NCERT Class 11 Physics • Chapter 5 Reprint 2026-27

Work, Energy and Power

Complete chapter notes, Scalar Product (Dot Product), Work-Energy Theorem for constant & variable forces, kinetic & potential energy, conservative forces, spring potential energy & Hooke's Law, vertical circular motion, power & horsepower, elastic & inelastic collisions in 1D/2D, stepwise solved examples 5.1–5.12, concept check quiz with per-option explanations, textbook exercises 5.1–5.23, tips & mnemonics, and 3-tier practice tests.
01 / Chapter Theory

Work, Energy and Power

Official NCERT syllabus: Vector Dot Product, Work definition & sign conventions, Work-Energy Theorem, Kinetic Energy, Variable Force integration, Gravitational & Elastic Potential Energy, Conservation of Mechanical Energy, Motion in a vertical circle, Power & Kilowatt-hour, and Collisions in 1D and 2D.

11 SECTIONS
Chapter Navigation
5.1Scalar (Dot) Product 5.2-5.3Work & Work-Energy Theorem 5.4Kinetic Energy 5.5-5.6Variable Force & WE Theorem 5.7Concept of Potential Energy 5.8Conservation of Mech Energy 5.9Spring Potential Energy 5.10Power & Kilowatt-Hour 5.11Collisions (1D & 2D)
5.1
NCERT Section

The Scalar Product (Dot Product)

Mathematical Prerequisite

5.1.1 Definition & Geometric Meaning

The scalar (or dot) product of two vectors $\mathbf{A}$ and $\mathbf{B}$ is a scalar defined as:

Scalar (Dot) Product Formula $$\mathbf{A} \cdot \mathbf{B} = A B \cos\theta$$ $$\mathbf{A} \cdot \mathbf{B} = A (B \cos\theta) = B (A \cos\theta)$$

Geometrical meaning: Product of magnitude of $\mathbf{A}$ and the projection of $\mathbf{B}$ onto $\mathbf{A}$ (or magnitude of $\mathbf{B}$ and projection of $\mathbf{A}$ onto $\mathbf{B}$).

Component Form & Unit Vector Rules $$\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z$$ $$\hat{\mathbf{i}}\cdot\hat{\mathbf{i}} = \hat{\mathbf{j}}\cdot\hat{\mathbf{j}} = \hat{\mathbf{k}}\cdot\hat{\mathbf{k}} = 1, \qquad \hat{\mathbf{i}}\cdot\hat{\mathbf{j}} = \hat{\mathbf{j}}\cdot\hat{\mathbf{k}} = \hat{\mathbf{k}}\cdot\hat{\mathbf{i}} = 0$$ $$\cos\theta = \frac{\mathbf{A}\cdot\mathbf{B}}{|\mathbf{A}||\mathbf{B}|} = \frac{A_x B_x + A_y B_y + A_z B_z}{\sqrt{A_x^2+A_y^2+A_z^2}\sqrt{B_x^2+B_y^2+B_z^2}}$$
θ A B B cos θ (Projection on A) (a) A • B = A (B cos θ) A B (b) A • B = B (A cos θ)
Figure 5.1: (a) $B \cos\theta$ is the projection of $\mathbf{B}$ onto $\mathbf{A}$. (b) $A \cos\theta$ is the projection of $\mathbf{A}$ onto $\mathbf{B}$.
Interactive 3D Work Done & Vector Dot Product ($W = \mathbf{F} \cdot \mathbf{d}$)
Dot Product Lab
Drag to rotate vector plane
Work Calculation & Sign Nature
W = F d cosθ = (10N) × (6.0m) × cos(45°) = 42.43 J • Positive Work
Key Physics Principles: Work is the scalar projection of applied force $\mathbf{F}$ (rose arrow) along the displacement $\mathbf{d}$ (emerald arrow). When $\theta < 90^\circ$, work is positive; when $\theta = 90^\circ$ (perpendicular force), $W = 0$; when $\theta > 90^\circ$ (e.g. friction or retarding force), work is negative.

Example 5.1 • Angle and Projection of Vectors

NCERT Solved

Find the angle between force $\mathbf{F} = (3\hat{\mathbf{i}} + 4\hat{\mathbf{j}} - 5\hat{\mathbf{k}})$ unit and displacement $\mathbf{d} = (5\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 3\hat{\mathbf{k}})$ unit. Also find the projection of $\mathbf{F}$ on $\mathbf{d}$.

Angle $\theta = \cos^{-1}(0.32) \approx 71.3^\circ$ • Projection of $\mathbf{F}$ on $\mathbf{d} \approx 2.26\text{ units}$
Step 1: Compute scalar dot product: $$\mathbf{F}\cdot\mathbf{d} = F_x d_x + F_y d_y + F_z d_z$$ $$\mathbf{F}\cdot\mathbf{d} = 3(5) + 4(4) + (-5)(3) = 15 + 16 - 15 = 16\text{ units}$$
Step 2: Compute vector magnitudes: $$|\mathbf{F}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$$ $$|\mathbf{d}| = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{25 + 16 + 9} = \sqrt{50} = 5\sqrt{2}$$
Step 3: Calculate angle and projection: $$\cos\theta = \frac{\mathbf{F}\cdot\mathbf{d}}{|\mathbf{F}||\mathbf{d}|} = \frac{16}{50} = 0.32 \implies \theta = \cos^{-1}(0.32) \approx 71.3^\circ$$ $$\text{Projection of }\mathbf{F}\text{ on }\mathbf{d} = |\mathbf{F}|\cos\theta = \frac{\mathbf{F}\cdot\mathbf{d}}{|\mathbf{d}|} = \frac{16}{\sqrt{50}} \approx 2.26\text{ units}$$
5.2 - 5.3
NCERT Sections

Work & The Work-Energy Theorem

Fundamental Work Definition

5.3.1 Work Done by a Constant Force

Work Equation $$W = \mathbf{F} \cdot \mathbf{d} = F d \cos\theta$$ $$\text{SI Unit: } 1\text{ Joule (J)} = 1\text{ N}\cdot\text{m} = 1\text{ kg}\cdot\text{m}^2\cdot\text{s}^{-2}, \qquad \text{Dimensions: } [\text{M L}^2 \text{T}^{-2}]$$
  • Positive Work ($0 \le \theta < 90^\circ$): Force has a component along displacement ($\cos\theta > 0$).
  • Zero Work ($\theta = 90^\circ$ or $d = 0$): Force is perpendicular to displacement (e.g. gravitational force on Moon in circular orbit, normal reaction on horizontal floor).
  • Negative Work ($90^\circ < \theta \le 180^\circ$): Force opposes displacement ($\cos\theta < 0$, e.g. friction $W = -f d$).

5.2.1 The Work-Energy Theorem

Work-Energy Theorem $$K_\text{f} - K_\text{i} = W_\text{net}$$ $$\Delta K = \frac{1}{2}m v_\text{f}^2 - \frac{1}{2}m v_\text{i}^2 = \int \mathbf{F}_\text{net}\cdot d\mathbf{r}$$

Example 5.2 • Raindrop Falling with Resistive Force

NCERT Solved

A raindrop of mass $1.00\text{ g}$ falls from height $h = 1.00\text{ km}$ hitting the ground with speed $v = 50.0\text{ m s}^{-1}$. (a) What is the work done by gravitational force? (b) What is the work done by the unknown resistive force? (Take $g = 10\text{ m s}^{-2}$).

(a) $W_\text{g} = +10.0\text{ J}$ • (b) $W_\text{r} = -8.75\text{ J}$
(a) Work done by gravity: $$W_\text{g} = mgh = (10^{-3}\text{ kg}) \times (10\text{ m s}^{-2}) \times (1000\text{ m}) = +10.0\text{ J}$$
(b) Kinetic energy change and resistive work: $$\Delta K = K_\text{f} - K_\text{i} = \frac{1}{2}m v^2 - 0 = \frac{1}{2} \times 10^{-3} \times (50)^2 = 1.25\text{ J}$$ By Work-Energy Theorem: $$\Delta K = W_\text{g} + W_\text{r} \implies W_\text{r} = \Delta K - W_\text{g}$$ $$W_\text{r} = 1.25\text{ J} - 10.0\text{ J} = -8.75\text{ J}$$

Example 5.3 • Skidding Cyclist Work Analysis

NCERT Solved

A cyclist comes to a skidding stop in $10\text{ m}$. The stopping force on the cycle due to road is $200\text{ N}$. (a) How much work does the road do on the cycle? (b) How much work does the cycle do on the road?

(a) $W_\text{road} = -2000\text{ J}$ • (b) $W_\text{cycle} = 0\text{ J}$
(a) Work by road on cycle: $$W = F d \cos\theta = 200 \times 10 \times \cos(180^\circ) = -2000\text{ J}$$
(b) Work by cycle on road: By Newton's Third Law, force on road is $200\text{ N}$ forward, but the displacement of the road is strictly zero ($d = 0$). $$W = F \times 0 = 0\text{ J}$$
5.4 - 5.6
NCERT Sections

Kinetic Energy & Variable Force Integration

Calculus Formulation

5.5.1 Work Done by Variable Force & Proof of WE Theorem

Variable Force Work Integral $$W = \int_{x_\text{i}}^{x_\text{f}} F(x)\, dx = \text{Area under the } F(x) \text{ vs } x \text{ curve}$$
Step-by-Step Calculus Proof of WE Theorem $$\frac{dK}{dt} = \frac{d}{dt}\left(\frac{1}{2}m v^2\right) = m v \frac{dv}{dt} = m a v = F v = F \frac{dx}{dt}$$ $$dK = F\, dx$$ $$\int_{K_\text{i}}^{K_\text{f}} dK = \int_{x_\text{i}}^{x_\text{f}} F(x)\, dx \implies K_\text{f} - K_\text{i} = W$$
x F(x) W = ∫ F(x) dx (Area under curve) x_i x_f
Figure 5.3: Work done by a variable force $F(x)$ equals the area under the curve between $x_\text{i}$ and $x_\text{f}$.

Example 5.4 • Bullet Fired Through Soft Plywood

NCERT Solved

A police officer fires a bullet of mass $50.0\text{ g}$ with speed $200\text{ m s}^{-1}$ into plywood of thickness $2.00\text{ cm}$. The bullet emerges with only $10\%$ of initial KE. Find emergent speed.

Emergent speed $v_\text{f} = 63.2\text{ m s}^{-1}$ (Speed reduced by ~68%)
Step 1: Calculate initial KE: $$K_\text{i} = \frac{1}{2} m v_\text{i}^2 = \frac{1}{2} (0.050\text{ kg}) (200\text{ m s}^{-1})^2 = 1000\text{ J}$$
Step 2: Calculate emergent speed: $$K_\text{f} = 0.10 \times 1000\text{ J} = 100\text{ J}$$ $$\frac{1}{2} m v_\text{f}^2 = 100 \implies v_\text{f} = \sqrt{\frac{2 \times 100}{0.050}} = \sqrt{4000} \approx 63.2\text{ m s}^{-1}$$

Example 5.5 • Pushing Trunk on Platform (Varying Force)

NCERT Solved

A woman pushes a trunk $20\text{ m}$: $100\text{ N}$ for the first $10\text{ m}$, then reducing linearly to $50\text{ N}$ at $20\text{ m}$. Opposing friction is constant $50\text{ N}$. Calculate work done by woman and friction.

$W_\text{woman} = +1750\text{ J}$ • $W_\text{friction} = -1000\text{ J}$ • $W_\text{net} = +750\text{ J}$
Step 1: Work done by woman: $$W_\text{woman} = \text{Area of Rectangle } (100 \times 10) + \text{Area of Trapezium } \left[\frac{100 + 50}{2} \times 10\right]$$ $$W_\text{woman} = 1000\text{ J} + 750\text{ J} = 1750\text{ J}$$
Step 2: Work done by friction: $$W_\text{friction} = (-50\text{ N}) \times 20\text{ m} = -1000\text{ J}$$

Example 5.6 • Block Entering 1/x Rough Patch

NCERT Solved

A $1\text{ kg}$ block moving with $v_\text{i} = 2\text{ m s}^{-1}$ enters a rough patch from $x = 0.10\text{ m}$ to $x = 2.01\text{ m}$ with retarding force $F_\text{r} = -k/x$ ($k = 0.5\text{ J}$). Find final KE and emergent speed.

$K_\text{f} = 0.5\text{ J}$ • $v_\text{f} = 1.0\text{ m s}^{-1}$
Step 1: Work done by retarding force: $$W = \int_{0.10}^{2.01} \left(-\frac{k}{x}\right) dx = -k \left[\ln x\right]_{0.10}^{2.01} = -0.5 \ln\left(\frac{2.01}{0.10}\right) = -0.5 \ln(20.1)$$ $$W \approx -0.5 (3.0) = -1.5\text{ J}$$
Step 2: Final KE and speed: $$K_\text{f} = K_\text{i} + W = \frac{1}{2}(1)(2)^2 - 1.5 = 2.0 - 1.5 = 0.5\text{ J}$$ $$v_\text{f} = \sqrt{\frac{2 K_\text{f}}{m}} = \sqrt{\frac{2 \times 0.5}{1}} = 1.0\text{ m s}^{-1}$$
5.7 - 5.8
NCERT Sections

Potential Energy & Conservation of Mechanical Energy

Conservative Forces

5.7.1 Mathematical Relationship Between Force & Potential Energy

Force and Potential Gradient $$F(x) = -\frac{dV}{dx} \iff \Delta V = -\int_{x_\text{i}}^{x_\text{f}} F(x)\, dx = -W_\text{c}$$ $$\Delta K + \Delta V = 0 \implies E = K + V = \text{Constant}$$ $$\oint \mathbf{F} \cdot d\mathbf{r} = 0 \quad (\text{Path Independent})$$

5.8.1 Motion in a Vertical Circle

A mass $m$ attached to a string of length $L$ whirled in a vertical plane:

Key Vertical Circle Speeds & Energies $$v_0 = \sqrt{5 g L} \quad (\text{Minimum speed at lowest point A to loop})$$ $$v_\text{B} = \sqrt{3 g L} \quad (\text{Speed at midpoint B, string horizontal})$$ $$v_\text{C} = \sqrt{g L} \quad (\text{Minimum speed at topmost point C, string slackens } T_\text{C}=0)$$ $$\frac{K_\text{B}}{K_\text{C}} = \frac{\frac{1}{2}m(3gL)}{\frac{1}{2}m(gL)} = 3$$
Interactive 3D Motion in a Vertical Circle & Energy Exchange
Loop-the-Loop Lab
Drag to change perspective
Instantaneous Speed ($v$)
17.1 m/s (h = 0.0m)
Energy Partition ($E = K + V = \text{const}$)
KE: 147 J | PE: 0 J
Conservation Law: As the bob ascends, Kinetic Energy converts continuously into Gravitational Potential Energy ($mg\Delta h$). At the topmost point, $v_\text{min} = \sqrt{gL}$ keeps string tension $T \ge 0$.

Example 5.7 • Bob in a Vertical Circle

NCERT Solved

A bob of mass $m$ on string length $L$ is given horizontal velocity $v_0$ at lowest point A to loop a vertical semi-circle with string becoming slack at top C. Find: (i) $v_0$, (ii) speeds at B and C, (iii) $K_\text{B}/K_\text{C}$.

(i) $v_0 = \sqrt{5gL}$ • (ii) $v_\text{B} = \sqrt{3gL}$, $v_\text{C} = \sqrt{gL}$ • (iii) $K_\text{B}/K_\text{C} = 3:1$
Step 1: Topmost Point C (String slackens $T_\text{C} = 0$): $$\frac{m v_\text{C}^2}{L} = mg \implies v_\text{C} = \sqrt{gL}$$ Total Energy at C: $$E = \frac{1}{2}m v_\text{C}^2 + mg(2L) = \frac{1}{2}mgL + 2mgL = \frac{5}{2}mgL$$
Step 2: Lowest Point A: $$\frac{1}{2}m v_0^2 = \frac{5}{2}mgL \implies v_0 = \sqrt{5gL}$$
Step 3: Midpoint B ($h = L$): $$\frac{1}{2}m v_\text{B}^2 + mgL = \frac{5}{2}mgL \implies \frac{1}{2}m v_\text{B}^2 = \frac{3}{2}mgL \implies v_\text{B} = \sqrt{3gL}$$
Step 4: Ratio of kinetic energies: $$\frac{K_\text{B}}{K_\text{C}} = \frac{\frac{3}{2}mgL}{\frac{1}{2}mgL} = 3$$
5.9
NCERT Section

The Potential Energy of a Spring

Hooke's Law Dynamics

5.9.1 Restoring Force & Elastic Energy

Hooke's Law & Potential Energy $$F_\text{s} = -k x \quad (k \text{ in } \text{N m}^{-1})$$ $$V(x) = -\int_0^x (-k x)\, dx = \frac{1}{2}k x^2$$ $$v_\text{m} = x_\text{m} \sqrt{\frac{k}{m}} \quad (\text{Max speed at equilibrium } x=0)$$
E = K + V V(x) = ½kx² K(x) = E − ½kx² −x_m 0 +x_m
Figure 5.8: Parabolic potential energy $V(x) = \frac{1}{2}kx^2$ and kinetic energy $K(x)$ curves for an ideal spring-mass system.
Interactive 3D Spring Potential Energy & Harmonic Oscillation
Hooke's Law Lab
Drag to change perspective
Instantaneous Displacement & Energy Partition
x = 0.00m • PE = 0.0 J | KE = 800.0 J | Total: 800.0 J
Conservation Law: At extreme positions ($x = \pm x_\text{m}$), speed drops to zero and total energy is 100% Potential Energy ($V = \frac{1}{2}kx_\text{m}^2$). At equilibrium ($x = 0$), $V=0$ and speed peaks with maximum Kinetic Energy ($K_\text{max} = \frac{1}{2}m v_\text{m}^2$).

Example 5.8 • Car Colliding with Horizontal Spring

NCERT Solved

A $1000\text{ kg}$ car moving at $18.0\text{ km/h}$ ($5\text{ m s}^{-1}$) collides with a spring of constant $k = 5.25 \times 10^3\text{ N m}^{-1}$ on a frictionless track. Find maximum compression $x_\text{m}$.

Maximum Compression $x_\text{m} = 2.00\text{ m}$
Step 1: Initial kinetic energy of car: $$K = \frac{1}{2} m v^2 = \frac{1}{2} (1000\text{ kg}) (5\text{ m s}^{-1})^2 = 1.25 \times 10^4\text{ J}$$
Step 2: Maximum compression by energy conservation: $$\frac{1}{2} k x_\text{m}^2 = K \implies x_\text{m} = \sqrt{\frac{2 K}{k}}$$ $$x_\text{m} = \sqrt{\frac{2 \times 12500}{5250}} = \sqrt{\frac{25000}{5250}} = \sqrt{4.76} \approx 2.00\text{ m}$$

Example 5.9 • Spring Collision with Friction ($\mu = 0.5$)

NCERT Solved

In Example 5.8, if surface friction has coefficient $\mu = 0.5$, calculate maximum compression $x_\text{m}$ (taking $g = 10\text{ m s}^{-2}$).

Maximum Compression $x_\text{m} = 1.35\text{ m}$
Step 1: Work-Energy Theorem with Friction: $$\Delta K = W_\text{spring} + W_\text{friction}$$ $$0 - \frac{1}{2}m v^2 = -\frac{1}{2}k x_\text{m}^2 - \mu m g x_\text{m}$$ $$\frac{1}{2}k x_\text{m}^2 + \mu m g x_\text{m} - \frac{1}{2}m v^2 = 0$$
Step 2: Substitute numerical values: $$\frac{1}{2}(5250)x_\text{m}^2 + (0.5 \times 1000 \times 10)x_\text{m} - 12500 = 0$$ $$2625 x_\text{m}^2 + 5000 x_\text{m} - 12500 = 0 \implies 21 x_\text{m}^2 + 40 x_\text{m} - 100 = 0$$ $$x_\text{m} = \frac{-40 \pm \sqrt{1600 - 4(21)(-100)}}{2(21)} = \frac{-40 \pm \sqrt{1600 + 8400}}{42} = \frac{-40 + 100}{42} = \frac{60}{42} \approx 1.35\text{ m}$$
5.10
NCERT Section

Power & Energy Units

Rate of Doing Work

5.10.1 Average and Instantaneous Power

Power Equations & Units $$P_\text{avg} = \frac{\Delta W}{\Delta t}, \qquad P = \frac{dW}{dt} = \mathbf{F} \cdot \mathbf{v} = F v \cos\theta$$ $$1\text{ Watt (W)} = 1\text{ J s}^{-1}, \qquad 1\text{ Horsepower (hp)} = 746\text{ W}$$ $$1\text{ Kilowatt-hour (kWh)} = (10^3\text{ W}) \times (3600\text{ s}) = 3.6 \times 10^6\text{ J} = 3.6\text{ MJ}$$

Example 5.10 • Elevator Motor Power

NCERT Solved

An $1800\text{ kg}$ elevator ascends at constant $2\text{ m s}^{-1}$ against $4000\text{ N}$ frictional force. Determine minimum motor power in Watts and horsepower ($g = 10\text{ m s}^{-2}$).

$P = 44,000\text{ W} = 44\text{ kW} \approx 59.0\text{ hp}$
Step 1: Total downward force: $$F = mg + F_\text{friction} = (1800 \times 10) + 4000 = 18000 + 4000 = 22,000\text{ N}$$
Step 2: Power delivered by motor: $$P = F v = (22,000\text{ N}) \times (2\text{ m s}^{-1}) = 44,000\text{ W} = 44\text{ kW}$$ $$P = \frac{44,000}{746} \approx 59.0\text{ hp}$$
5.11
NCERT Section

Collisions in One and Two Dimensions

Impact Mechanics

5.11.1 Types of Collisions

Collision Type Linear Momentum Kinetic Energy Coefficient of Restitution ($e$)
Elastic Collision Conserved Conserved $e = 1$
Inelastic Collision Conserved Not Conserved (Partially Lost) $0 < e < 1$
Completely Inelastic Conserved Maximum KE Lost (Bodies Stick) $e = 0$

5.11.2 1D Elastic Collision Velocities

For target $m_2$ initially at rest ($u_2 = 0$):

Final Velocities (1D Elastic Collision) $$v_{1\text{f}} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) v_{1\text{i}}$$ $$v_{2\text{f}} = \left(\frac{2 m_1}{m_1 + m_2}\right) v_{1\text{i}}$$
Special Physical Cases $$\text{Equal Masses } (m_1 = m_2): \quad v_{1\text{f}} = 0, \quad v_{2\text{f}} = v_{1\text{i}} \quad (\text{Complete Velocity Swap})$$ $$\text{Massive Target } (m_2 \gg m_1): \quad v_{1\text{f}} \approx -v_{1\text{i}}, \quad v_{2\text{f}} \approx 0 \quad (\text{Light Body Rebounds})$$ $$\text{2D Glancing Elastic Collision } (m_1 = m_2, u_2 = 0): \quad \theta_1 + \theta_2 = 90^\circ$$
Interactive 3D 1D/2D Collision & Restitution Simulator
Impact Lab
Drag to change perspective
Post-Collision Velocities & Restitution Coefficient
Impact pending • Press "Fire Collision" to launch mass $m_1$
Key Physics Principles: When $e = 1.0$ and $m_1 = m_2$, the incident particle stops dead and completely transfers its momentum and KE to the target. When $e = 0.0$ (completely inelastic), the two bodies coalesce and move together with maximum loss in mechanical kinetic energy.

Example 5.11 • Slowing Down of Fast Neutrons in Reactors

NCERT Solved

Show that a fast neutron ($m_1$) transfers $\approx 90\%$ of its KE to a stationary deuterium nucleus ($m_2 = 2 m_1$) and $\approx 28\%$ to carbon ($m_2 = 12 m_1$) in an elastic head-on collision.

Fraction of KE to Deuterium = $\frac{8}{9} \approx 88.9\%$ • Fraction to Carbon = $\frac{48}{169} \approx 28.4\%$
Step 1: Fractional kinetic energy transfer formula: $$f_2 = \frac{K_{2\text{f}}}{K_{1\text{i}}} = \frac{\frac{1}{2}m_2 v_{2\text{f}}^2}{\frac{1}{2}m_1 v_{1\text{i}}^2} = \frac{4 m_1 m_2}{(m_1 + m_2)^2}$$
Step 2: For Deuterium ($m_2 = 2m_1$): $$f_2 = \frac{4(1)(2)}{(1 + 2)^2} = \frac{8}{9} \approx 0.889 = 88.9\% \approx 90\%$$
Step 3: For Carbon ($m_2 = 12m_1$): $$f_2 = \frac{4(1)(12)}{(1 + 12)^2} = \frac{48}{169} \approx 0.284 = 28.4\%$$

Example 5.12 • 2D Elastic Billiard Ball Collision

NCERT Solved

Two identical billiard balls ($m_1 = m_2$) collide elastically. The target ball is initially at rest and recoils at angle $\theta_2 = 37^\circ$. Find the deflection angle $\theta_1$ of the cue ball.

Angle $\theta_1 = 90^\circ - 37^\circ = 53^\circ$
Step 1: Conservation of linear momentum: $$\mathbf{p}_{1\text{i}} = \mathbf{p}_{1\text{f}} + \mathbf{p}_{2\text{f}} \implies p_{1\text{i}}^2 = p_{1\text{f}}^2 + p_{2\text{f}}^2 + 2 (\mathbf{p}_{1\text{f}} \cdot \mathbf{p}_{2\text{f}})$$
Step 2: Conservation of kinetic energy ($m_1 = m_2 = m$): $$\frac{p_{1\text{i}}^2}{2m} = \frac{p_{1\text{f}}^2}{2m} + \frac{p_{2\text{f}}^2}{2m} \implies p_{1\text{i}}^2 = p_{1\text{f}}^2 + p_{2\text{f}}^2$$
Step 3: Equating equations gives orthogonality: $$2 (\mathbf{p}_{1\text{f}} \cdot \mathbf{p}_{2\text{f}}) = 0 \implies \mathbf{p}_{1\text{f}} \cdot \mathbf{p}_{2\text{f}} = 0 \implies \cos(\theta_1 + \theta_2) = 0$$ $$\theta_1 + \theta_2 = 90^\circ \implies \theta_1 = 90^\circ - 37^\circ = 53^\circ$$
02 / Self-Assessment Lab

Concept Check Questions

Instant per-option explanations for every choice, illustrating fundamental dynamic principles and pinpointing common misconceptions.

20 QUESTIONS
Progress: 0 / 20 Answered
Score: 0
Q01UNANSWERED

When a body is moving in a circular orbit under a central gravitational force, work done by the gravitational force over one full orbit is:

Option A is incorrect. The force is perpendicular to tangential displacement at every instant.
Option B is correct. At every point θ = 90° (cos 90° = 0), and gravity is conservative (∮ F • dr = 0).
Option C is incorrect. No net kinetic energy is drained.
Option D is incorrect. Work is identically zero for all radii.
Core Rule
Centripetal forces perpendicular to instantaneous velocity do zero work.
Q02UNANSWERED

When a conservative force does positive work on a body, the potential energy of the body:

Option A is incorrect. Positive work converts stored potential energy into kinetic energy.
Option B is correct. ΔV = −Wc. If Wc > 0, ΔV < 0 (potential energy decreases).
Option C is incorrect. Potential energy changes whenever work is done.
Option D is incorrect. It simply decreases.
Core Rule
ΔV = −Wc. Positive conservative work decreases potential energy.
Q03UNANSWERED

For an ideal spring obeying Hooke's Law (F = −kx), if extension is doubled from x to 2x, stored elastic potential energy increases by a factor of:

Option A is incorrect. Potential energy is quadratic with displacement.
Option B is correct. V(x) = ½kx2 → V(2x) = ½k(2x)2 = 4(½kx2) = 4 V(x).
Option C is incorrect. Power is 2, not 3.
Option D is incorrect. Energy scales as x2.
Core Rule
Spring energy scales quadratically: V ∝ x2.
Q04UNANSWERED

The minimum speed at the lowest point of a vertical circle of radius L to complete a full revolution without the string slacking is:

Option A is incorrect. √(gL) is the critical speed at the top point C.
Option B is incorrect. √(3gL) is the speed at the horizontal mid-point B.
Option C is correct. Conservation of energy requires ½m v02 = ½m(√gL)2 + mg(2L) = 5/2 mgL ⇒ v0 = √(5gL).
Option D is incorrect. Insufficient to reach the top.
Core Rule
Vertical circle bottom threshold: v0 = √(5gL).
Q05UNANSWERED

One kilowatt-hour (1 kWh) is equal to:

Option A is incorrect. Missed the factor of 1000 from kilo.
Option B is correct. 1000 W × 3600 s = 3,600,000 J = 3.6 × 106 J.
Option C is incorrect. 746 W is 1 horsepower.
Option D is incorrect. 1 kWh includes 3600 seconds.
Core Rule
1 kWh = 3.6 × 106 J.
Q06UNANSWERED

When two equal mass billiard balls collide elastically in 1D with one ball initially at rest, after collision:

Option A is incorrect. That occurs only in completely inelastic collisions.
Option B is correct. When m1 = m2 in 1D elastic collision, velocities completely exchange (v1f = 0, v2f = u1).
Option C is incorrect. Rebound only occurs when target mass is heavier.
Option D is incorrect. Sticking means completely inelastic.
Core Rule
Equal masses in 1D elastic collision completely exchange velocities.
Q07UNANSWERED

In an inelastic collision of two bodies, which physical quantities are conserved throughout and after collision?

Option A is incorrect. KE is not conserved in inelastic collisions.
Option B is correct. Linear momentum is conserved due to Newton's 3rd law, and total energy (including heat/sound) is universally conserved.
Option C is incorrect. Potential energy changes dynamically during impact.
Option D is incorrect. Velocities change upon collision.
Core Rule
Momentum and total energy are conserved in all isolated collisions.
Q08UNANSWERED

A body initially at rest undergoes 1D motion with constant acceleration a. The power delivered to it at time t is proportional to:

Option A is incorrect.
Option B is correct. P = F v = (ma)(at) = m a2 t ⇒ P ∝ t.
Option C is incorrect.
Option D is incorrect. Work is proportional to t2, power is proportional to t.
Core Rule
Under constant acceleration: P = m a2 t ∝ t.
Q09UNANSWERED

A body is moving unidirectionally under constant power P. Its displacement x in time t is proportional to:

Option A is incorrect.
Option B is incorrect. Speed is ∝ t1/2.
Option B is correct. ½mv2 = Pt → v = √(2P/m) t1/2 → x = ∫ v dt ∝ t3/2.
Option D is incorrect.
Core Rule
Under constant power: v ∝ t1/2 and x ∝ t3/2.
Q10UNANSWERED

In a 2D glancing elastic collision between two equal masses where one mass is initially at rest, the angle between their final velocity vectors is always:

Option A is incorrect.
Option B is correct. Dot product of final velocities v1f • v2f = 0 by momentum and energy conservation, so θ1 + θ2 = 90°.
Option C is incorrect. 180° is head-on rebound.
Option D is incorrect.
Core Rule
Glancing 2D elastic collision of equal masses: θ1 + θ2 = 90°.
Q11UNANSWERED

Work done by a body against friction always results in:

Option A is correct. Friction is a dissipative force that drains mechanical energy and converts kinetic energy into thermal energy (heat).
Option B is incorrect. Friction cannot store energy into potential energy.
Option C is incorrect. Kinetic energy is diminished.
Option D is incorrect. Total mechanical energy decreases.
Core Rule
Friction converts kinetic energy into thermal energy.
Q12UNANSWERED

A force F = 2 î + 3 ĵ + 4 k̂ N moves a body along the z-axis by 4 m. The work done is:

Option A is incorrect. 8 J is Fx × 4.
Option B is correct. d = 0 î + 0 ĵ + 4 k̂ m → W = F • d = 4 × 4 = 16 J.
Option C is incorrect. 12 J is Fy × 4.
Option D is incorrect. Summing all components incorrectly.
Core Rule
W = F • d = Fz dz for motion along the z-axis.
Q13UNANSWERED

During the brief instant of contact in an elastic collision between two balls, the total kinetic energy:

Option A is incorrect. During contact, KE is momentarily converted into elastic potential energy of deformation.
Option B is correct. At maximum deformation, kinetic energy is at a minimum. KE is fully recovered only after separation.
Option C is incorrect. Kinetic energy cannot exceed initial energy.
Option D is incorrect. Unphysical.
Core Rule
Kinetic energy conservation applies before and after collision, not during contact deformation.
Q14UNANSWERED

An electron has kinetic energy 10 keV and a proton has 100 keV. Which particle has greater speed?

Option A is correct. v = √(2K/m). Since mp ≈ 1836 me, ve/vp = √[(10/100) × 1836] = √183.6 ≈ 13.5 > 1. Electron is ~13.5 times faster.
Option B is incorrect. Proton is much heavier, resulting in a lower speed despite 10× more energy.
Option C is incorrect. Masses differ by a factor of 1836.
Option D is incorrect. Readily calculated from masses.
Core Rule
v ∝ √(K/m). Lighter particles can have much higher speeds even with lower kinetic energy.
Q15UNANSWERED

A particle moves in potential V(x) = ½ k x2 with k = 0.5 N m−1 and total energy E = 1 J. Its turning points are:

Option A is incorrect. At 1 m, V = 0.25 J ≠ 1 J.
Option B is correct. At turning points K = 0 ⇒ E = V(x) ⇒ 1 = ½(0.5)x2 = 0.25 x2 ⇒ x2 = 4 ⇒ x = ±2 m.
Option C is incorrect. V(4) = 4 J > 1 J (forbidden region).
Option D is incorrect.
Core Rule
Turning points occur where K = 0 ⇒ V(x) = E.
Q16UNANSWERED

A sandbag on a frictionless trolley leaks sand steadily onto the ground. The speed of the trolley:

Option A is correct. Leaking sand drops vertically with horizontal velocity identical to the trolley; no horizontal external force acts on the remaining system, so speed remains constant.
Option B is incorrect. Mass reduction does not accelerate the trolley because leaking sand carries away momentum.
Option C is incorrect. There is no retarding force on frictionless track.
Option D is incorrect.
Core Rule
Leaking mass with identical forward velocity produces zero thrust ⇒ constant velocity.
Q17UNANSWERED

A pump lifts 30 m3 of water to a 40 m high tank in 15 min with 30% efficiency. Electric power consumed is:

Option A is incorrect. 13.3 kW is the output mechanical power.
Option B is correct. Wout = mgh = (30000 × 9.8 × 40) = 1.176 × 107 J. Pout = 1.176×107 / 900 = 13.07 kW. Pin = 13.07 / 0.30 = 43.6 kW.
Option C is incorrect.
Option D is incorrect.
Core Rule
Pinput = Poutput / η.
Q18UNANSWERED

A pendulum bob A released from 30° strikes an identical resting bob B at the lowest point in an elastic collision. Bob A rises to:

Option A is incorrect. Bob A transfers all its kinetic energy to Bob B.
Option B is correct. Equal masses in 1D elastic collision exchange velocities: Bob A comes to rest and Bob B swings up to 30°.
Option C is incorrect. Velocity is not split.
Option D is incorrect. Violates energy conservation.
Core Rule
Equal mass elastic collision: Incident bob transfers 100% velocity and comes to rest.
Q19UNANSWERED

If the kinetic energy of a moving body is increased by 300%, its momentum increases by:

Option A is incorrect. p ∝ √K, not linear.
Option B is correct. K' = K + 3K = 4K. Since p = √(2mK), p' = √(4) p = 2p → Increase = (2p − p)/p × 100% = 100%.
Option C is incorrect. Factor of 2 in magnitude means 100% increase.
Option D is incorrect.
Core Rule
p = √(2mK). Quadrupling energy doubles momentum (+100%).
Q20UNANSWERED

Why does a satellite speeding in a decaying orbit around Earth speed up as it loses mechanical energy to atmospheric resistance?

Option A is correct. By virial theorem, orbital speed v = √(GM/r). As radius r shrinks, potential energy drops twice as fast as KE rises, so satellite accelerates despite losing total energy.
Option B is incorrect. Drag always opposes motion.
Option C is incorrect. In a decaying spiral, displacement has a radial component along gravity.
Option D is incorrect. Mass is constant.
Core Rule
Orbital decay converts gravitational potential energy into both increased kinetic energy and frictional heat.
03 / Textbook Solutions

NCERT Exercises 5.1 – 5.23

Complete stepwise solutions for every textbook exercise question in NCERT Class 11 Physics Chapter 5 Reprint 2026-27 (take g = 9.8 or 10 m s−2 as specified).

23 QUESTIONS
Ex 5.1Sign of Work Done

5.1 The sign of work done by a force on a body is important to understand. State clearly whether the following quantities are positive or negative:

(a)
Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b)
Work done by gravitational force in the above case.
(c)
Work done by friction on a body sliding down an inclined plane.
(d)
Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
(e)
Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
(a) Positive • (b) Negative • (c) Negative • (d) Positive • (e) Negative
(a) Positive: The lifting force $\mathbf{F}$ and displacement $\mathbf{d}$ are in the same direction ($\theta = 0^\circ \implies \cos 0^\circ = +1$).
(b) Negative: Gravitational force acts downward while displacement is upward ($\theta = 180^\circ \implies \cos 180^\circ = -1$).
(c) Negative: Friction opposes the relative sliding motion along the incline ($\theta = 180^\circ \implies W = -f d$).
(d) Positive: The applied force acts in the direction of forward motion ($\theta = 0^\circ$).
(e) Negative: Air resistance opposes the velocity of the bob at every instant ($\theta = 180^\circ$).
Ex 5.2Work Done on Rough Table in 10 s

5.2 A body of mass $2\text{ kg}$ initially at rest moves under the action of an applied horizontal force of $7\text{ N}$ on a table with coefficient of kinetic friction $\mu_\text{k} = 0.1$. Compute the: (a) work done by the applied force in $10\text{ s}$, (b) work done by friction in $10\text{ s}$, (c) work done by the net force on the body in $10\text{ s}$, (d) change in kinetic energy of the body in $10\text{ s}$, and interpret your results. ($g = 9.8\text{ m s}^{-2}$).

(a) $W_\text{applied} = 882\text{ J}$ • (b) $W_\text{friction} = -246.96\text{ J} \approx -247\text{ J}$ • (c) $W_\text{net} = 635.04\text{ J} \approx 635\text{ J}$ • (d) $\Delta K = 635.04\text{ J}$
Step 1: Dynamics and kinematics in $10\text{ s}$: $$f_\text{k} = \mu_\text{k} m g = 0.1 \times 2 \times 9.8 = 1.96\text{ N}$$ $$F_\text{net} = F_\text{applied} - f_\text{k} = 7 - 1.96 = 5.04\text{ N}$$ $$a = \frac{F_\text{net}}{m} = \frac{5.04}{2} = 2.52\text{ m s}^{-2}$$ $$d = \frac{1}{2} a t^2 = \frac{1}{2} (2.52\text{ m s}^{-2}) (10\text{ s})^2 = 126\text{ m}$$
Step 2: Work calculations: $$(a) \quad W_\text{applied} = F_\text{applied} d = 7\text{ N} \times 126\text{ m} = 882\text{ J}$$ $$(b) \quad W_\text{friction} = -f_\text{k} d = -1.96\text{ N} \times 126\text{ m} = -246.96\text{ J}$$ $$(c) \quad W_\text{net} = F_\text{net} d = 5.04\text{ N} \times 126\text{ m} = 635.04\text{ J}$$
Step 3: Kinetic energy change & interpretation: $$v = u + at = 0 + (2.52)(10) = 25.2\text{ m s}^{-1}$$ $$\Delta K = \frac{1}{2} m v^2 - 0 = \frac{1}{2} (2\text{ kg}) (25.2\text{ m s}^{-1})^2 = 635.04\text{ J}$$ Interpretation: $W_\text{net} = \Delta K$, directly verifying the Work-Energy Theorem.
Ex 5.3Potential Energy Graphs Forbidden Regions

5.3 In Fig. 5.11, four 1D potential energy functions V(x) are shown with total energy E marked. Specify regions where the particle cannot be found and find minimum total energy in each case.

(a) Step V0 a E V0 (b) Well -a to b E (c) Finite Well -V1 E (d) Incline Barrier E
Figure 5.11: Potential energy curves V(x) for Exercise 5.3.
General Principle: Kinetic energy $K = E - V(x) \ge 0$. A classical particle cannot exist in regions where $V(x) > E$ (negative kinetic energy is unphysical).
(a) For $x > a$, $V(x) = V_0 > E$. Particle cannot exist in $x > a$. Minimum total energy $E_\text{min} = 0$.
(b) For all $x$ except $-a/2 < x < a/2$, $V(x) > E$. Particle cannot exist in $|x| > a/2$. $E_\text{min} = 0$.
(c) Particle cannot exist for $x < -a/2$ and $x > a/2$ (where $V(x) = 0 > E$). $E_\text{min} = -V_1$.
(d) Particle cannot exist in $-b/2 < x < b/2$ (where $V(x) > E$). $E_\text{min} = -V_1$.
Ex 5.4SHM Turning Points

5.4 The potential energy function for a particle executing simple harmonic motion is given by $V(x) = \frac{1}{2} k x^2$, where $k = 0.5\text{ N m}^{-1}$. Show that a particle of total energy $1\text{ J}$ moving under this potential must turn back when it reaches $x = \pm 2\text{ m}$.

Proof: $$E = K + V(x) = 1\text{ J}$$ At turning points, instantaneous velocity $v = 0 \implies K = 0$: $$V(x) = E \implies \frac{1}{2} k x^2 = 1$$ $$\frac{1}{2}(0.5) x^2 = 1 \implies 0.25 x^2 = 1 \implies x^2 = 4 \implies x = \pm 2\text{ m}$$ Since $K \ge 0$, for $|x| > 2\text{ m}$, $V(x) > 1\text{ J}$ which would make $K < 0$ (forbidden). Hence the particle must turn back at $x = \pm 2\text{ m}$.
Ex 5.5Conceptual Physics Questions

5.5 Answer the following:

(a)
A rocket casing burns up due to friction in flight. At whose expense is this heat energy obtained (rocket or atmosphere)?
(b)
Comets move in highly elliptical orbits. Why is work done by gravitational force zero over each complete orbit?
(c)
Why does a satellite's speed increase as its orbit decays due to atmospheric resistance?
(d)
Fig 5.13: Man walks $2\text{ m}$ holding $15\text{ kg}$ vs pulling $15\text{ kg}$ hanging over a pulley. In which case is work done greater?
(a) Rocket: Heat energy is obtained at the expense of the rocket's own mechanical energy (kinetic and potential energy), causing its total energy to decrease.
(b) Conservative Field: Gravitational force is conservative ($\oint \mathbf{F}\cdot d\mathbf{r} = 0$). Over any complete closed orbit, the net displacement is zero and work done is strictly zero.
(c) Orbital Decay: As orbital radius $r$ decreases, gravitational potential energy $V = -\frac{GMm}{r}$ decreases twice as fast as kinetic energy $K = \frac{GMm}{2r}$ increases ($\Delta K = \frac{1}{2}|\Delta V|$). The remaining mechanical energy is lost as heat.
(d) Second case (Pulling via pulley): In Fig (i), $\theta = 90^\circ \implies W = F d \cos 90^\circ = 0\text{ J}$. In Fig (ii), upward pulling force equals weight $mg$: $$W = F d = m g d = (15\text{ kg}) \times (9.8\text{ m s}^{-2}) \times (2\text{ m}) = 294\text{ J}$$ Work done is much greater in Case (ii).
Ex 5.6Choose the Correct Alternative

5.6 Select the correct alternative:
(a) When a conservative force does positive work, PE increases/decreases/remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
(c) Rate of change of total momentum is proportional to external force/sum of internal forces.
(d) In an inelastic collision, quantities which do not change after collision are total KE/total linear momentum/total energy.

(a) Decreases • (b) Kinetic energy • (c) External force • (d) Total linear momentum and total energy
(a) $\Delta V = -W_\text{c}$. Since $W_\text{c} > 0$, $\Delta V < 0 \implies$ PE decreases.
(b) Work against friction directly dissipates kinetic energy into heat.
(c) $\frac{d\mathbf{p}}{dt} = \mathbf{F}_\text{ext}$ (internal forces cancel in pairs $\Sigma \mathbf{F}_\text{int} = 0$).
(d) In any collision (elastic or inelastic), total linear momentum and total energy remain conserved.
Ex 5.7True / False Statements

5.7 State if True or False with reasons:
(a) In an elastic collision, momentum and energy of each individual body is conserved.
(b) Total energy of a system is always conserved, regardless of internal/external forces.
(c) Work done over a closed loop is zero for every force in nature.
(d) In an inelastic collision, final KE is always less than initial KE.

(a) False: Only the TOTAL momentum and energy of the composite system is conserved; individual bodies exchange momentum and energy.
(b) False: True only for an isolated system. If an external agency does work or heat is transferred, total system energy changes.
(c) False: Only true for conservative forces (e.g. friction does negative work over a closed loop).
(d) True (for passive collisions): In classical inelastic collisions without internal chemical/nuclear release, final kinetic energy is always less than initial KE.
Ex 5.8Collision Duration Questions

5.8 Answer carefully with reasons:
(a) Is total KE conserved during the short contact time of elastic collision?
(b) Is total linear momentum conserved during contact time of elastic collision?
(c) What are answers for inelastic collision?
(d) If PE of balls depends only on center-to-center distance, is collision elastic or inelastic?

(a) No: Kinetic energy is temporarily converted into elastic potential energy of deformation during contact.
(b) Yes: Internal forces are equal and opposite at every instant (Δp = 0).
(c) For inelastic collision: (a) No (KE lost), (b) Yes (momentum conserved at all times).
(d) Elastic: If potential energy depends solely on distance, the force is conservative and total mechanical energy is conserved → Elastic collision.
Ex 5.9Power under Constant Acceleration

5.9 A body initially at rest undergoes 1D motion with constant acceleration. The power delivered to it at time $t$ is proportional to: (i) $t^{1/2}$, (ii) $t$, (iii) $t^{3/2}$, (iv) $t^2$.

Correct Alternative: (ii) $t$
Derivation: $$v = u + at = 0 + at = at$$ $$P = F v = (m a)(at) = m a^2 t \implies P \propto t$$
Ex 5.10Displacement under Constant Power

5.10 A body moves unidirectionally under a source of constant power. Its displacement in time $t$ is proportional to: (i) $t^{1/2}$, (ii) $t$, (iii) $t^{3/2}$, (iv) $t^2$.

Correct Alternative: (iii) $t^{3/2}$
Derivation: $$P = F v = \left(m \frac{dv}{dt}\right) v = \text{Constant}$$ $$v\, dv = \left(\frac{P}{m}\right) dt \implies \int_0^v v\, dv = \frac{P}{m} \int_0^t dt$$ $$\frac{1}{2} v^2 = \frac{P}{m} t \implies v = \sqrt{\frac{2P}{m}}\, t^{1/2}$$ $$x = \int_0^t v\, dt = \sqrt{\frac{2P}{m}} \int_0^t t^{1/2}\, dt = \frac{2}{3} \sqrt{\frac{2P}{m}}\, t^{3/2} \implies x \propto t^{3/2}$$
Ex 5.11Vector Work along Z-Axis

5.11 A body constrained to move along the $z$-axis of a coordinate system is subjected to a constant force $\mathbf{F} = (-\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 3\hat{\mathbf{k}})\text{ N}$. What is the work done by this force in moving the body a distance of $4\text{ m}$ along the $z$-axis?

Work done $W = 12\text{ J}$
Calculation: $$\mathbf{d} = 0\hat{\mathbf{i}} + 0\hat{\mathbf{j}} + 4\hat{\mathbf{k}}\text{ m}$$ $$W = \mathbf{F} \cdot \mathbf{d} = (-1)(0) + (2)(0) + (3)(4) = 12\text{ J}$$
Ex 5.12Electron vs Proton Speed Ratio

5.12 An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy $10\text{ keV}$, and the second with $100\text{ keV}$. Which is faster, the electron or the proton? Obtain the ratio of their speeds. ($m_\text{e} = 9.11 \times 10^{-31}\text{ kg}, m_\text{p} = 1.67 \times 10^{-27}\text{ kg}$).

The Electron is faster • Speed ratio $\frac{v_\text{e}}{v_\text{p}} \approx 13.5$
Calculation: $$K = \frac{1}{2} m v^2 \implies v = \sqrt{\frac{2K}{m}}$$ $$\frac{v_\text{e}}{v_\text{p}} = \sqrt{\left(\frac{K_\text{e}}{K_\text{p}}\right) \left(\frac{m_\text{p}}{m_\text{e}}\right)}$$ $$\frac{v_\text{e}}{v_\text{p}} = \sqrt{\left(\frac{10\text{ keV}}{100\text{ keV}}\right) \times \left(\frac{1.67 \times 10^{-27}\text{ kg}}{9.11 \times 10^{-31}\text{ kg}}\right)} = \sqrt{0.10 \times 1833.15} = \sqrt{183.3} \approx 13.54$$
Ex 5.13Raindrop Falling 500 m with Viscous Drag

5.13 A rain drop of radius $2\text{ mm}$ falls from a height of $500\text{ m}$ above the ground. It falls with decreasing acceleration (due to viscous resistance) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is $10\text{ m s}^{-1}$? ($g = 9.8\text{ m s}^{-2}$).

$W_{\text{g},1} = W_{\text{g},2} = +0.082\text{ J}$ (Total $W_\text{g} = 0.164\text{ J}$) • $W_\text{resistive} = -0.162\text{ J}$
Step 1: Mass of raindrop: $$r = 2 \times 10^{-3}\text{ m}, \quad \rho = 1000\text{ kg m}^{-3}$$ $$m = \rho V = 1000 \times \left[\frac{4}{3}\pi (2 \times 10^{-3})^3\right] \approx 3.35 \times 10^{-5}\text{ kg}$$
Step 2: Work done by gravity in each half ($h = 250\text{ m}$): $$W_{\text{g},1} = W_{\text{g},2} = m g h = (3.35 \times 10^{-5}) \times 9.8 \times 250 = 0.082\text{ J}$$ $$W_{\text{g},\text{total}} = 2 \times 0.082 = 0.164\text{ J}$$
Step 3: Work done by resistive force: $$\Delta K = \frac{1}{2}m v^2 - 0 = \frac{1}{2} (3.35 \times 10^{-5})(10)^2 = 1.675 \times 10^{-3}\text{ J} \approx 0.0017\text{ J}$$ $$W_\text{r} = \Delta K - W_{\text{g},\text{total}} = 0.0017 - 0.164 = -0.1623\text{ J} \approx -0.162\text{ J}$$
Ex 5.14Gas Molecule Collision with Wall

5.14 A molecule in a gas container hits a horizontal wall with speed $200\text{ m s}^{-1}$ and angle $30^\circ$ with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

Total momentum is conserved • Collision is elastic
Momentum: The total momentum of the isolated system (molecule + entire container wall) is conserved. The wall experiences recoil momentum $\Delta p = 2 m v \cos 30^\circ$.
Elasticity: The kinetic energy before impact ($\frac{1}{2}mv^2$) equals kinetic energy after impact ($\frac{1}{2}mv^2$) since speed remains unchanged ($200\text{ m s}^{-1}$), confirming an elastic collision.
Ex 5.15Water Pump Power Consumption

5.15 A pump on the ground floor of a building can pump up water to fill a tank of volume $30\text{ m}^3$ in $15\text{ min}$. If the tank is $40\text{ m}$ above the ground, and the efficiency of the pump is $30\%$, how much electric power is consumed by the pump? ($g = 9.8\text{ m s}^{-2}$).

Electric Power Consumed $P_\text{in} = 43.56\text{ kW} \approx 43.6\text{ kW}$
Step 1: Output work and useful power: $$m = \rho V = (1000\text{ kg m}^{-3})(30\text{ m}^3) = 30,000\text{ kg}$$ $$W_\text{out} = mgh = 30,000 \times 9.8 \times 40 = 1.176 \times 10^7\text{ J}$$ $$P_\text{out} = \frac{W_\text{out}}{t} = \frac{1.176 \times 10^7\text{ J}}{15 \times 60\text{ s}} = \frac{1.176 \times 10^7}{900} = 13.067\text{ kW}$$
Step 2: Input electric power: $$\eta = \frac{P_\text{out}}{P_\text{in}} = 0.30 \implies P_\text{in} = \frac{13.067\text{ kW}}{0.30} = 43.56\text{ kW}$$
Ex 5.16Ball Bearing Collision (Fig 5.14)

5.16 Two identical ball bearings in contact with each other and at rest on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed $V$. If the collision is elastic, which of the situations shown in Fig 5.14 is a possible result after collision?

Correct Result: Case (ii) – The third ball moves off with speed $V$, and first two balls remain at rest.
Test Case (i) (Two balls moving with speed $V/2$): $$p_\text{f} = (2m)\left(\frac{V}{2}\right) = mV \quad (\text{Momentum conserved})$$ $$K_\text{f} = \frac{1}{2}(2m)\left(\frac{V}{2}\right)^2 = \frac{1}{4}mV^2 \neq \frac{1}{2}mV^2 \quad (\text{KE is halved, Violates Elasticity})$$
Test Case (ii) (Third ball moving with speed $V$): $$p_\text{f} = mV, \qquad K_\text{f} = \frac{1}{2}mV^2$$ Both momentum and kinetic energy are conserved simultaneously.
Ex 5.17Pendulum Bob Elastic Impact (Fig 5.15)

5.17 The bob A of a pendulum released from $30^\circ$ to the vertical hits another bob B of the same mass at rest on a table as shown in Fig 5.15. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Bob A rises to zero height (comes to complete rest).
Step-by-step reasoning: In an elastic head-on collision between two equal masses ($m_1 = m_2$) with target initially at rest: $$v_{1\text{f}} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) v_{1\text{i}} = 0$$ $$v_{2\text{f}} = \left(\frac{2m_1}{m_1 + m_2}\right) v_{1\text{i}} = v_{1\text{i}}$$ Bob A transfers $100\%$ of its momentum and kinetic energy to Bob B and comes to rest at the lowest point. Hence Bob A rises to zero height.
Ex 5.18Pendulum with 5% Air Dissipation

5.18 The bob of a pendulum is released from a horizontal position. If the length of the pendulum is $1.5\text{ m}$, what is the speed with which the bob arrives at the lowermost point, given that it dissipated $5\%$ of its initial energy against air resistance? ($g = 9.8\text{ m s}^{-2}$).

Speed $v = 5.28\text{ m s}^{-1} \approx 5.3\text{ m s}^{-1}$
Calculation: Initial PE at horizontal position: $E_\text{i} = mgL$. Since $5\%$ energy is dissipated, final KE at bottom is $95\%$ of $E_\text{i}$: $$\frac{1}{2} m v^2 = 0.95 \times m g L$$ $$v = \sqrt{2 \times 0.95 \times g L} = \sqrt{2 \times 0.95 \times 9.8 \times 1.5} = \sqrt{27.93} \approx 5.285\text{ m s}^{-1}$$
Ex 5.19Trolley with Leaking Sandbag

5.19 A trolley of mass $300\text{ kg}$ carrying a sandbag of $25\text{ kg}$ is moving uniformly with a speed of $27\text{ km/h}$ on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of $0.05\text{ kg s}^{-1}$. What is the speed of the trolley after the entire sand bag is empty?

Speed remains unchanged: $27\text{ km/h} = 7.5\text{ m s}^{-1}$
Reasoning: The sand leaks vertically downwards and carries the same forward horizontal speed as the trolley. There is no external horizontal force acting on the trolley-sand system ($\Sigma F_x = 0$). By Newton's First Law, the horizontal acceleration of the trolley is zero ($a_x = 0$), so its speed remains constant at $27\text{ km/h}$.
Ex 5.20Work for v = a x3/2

5.20 A body of mass $0.5\text{ kg}$ travels in a straight line with velocity $v = a x^{3/2}$ where $a = 5\text{ m}^{-1/2}\text{ s}^{-1}$. What is the work done by the net force during its displacement from $x = 0$ to $x = 2\text{ m}$?

Work done $W = 50\text{ J}$
By Work-Energy Theorem: $$\text{At } x = 0: \quad v_\text{i} = 0$$ $$\text{At } x = 2\text{ m}: \quad v_\text{f} = a (2)^{3/2} = 5 \times 2^{3/2} \implies v_\text{f}^2 = 25 \times 8 = 200\text{ m}^2\text{ s}^{-2}$$ $$W = \Delta K = \frac{1}{2} m v_\text{f}^2 - 0 = \frac{1}{2} (0.5\text{ kg}) (200\text{ m}^2\text{ s}^{-2}) = 50\text{ J}$$
Ex 5.21Windmill Power Generation

5.21 The blades of a windmill sweep out a circle of area $A = 30\text{ m}^2$. (a) If the wind flows at a velocity $v = 36\text{ km/h}$ ($10\text{ m/s}$) perpendicular to the circle, what is the mass of the air passing through it in time $t$? (b) What is the kinetic energy of the air? (c) Assume that the windmill converts $25\%$ of the wind’s energy into electrical energy, and that $\rho = 1.2\text{ kg m}^{-3}$. What is the electric power produced?

Electrical Power $P = 4500\text{ W} = 4.5\text{ kW}$
(a) Mass of air passing in time $t$: $$m = \rho V = \rho (A v t) = 1.2 \times 30 \times 10 \times t = 360 t\text{ kg}$$
(b) Kinetic energy of the air: $$K = \frac{1}{2} m v^2 = \frac{1}{2} (\rho A v t) v^2 = \frac{1}{2} \rho A v^3 t = \frac{1}{2}(1.2)(30)(10)^3 t = 18,000 t\text{ J}$$
(c) Electrical power generated: $$P = 0.25 \times \frac{K}{t} = 0.25 \times 18,000\text{ W} = 4500\text{ W} = 4.5\text{ kW}$$
Ex 5.22Dieter Fat Loss Calculation

5.22 A person trying to lose weight (dieter) lifts a $10\text{ kg}$ mass, one thousand times, to a height of $0.5\text{ m}$ each time. Assume that the potential energy lost each time she lowers the mass is dissipated. (a) How much work does she do against the gravitational force? (b) Fat supplies $3.8 \times 10^7\text{ J}$ of energy per kilogram which is converted to mechanical energy with a $20\%$ efficiency rate. How much fat will the dieter use up? ($g = 9.8\text{ m s}^{-2}$).

(a) Work done $W = 4.9 \times 10^4\text{ J} = 49\text{ kJ}$ • (b) Fat burned $m_\text{fat} = 6.45 \times 10^{-3}\text{ kg} = 6.45\text{ g}$
(a) Work against gravity: $$W = n (mgh) = 1000 \times (10\text{ kg} \times 9.8\text{ m s}^{-2} \times 0.5\text{ m}) = 49,000\text{ J} = 4.9 \times 10^4\text{ J}$$
(b) Fat consumed: $$\text{Total chemical energy needed} = \frac{W}{\eta} = \frac{49,000\text{ J}}{0.20} = 2.45 \times 10^5\text{ J}$$ $$m_\text{fat} = \frac{2.45 \times 10^5\text{ J}}{3.8 \times 10^7\text{ J kg}^{-1}} \approx 6.45 \times 10^{-3}\text{ kg} = 6.45\text{ g}$$
Ex 5.23Solar Panel Area for 8 kW Family

5.23 A family uses $8\text{ kW}$ of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of $200\text{ W per square metre}$. If $20\%$ of this energy can be converted to useful electrical energy, how large an area is needed to supply $8\text{ kW}$? (b) Compare this area to that of the roof of a typical house.

(a) Required Solar Collector Area $A = 200\text{ m}^2$ • (b) Comparable to a typical roof ($14\text{ m} \times 14\text{ m} \approx 196\text{ m}^2$).
Calculation: $$\text{Effective power output per unit area} = 0.20 \times 200\text{ W m}^{-2} = 40\text{ W m}^{-2}$$ $$A = \frac{\text{Total Power Required}}{\text{Power Output per }\text{m}^2} = \frac{8000\text{ W}}{40\text{ W m}^{-2}} = 200\text{ m}^2$$
04 / Rapid Reference

Chapter Summary & Formulas

Work-energy definitions, spring physics, conservation laws, vertical circular motion limits, 1D/2D collision equations, and high-yield exam shortcuts.

100% SYLLABUS
5.1 - 5.3 • Work & Power

Work Done by Constant / Variable Forces & Power

Work is the scalar product of force and displacement.

  • Work Done (Constant Force): $W = \vec{F} \cdot \vec{d} = Fd\cos\theta$
    • Positive: $\theta < 90^\circ$ (e.g. pulling a cart).
    • Negative: $\theta > 90^\circ$ (e.g. friction).
    • Zero: $\theta = 90^\circ$ (e.g. centripetal force).
  • Work Done (Variable Force): $W = \int_{x_i}^{x_f} F(x) dx$ (Equal to the **area under $F-x$ curve**).

Power (Rate of doing work)

  • Average Power: $P_{\text{avg}} = \frac{\Delta W}{\Delta t}$
  • Instantaneous Power: $P = \frac{dW}{dt} = \vec{F} \cdot \vec{v}$ (Unit: Watt, $\text{W}$; $1\text{ hp} = 746\text{ W}$).
5.4 & 5.5 • Energy Physics

Kinetic vs. Potential Energy & WE Theorem

Energy is a scalar quantity measuring the capacity to do work.

Work-Energy Theorem: The work done by the net force acting on a body is equal to the change in its kinetic energy: $$W_{\text{net}} = \Delta K = K_f - K_i$$ Valid for both constant and variable forces.

Energy Formulas & Relationships

Energy Term Formula Conditions & Tricks
Kinetic Energy ($K$) $K = \frac{1}{2}mv^2 = \frac{p^2}{2m}$ Always positive. Trick: For small changes ($\Delta \le 5\%$), $\frac{\Delta K}{K} \approx 2\frac{\Delta p}{p}$.
Conservative Force $F(x) = -\frac{dV}{dx}$ Work done depends only on endpoints: $W_c = -\Delta V$. Associated with Potential Energy.
Gravitational PE $V(h) = mgh$ Relative to a chosen reference height $h=0$.
Spring Elastic PE $V_s = \frac{1}{2}kx^2$ Associated with spring force $F_s = -kx$. $k$ is spring constant.
5.7 • Mechanical Energy Conservation

Conservation of Mechanical Energy & Vertical Circle

If only conservative forces do work, the total mechanical energy ($E = K+V$) of a system is conserved.

Critical Limits for Vertical Circular Motion (String of length $L$)

Position in Loop Critical Velocity Limit String Tension Limit
Lowest Point (Bottom, A) $$v_A \ge \sqrt{5gL}$$ $$T_A \ge 6mg$$
Horizontal Point (Middle, B) $$v_B \ge \sqrt{3gL}$$ $$T_B \ge 3mg$$
Highest Point (Top, C) $$v_C \ge \sqrt{gL}$$ $$T_C \ge 0$$
Mnemonic: "Vertical Circle 5-3-1 Rule": Under the square root, the coefficient descends as odd numbers **5 (bottom) $\to$ 3 (middle) $\to$ 1 (top)**.
5.8 • Collision Dynamics

Collisions: Elastic & Inelastic

In all collisions, linear momentum is conserved at every instant. Kinetic energy may or may not be conserved.

Collision Type Coefficient of Restitution ($e$) KE Conservation & Velocity Rules
Completely Elastic $$e = 1$$ KE is conserved.
Trick: Identical masses ($m_1=m_2$) swap velocities after 1D collision.
Inelastic $$0 < e < 1$$ KE is lost as heat/noise. Deformation occurs.
Perfect Inelastic $$e = 0$$ Maximum KE loss. Objects stick together post-collision: $v_f = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$.

Coeff of Restitution ($e$) & Glancing Collisions

$$e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{v_2 - v_1}{u_1 - u_2}$$

Glancing 2D Collision Rule: For two identical masses colliding elastically in 2D with one initially at rest, their post-collision trajectories are always mutually perpendicular ($\theta_1 + \theta_2 = 90^\circ$).
NCERT Official

Points to Ponder & Exam Tips

  1. Internal forces can do work and change the kinetic energy of a system, but they cannot change the total momentum of the system.
  2. Potential energy $V(x)$ is defined up to an arbitrary constant. Adding a constant to $V(x)$ does not change the physical force ($F = -dV/dx$).
  3. In inelastic collisions, KE is not conserved *during* or *after* the collision, but total momentum is conserved at all times.
  4. Work done by a force depends on the frame of reference because displacement is frame-dependent.
  5. The work-energy theorem represents a scalar equation, which is often easier to apply than the vector equations of Newton's laws.
05 / Practice Tests

3-Tier Practice Tests

Level 1 (Foundation), Level 2 (Application), and Level 3 (Challenge). Select options and submit to calculate score and review answers.

24 QUESTIONS
1. Dimensional formula of work and energy is:
2. One kilowatt-hour (1 kWh) in joules is:
3. If force is perpendicular to displacement (θ = 90°), work done is:
4. The work-energy theorem states that ΔK equals:
5. 1 horsepower (hp) is equal to:
6. Potential energy of an ideal spring of constant k extended by x is:
7. Minimum speed at bottom of vertical circle of length L to loop is:
8. In completely inelastic collision, the colliding bodies:
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