Concept Card: Mechanics Master Formula Blueprint (Units 1 to 6)
1. Structural Integration of the Mechanics Curriculum:
The comprehensive Mechanics syllabus for JEE Main and JEE Advanced synthesizes six foundational units:
- Unit 1 (Units & Measurements): Dimensional formulas, Principle of Homogeneity, error propagation rules, and vernier/screw gauge instrument calibrations.
- Units 2 & 3 (Kinematics in 1D & 2D): Differential calculus equations of motion, parabolic trajectory geometry, and relative reference frames.
- Unit 4 (Laws of Motion & Friction): Newton’s laws, inertial vs non-inertial pseudo forces, string and wedge constraints, and static/kinetic friction boundaries.
- Unit 5 (Work, Energy & Power): Work-energy theorem, conservative potential functions, equilibrium stability, and vertical circular motion thresholds.
- Unit 6 (System of Particles & Rotational Motion): Center of mass, parallel/perpendicular axis theorems, angular momentum conservation, and pure rolling dynamics.
- Unit 6 / Gravitation: Inverse-square law, spherical mass potentials, orbital satellite mechanics, Kepler’s laws, and escape dynamics.
2. High-Yield Master Formula Sheet across Mechanics
| Domain / Topic | Core Governing Formula | Physical Context & Constraints |
|---|---|---|
| Error Propagation | $\frac{\Delta Z}{Z} = p \frac{\Delta A}{A} + q \frac{\Delta B}{B} + r \frac{\Delta C}{C}$ | For power formulations $Z = \frac{A^p B^q}{C^r}$; valid for small fractional errors ($\le 10\%$). |
| Calculus Kinematics | $a = \frac{dv}{dt} = v \frac{dv}{dx} = \frac{d^2 x}{dt^2}$ | Integrate $v dv = a(x) dx$ for position-dependent acceleration. |
| Projectile Trajectory | $y = x \tan\theta \left(1 – \frac{x}{R}\right)$ | Range $R = \frac{u^2 \sin 2\theta}{g}$, maximum height $H = \frac{u^2 \sin^2\theta}{2g}$. |
| Virtual Work Constraint | $\sum \vec{T} \cdot \vec{a}_i = 0$ | For ideal, inextensible string-pulley systems. |
| Vertical Circle Threshold | $u_{\min} = \sqrt{5 g L}, \quad \Delta T = 6 m g$ | Minimum launch speed for complete loop with string; $\Delta T = T_{\text{bottom}} – T_{\text{top}}$. |
| Pure Rolling Acceleration | $a_{\text{cm}} = \frac{g \sin\theta}{1 + k^2 / R^2}$ | Down an incline $\theta$; friction $f = \frac{M g \sin\theta}{1 + R^2 / k^2} \le \mu_s N$. |
| Angular Momentum | $\vec{L}_O = \vec{r}_{\text{cm}} \times M \vec{v}_{\text{cm}} + I_{\text{cm}} \vec{\omega}$ | Conserved about pivot $O$ if $\sum \vec{\tau}_{\text{ext}, O} = \vec{0}$. |
| Solid Sphere Potential | $V_{\text{in}}(r) = -\frac{G M}{2 R^3}(3 R^2 – r^2)$ | At the center $r = 0$: $V_c = -1.5 \frac{G M}{R} = 1.5 V_{\text{surface}}$. |
| Satellite Mechanics | $E = -K = \frac{U}{2} = -\frac{G M m}{2 r}$ | Binding energy $BE = -E = K$; escape velocity $v_e = \sqrt{2} v_o$. |
3. Critical Insights & Exam Traps in JEE Mechanics:
- Trap 1 (Average Speed vs. Magnitude of Average Velocity): Never assume $\langle v \rangle = |\langle \vec{v} \rangle|$ unless motion is strictly 1D without reversal. Average speed is total distance divided by total time.
- Trap 2 (Direction of Friction in Rolling): In pure rolling on a flat surface under applied force $F$ at height $h$ above CM, friction vanishes when $h = I_{\text{cm}} / (M R)$ (for a solid cylinder, $h = R/2$).
- Trap 3 (Atwood Machine Pulley Inertia): When a pulley has mass $M$, rotational inertia reduces system acceleration: $a = \frac{(m_1 – m_2)g}{m_1 + m_2 + I/R^2}$.
- Trap 4 (Gravitational Potential at Center): Even though field $E = 0$ at the center of a solid sphere, potential is non-zero and at its minimum negative value: $V_c = -1.5 G M / R$.
Solved Examples
Example 1 (Calculus Kinematics & Motion Reversal – 1D Kinematics):
A particle moves along the $x$-axis with time-dependent acceleration $a(t) = (6t – 4)\text{ m/s}^2$. At $t = 0$, its initial position is $x_0 = 2.0\text{ m}$ and initial velocity is $v_0 = -4.0\text{ m/s}$.
(a) Determine the velocity $v(t)$ and position $x(t)$ as functions of time.
(b) Find the time $t_1 > 0$ when the particle momentarily comes to rest.
(c) Calculate the net displacement and the total distance travelled by the particle between $t = 0$ and $t = 3.0\text{ s}$.
Solution:
(a) Velocity and Position Functions:
$$v(t) = v_0 + \int_0^t a(t’) dt’ = -4 + \int_0^t (6t’ – 4) dt’ = -4 + 3t^2 – 4t = \mathbf{3t^2 – 4t – 4\text{ m/s}}$$
$$x(t) = x_0 + \int_0^t v(t’) dt’ = 2 + \int_0^t (3t’^2 – 4t’ – 4) dt’ = \mathbf{t^3 – 2t^2 – 4t + 2\text{ m}}$$
(b) Instantaneous Rest ($v = 0$):
$$3t^2 – 4t – 4 = 0 \implies (3t + 2)(t – 2) = 0$$
Since time $t > 0$, the particle comes to rest at $\mathbf{t_1 = 2.0\text{ s}}$.
(c) Displacement and Total Distance ($t = 0$ to $t = 3.0\text{ s}$):
At $t = 0$: $x(0) = 2.0\text{ m}$.
At $t = 2.0\text{ s}$: $x(2) = (2)^3 – 2(2)^2 – 4(2) + 2 = 8 – 8 – 8 + 2 = -6.0\text{ m}$.
At $t = 3.0\text{ s}$: $x(3) = (3)^3 – 2(3)^2 – 4(3) + 2 = 27 – 18 – 12 + 2 = -1.0\text{ m}$.
– Net Displacement:
$$\Delta x = x(3) – x(0) = -1.0 – 2.0 = -\mathbf{3.0\text{ m}}$$
– Total Distance Travelled:
From $t = 0$ to $2\text{ s}$: $|x(2) – x(0)| = |-6.0 – 2.0| = 8.0\text{ m}$ (moving in $-x$ direction).
From $t = 2$ to $3\text{ s}$: $|x(3) – x(2)| = |-1.0 – (-6.0)| = 5.0\text{ m}$ (moving in $+x$ direction).
Total distance = $8.0 + 5.0 = \mathbf{13.0\text{ m}}$.
Example 2 (Accelerating Wedge & Friction Range – Newton’s Laws & Friction):
A block of mass $m = 2.0\text{ kg}$ is placed on the inclined surface of a wedge of angle $\theta = 37^\circ$ ($\sin 37^\circ = 0.6, \cos 37^\circ = 0.8$). The coefficient of static friction between the block and the incline is $\mu_s = 0.50$ ($g = 10\text{ m/s}^2$). The wedge is given a rightward horizontal acceleration $a_0$.
Determine the allowable range of acceleration $a_0$ such that the block remains stationary relative to the inclined plane.
Solution:
In the non-inertial reference frame of the wedge accelerating rightward with $a_0$, the block experiences a leftward pseudo force $m a_0$.
Normal force perpendicular to the incline:
$$N = m g \cos\theta + m a_0 \sin\theta = m(10 \times 0.8 + a_0 \times 0.6) = m(8.0 + 0.6 a_0)$$
Maximum static friction available:
$$f_{s, \max} = \mu_s N = 0.50 m(8.0 + 0.6 a_0) = m(4.0 + 0.3 a_0)$$
Forces parallel to the incline:
– Downward component of gravity: $m g \sin\theta = 0.6 m g = 6.0 m$.
– Upward component of pseudo force: $m a_0 \cos\theta = 0.8 m a_0$.
Case 1: Minimum Acceleration $a_{0, \min}$ (Block tends to slip down the incline):
Friction acts up the incline to prevent downward sliding:
$$m g \sin\theta – m a_0 \cos\theta \le f_{s, \max} \implies 6.0 m – 0.8 m a_0 \le m(4.0 + 0.3 a_0)$$
$$6.0 – 4.0 \le (0.8 + 0.3) a_0 \implies 2.0 \le 1.1 a_0 \implies \mathbf{a_{0, \min} = \frac{20}{11}\text{ m/s}^2 \approx 1.82\text{ m/s}^2}$$
Case 2: Maximum Acceleration $a_{0, \max}$ (Block tends to slip up the incline):
Friction acts down the incline to prevent upward sliding:
$$m a_0 \cos\theta – m g \sin\theta \le f_{s, \max} \implies 0.8 m a_0 – 6.0 m \le m(4.0 + 0.3 a_0)$$
$$(0.8 – 0.3) a_0 \le 6.0 + 4.0 \implies 0.5 a_0 \le 10.0 \implies \mathbf{a_{0, \max} = 20.0\text{ m/s}^2}$$
Allowable Range:
$$\mathbf{\frac{20}{11}\text{ m/s}^2 \le a_0 \le 20.0\text{ m/s}^2}$$
Example 3 (Rotational Dynamics of Rolling Body with Hanging Load):
A uniform solid cylinder of mass $M = 4.0\text{ kg}$ and radius $R$ is placed on a rough horizontal floor. A light string wrapped around an axle through its center of mass passes over a light frictionless pulley and supports a suspended mass $m = 2.0\text{ kg}$. The cylinder rolls without slipping on the floor ($g = 10\text{ m/s}^2$).
(a) Derive the linear acceleration $a$ of the system.
(b) Calculate the string tension $T$ and the friction force $f$ acting on the cylinder.
Solution:
Let $a$ be the forward linear acceleration of the center of mass of the cylinder, and $\alpha = a/R$ be its angular acceleration (clockwise).
Because the string is connected at the center of mass, the suspended block moves downward with the same acceleration $a$.
Equations of motion:
– For the suspended block ($m$):
$$m g – T = m a \implies T = m(g – a)$$
– For the cylinder ($M$):
Translational equation along horizontal:
$$T – f = M a$$
Rotational equation about center of mass:
$$\tau_{\text{cm}} = f R = I_{\text{cm}} \alpha = \left(\frac{1}{2} M R^2\right) \frac{a}{R} = \frac{1}{2} M R a \implies f = \frac{1}{2} M a$$
Substitute $f$ into translation:
$$T – \frac{1}{2} M a = M a \implies T = \frac{3}{2} M a$$
Equating the two expressions for tension $T$:
$$m(g – a) = \frac{3}{2} M a \implies m g = \left(m + \frac{3}{2} M\right) a$$
(a) Acceleration of the System:
$$a = \frac{m g}{m + 1.5 M} = \frac{(2.0)(10)}{2.0 + 1.5(4.0)} = \frac{20.0}{2.0 + 6.0} = \frac{20.0}{8.0} = \mathbf{2.5\text{ m/s}^2}$$
(b) Tension and Friction:
$$T = \frac{3}{2} M a = 1.5(4.0)(2.5) = \mathbf{15.0\text{ N}}$$
$$f = \frac{1}{2} M a = 0.5(4.0)(2.5) = \mathbf{5.0\text{ N}}$$
(Friction acts in the backward direction opposing translation).
Example 4 (Satellite Orbit Elevation & Work-Energy Balance – Gravitation):
A satellite of mass $m = 500\text{ kg}$ revolves around Earth in a circular orbit at altitude $h_1 = R$ ($r_1 = 2R$). Rocket thrusters are fired to boost the satellite into a higher circular orbit at altitude $h_2 = 2R$ ($r_2 = 3R$). (Take Earth’s radius $R = 6400\text{ km}$, surface gravity $g_0 = 9.8\text{ m/s}^2$).
(a) Determine the initial and final orbital speeds $v_1$ and $v_2$.
(b) Calculate the changes in kinetic energy $\Delta K$, potential energy $\Delta U$, and total energy $\Delta E$.
(c) Determine the work $W$ performed by the rocket engines.
Solution:
Recall $G M = g_0 R^2$.
(a) Orbital Speeds:
$$v_1 = \sqrt{\frac{G M}{r_1}} = \sqrt{\frac{g_0 R^2}{2R}} = \sqrt{\frac{g_0 R}{2}} = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{2}} = \mathbf{5600\text{ m/s} = 5.60\text{ km/s}}$$
$$v_2 = \sqrt{\frac{G M}{r_2}} = \sqrt{\frac{g_0 R^2}{3R}} = \sqrt{\frac{g_0 R}{3}} = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{3}} \approx \mathbf{4572\text{ m/s} = 4.57\text{ km/s}}$$
(b) Energy Calculations:
Total energy in orbit $r$: $E = -\frac{G M m}{2r} = -\frac{m g_0 R^2}{2r}$.
$$\mathbf{\Delta E = E_2 – E_1 = \frac{m g_0 R^2}{2}\left(\frac{1}{r_1} – \frac{1}{r_2}\right) = \frac{m g_0 R^2}{2}\left(\frac{1}{2R} – \frac{1}{3R}\right) = \frac{1}{12} m g_0 R}$$
$$\Delta E = \frac{1}{12}(500)(9.8)(6.4 \times 10^6) = \mathbf{+2.613 \times 10^9\text{ J} = +2.613\text{ GJ}}$$
Kinetic energy change:
$$\mathbf{\Delta K = -\Delta E = -2.613 \times 10^9\text{ J} = -2.613\text{ GJ}}$$
Potential energy change:
$$\mathbf{\Delta U = 2 \Delta E = +5.227 \times 10^9\text{ J} = +5.227\text{ GJ}}$$
(c) Work Performed by Engines:
$$W = \Delta E = \mathbf{+2.613 \times 10^9\text{ J} = +2.613\text{ GJ}}$$
Practice Worksheet: Full Mechanics Revision Mock Exam
Section A: Single Correct Questions (Q1 to Q5)
Q1. The dimensional formula of the characteristic impedance of free space $Z_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}}$ is identical to that of electrical resistance and is given by:
(A) $[M L^2 T^{-3} A^{-2}]$
(B) $[M L^2 T^{-2} A^{-1}]$
(C) $[M L^3 T^{-3} A^{-2}]$
(D) $[M^0 L^0 T^0 A^0]$
Q2. A projectile is projected horizontally with initial speed $u$ from the top of a high cliff. The time $t$ after launch at which its instantaneous velocity vector makes an angle of $45^\circ$ with the horizontal is:
(A) $\frac{u}{g}$
(B) $\frac{2u}{g}$
(C) $\frac{u}{2g}$
(D) $\frac{\sqrt{2}u}{g}$
Q3. In an Atwood machine, two blocks of masses $m_1 = 3.0\text{ kg}$ and $m_2 = 1.0\text{ kg}$ are connected by a light string running over a pulley of mass $M = 2.0\text{ kg}$ and radius $R$. Treating the pulley as a uniform solid disc ($I = \frac{1}{2} M R^2$) with no string slipping, the acceleration of the blocks is:
(A) $\frac{g}{3}$
(B) $\frac{2g}{5}$
(C) $\frac{g}{2}$
(D) $\frac{3g}{7}$
Q4. A position-dependent force $\vec{F} = -k(y\hat{i} + x\hat{j})$ acts on a particle moving in the $xy$-plane. The work done by this force as the particle traverses a circular arc of radius $R$ centered at the origin from $(R, 0)$ to $(0, R)$ in the first quadrant is:
(A) Zero
(B) $-k R^2$
(C) $-\frac{1}{2} k R^2$
(D) $\pi k R^2$
Q5. The escape velocity from the surface of a spherical planet of radius $R$ and uniform mean mass density $\rho$ is directly proportional to:
(A) $R \sqrt{\rho}$
(B) $R^2 \rho$
(C) $\sqrt{R \rho}$
(D) $\frac{R}{\sqrt{\rho}}$
Section B: One or More than One Correct Questions (Q6 to Q8)
Q6. Two blocks of masses $m_1$ and $m_2$ rest on a smooth horizontal floor, connected by an ideal massless spring of force constant $k$. A constant horizontal pull $F$ is applied to block $m_2$ directed away from $m_1$. Which of the following statements is/are correct?
(A) The acceleration of the center of mass of the system is $a_{\text{cm}} = \frac{F}{m_1 + m_2}$ at all times.
(B) The maximum extension attained by the spring during the motion is $x_{\max} = \frac{2 m_1 F}{k (m_1 + m_2)}$.
(C) In the center of mass frame, both blocks oscillate with angular frequency $\omega = \sqrt{\frac{k}{\mu}}$ where $\mu = \frac{m_1 m_2}{m_1 + m_2}$.
(D) At the instant of maximum spring extension, both blocks move with the same instantaneous velocity equal to $v_{\text{cm}}$.
Q7. Four symmetric bodies – a solid sphere ($S_1$), a solid cylinder ($S_2$), a thin spherical shell ($S_3$), and a thin ring ($S_4$) – each of mass $M$ and radius $R$, are released simultaneously from rest from the same height at the top of a rough inclined plane of inclination $\theta$. Assuming pure rolling without slipping for all bodies:
(A) The solid sphere reaches the bottom first with the highest linear speed.
(B) The ring reaches the bottom last with the lowest linear speed.
(C) The friction force acting on the ring is greater than that acting on the solid sphere.
(D) At any given vertical height down the incline, all four bodies have identical total mechanical energy.
Q8. A satellite describes an elliptical orbit around Earth with perigee distance $r_p$ and apogee distance $r_a$. Which of the following statements is/are correct?
(A) The areal velocity of the satellite remains constant throughout its orbit.
(B) The ratio of its speed at perigee to that at apogee is $\frac{v_p}{v_a} = \frac{r_a}{r_p}$.
(C) The total mechanical energy of the satellite is $E = -\frac{G M m}{r_p + r_a}$.
(D) The speed at perigee is $v_p = \sqrt{\frac{2 G M r_a}{r_p (r_p + r_a)}}$.
Section C: Numerical Value Type Questions (Q9 to Q10)
Q9. A particle of mass $m$ attached to an inextensible light string of length $L$ is projected horizontally from the lowest point of a vertical circle. The minimum projection speed required for the string to stay taut throughout the complete vertical circular loop is $u_{\min} = \sqrt{n g L}$. Determine the integer value of $n$.
Q10. A uniform thin rod of mass $M$ and length $L$ has a moment of inertia about a transverse axis passing through a point at distance $L/4$ from its center of mass given by $I = \frac{k}{48} M L^2$. Determine the integer value of $k$.
Comprehensive Solutions and Answer Key
Answer Key
| Question | Correct Answer | Core Physics Concept |
|---|---|---|
| Q1 | (A) | Impedance of Free Space Dimensions: $[M L^2 T^{-3} A^{-2}]$ (Resistance) |
| Q2 | (A) | Horizontal Projectile Velocity Angle: $\tan 45^\circ = g t / u = 1 \implies t = u/g$ |
| Q3 | (B) | Atwood Machine with Pulley Inertia: $a = \frac{(3-1)g}{3+1+1} = \frac{2g}{5}$ |
| Q4 | (A) | Exact Differential Conservative Work: $dW = -k d(xy) \implies W = 0$ |
| Q5 | (A) | Density-Radius Scaling of Escape Speed: $v_e = R \sqrt{\frac{8}{3}\pi G \rho} \propto R\sqrt{\rho}$ |
| Q6 | (A, B, C, D) | Coupled Two-Body Spring System Dynamics & Maximum Extension |
| Q7 | (A, B, C, D) | Rolling Race Down an Incline: Radius of Gyration Hierarchy |
| Q8 | (A, B, C, D) | Keplerian Elliptical Orbit Conservation Laws & Vis-Viva Linkage |
| Q9 | 5 | Vertical Circle Inextensible String Minimum Speed: $u = \sqrt{5 g L} \implies n = 5$ |
| Q10 | 7 | Parallel Axis Theorem: $I = \frac{1}{12} + \frac{1}{16} = \frac{7}{48} M L^2 \implies k = 7$ |
Detailed Explanations
Solution 1:
The characteristic impedance of free space is:
$$Z_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}}$$
Recall the dimensional formulas:
$[\mu_0] = [M L T^{-2} A^{-2}]$
$[\varepsilon_0] = [M^{-1} L^{-3} T^4 A^2]$
Dividing the two:
$$\frac{[\mu_0]}{[\varepsilon_0]} = \frac{[M L T^{-2} A^{-2}]}{[M^{-1} L^{-3} T^4 A^2]} = [M^2 L^4 T^{-6} A^{-4}]$$
Taking the square root:
$$[Z_0] = \sqrt{[M^2 L^4 T^{-6} A^{-4}]} = \mathbf{[M L^2 T^{-3} A^{-2}]}$$
This matches the dimensional formula for electrical resistance ($R = V / I$).
Correct Answer: (A)
Solution 2:
For a horizontal projectile with initial velocity $\vec{u} = u \hat{i}$:
– Horizontal velocity at time $t$: $v_x = u$ (constant).
– Vertical downward velocity at time $t$: $v_y = g t$.
When the velocity vector makes an angle of $45^\circ$ with the horizontal:
$$\tan 45^\circ = \frac{v_y}{v_x} = \frac{g t}{u} = 1 \implies \mathbf{t = \frac{u}{g}}$$
Correct Answer: (A)
Solution 3:
For an Atwood machine with pulley of mass $M$ and radius $R$ ($I = \frac{1}{2} M R^2$):
Equations of motion:
1. $m_1 g – T_1 = m_1 a$
2. $T_2 – m_2 g = m_2 a$
3. $(T_1 – T_2) R = I \alpha = \left(\frac{1}{2} M R^2\right) \frac{a}{R} \implies T_1 – T_2 = \frac{1}{2} M a$.
Adding (1) and (2):
$$(m_1 – m_2) g – (T_1 – T_2) = (m_1 + m_2) a$$
Substitute $T_1 – T_2 = \frac{1}{2} M a$:
$$(m_1 – m_2) g = \left(m_1 + m_2 + \frac{1}{2} M\right) a \implies a = \frac{(m_1 – m_2) g}{m_1 + m_2 + M / 2}$$
Given $m_1 = 3.0\text{ kg}, m_2 = 1.0\text{ kg}, M = 2.0\text{ kg}$:
$$a = \frac{(3.0 – 1.0) g}{3.0 + 1.0 + 2.0 / 2} = \frac{2.0 g}{4.0 + 1.0} = \mathbf{\frac{2 g}{5} = 0.4 g}$$
Correct Answer: (B)
Solution 4:
The work done by $\vec{F} = -k(y\hat{i} + x\hat{j})$ is:
$$W = \int \vec{F} \cdot d\vec{r} = \int -k (y dx + x dy) = -k \int d(x y) = -k [x y]_{(R, 0)}^{(0, R)}$$
At the initial point $(R, 0)$: $x y = R \times 0 = 0$.
At the final point $(0, R)$: $x y = 0 \times R = 0$.
$$W = -k (0 – 0) = \mathbf{0}$$
Because $\vec{F} = -\vec{\nabla}(k x y)$ is a conservative force and $U(R, 0) = U(0, R) = 0$, the net work done along any path connecting these points is zero.
Correct Answer: (A)
Solution 5:
The escape velocity from the surface of a spherical planet is:
$$v_e = \sqrt{\frac{2 G M}{R}}$$
Expressing mass $M$ in terms of uniform density $\rho$:
$$M = \frac{4}{3} \pi R^3 \rho$$
$$v_e = \sqrt{\frac{2 G \left(\frac{4}{3} \pi R^3 \rho\right)}{R}} = \sqrt{\frac{8}{3} \pi G \rho R^2} = R \sqrt{\frac{8}{3} \pi G \rho}$$
Therefore, $v_e \propto \mathbf{R \sqrt{\rho}}$.
Correct Answer: (A)
Solution 6:
– (A) True: The external horizontal force on the system is $F$. Hence $a_{\text{cm}} = \frac{F}{m_1 + m_2}$ at all times.
– (B) True: In the center of mass frame, reduced mass is $\mu = \frac{m_1 m_2}{m_1 + m_2}$. Effective force on $m_1$ is $F_1′ = \mu F / m_2$. Work-energy analysis gives $x_{\max} = \frac{2 m_1 F}{k (m_1 + m_2)}$.
– (C) True: In the CM frame, the two-body oscillation reduces to a single harmonic oscillator of effective mass $\mu$: $\omega = \sqrt{k / \mu}$.
– (D) True: At maximum extension, relative velocity between the two blocks is zero; therefore, both blocks move with the common velocity $v_{\text{cm}}$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
For pure rolling down an incline of height $h$:
$$a_{\text{cm}} = \frac{g \sin\theta}{1 + k^2 / R^2}, \quad v = \sqrt{\frac{2 g h}{1 + k^2 / R^2}}, \quad f = \frac{M g \sin\theta}{1 + R^2 / k^2}$$
Values of $k^2 / R^2$:
– Solid Sphere: $k^2 / R^2 = 2/5 = 0.40$ (least $k^2/R^2 \implies$ largest $a_{\text{cm}}, v$).
– Solid Cylinder: $k^2 / R^2 = 1/2 = 0.50$.
– Spherical Shell: $k^2 / R^2 = 2/3 \approx 0.67$.
– Ring: $k^2 / R^2 = 1.00$ (highest $k^2/R^2 \implies$ least $a_{\text{cm}}, v$).
– (A) True: Solid sphere has the highest acceleration and velocity; it arrives first.
– (B) True: Ring has the lowest acceleration and velocity; it arrives last.
– (C) True: $f = \frac{M g \sin\theta}{1 + R^2/k^2}$. For ring ($R^2/k^2 = 1$), $f = \frac{1}{2} M g \sin\theta$. For sphere ($R^2/k^2 = 2.5$), $f = \frac{2}{7} M g \sin\theta$. Since $1/2 > 2/7$, friction is greater on the ring.
– (D) True: Since static friction does zero work in pure rolling, mechanical energy is conserved for all four bodies: $E_{\text{total}} = M g h$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
– (A) True: Gravitational force is central; $\vec{\tau} = \vec{0} \implies$ areal velocity $\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$.
– (B) True: Angular momentum conservation: $m v_p r_p = m v_a r_a \implies \frac{v_p}{v_a} = \frac{r_a}{r_p}$.
– (C) True: Semi-major axis $a = \frac{r_p + r_a}{2}$. Total mechanical energy of an elliptical orbit is $E = -\frac{G M m}{2a} = -\frac{G M m}{r_p + r_a}$.
– (D) True: Combining energy conservation $E = \frac{1}{2} m v_p^2 – \frac{G M m}{r_p} = -\frac{G M m}{r_p + r_a}$ gives $v_p = \sqrt{\frac{2 G M r_a}{r_p (r_p + r_a)}}$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 9:
For a particle tied to an inextensible string of length $L$ moving in a vertical circle, tension at the highest point must be non-negative:
$$T_{\text{top}} \ge 0 \implies v_{\text{top}}^2 \ge g L$$
By conservation of mechanical energy between the lowest and highest points:
$$\frac{1}{2} m u^2 = \frac{1}{2} m v_{\text{top}}^2 + m g (2L) \implies u^2 = v_{\text{top}}^2 + 4 g L$$
For minimum speed:
$$u_{\min}^2 = g L + 4 g L = 5 g L \implies u_{\min} = \sqrt{5 g L}$$
Comparing with $\sqrt{n g L}$:
$$\mathbf{n = 5}$$
Correct Answer: 5
Solution 10:
Moment of inertia of a uniform thin rod about an axis through its center of mass perpendicular to its length:
$$I_{\text{cm}} = \frac{1}{12} M L^2$$
Distance of the new axis from the center of mass: $d = \frac{L}{4}$.
By the Parallel Axis Theorem:
$$I = I_{\text{cm}} + M d^2 = \frac{1}{12} M L^2 + M \left(\frac{L}{4}\right)^2 = \frac{1}{12} M L^2 + \frac{1}{16} M L^2$$
Finding a common denominator of 48:
$$I = \left(\frac{4}{48} + \frac{3}{48}\right) M L^2 = \mathbf{\frac{7}{48} M L^2}$$
Comparing with $\frac{k}{48} M L^2$:
$$\mathbf{k = 7}$$
Correct Answer: 7