Concept Card: Mechanics Core Revision – Units, Kinematics & Dynamics
1. Architectural Overview of Fundamental Mechanics:
The classical mechanics framework for JEE Main and JEE Advanced rests upon three interconnected structural pillars:
- Measurement & Dimensions: Establishing physical quantities, dimensional homogeneity, and error bounds.
- Kinematics (1D & 2D): Describing trajectories, rates of change, and relative frames of reference without reference to the forces causing them.
- Dynamics (Newton’s Laws & Friction): Formulating causal equations of motion, constraint geometry, and contact interaction thresholds.
2. Module 1: Units, Dimensions & Precision Measurement
(A) Dimensional Analysis & Principle of Homogeneity:
- Any physically valid equation $A + B = C$ requires that terms added, subtracted, or equated must possess identical dimensional formulas: $[A] = [B] = [C]$.
- Arguments of transcendental functions (trigonometric $\sin\theta$, exponential $e^x$, logarithmic $\ln x$) must be strictly dimensionless ($[M^0 L^0 T^0]$).
- High-Yield Universal Constants:
- Universal Gravitational Constant: $[G] = [M^{-1} L^3 T^{-2}]$
- Planck’s Constant: $[h] = [M L^2 T^{-1}]$
- Coefficient of Viscosity: $[\eta] = [M L^{-1} T^{-1}]$
- Permittivity of Free Space: $[\varepsilon_0] = [M^{-1} L^{-3} T^4 A^2]$
(B) Error Propagation & Precision Instruments:
- For a derived quantity $Z = \frac{A^p B^q}{C^r}$, the maximum fractional error is:
$$\mathbf{\frac{\Delta Z}{Z} = p \frac{\Delta A}{A} + q \frac{\Delta B}{B} + r \frac{\Delta C}{C}}$$
Caveat: This differential rule applies exclusively when fractional errors are small (typically $\le 10\%$). - Vernier Callipers:
$$\text{Least Count (LC)} = 1\text{ MSD} – 1\text{ VSD} = \left(1 – \frac{n-1}{n}\right)\text{MSD} = \frac{1}{n}\text{MSD}$$ - Screw Gauge:
$$\text{Least Count (LC)} = \frac{\text{Pitch}}{\text{Total Circular Scale Divisions}}$$
$$\text{True Reading} = \text{Main Scale Reading (MSR)} + [\text{Circular Scale Reading (CSR)} \times \text{LC}] – (\pm\text{Zero Error})$$
3. Module 2: Kinematics (Calculus, 1D Motion & Projectiles)
(A) Calculus-Based Differential Equations:
$$v = \frac{dx}{dt}, \quad a = \frac{dv}{dt} = v \frac{dv}{dx} = \frac{d^2 x}{dt^2}$$
To integrate variable acceleration:
- When $a = f(t)$: $\int_{v_0}^v dv = \int_0^t f(t) dt$.
- When $a = f(x)$: $\int_{v_0}^v v dv = \int_{x_0}^x f(x) dx$.
- When $a = f(v)$: $\int_{x_0}^x dx = \int_{v_0}^v \frac{v dv}{f(v)} \quad \text{and} \quad \int_0^t dt = \int_{v_0}^v \frac{dv}{f(v)}$.
(B) Projectile Motion & Relative Trajectories:
- Ground-to-Ground Projectile ($u, \theta$):
$$T = \frac{2 u \sin\theta}{g}, \quad H = \frac{u^2 \sin^2\theta}{2g}, \quad R = \frac{u^2 \sin 2\theta}{g}$$
Trajectory equation:
$$y = x \tan\theta \left(1 – \frac{x}{R}\right) = x \tan\theta – \frac{g x^2}{2 u^2 \cos^2\theta}$$ - Relative Motion Vector Formulation:
$$\vec{v}_{A/B} = \vec{v}_A – \vec{v}_B, \quad \vec{a}_{A/B} = \vec{a}_A – \vec{a}_B$$
River-Boat Shortest Path: Swimmer heads upstream at angle $\sin\alpha = \frac{u}{v}$ to drift zero, crossing speed $v_{\text{cross}} = \sqrt{v^2 – u^2}$.
4. Module 3: Newton’s Laws, Constraint Relations & Friction Dynamics
(A) Constraint Equations:
- String Constraint: For inextensible ropes passing over ideal pulleys, the virtual work done by tension vanishes:
$$\sum \vec{T} \cdot \vec{v}_i = 0 \implies \sum \vec{T} \cdot \vec{a}_i = 0$$ - Wedge Contact Constraint: Contacting surfaces cannot interpenetrate; therefore, accelerations and velocities perpendicular to the contact interface must be identical:
$$a_{1, \perp} = a_{2, \perp}$$
(B) Friction Mechanics (Static vs. Kinetic):
- Static Friction ($f_s$): Self-adjusting in magnitude and direction to oppose incipient motion up to a maximum threshold:
$$0 \le f_s \le f_{s, \max} = \mu_s N$$ - Kinetic Friction ($f_k$): Acts opposite to actual relative surface velocity:
$$f_k = \mu_k N$$ - Pseudo Force in Non-Inertial Frames: For an observer accelerating with $\vec{a}_0$:
$$\vec{F}_{\text{pseudo}} = -m \vec{a}_0$$
5. Critical Insights & Exam Traps in JEE:
- Trap 1 (Average Speed vs. Velocity): Average speed $\langle v \rangle = \frac{\text{Total Distance}}{\text{Total Time}}$, whereas average velocity $\langle \vec{v} \rangle = \frac{\Delta \vec{r}}{\Delta t}$. $\langle v \rangle \ge |\langle \vec{v} \rangle|$ always; equality holds strictly for rectilinear unidirectional motion.
- Trap 2 (Direction of Friction): Friction does not necessarily oppose motion; it strictly opposes relative slipping between contacting surfaces. Friction on the driving wheel of a car or on a walking person acts in the forward direction.
- Trap 3 (Tension & Normal Reaction Realizability): Tensions in flexible strings cannot be negative ($T \ge 0$). If a dynamic equation predicts $T < 0$, the string has gone slack. Similarly, normal forces must satisfy $N \ge 0$; $N = 0$ indicates loss of contact.
Solved Examples
Example 1 (Error Analysis & Instrument Least Count – JEE Main Level):
A physical quantity $P$ is calculated from measured parameters $A, B, C,$ and $D$ via the relation:
$$P = \frac{A^3 B^{1/2}}{C^2 D}$$
The percentage errors in the measurements of $A, B, C,$ and $D$ are $1.0\%, 2.0\%, 1.5\%,$ and $0.5\%$ respectively.
(a) Determine the maximum percentage error in the calculated value of $P$.
(b) A Vernier calliper has $1\text{ MSD} = 1\text{ mm}$, and $10$ vernier scale divisions coincide with $9$ main scale divisions. Find its least count in $\text{mm}$ and $\text{cm}$.
Solution:
(a) Maximum Percentage Error in $P$:
Taking the natural logarithm and differentiating:
$$\frac{\Delta P}{P} = 3\left(\frac{\Delta A}{A}\right) + \frac{1}{2}\left(\frac{\Delta B}{B}\right) + 2\left(\frac{\Delta C}{C}\right) + 1\left(\frac{\Delta D}{D}\right)$$
Multiplying by $100\%$ for worst-case additive errors:
$$\% \text{ error in } P = 3(1.0\%) + 0.5(2.0\%) + 2(1.5\%) + 1(0.5\%)$$
$$\% \text{ error in } P = 3.0\% + 1.0\% + 3.0\% + 0.5\% = \mathbf{7.5\%}$$
(b) Least Count of Vernier Calliper:
Given $10\text{ VSD} = 9\text{ MSD} \implies 1\text{ VSD} = 0.9\text{ MSD}$.
$$\text{Least Count} = 1\text{ MSD} – 1\text{ VSD} = 1\text{ MSD} – 0.9\text{ MSD} = 0.1\text{ MSD}$$
With $1\text{ MSD} = 1\text{ mm}$:
$$\mathbf{\text{LC} = 0.1\text{ mm} = 0.01\text{ cm}}$$
Example 2 (Calculus-Based Non-Uniform Deceleration – Kinematics):
A particle moves along the positive $x$-axis subject to a velocity-dependent retarding acceleration:
$$a(v) = -(4v + 2)\text{ m/s}^2$$
At $t = 0$, the particle is at $x = 0$ with an initial velocity $v_0 = 10\text{ m/s}$.
(a) Find the time $t_1$ required for the velocity to decrease to $v_1 = 2\text{ m/s}$.
(b) Find the maximum displacement $x_{\max}$ attained by the particle as it comes to rest ($v \to 0$).
Solution:
(a) Time Calculation ($a = \frac{dv}{dt}$):
$$\frac{dv}{dt} = -(4v + 2) \implies dt = -\frac{dv}{4v + 2}$$
Integrating from $t = 0$ ($v = 10\text{ m/s}$) to $t = t_1$ ($v = 2\text{ m/s}$):
$$t_1 = -\int_{10}^2 \frac{dv}{4v + 2} = -\frac{1}{4} [\ln(4v + 2)]_{10}^2 = -\frac{1}{4} [\ln(10) – \ln(42)] = \frac{1}{4} \ln\left(\frac{42}{10}\right) = \frac{1}{4} \ln(4.2)$$
Since $\ln(4.2) \approx 1.4351$:
$$\mathbf{t_1 = \frac{1.4351}{4} \approx 0.3588\text{ seconds}}$$
(b) Displacement Calculation ($a = v \frac{dv}{dx}$):
$$v \frac{dv}{dx} = -(4v + 2) \implies dx = -\frac{v dv}{4v + 2}$$
Rewrite the integrand:
$$\frac{v}{4v + 2} = \frac{1}{4}\left(\frac{4v + 2 – 2}{4v + 2}\right) = \frac{1}{4} – \frac{1}{2(4v + 2)}$$
Integrating from $x = 0$ ($v = 10\text{ m/s}$) to $x_{\max}$ ($v = 0$):
$$x_{\max} = -\int_{10}^0 \left[\frac{1}{4} – \frac{1}{2(4v + 2)}\right] dv = \int_0^{10} \left[\frac{1}{4} – \frac{1}{2(4v + 2)}\right] dv$$
$$x_{\max} = \frac{1}{4}[v]_0^{10} – \frac{1}{8}[\ln(4v + 2)]_0^{10} = \frac{10}{4} – \frac{1}{8}[\ln(42) – \ln(2)] = 2.5 – \frac{1}{8}\ln(21)$$
With $\ln(21) \approx 3.0445$:
$$\mathbf{x_{\max} = 2.5 – \frac{3.0445}{8} = 2.5 – 0.3806 = 2.1194\text{ m}}$$
Example 3 (Two-Block Friction Threshold Dynamics – JEE Advanced):
A block $A$ of mass $m = 2.0\text{ kg}$ rests on top of block $B$ of mass $M = 4.0\text{ kg}$, which in turn rests on a frictionless horizontal floor. The coefficient of static and kinetic friction between $A$ and $B$ is $\mu_s = \mu_k = 0.40$ ($g = 10\text{ m/s}^2$). A horizontal driving force $F$ is applied to the bottom block $B$.
(a) Determine the maximum force $F_{\max}$ that can be applied to block $B$ such that block $A$ accelerates together with $B$ without relative slipping.
(b) If an applied force $F = 36.0\text{ N}$ is exerted on block $B$, calculate the resulting accelerations $a_A$ and $a_B$ of both blocks.
Solution:
(a) Maximum Force for No Relative Slipping:
Normal force between $A$ and $B$: $N_{AB} = m g = 2.0 \times 10 = 20.0\text{ N}$.
Maximum static friction available to accelerate block $A$:
$$f_{s, \max} = \mu_s N_{AB} = 0.40 \times 20.0\text{ N} = 8.0\text{ N}$$
The sole horizontal force accelerating block $A$ is friction $f_s$:
$$f_s = m a_A \implies a_{A, \max} = \frac{f_{s, \max}}{m} = \frac{8.0\text{ N}}{2.0\text{ kg}} = 4.0\text{ m/s}^2$$
For both blocks to move as a unified system with maximum acceleration $a_{\max} = 4.0\text{ m/s}^2$:
$$\mathbf{F_{\max} = (M + m) a_{\max} = (4.0 + 2.0)(4.0) = 24.0\text{ N}}$$
(b) Accelerations for $F = 36.0\text{ N} > 24.0\text{ N}$ (Slipping Regime):
Since $F > F_{\max}$, relative sliding occurs. Kinetic friction acts between $A$ and $B$:
$$f_k = \mu_k N_{AB} = 8.0\text{ N}$$
– For Block $A$: Accelerated forward by $f_k$:
$$a_A = \frac{f_k}{m} = \frac{8.0\text{ N}}{2.0\text{ kg}} = \mathbf{4.0\text{ m/s}^2}$$
– For Block $B$: Driven forward by $F$ and opposed by friction $f_k$:
$$F – f_k = M a_B \implies 36.0 – 8.0 = 4.0 a_B \implies 28.0 = 4.0 a_B \implies \mathbf{a_B = 7.0\text{ m/s}^2}$$
Example 4 (Smooth Wedge Constraint Dynamics – Multi-Body Systems):
A smooth wedge of mass $M$ with an incline angle $\theta = 30^\circ$ is free to slide on a frictionless horizontal floor. A small block of mass $m$ is released from rest on the incline.
(a) Derive the acceleration $A$ of the wedge across the floor in terms of $M, m, g,$ and $\theta$.
(b) If $M = 3m$, evaluate the numerical acceleration of the wedge in units of $g$.
Solution:
(a) Wedge Acceleration Derivation:
Let $A$ be the acceleration of the wedge to the left. In the non-inertial frame of the wedge (accelerating left with $A$):
– The block experiences a pseudo force $m A$ directed to the right.
– Normal reaction $N$ perpendicular to the incline satisfies:
$$N = m g \cos\theta – m A \sin\theta$$
For the wedge in the horizontal direction (inertial frame):
$$N \sin\theta = M A$$
Substitute $N$:
$$(m g \cos\theta – m A \sin\theta) \sin\theta = M A$$
$$m g \sin\theta \cos\theta – m A \sin^2\theta = M A \implies A (M + m \sin^2\theta) = m g \sin\theta \cos\theta$$
$$\mathbf{A = \frac{m g \sin\theta \cos\theta}{M + m \sin^2\theta}}$$
(b) Numerical Evaluation for $M = 3m$ and $\theta = 30^\circ$:
$$\sin 30^\circ = \frac{1}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin^2 30^\circ = \frac{1}{4}$$
Numerator: $m g \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4} m g$.
Denominator: $3m + m\left(\frac{1}{4}\right) = \frac{13}{4} m$.
$$A = \frac{\frac{\sqrt{3}}{4} m g}{\frac{13}{4} m} = \mathbf{\frac{\sqrt{3}}{13} g \approx 0.1332 g}$$
Practice Worksheet: Mechanics Comprehensive Revision
Section A: Single Correct Questions (Q1 to Q5)
Q1. If the Planck constant ($h$), speed of light in vacuum ($c$), and universal gravitational constant ($G$) are selected as fundamental dimensions, the dimensional formula for Length is:
(A) $\sqrt{\frac{h G}{c^3}}$
(B) $\sqrt{\frac{h c}{G}}$
(C) $\sqrt{\frac{G c^3}{h}}$
(D) $\frac{h G}{c^2}$
Q2. A particle starts from rest at the origin ($x = 0$) and moves along the positive $x$-axis such that its instantaneous velocity varies with position as $v = \alpha \sqrt{x}$, where $\alpha$ is a positive constant. The displacement $x$ of the particle as a function of time $t$ is:
(A) $\frac{1}{4} \alpha^2 t^2$
(B) $\frac{1}{2} \alpha^2 t^2$
(C) $\alpha^2 t^2$
(D) $\frac{1}{4} \alpha t^2$
Q3. A projectile is launched from ground level with speed $u$ at an angle $\theta$ ($0 < \theta < 90^\circ$) above the horizontal. The time $t$ after launch at which its velocity vector becomes perpendicular to its initial velocity vector is:
(A) $\frac{u}{g \sin\theta}$
(B) $\frac{u}{g \cos\theta}$
(C) $\frac{u \sin\theta}{g}$
(D) $\frac{2 u \sin\theta}{g}$
Q4. In an ideal pulley-string system, block $A$ of mass $m$ hangs from a movable pulley, while block $B$ of mass $m$ rests on a frictionless horizontal table. The string from $B$ runs over a fixed pulley and supports the movable pulley carrying $A$. The downward acceleration of block $A$ is:
(A) $\frac{g}{5}$
(B) $\frac{g}{3}$
(C) $\frac{2g}{5}$
(D) $\frac{g}{2}$
Q5. A block of mass $m$ rests on a rough plane inclined at an angle $\theta = 45^\circ$ to the horizontal. The coefficient of static friction between the block and the incline is $\mu_s = 0.50$. The minimum force $F$ applied parallel to the inclined plane required to prevent the block from sliding down is:
(A) $\frac{m g}{2\sqrt{2}}$
(B) $\frac{m g}{\sqrt{2}}$
(C) $\frac{3 m g}{2\sqrt{2}}$
(D) Zero
Section B: One or More than One Correct Questions (Q6 to Q8)
Q6. A particle moves along a straight line with velocity $v(t) = 4t – t^2\text{ m/s}$ for $0 \le t \le 6\text{ s}$. Which of the following statements is/are correct?
(A) The acceleration of the particle is zero at $t = 2\text{ s}$.
(B) The particle reverses its direction of motion at $t = 4\text{ s}$.
(C) The net displacement of the particle from $t = 0$ to $t = 6\text{ s}$ is zero.
(D) The total distance travelled by the particle from $t = 0$ to $t = 6\text{ s}$ is $\frac{64}{3}\text{ m}$.
Q7. A person of mass $m$ stands on a spring weighing machine placed on the floor of an elevator. Which of the following statements is/are correct?
(A) The reading of the weighing machine is greater than $m g$ when the elevator accelerates upwards.
(B) The reading is less than $m g$ when the elevator ascends with decreasing speed.
(C) The reading is zero when the elevator cable snaps and it falls freely under gravity.
(D) The reading is greater than $m g$ when the elevator moves downwards with decreasing speed.
Q8. Two blocks $A$ ($m_A = 3\text{ kg}$) and $B$ ($m_B = 2\text{ kg}$) are placed in contact on a frictionless horizontal floor. A horizontal force $F = 10\text{ N}$ is applied to block $A$ directed towards block $B$. Which of the following statements is/are correct?
(A) The common acceleration of the two blocks is $2\text{ m/s}^2$.
(B) The contact force between blocks $A$ and $B$ is $4\text{ N}$.
(C) If the force $F = 10\text{ N}$ is applied instead to block $B$ directed towards block $A$, the contact force between them becomes $6\text{ N}$.
(D) The net horizontal force acting on block $A$ is $6\text{ N}$ when $F$ acts on $A$.
Section C: Numerical Value Type Questions (Q9 to Q10)
Q9. A river of width $d = 400\text{ m}$ flows with a uniform current speed $u = 3\text{ m/s}$. A swimmer can swim with a speed $v = 5\text{ m/s}$ in still water. If the swimmer steers so as to cross the river along the shortest possible path (perpendicular to the river banks without any drift), the time taken to cross the river (in seconds) is $T$. Find the value of $T$.
Q10. The electrical resistance $R$ of a circuit component is determined by measuring the potential difference $V$ across it and the current $I$ passing through it ($R = V / I$). The experimental measurements yield $V = (100 \pm 2)\text{ V}$ and $I = (10 \pm 0.2)\text{ A}$. The percentage error in the calculated resistance $R$ is $x\%$. Find the integer value of $x$.
Comprehensive Solutions and Answer Key
Answer Key
| Question | Correct Answer | Core Physics Concept |
|---|---|---|
| Q1 | (A) | Planck Length Dimensional Derivation: $l_p = \sqrt{h G / c^3}$ |
| Q2 | (A) | Separation of Variables Kinematics: $x = \frac{1}{4} \alpha^2 t^2$ |
| Q3 | (A) | Orthogonal Velocity Condition: $\vec{u} \cdot \vec{v} = 0 \implies t = \frac{u}{g \sin\theta}$ |
| Q4 | (A) | Movable Pulley Constraint: $a_B = 2 a_A \implies a_A = g/5$ |
| Q5 | (A) | Incline Equilibrium with Friction: $F = mg(\sin 45^\circ – \mu \cos 45^\circ) = \frac{mg}{2\sqrt{2}}$ |
| Q6 | (A, B, C, D) | Kinematics Calculus: Direction Reversals, Distance vs Displacement |
| Q7 | (A, B, C, D) | Apparent Weight in Non-Inertial Elevators ($N = m(g \pm a)$) |
| Q8 | (A, B, C, D) | Action-Reaction Contact Dynamics in Multi-Body Systems |
| Q9 | 100 | Shortest River Crossing: $v_{\text{cross}} = \sqrt{5^2 – 3^2} = 4\text{ m/s} \implies T = 400/4 = 100\text{ s}$ |
| Q10 | 4 | Relative Error Combination: $\frac{\Delta R}{R} = 2\% + 2\% = 4\%$ |
Detailed Explanations
Solution 1:
Let Length $L = h^a c^b G^c$.
Dimensions:
$[h] = [M L^2 T^{-1}]$
$[c] = [L T^{-1}]$
$[G] = [M^{-1} L^3 T^{-2}]$
Substitute into $[L^1] = [M L^2 T^{-1}]^a [L T^{-1}]^b [M^{-1} L^3 T^{-2}]^c = M^{a-c} L^{2a+b+3c} T^{-a-b-2c}$.
Equating exponents:
1. $a – c = 0 \implies a = c$
2. $-a – b – 2c = 0 \implies b = -3a$
3. $2a + b + 3c = 1 \implies 2a – 3a + 3a = 1 \implies 2a = 1 \implies a = 1/2$.
Hence $c = 1/2$ and $b = -3/2$.
$$L = h^{1/2} G^{1/2} c^{-3/2} = \mathbf{\sqrt{\frac{h G}{c^3}}}$$
Correct Answer: (A)
Solution 2:
Given $v = \frac{dx}{dt} = \alpha \sqrt{x}$.
Separating variables:
$$\frac{dx}{\sqrt{x}} = \alpha dt$$
Integrating from $x = 0$ at $t = 0$:
$$\int_0^x x^{-1/2} dx = \alpha \int_0^t dt \implies [2 \sqrt{x}]_0^x = \alpha t \implies 2 \sqrt{x} = \alpha t$$
Squaring both sides:
$$4 x = \alpha^2 t^2 \implies \mathbf{x = \frac{1}{4} \alpha^2 t^2}$$
Correct Answer: (A)
Solution 3:
Initial velocity vector: $\vec{u} = u \cos\theta \hat{i} + u \sin\theta \hat{j}$.
Velocity at time $t$: $\vec{v} = u \cos\theta \hat{i} + (u \sin\theta – g t) \hat{j}$.
For $\vec{v} \perp \vec{u}$, the dot product vanishes ($\vec{u} \cdot \vec{v} = 0$):
$$(u \cos\theta)(u \cos\theta) + (u \sin\theta)(u \sin\theta – g t) = 0$$
$$u^2 \cos^2\theta + u^2 \sin^2\theta – u g t \sin\theta = 0$$
$$u^2 – u g t \sin\theta = 0 \implies \mathbf{t = \frac{u}{g \sin\theta}}$$
Correct Answer: (A)
Solution 4:
Let $y_A$ be the downward displacement of movable pulley $A$, and $x_B$ be the rightward displacement of block $B$.
The total length of the string is $L = x_B + 2 y_A + \text{const}$.
Differentiating twice: $a_B + 2 a_A = 0 \implies a_B = 2 a_A$.
Equations of motion:
– For block $B$: $T = m a_B = m (2 a_A) = 2 m a_A$.
– For block $A$: $m g – 2 T = m a_A$.
Substitute $T = 2 m a_A$:
$$m g – 2(2 m a_A) = m a_A \implies m g – 4 m a_A = m a_A \implies 5 m a_A = m g \implies \mathbf{a_A = \frac{g}{5}}$$
Correct Answer: (A)
Solution 5:
The component of gravity pulling the block down the incline is:
$$F_g = m g \sin 45^\circ = \frac{m g}{\sqrt{2}}$$
The normal reaction is $N = m g \cos 45^\circ = \frac{m g}{\sqrt{2}}$.
The maximum static friction opposing downward slip (directed up the plane) is:
$$f_{s, \max} = \mu_s N = 0.50 \times \frac{m g}{\sqrt{2}} = \frac{m g}{2\sqrt{2}}$$
To just prevent sliding down, the applied force $F$ directed up the incline plus maximum static friction must balance $F_g$:
$$F + f_{s, \max} = F_g \implies F = \frac{m g}{\sqrt{2}} – \frac{m g}{2\sqrt{2}} = \mathbf{\frac{m g}{2\sqrt{2}}}$$
Correct Answer: (A)
Solution 6:
Given $v(t) = 4t – t^2$.
– (A) Acceleration $a(t) = \frac{dv}{dt} = 4 – 2t$. At $t = 2\text{ s}$, $a = 4 – 4 = 0$. (True)
– (B) Velocity vanishes at $t = 0$ and $t = 4\text{ s}$. For $t < 4\text{ s}$, $v > 0$; for $t > 4\text{ s}$, $v < 0$. Thus, direction reverses at $t = 4\text{ s}$. (True)
- (C) Displacement from $0$ to $6\text{ s}$:
$$s = \int_0^6 (4t - t^2) dt = \left[2t^2 - \frac{t^3}{3}\right]_0^6 = 2(36) - \frac{216}{3} = 72 - 72 = 0\text{ m}.$$ (True)
- (D) Distance is the sum of magnitudes of displacement in each direction:
From $t = 0$ to $4$: $s_1 = \left[2(16) - \frac{64}{3}\right] = \frac{32}{3}\text{ m}$.
From $t = 4$ to $6$: $s_2 = \left|\left[2t^2 – \frac{t^3}{3}\right]_4^6\right| = |0 – 32/3| = \frac{32}{3}\text{ m}$.
Total distance = $s_1 + s_2 = \frac{32}{3} + \frac{32}{3} = \frac{64}{3}\text{ m}$. (True)
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
The apparent weight recorded by the scale is the normal reaction $N$:
– (A) Upward acceleration: $N – m g = m a \implies N = m(g + a) > m g$. (True)
– (B) Upward deceleration: $a$ is downward $\implies N = m(g – a) < m g$. (True)
- (C) Free fall ($a = g$ downward): $N = m(g - g) = 0$. (True)
- (D) Downward deceleration: upward acceleration $\implies N = m(g + a) > m g$. (True)
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
– (A) Common acceleration: $a = \frac{F}{m_A + m_B} = \frac{10}{3 + 2} = 2.0\text{ m/s}^2$. (True)
– (B) When $F$ acts on $A$, the normal force $N$ accelerates block $B$: $N = m_B a = 2 \times 2 = 4.0\text{ N}$. (True)
– (C) When $F$ acts on $B$, normal force $N’$ accelerates block $A$: $N’ = m_A a = 3 \times 2 = 6.0\text{ N}$. (True)
– (D) Net horizontal force on block $A$: $F_{\text{net}, A} = m_A a = 3 \times 2 = 6.0\text{ N}$. (True)
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 9:
For the shortest crossing path perpendicular to the river flow, the resultant velocity $\vec{v}_{\text{res}}$ must be directed straight across:
$$v_{\text{res}} = \sqrt{v^2 – u^2} = \sqrt{5^2 – 3^2} = \sqrt{25 – 9} = \sqrt{16} = 4.0\text{ m/s}$$
The time taken to traverse width $d = 400\text{ m}$:
$$T = \frac{d}{v_{\text{res}}} = \frac{400\text{ m}}{4.0\text{ m/s}} = \mathbf{100\text{ seconds}}$$
Correct Answer: 100
Solution 10:
Given $R = \frac{V}{I}$.
The fractional error in $R$ is:
$$\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$$
Substitute measured values:
$$\frac{\Delta R}{R} = \frac{2}{100} + \frac{0.2}{10} = 0.02 + 0.02 = 0.04$$
Expressed as a percentage:
$$\% \text{ error} = 0.04 \times 100\% = \mathbf{4\%}$$
Correct Answer: 4