the curve's slopes pointing DOWNHILL, each labelled "F = −dU/dx (points toward lower U)". Style: flat minimalist vector textbook graph, thin black axes, blue curve, dashed green energy line with light green shading, red slope arrows, clean sans-serif labels, no gradients, no shadows, pure white background.

This one graph carries four JEE question types: equilibria, F from slope, turning points, allowed regions.

✍ IN-TEXT PRACTICE 4.1 — JEE Main (U to F)

Q. The potential energy of a particle is $U = (x^2 - 4x + 3)$ J. Find the equilibrium position and its type, and the force at x = 0.

Force: $F = -\dfrac{dU}{dx} = -(2x - 4) = 4 - 2x$

Equilibrium: F = 0 at x = 2 m; $\dfrac{d^2U}{dx^2} = 2 > 0$ → stable (valley).

At x = 0: $F = \mathbf{+4 \text{ N}}$ (toward the valley, as expected).

5. The Spring Force & Elastic Potential Energy

Hooke's law: $F_{spring} = -kx$ (restoring, opposite to stretch/compression x; k = spring constant, N/m; stiffer spring → larger k).

Derivation of spring PE. Work done by the agent in stretching slowly from 0 to x:

$$W = \int_0^x kx'\,dx' = \tfrac12kx^2 \;\Rightarrow\; U_{spring} = \tfrac12kx^2$$

Work done by the spring in the same stretch = −½kx². From x₁ to x₂: $W_{spring} = \tfrac12k(x_1^2 - x_2^2)$ — depends only on end deformations (conservative). U is the area of the triangle under the F–x line.

Energy is the same for stretch or compression of equal |x| (U ∝ x²).
Series/parallel k (JEE): series $\frac{1}{k_{eq}} = \frac{1}{k_1}+\frac{1}{k_2}$; parallel $k_{eq} = k_1 + k_2$. Cutting a spring into n equal parts makes each part n times stiffer.
Block hits a spring: maximum compression when the block momentarily stops: $\tfrac12mv^2 = \tfrac12kx_{max}^2 \Rightarrow x_{max} = v\sqrt{m/k}$ (smooth floor).

Image Placeholder — Fig. 3

Spring F–x graph, energy triangle, and the block–spring collision (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a two-panel physics textbook diagram side by side, LANDSCAPE orientation (16:9), pure white background (#ffffff). LEFT PANEL titled "Spring force and energy": a first-quadrant graph, x-axis "extension x" with arrowhead, y-axis "applied force F" with arrowhead; a straight blue (#1e88e5) line from the origin with slope labelled "slope = k (Hooke's law F = kx)"; shade the triangle under the line up to extension x₀ in light blue (#e3f2fd) with centred label "area = ½kx₀² = elastic PE stored". Above the graph draw a small horizontal spring sketch (a zigzag coil between a hatched wall on the left and a small block on the right) with a green (#43a047) rightward arrow on the block labelled "stretch x" and a red (#d32f2f) leftward arrow labelled "F_spring = −kx (restoring)". RIGHT PANEL titled "Block hits a spring": a horizontal smooth floor line; on the left a grey block labelled "m" sliding right with a green velocity arrow "v"; on the right a zigzag spring attached to a hatched wall; below, a second snapshot of the same scene showing the spring COMPRESSED with the block touching it at rest, compression marked by a dimension arrow labelled "x_max"; boxed energy equation between the snapshots: "½mv² = ½k x_max² ⟹ x_max = v√(m/k)". Style: flat minimalist vector textbook figure, thin black outlines, zigzag spring coils, hatched walls, blue/green/red elements exactly as specified, clean sans-serif labels, no gradients, no shadows, pure white background.

The shaded triangle IS the formula ½kx² — read energy straight off the graph.

✍ IN-TEXT PRACTICE 5.1 — NEET (Two stretches)

Q. A spring (k = 200 N/m) is stretched by 5 cm. Find the work needed to stretch it by a further 5 cm.

Not equal to the first work! $W = \tfrac12k(x_2^2 - x_1^2) = \tfrac12(200)(0.10^2 - 0.05^2)$

$= 100(0.01 - 0.0025) = \mathbf{0.75 \text{ J}}$ — three times the first-5-cm work (0.25 J). U ∝ x² is the trap.

6. Conservation of Mechanical Energy

Statement: If only conservative forces do work, the total mechanical energy E = K + U of a system remains constant.

Proof for a freely falling body (dropped from height H; take ground as U = 0):

At top A (h = H, v = 0): $E = 0 + mgH = mgH$
At B after falling x: $v^2 = 2gx$ → $E = \tfrac12m(2gx) + mg(H-x) = mgH$ ✓
At ground C: $v^2 = 2gH$ → $E = \tfrac12m(2gH) + 0 = mgH$ ✓

E is the same everywhere; K and U merely interconvert. With friction: $\Delta E = -W_{friction}$ (energy lost = heat) — use the modified theorem $K_f + U_f = K_i + U_i - |W_{fric}|$.

Pendulum released from horizontal: speed at the bottom $v = \sqrt{2gL}$.
Sliding down any smooth surface (incline, curve, chute): $v = \sqrt{2gh}$, independent of the path shape — energy methods beat FBDs on curved tracks.

Image Placeholder — Fig. 4

Free-fall energy conservation with K/U bar charts at three levels (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a professional physics textbook diagram, LANDSCAPE orientation (16:9), pure white background (#ffffff). LEFT HALF: a vertical dashed drop line from a point A at the top to the ground line at the bottom; a small blue ball shown at three positions: A (top, labelled "A: v = 0, h = H"), B (middle, labelled "B: h = H/2") and C (just above the ground, labelled "C: h = 0, v = √(2gH)"); a curly bracket on the left spanning the full drop labelled "H"; a red (#d32f2f) downward arrow labelled "g". RIGHT HALF: three vertical stacked-bar charts side by side, one for each point, each bar of the SAME total height labelled "E = mgH (constant)": bar A entirely orange (#fbc02d) labelled "U = mgH, K = 0"; bar B half orange (top, "U = mgH/2") and half green (#43a047) (bottom, "K = mgH/2"); bar C entirely green labelled "K = mgH, U = 0". Above the bars a heading "K + U = constant"; below them the line "PE converts to KE; total stays mgH". Style: flat minimalist vector textbook figure, thin black outlines and axes, orange = potential energy, green = kinetic energy consistently, equal-height bars, clean sans-serif labels, no gradients, no shadows, pure white background.

Equal-height bars are the visual proof of conservation — reproduce them in Board answers for full marks.

✍ IN-TEXT PRACTICE 6.1 — Boards / NEET (Height–speed exchange)

Q. A ball dropped from 80 m: at what height is its KE equal to its PE? What is its speed there? (g = 10 m/s²)

K = U ⟹ each = E/2: $mgh = \tfrac12mgH \Rightarrow h = H/2 = \mathbf{40 \text{ m}}$

Speed: $\tfrac12mv^2 = mg(40) \Rightarrow v = \sqrt{800} = \mathbf{20\sqrt2 \approx 28.3 \text{ m/s}}$

7. Motion in a Vertical Circle (Energy + Circular Dynamics)

Setup: a bob on a string of length L whirled in a vertical circle. At the TOP, string tension and gravity both point toward the centre: $T_{top} + mg = \dfrac{mv_{top}^2}{L}$. The critical condition is T_top = 0:

$$v_{top(min)} = \sqrt{gL}$$

Energy conservation top → bottom (height difference 2L):

$$\tfrac12mv_{bottom}^2 = \tfrac12mv_{top}^2 + mg(2L)$$

Vertical Circle — Critical Speeds $$v_{top} \ge \sqrt{gL} \qquad v_{bottom} \ge \sqrt{5gL} \qquad v_{side} \ge \sqrt{3gL}$$ $$T_{bottom} - T_{top} = 6mg \ (\text{always, at any speed})$$

At the bottom: $T_{bottom} = \dfrac{mv_b^2}{L} + mg$ (minimum 6mg for a full loop). If $v_{bottom} < \sqrt{2gL}$: oscillates like a pendulum (never reaches horizontal); between $\sqrt{2gL}$ and $\sqrt{5gL}$: leaves the circle somewhere above the horizontal and becomes a projectile. For a bead on a rigid rod/track (can push), the top condition relaxes to v_top ≥ 0 ⟹ v_bottom ≥ $\sqrt{4gL} = 2\sqrt{gL}$.

Image Placeholder — Fig. 5

Vertical circle with forces at top, bottom and side, and the three speed regimes (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a professional physics textbook diagram, LANDSCAPE orientation (16:9), pure white background (#ffffff). CENTRE-LEFT: a large thin black circle with centre C; a string drawn as a thin black line with rope-texture ticks from C to a small blue bob shown at THREE positions: TOP (bob at 12 o'clock: green (#43a047) arrow from bob toward C labelled "T_top" and red (#d32f2f) downward arrow "mg", caption beside it "T + mg = mv²/L; critical: T = 0 ⟹ v_top = √(gL)"); BOTTOM (bob at 6 o'clock: green arrow pointing UP toward C labelled "T_bottom", red arrow down "mg", caption "T − mg = mv²/L ⟹ v_bottom(min) = √(5gL), T_min = 6mg"); SIDE (bob at 3 o'clock: green arrow pointing horizontally toward C labelled "T = mv²/L", red arrow down "mg tangential here", caption "v_side(min) = √(3gL)"). Mark the vertical diameter with a dimension arrow labelled "2L". RIGHT PANEL: a small table drawn as three text rows with tick/cross icons: "v_b < √(2gL) → oscillates (pendulum)"; "√(2gL) < v_b < √(5gL) → leaves circle, becomes projectile"; "v_b ≥ √(5gL) → completes the loop ✓". Style: flat minimalist vector textbook figure, thin black circle, rope-textured string, green tension and red weight arrows exactly as specified, clean sans-serif labels, no gradients, no shadows, pure white background.

Three positions + three regimes = the complete vertical circle story of Section 7.

✍ IN-TEXT PRACTICE 7.1 — NEET / JEE (Vertical circle numbers)

Q. A 0.5 kg stone on a 2 m string must just complete a vertical circle (g = 10 m/s²). Find the minimum speeds at the top and bottom and the tension at the bottom in that case.

Top: $v = \sqrt{gL} = \sqrt{20} \approx \mathbf{4.47 \text{ m/s}}$

Bottom: $v = \sqrt{5gL} = \sqrt{100} = \mathbf{10 \text{ m/s}}$

Tension: $T = 6mg = 6\times0.5\times10 = \mathbf{30 \text{ N}}$

8. Power

Power $$P_{avg} = \frac{W}{t} \qquad P_{inst} = \frac{dW}{dt} = \vec{F}\cdot\vec{v} = Fv\cos\theta$$

Scalar. SI unit watt (W) = J/s; dimension $[ML^2T^{-3}]$. 1 hp = 746 W. kWh is a unit of ENERGY, not power: 1 kWh = 3.6 × 10⁶ J (the "unit" on an electricity bill).

Pump lifting water: power to lift mass m per second to height h and eject at speed v: $P = \dfrac{mgh + \tfrac12mv^2}{t}$; with efficiency η, input power = output/η.
Vehicle at constant speed: engine power P = (resistive force) × v; maximum speed when the whole power fights resistance: $v_{max} = P/F_{res}$.
Constant-power machine (JEE): $P = Fv = m v\dfrac{dv}{dt}$ ⟹ $v = \sqrt{2Pt/m}$, and displacement $x \propto t^{3/2}$ — a standard derivation.

✍ IN-TEXT PRACTICE 8.1 — Boards / NDA (Pump with efficiency)

Q. A pump lifts 600 kg of water per minute from a 10 m deep well and ejects it at 5 m/s. Find the power of the pump if its efficiency is 80%. (g = 10 m/s²)

Useful work per second: mass/s = 10 kg/s → $P_{out} = 10(10)(10) + \tfrac12(10)(25) = 1000 + 125 = 1125$ W

Input power: $P_{in} = \dfrac{1125}{0.8} = \mathbf{1406\ \text{W} \approx 1.4 \text{ kW}}$

✍ IN-TEXT PRACTICE 8.2 — NEET (P = Fv)

Q. A car of mass 1000 kg moves up at a constant 15 m/s against total resistance 500 N on a level road. Find the engine power.

Constant speed → F_engine = F_res: $P = Fv = 500\times15 = \mathbf{7500 \text{ W} = 7.5 \text{ kW}}$

9. Collisions — Complete Theory

In every collision (no external force during the brief impact) momentum is conserved. Kinetic energy may or may not be conserved — that defines the type.

Type	Momentum	Kinetic Energy	Coefficient of restitution e
Elastic	Conserved	Conserved	e = 1
Inelastic	Conserved	Partly lost (heat, sound, deformation)	0 < e < 1
Perfectly inelastic	Conserved	Maximum possible loss; bodies stick together	e = 0

Coefficient of Restitution $$e = \frac{\text{speed of separation}}{\text{speed of approach}} = \frac{v_2 - v_1}{u_1 - u_2}$$

Ball dropped from height h rebounds to $h' = e^2h$; after n bounces $h_n = e^{2n}h$ (a NEET regular).

9.1 Elastic Collision in 1-D (Full Derivation)

Given: m₁ at u₁ hits m₂ at u₂ (u₁ > u₂), head-on, elastic.

Conservation equations:

$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$ $$\tfrac12m_1u_1^2 + \tfrac12m_2u_2^2 = \tfrac12m_1v_1^2 + \tfrac12m_2v_2^2$$

Rearrange each and divide (the classic trick): $u_1 + v_1 = u_2 + v_2$, i.e. relative velocity reverses: $u_1 - u_2 = -(v_1 - v_2)$ (this IS e = 1). Solving the linear pair:

$$v_1 = \frac{(m_1-m_2)u_1 + 2m_2u_2}{m_1+m_2} \qquad v_2 = \frac{(m_2-m_1)u_2 + 2m_1u_1}{m_1+m_2}$$

9.2 Special Cases — The MCQ Goldmine

Case (elastic, u₂ = 0 unless stated)	Result
Equal masses (m₁ = m₂)	They exchange velocities: v₁ = u₂, v₂ = u₁ (Newton's cradle!)
Heavy hits light (m₁ ≫ m₂)	v₁ ≈ u₁ (barely slows); v₂ ≈ 2u₁ (light one shoots off at double speed)
Light hits heavy (m₁ ≪ m₂)	v₁ ≈ −u₁ (bounces back with same speed); v₂ ≈ 0 (ball off a wall)
Energy transfer	Fraction transferred to m₂: $\dfrac{4m_1m_2}{(m_1+m_2)^2}$ — maximum (100%) when m₁ = m₂ (why moderators in reactors use light nuclei!)

9.3 Perfectly Inelastic Collision & a Word on 2-D

Sticking together: common velocity $v = \dfrac{m_1u_1 + m_2u_2}{m_1+m_2}$.

KE loss (u₂ = 0): $\Delta K = \dfrac{m_1m_2}{2(m_1+m_2)}u_1^2 = \dfrac{1}{2}\mu_{red}\,u_{rel}^2$ where $\mu_{red} = \dfrac{m_1m_2}{m_1+m_2}$ (reduced mass form — elegant and exam-usable).

Ballistic pendulum (bullet embeds in a hanging block): first momentum conservation for the impact, then energy conservation for the swing: $h = \dfrac{v_{common}^2}{2g}$. Never apply energy conservation ACROSS the impact! In 2-D elastic collision of equal masses (one at rest), the two final velocities are perpendicular (θ₁ + θ₂ = 90°) — billiards physics, a JEE favourite.

Image Placeholder — Fig. 6

Collision gallery: elastic special cases and the perfectly inelastic stick-together (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a four-row physics textbook diagram stacked vertically, LANDSCAPE orientation (16:9), pure white background (#ffffff), thin grey dividers between rows; each row has a "BEFORE" scene on the left, a large grey arrow "⟶" in the middle, and an "AFTER" scene on the right, all on a common horizontal floor line. ROW 1 titled "Equal masses (elastic): velocities exchange": BEFORE — blue ball labelled "m, u" moving right (green velocity arrow), red ball labelled "m, at rest"; AFTER — blue ball at rest, red ball moving right with green arrow labelled "u". ROW 2 titled "Heavy hits light (elastic)": BEFORE — big blue ball "M, u" moving right, small red ball "m ≪ M, rest"; AFTER — big ball still moving right "≈ u", small ball shooting right with a LONGER green arrow "≈ 2u". ROW 3 titled "Light hits heavy (elastic)": BEFORE — small blue ball "m, u" moving right toward a big red ball "M ≫ m, rest"; AFTER — small ball moving LEFT with green arrow "≈ −u (rebounds)", big ball "≈ 0 (almost still)". ROW 4 titled "Perfectly inelastic (e = 0): stick together": BEFORE — blue ball "m₁, u₁" moving right, red ball "m₂, at rest"; AFTER — the two balls drawn TOUCHING as one combined lump moving right with green arrow labelled "v = m₁u₁/(m₁+m₂)", caption "KE lost = m₁m₂u₁² / 2(m₁+m₂)". Style: flat minimalist vector textbook figure, thin black floor lines, blue and red balls sized by mass, green velocity arrows with lengths proportional to speeds, clean sans-serif labels, no gradients, no shadows, pure white background.

Arrow lengths encode the answers — the whole of Section 9.2 in one glance.

✍ IN-TEXT PRACTICE 9.1 — NEET (Perfectly inelastic + KE loss)

Q. A 2 kg body moving at 6 m/s collides with a stationary 4 kg body and sticks to it. Find the common velocity and the kinetic energy lost.

Momentum: $v = \dfrac{2\times6}{6} = \mathbf{2 \text{ m/s}}$

KE loss: $K_i - K_f = \tfrac12(2)(36) - \tfrac12(6)(4) = 36 - 12 = \mathbf{24 \text{ J}}$ (or directly $\frac{m_1m_2}{2(m_1+m_2)}u^2 = \frac{8}{12}\times36 = 24$ ✓)

✍ IN-TEXT PRACTICE 9.2 — Boards / JEE (Ballistic pendulum)

Q. A 10 g bullet at 400 m/s embeds in a 1.99 kg block hanging from long strings. To what height does the block rise? (g = 10 m/s²)

Impact (momentum only!): $v = \dfrac{0.01\times400}{2} = 2$ m/s.

Swing (energy): $h = \dfrac{v^2}{2g} = \dfrac{4}{20} = \mathbf{0.2 \text{ m}}$

Trap: applying energy conservation across the embedding step gives a wrong (huge) height — most KE is lost as heat there.

10. Solved Examples — Every Question Type in the Chapter

One worked model of each distinct question type asked from this chapter, tagged by exam. Attempt each yourself first, then check.

✍ TYPE 1 — Boards · State, Prove & Derive

Q. State and prove the work–energy theorem for a variable force (5 marks). / Prove conservation of mechanical energy for a freely falling body (5 marks). / Derive the velocities after a 1-D elastic collision and discuss the equal-mass case (5 marks).

Where to answer from: Section 3.1 (∫mv dv derivation), Section 6 (three-point A/B/C proof, reproduce Fig. 4's bar chart), Section 9.1 (divide-the-equations trick, then Section 9.2 cases).

✍ TYPE 2 — NEET / NDA · p–K Percentage Games

Q. The momentum of a body increases by 50%. By what percentage does its kinetic energy increase?

K ∝ p²: $\dfrac{K_2}{K_1} = (1.5)^2 = 2.25$

Increase = 125%. (Reverse: K up 44% → p up 20%. Always square / square-root the ratio.)

✍ TYPE 3 — JEE Main · Work from an F–x Graph

Q. A force on a particle varies as: F = 10 N (constant) from x = 0 to 2 m, then decreases linearly to 0 at x = 4 m, then is −5 N (constant) from 4 to 6 m. Find the total work.

Areas: rectangle 10×2 = 20 J; triangle ½×2×10 = 10 J; negative rectangle −5×2 = −10 J.

Total: $20 + 10 - 10 = \mathbf{20 \text{ J}}$ — sign the areas, then add.

✍ TYPE 4 — JEE · Work–Energy on an Incline with Friction

Q. A 2 kg block slides from rest down a rough incline of height 5 m and reaches the bottom at 8 m/s (g = 10 m/s²). Find the energy lost to friction.

Energy audit: $|W_{fric}| = mgh - \tfrac12mv^2 = 100 - 64 = \mathbf{36 \text{ J}}$

No geometry of the slope needed — the power of energy methods.

✍ TYPE 5 — NEET · Chain / Hanging-Part Problems

Q. A uniform chain of mass M and length L lies on a table with one-third of its length hanging over the edge. Find the work required to pull the hanging part back onto the table.

Hanging part: mass M/3, its centre of mass hangs L/6 below the table top.

Work = raise that CM to table level: $W = \dfrac{M}{3}g\cdot\dfrac{L}{6} = \mathbf{\dfrac{MgL}{18}}$

General: fraction 1/n hanging → W = MgL/2n².

✍ TYPE 6 — JEE Main · Block–Spring on a Rough Floor

Q. A 1 kg block at 4 m/s hits a spring (k = 100 N/m) on a floor with μ = 0.2 (g = 10). Find the maximum compression. (Quadratic expected!)

Energy with friction: $\tfrac12mv^2 = \tfrac12kx^2 + \mu mg x$

$8 = 50x^2 + 2x \Rightarrow 50x^2 + 2x - 8 = 0 \Rightarrow 25x^2 + x - 4 = 0$

x: $x = \dfrac{-1 + \sqrt{1+400}}{50} = \dfrac{-1+20.02}{50} \approx \mathbf{0.38 \text{ m}}$

✍ TYPE 7 — NEET · Rebounding Ball (Restitution)

Q. A ball dropped from 20 m rebounds to 5 m. Find e and the height after the second bounce.

e: $h' = e^2h \Rightarrow e^2 = 5/20 = 0.25 \Rightarrow e = \mathbf{0.5}$

Second bounce: $h_2 = e^4h = (0.0625)(20) = \mathbf{1.25 \text{ m}}$

✍ TYPE 8 — JEE Main · Constant Power Machine

Q. A 2 kg body starts from rest and is driven by a machine delivering a constant power of 16 W. Find its speed and displacement after 4 s.

Energy = Pt: $\tfrac12mv^2 = Pt \Rightarrow v = \sqrt{2Pt/m} = \sqrt{64} = \mathbf{8 \text{ m/s}}$

Displacement: $x = \displaystyle\int_0^4 \sqrt{\tfrac{2P}{m}}\, t^{1/2}dt = \sqrt{16}\cdot\tfrac{2}{3}(4)^{3/2} = 4\cdot\tfrac{2}{3}\cdot8 = \mathbf{21.3 \text{ m}}$ ($x \propto t^{3/2}$)

✍ TYPE 9 — JEE · Energy Transfer Fraction in Elastic Collision

Q. A neutron (mass m) collides elastically head-on with a stationary carbon nucleus (mass 12m). What fraction of its kinetic energy does it transfer?

Fraction: $\dfrac{4m_1m_2}{(m_1+m_2)^2} = \dfrac{4\times12}{169} = \mathbf{\dfrac{48}{169} \approx 28\%}$

Maximum transfer (100%) needs equal masses — that's why hydrogen-rich moderators slow neutrons best.

✍ TYPE 10 — JEE Advanced Level · Leaving a Smooth Sphere

Q. A particle slides from rest from the top of a smooth hemisphere of radius R. At what height (from the ground) does it leave the surface?

Leaving condition (N = 0): $mg\cos\theta = \dfrac{mv^2}{R}$ where θ is measured from the vertical.

Energy: $v^2 = 2gR(1-\cos\theta)$. Combine: $\cos\theta = 2(1-\cos\theta) \Rightarrow \cos\theta = \dfrac{2}{3}$

Height: $h = R\cos\theta = \mathbf{\dfrac{2R}{3}}$ from the ground (i.e., it falls R/3 along the sphere before flying off).

11. Common Mistakes & Misconceptions — Final Checklist

Night Before Exam

Read this the night before the exam:

Work is a scalar but has a SIGN — friction usually does negative work; centripetal force always does zero work.
Holding a weight stationary or carrying it horizontally = zero work by the person (physics definition).
The work–energy theorem uses the net work of ALL forces (including friction) — it is not restricted to conservative forces.
Mechanical energy conservation IS restricted to conservative forces; with friction, write $\Delta E = -|W_{fric}|$.
Spring energy ∝ x²: the second centimetre of stretch costs more work than the first.
K ∝ p²: double the momentum → four times the KE. Square/root the ratios, never scale linearly.
kWh is energy, not power; watt is power, not energy.
In ANY collision momentum is conserved; KE is conserved only in elastic ones. Never conserve KE across an embedding/sticking step (ballistic pendulum!).
Equal-mass elastic collision (target at rest): velocities exchange; energy transfer is maximum.
Vertical circle: minimum condition comes from T = 0 at the TOP, not v = 0; and $T_{bottom} - T_{top} = 6mg$ always.
For a rod (can push) the top-speed condition relaxes to v = 0 → v_bottom = 2√(gL), not √(5gL).
Potential energy zero level is arbitrary — only ΔU matters; F = −dU/dx (mind the minus sign).
Stable equilibrium = U minimum; particles oscillate about minima, never maxima.
Rebound height h' = e²h (not eh); after n bounces e²ⁿh.

12. Rapid Revision — One-Page Formula Sheet

Concept	Formula	Condition / Note
Work (constant F)	$W = Fs\cos\theta = \vec{F}\cdot\vec{s}$	Zero at θ = 90°; 1 kWh = 3.6×10⁶ J
Work (variable F)	$W = \int F\,dx$	= signed area under F–x graph
Kinetic energy	$K = \tfrac12mv^2 = p^2/2m$	$p = \sqrt{2mK}$; %ΔK ≈ 2%Δp (small)
Work–energy theorem	$W_{net} = \Delta K$	All forces, any frame (inertial)
PE ↔ force	$F = -dU/dx$	Stable eq.: U min; unstable: U max
Gravitational PE	$U = mgh$	Near surface; zero level arbitrary
Spring PE	$U = \tfrac12kx^2$	$W_{x_1\to x_2} = \tfrac12k(x_2^2-x_1^2)$
Springs combined	Series: $\tfrac1{k} = \tfrac1{k_1}+\tfrac1{k_2}$	Parallel: $k = k_1+k_2$; n cuts → each nk
Energy conservation	$K + U = $ constant	Conservative forces only; else ΔE = −\|W_fric\|
Vertical circle	$v_{top}=\sqrt{gL},\ v_{bot}=\sqrt{5gL}$	$T_{bot}-T_{top} = 6mg$; rod: $v_{bot} = 2\sqrt{gL}$
Leaving a smooth sphere	$\cos\theta = 2/3$	Leaves at height 2R/3 from ground
Power	$P = W/t = \vec{F}\cdot\vec{v}$	1 hp = 746 W; pump: P = (mgh+½mv²)/t ÷ η
Constant power motion	$v = \sqrt{2Pt/m}$	$x \propto t^{3/2}$
Elastic collision (1-D)	$v_1 = \frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2}$	Swap indices for v₂; equal masses exchange v
Energy transfer fraction	$\frac{4m_1m_2}{(m_1+m_2)^2}$	Max (=1) when m₁ = m₂
Perfectly inelastic	$v = \frac{m_1u_1+m_2u_2}{m_1+m_2}$	KE lost $= \frac{m_1m_2}{2(m_1+m_2)}u_{rel}^2$
Restitution	$e = \frac{v_2-v_1}{u_1-u_2}$	Bounce: $h' = e^2h$; n-th: $e^{2n}h$
Chain pull-up	$W = \frac{MgL}{2n^2}$	Fraction 1/n hanging over the edge

Study Plan

How to use these notes: Day 1: Sections 1–3 (work, W–E theorem) + Fig 1; redo both theorem derivations. Day 2: Sections 4–6 (PE, springs, conservation) + Figs 2–4; sketch the U–x curve analysis from memory. Day 3: Sections 7–8 (vertical circle, power) + Fig 5; derive √(5gL) unaided. Day 4: Section 9 (collisions) + Fig 6, then all ten Types of Section 10 without looking, followed by the mistakes checklist. Finish every session by writing the formula sheet from memory.