Unit 4: Work, Energy, Power & Circular Motion

This unit is the conceptual hinge of JEE Mechanics: energy methods often solve problems that would be intractable by force methods. Expect 2–3 direct questions in JEE Main on collisions, work-energy, vertical circle; in JEE Advanced, energy and collisions almost always appear in multi-step problems (collision + circular motion, or spring + collision).

Typical question types:

• JEE Main: work done by a variable force, KE at the bottom of an incline, collision with a wall, conical pendulum, banking angle.
• JEE Advanced: energy diagrams (turning points, stable/unstable equilibrium), inelastic collisions with constraints (block on spring), vertical-circle minimum speed, non-uniform circular motion with friction.

Concept Map

• Work
  • Constant force, variable force, line integral
  • Work-energy theorem
• Energy
  • Kinetic energy
  • Conservative vs non-conservative forces
  • Potential energy: gravitational (near earth + general), spring, elastic
  • Conservation of mechanical energy
  • Energy diagrams: turning points, equilibrium types
• Power
  • Average and instantaneous; $P = \vec F\cdot\vec v$
• Collisions
  • 1D and 2D elastic & inelastic
  • Coefficient of restitution
  • Ballistic pendulum
• Centre of mass (intro)
• Circular Motion
  • Kinematics: $\omega, \alpha, v, a_c$
  • Dynamics: centripetal force, banking
  • Conical pendulum
  • Vertical circle (full analysis)
  • Non-uniform circular motion

Topic 1: Work

Sub-topic A: Constant force

For a constant force $\vec F$ acting on a particle that undergoes displacement $\vec d$:

$W = \vec F\cdot\vec d = F d\cos\theta.$

Units: joule (J). Work is a scalar, can be positive, negative, or zero.

Sub-topic B: Variable force — line integral

For a force $\vec F(\vec r)$ depending on position, work along path $C$ is

$W = \int_C \vec F\cdot d\vec r.$

In 1D: $W = \int F(x)\,dx$ — area under $F$-$x$ graph.

Sub-topic C: Examples

Spring work. For a spring stretched from $x_1$ to $x_2$, the work done by the spring on the mass is

$W = \int_{x_1}^{x_2} (-kx)\,dx = -\tfrac12 k(x_2^2 - x_1^2).$

Gravity, near earth. $W_g = -mg\Delta y$ (negative if $\Delta y>0$, i.e. going up).

Worked Examples (JEE Main level)

Example 1.1. A force $F = 3x^2$ N acts on a particle moving from $x=0$ to $x=2$ m. Work done?

$W = \int_0^2 3x^2\,dx = x^3\big\vert _0^2 = 8$ J.

Example 1.2. A spring of $k=200$ N/m is stretched from natural length by $0.1$ m and then by another $0.1$ m. Work done by external agent in the second stretch?

$W_{ext} = -W_{spring} = \tfrac12 k(x_2^2 - x_1^2) = \tfrac12\cdot 200\cdot (0.04 - 0.01) = 100\cdot 0.03 = 3$ J.

Topic 2: Work-Energy Theorem and Energy

Sub-topic A: Kinetic energy

$K = \tfrac12 m v^2.$

Sub-topic B: Work-energy theorem

For any particle, net work done = change in KE:

$W_{net} = \Delta K = \tfrac12 m v_f^2 - \tfrac12 m v_i^2.$

Derivation. $\vec F = m\,d\vec v/dt$. $W = \int \vec F\cdot d\vec r = \int m\,d\vec v/dt\cdot\vec v\,dt = m\int \vec v\cdot d\vec v = \tfrac12 m v^2\big\vert _i^f.$

Sub-topic C: Conservative vs non-conservative

A force $\vec F$ is conservative if the work it does is path-independent — equivalently, if $\oint \vec F\cdot d\vec r=0$ for every closed loop, or if there exists a scalar $U$ such that

$\vec F = -\nabla U,\quad \text{1D:}\;F = -\frac{dU}{dx}.$

Examples: gravity, spring, electrostatic. Non-conservative: friction, viscous drag.

Sub-topic D: Potential energy

• Gravity near earth: $U = mgy$ (choosing $y=0$ as reference).
• Gravity general (point masses): $U = -GMm/r$ (zero at infinity).
• Spring: $U = \tfrac12 k x^2$.
• Elastic (e.g. stretched wire): $U = \tfrac12\cdot\text{stress}\cdot\text{strain}\cdot V = \tfrac12 Y\varepsilon^2 V$.

Sub-topic E: Conservation of mechanical energy

If only conservative forces do work, $K + U =$ const.

If non-conservative forces are present:

$\Delta K + \Delta U = W_{nc}.$

Sub-topic F: Energy diagrams

Given $U(x)$, the particle's motion is bounded between turning points where $E = U$ (so $K=0$). Equilibrium points are where $dU/dx=0$:

• Stable: $d^2U/dx^2 > 0$ (local minimum).
• Unstable: $d^2U/dx^2 < 0$ (local maximum).
• Neutral: $d^2U/dx^2 = 0$.

Worked Examples (JEE Main level)

Example 2.1. A $0.5$ kg ball is dropped from $20$ m height. Speed just before hitting ground (ignore air resistance):

$\tfrac12 mv^2 = mgh\Rightarrow v=\sqrt{2gh}=20$ m/s.

Example 2.2. A block of $2$ kg moves at $5$ m/s on a rough floor ($\mu=0.25$). Distance before stopping ($g=10$)?

By WE theorem: $-\mu mg\cdot d = -\tfrac12 mv^2\Rightarrow d = v^2/(2\mu g) = 25/5 = 5$ m.

Example 2.3. Spring of $k=400$ N/m. Block of $1$ kg compresses it $0.1$ m and is released on smooth floor. Speed at natural length?

$\tfrac12 k x^2 = \tfrac12 m v^2\Rightarrow v = x\sqrt{k/m} = 0.1\sqrt{400}=2$ m/s.

Worked Examples (JEE Advanced level)

Example 2.A1. $U(x) = U_0(x^2/a^2 - 1)^2$. Find equilibrium positions and classify.

$dU/dx = U_0\cdot 2(x^2/a^2-1)\cdot 2x/a^2 = 0$ at $x=0$ or $x=\pm a$. $d^2 U/dx^2$: messy; alternatively note $U$ has minima at $x=\pm a$ (where $U=0$) and maximum at $x=0$ (where $U=U_0$). So $x=\pm a$ are stable, $x=0$ is unstable (double-well potential).

Example 2.A2. A ball is released from height $h$ above a vertical spring of constant $k$ (rest position at floor). Find max compression $x$.

Energy: $mg(h+x) = \tfrac12 k x^2$. Quadratic: $kx^2 - 2mgx - 2mgh = 0$. Solve.

Topic 3: Power

Sub-topic A: Definitions

• Average power: $\bar P = W/\Delta t$.
• Instantaneous power: $P = dW/dt = \vec F\cdot\vec v$.

Unit: watt = J/s = kg·m²/s³.

1 hp = 746 W.

Worked Examples (JEE Main level)

Example 3.1. A pump lifts water at $20$ kg/s to a height of $10$ m. Power ($g=10$)?

$P = (dm/dt)\cdot g h = 20\cdot10\cdot10 = 2000$ W $=2$ kW.

Example 3.2. Engine delivers constant power $P$ to a car of mass $m$ starting from rest. Find velocity as a function of $x$.

$P = F v = m v\,dv/dx\cdot v = m v^2\,dv/dx$. So $\int v^2\,dv = (P/m)\,dx\Rightarrow v^3/3 = Px/m\Rightarrow v = (3Px/m)^{1/3}$.

Topic 4: Collisions

Sub-topic A: Types

• Elastic: KE conserved AND momentum conserved.
• Inelastic: momentum conserved, KE not conserved.
• Perfectly inelastic: bodies stick together after collision.

Sub-topic B: Coefficient of restitution $e$

For collision along a line,

$e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}} = \frac{v_2'-v_1'}{v_1-v_2}.$

$e=1$: elastic; $0\le e< 1$: inelastic; $e=0$: perfectly inelastic.

Sub-topic C: 1D collision formulas

Two bodies $m_1, m_2$ with initial velocities $u_1, u_2$. Coefficient $e$.

Momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.

Restitution: $v_2 - v_1 = e(u_1 - u_2)$.

Solving:

$v_1 = \frac{(m_1 - e m_2) u_1 + (1+e) m_2 u_2}{m_1 + m_2},$

$v_2 = \frac{(m_2 - e m_1) u_2 + (1+e) m_1 u_1}{m_1 + m_2}.$

Elastic case $e=1$:

$v_1 = \frac{m_1-m_2}{m_1+m_2} u_1 + \frac{2 m_2}{m_1+m_2} u_2,$

$v_2 = \frac{m_2-m_1}{m_1+m_2} u_2 + \frac{2 m_1}{m_1+m_2} u_1.$

Special cases:

• $m_1 = m_2$: velocities are exchanged ($v_1 = u_2$, $v_2 = u_1$).
• $m_2$ at rest, $m_1\ll m_2$: $v_1 \approx -u_1$ (bounces back), $v_2\approx 0$ (heavy stays).
• $m_2$ at rest, $m_1\gg m_2$: $v_1\approx u_1$ (continues), $v_2\approx 2u_1$.

Sub-topic D: KE loss in inelastic collision

For 1D collision:

$\Delta K = -\frac{m_1 m_2}{2(m_1+m_2)}(1 - e^2)(u_1 - u_2)^2.$

For perfectly inelastic ($e=0$):

$\Delta K_{max,\text{loss}} = -\frac{m_1 m_2}{2(m_1+m_2)}(u_1 - u_2)^2.$

Sub-topic E: 2D collisions & ballistic pendulum

In 2D, momentum conserves componentwise. Often: one body initially at rest, the other moves; after collision, they move at angles $\theta_1, \theta_2$. Three unknowns ($v_1', v_2', \theta_2$ say) — need three equations: two momentum (x, y) and either KE conservation (elastic) or restitution (inelastic).

Ballistic pendulum. A bullet of mass $m$ with speed $u$ embeds in a block of mass $M$ on a string. After collision, the block-bullet rises to height $h$.

Inelastic collision: $mu = (m+M) v\Rightarrow v = mu/(m+M)$.

Then energy: $\tfrac12 (m+M) v^2 = (m+M) g h\Rightarrow h = v^2/(2g) = m^2 u^2/[2g(m+M)^2]$.

So $u = \dfrac{m+M}{m}\sqrt{2gh}$.

Worked Examples (JEE Main level)

Example 4.1. A $2$ kg ball moving at $10$ m/s collides elastically with a stationary $3$ kg ball. Final velocities?

$m_1 = 2, m_2 = 3$. $v_1 = (2-3)/5\cdot 10 = -2$ m/s. $v_2 = 2\cdot 2/5\cdot 10 = 8$ m/s.

Example 4.2. A bullet of $20$ g with speed $400$ m/s embeds in a $2$ kg block on a smooth surface. Final speed?

$v = 0.02\cdot 400/2.02 \approx 3.96$ m/s.

Example 4.3. A ball is dropped from $5$ m onto a horizontal floor; $e = 0.6$. Height of first bounce?

After bounce, speed = $e\sqrt{2gh} = 0.6\sqrt{100}=6$ m/s. Height $=v^2/(2g) = 36/20 = 1.8$ m.

Worked Examples (JEE Advanced level)

Example 4.A1. A ball of mass $m$ hits a stationary ball of mass $2m$ obliquely; the collision is elastic. After the collision, the first ball moves at $90^\circ$ to its original direction. Find the angle the second ball makes.

Momentum (along initial): $m u = m v_1\cos90^\circ + 2m v_2\cos\theta_2 = 2m v_2\cos\theta_2$.

Perpendicular: $0 = m v_1\sin90^\circ - 2m v_2\sin\theta_2\Rightarrow v_1 = 2 v_2\sin\theta_2$.

Elastic: $\tfrac12 m u^2 = \tfrac12 m v_1^2 + \tfrac12(2m) v_2^2\Rightarrow u^2 = v_1^2 + 2 v_2^2$.

From first: $u = 2 v_2\cos\theta_2$. From second: $v_1 = 2 v_2\sin\theta_2$. Substitute:

$4 v_2^2\cos^2\theta_2 = 4 v_2^2\sin^2\theta_2 + 2 v_2^2\Rightarrow 4\cos 2\theta_2 = 2\Rightarrow \cos 2\theta_2 = 1/2\Rightarrow \theta_2 = 30^\circ$.

Example 4.A2. A block of $1$ kg moving at $6$ m/s on a smooth floor strikes a spring (constant $k = 100$ N/m) attached to a wall. Maximum compression?

$\tfrac12 m v^2 = \tfrac12 k x^2\Rightarrow x = v\sqrt{m/k} = 6/\sqrt{100} = 0.6$ m.

Example 4.A3 (Chained collisions). Three balls of equal mass on a smooth line. Ball 1 moves at $v_0$, balls 2 and 3 are at rest. Ball 1 collides elastically with ball 2, then ball 2 with ball 3. Final velocities?

After 1st (equal masses, elastic): ball 1 stops, ball 2 moves at $v_0$.

After 2nd: ball 2 stops, ball 3 moves at $v_0$.

Net: ball 1, 2 at rest; ball 3 moves at $v_0$. (Classic Newton's cradle behaviour.)

Topic 5: Centre of Mass (Introduction)

Sub-topic A: Definition

For a system of particles:

$\vec R_{cm} = \frac{\sum m_i \vec r_i}{\sum m_i},\quad \vec v_{cm} = \frac{\sum m_i \vec v_i}{\sum m_i}.$

For a continuous body: $\vec R_{cm} = \dfrac{1}{M}\int \vec r\,dm$.

Sub-topic B: Newton's law for CM

$\vec F_{ext} = M \vec a_{cm}.$

Internal forces (between particles) do not change $\vec v_{cm}$. Hence in any internal explosion/collision, $\vec v_{cm}$ remains the same as before.

(More detailed treatment of CM for extended bodies in Unit 5.)

Topic 6: Circular Motion

Sub-topic A: Kinematics of uniform circular motion

A particle moves on a circle of radius $r$ at constant speed $v$. Angular speed $\omega = v/r$. Period $T = 2\pi/\omega = 2\pi r/v$. Frequency $\nu = 1/T$.

Position: $\vec r = r(\cos\omega t,\sin\omega t)$.

Velocity: $\vec v = r\omega(-\sin\omega t,\cos\omega t)$ — tangential, magnitude $v=r\omega$.

Acceleration: $\vec a = -r\omega^2(\cos\omega t,\sin\omega t) = -\omega^2\vec r$ — centripetal, magnitude $a_c=v^2/r = r\omega^2$.

Sub-topic B: Dynamics — centripetal force

Net inward (centripetal) force = $mv^2/r$. This is NOT a new force; it is provided by tension, gravity, normal, friction etc.

Sub-topic C: Banking of roads

A car turns on a circular road of radius $r$, banked at angle $\theta$.

No friction. Normal $N$ provides centripetal $N\sin\theta = mv^2/r$, and $N\cos\theta = mg$. So

$\tan\theta = \frac{v^2}{rg}\Rightarrow v = \sqrt{rg\tan\theta}.$

With friction, max speed (about to slip outward, friction inward, down the bank):

$N\sin\theta + f\cos\theta = mv^2/r$, $N\cos\theta - f\sin\theta = mg$, $f = \mu N$.

$v_{\max} = \sqrt{rg\,\frac{\tan\theta + \mu}{1 - \mu\tan\theta}}.$

Min speed (about to slip inward, friction up the bank):

$v_{\min} = \sqrt{rg\,\frac{\tan\theta - \mu}{1 + \mu\tan\theta}}.$

(For $\tan\theta>\mu$; else $v_{\min}=0$.)

Unbanked road. $\theta=0$: $v_{\max}=\sqrt{\mu rg}$.

Sub-topic D: Conical pendulum

A bob of mass $m$ on string of length $\ell$ moves in a horizontal circle of radius $r$, string making angle $\theta$ with vertical.

$T\cos\theta = mg$, $T\sin\theta = m\omega^2 r = m\omega^2\ell\sin\theta$.

So $\omega = \sqrt{g/(\ell\cos\theta)}$, period $\mathcal T = 2\pi\sqrt{\ell\cos\theta/g}$.

Sub-topic E: Vertical circle

A bob on a string of length $\ell$ swings in a vertical circle. At angle $\theta$ from the lowest point:

Tangential: $-mg\sin\theta$ (back to lowest).

Radial: $T - mg\cos\theta = mv^2/\ell$ (centripetal).

Energy: $\tfrac12 m v_L^2 = \tfrac12 m v^2 + mg\ell(1-\cos\theta)$, where $v_L$ is speed at the lowest.

At top ($\theta=180^\circ$): $T_{top} + mg = mv_{top}^2/\ell$. For minimum speed to complete loop, $T_{top}\ge 0\Rightarrow v_{top}^2\ge g\ell$, i.e. $v_{top,\min} = \sqrt{g\ell}$.

Energy gives $v_{L,\min}^2 = v_{top}^2 + 4g\ell = 5g\ell$, so $v_{L,\min} = \sqrt{5g\ell}$.

At horizontal ($\theta=90^\circ$): $T_h = mv_h^2/\ell$ and $v_h^2 = v_L^2 - 2g\ell$.

For minimum case: $v_h^2 = 5g\ell - 2g\ell = 3g\ell$, $T_h = 3mg$.

Tensions at minimum case: $T_L = 6mg$, $T_h = 3mg$, $T_{top} = 0$.

Death well (mauka motorcycle in a vertical drum). Normal provides centripetal, friction provides weight support. Min speed: $\mu N \ge mg$ and $N=mv^2/r$. So $v_{\min}=\sqrt{gr/\mu}$.

Sub-topic F: Non-uniform circular motion

If speed varies with time, total acceleration has two components:

• Tangential $a_t = dv/dt = r\alpha$ (along velocity).
• Centripetal $a_c = v^2/r$ (toward centre).

Total $\vert \vec a\vert = \sqrt{a_t^2 + a_c^2}$.

Worked Examples (JEE Main level)

Example 6.1. A car moves at $20$ m/s on a circular road of radius $50$ m. Min coefficient of friction (unbanked)?

$\mu\ge v^2/(rg) = 400/500 = 0.8$.

Example 6.2. A pendulum bob of $0.2$ kg moves in a horizontal circle (conical pendulum) with string length $1$ m at $\theta=60^\circ$ with vertical. Find period and tension.

$\mathcal T = 2\pi\sqrt{\ell\cos\theta/g} = 2\pi\sqrt{0.5/10} \approx 1.4$ s. $T = mg/\cos\theta = 2/0.5 = 4$ N.

Example 6.3. A ball is tied to a string of length $1$ m and revolved in vertical circle. Minimum speed at the lowest point ($g=10$)?

$v_{\min} = \sqrt{5g\ell}=\sqrt{50}\approx 7.07$ m/s.

Worked Examples (JEE Advanced level)

Example 6.A1. A particle of mass $m$ moves on the inside of a smooth vertical circular track of radius $R$, starting from rest at a height $h$ above the bottom. Find min $h$ such that the particle completes the loop.

Speed at top: from energy, $\tfrac12 m v_{top}^2 = mg h - mg(2R)\Rightarrow v_{top}^2 = 2g(h-2R)$. Min condition: $v_{top}^2 = gR\Rightarrow h = 5R/2$.

Example 6.A2. A small bead is threaded on a smooth vertical circular wire of radius $R$. It is given a small push from the top. Where on the circle does it leave the wire?

Energy at angle $\theta$ from the top (along the inside of the circle): $\tfrac12 m v^2 = mg R(1 - \cos\theta)$ (using top as reference).

Normal (outward from centre): $N + mg\cos\theta = mv^2/R = 2mg(1-\cos\theta)\Rightarrow N = mg(2 - 3\cos\theta)$.

For the bead inside the wire, $N$ can be both inward (from outer wire) or outward (from inner wire). On a circular wire (groove), the bead doesn't leave at all — it's constrained. But on the outside of a sphere/sphere surface (ball on outer dome), the bead leaves when $N=0\Rightarrow\cos\theta = 2/3$, height fallen $= R(1-\cos\theta) = R/3$.

Example 6.A3 (Non-uniform CM). A car moves on a circular track of radius $R$ such that its speed increases at constant rate $a_t$. Find the angle through which it travels before the magnitude of total acceleration equals $\sqrt 2\,a_t$.

$\vert \vec a\vert = \sqrt{a_t^2 + a_c^2}=\sqrt 2 a_t\Rightarrow a_c=a_t\Rightarrow v^2/R = a_t$.

Starting from rest, $v^2 = 2 a_t s$ where $s = R\phi$ ($\phi$ = angle subtended). So $2 a_t R \phi/R = a_t\Rightarrow \phi = 1/2$ rad.

Problem-Solving Heuristics

1. Identify "energy" type problems: if forces depend on position, conservation of energy is far simpler than F=ma.
2. Always check whether the relevant forces are conservative before invoking $\Delta KE + \Delta PE = 0$.
3. For collisions, write momentum before deciding whether to use $e$ or KE conservation.
4. The CM moves under external forces only: $\vec F_{ext} = M\vec a_{cm}$. Internal forces (collisions, explosions) don't change $\vec v_{cm}$.
5. Circular motion: always identify the net inward force equation; never write "centrifugal" force as a real force in an inertial frame.
6. For vertical circle, the minimum speed at top is $\sqrt{gR}$ (string), but for a rod or rail it's $0$ (the rod can push as well as pull).
7. Banking with friction: when speed is too low, friction acts up; too high, friction acts down. Use this to decide signs.
8. For non-uniform circular motion, separate tangential and radial equations; never mix.
9. Energy diagrams: read off turning points from $E = U(x)$; equilibria from $dU/dx = 0$; stability from $d^2 U/dx^2$.

Common Traps & Mistakes

• Work done by friction on a moving block: $-fd$, not $-f\Delta x_{rel}$. (For block alone, $d$ is its displacement; for block on a moving surface, you must use rel. displacement to compute internal heat = $f\cdot d_{\text{rel}}$.)
• Confusing work by all forces with work by net force — they're equal.
• Spring PE is $\tfrac12 kx^2$, not $\tfrac12 k\vert x\vert$.
• Min speed at top of vertical circle: $\sqrt{gR}$ ONLY for a string; for a rod/rail/track, the min speed is $0$ — the constraint can push.
• "Velocity exchanged" in elastic collision is only for equal masses.
• Coefficient of restitution applies to components along the line of impact in 2D — not to the full velocity vectors.
• In banking + friction: when the road is banked, friction's direction depends on whether speed is above/below the "no-friction" speed $\sqrt{gr\tan\theta}$.
• "Centripetal" is the direction, not a separate force. Don't add centripetal as a force in the FBD; rather, recognize that the net inward force = $mv^2/r$.
• For the conical pendulum, $\theta\to 90^\circ$ requires $\omega\to\infty$.

Quick Revision Card

• $W = \int \vec F\cdot d\vec r$, work-energy theorem $W_{net} = \Delta K$.
• $F = -dU/dx$ for conservative forces.
• Spring PE $= \tfrac12 k x^2$. Gravity (near earth) $U=mgy$.
• $P = \vec F\cdot\vec v$.
• Elastic 1D: equal masses exchange velocities.
• KE loss in inelastic: $\Delta K = -\tfrac{m_1 m_2}{2(m_1+m_2)}(1-e^2)(u_1-u_2)^2$.
• Banking (no friction): $\tan\theta = v^2/(rg)$.
• Conical pendulum: $\mathcal T = 2\pi\sqrt{\ell\cos\theta/g}$.
• Vertical circle (string): $v_L^{min} = \sqrt{5gR}$, $v_{top}^{min}=\sqrt{gR}$; tensions $6mg, 3mg, 0$.
• Centripetal: $a_c = v^2/R = \omega^2 R$.

Formula Sheet