Concept Card: Unit 3 (Laws of Motion) Comprehensive Mastery Cheat Sheet
1. Fundamental Laws & Momentum Dynamics:
- Newton’s First Law: $\sum \vec{F}_{\text{ext}} = \vec{0} \iff \vec{a} = \vec{0} \iff \vec{v} = \text{constant}$. Establishes mass $m$ as the quantitative measure of translational inertia and defines inertial frames of reference.
- Newton’s Second Law: $\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt} = m\vec{a}$ (for constant mass). Linear momentum $\vec{p} = m\vec{v}$; kinetic energy relation $K = \frac{p^2}{2m} \iff p = \sqrt{2mK}$. Variable mass: $\vec{F}_{\text{ext}} + \vec{v}_{\text{rel}}\frac{dm}{dt} = m\frac{d\vec{v}}{dt}$.
- Impulse-Momentum Theorem: $\vec{J} = \int_{t_1}^{t_2} \vec{F}\,dt = \Delta\vec{p} = \text{Area under } F-t \text{ curve}$. Average force $\vec{F}_{\text{avg}} = \frac{\Delta\vec{p}}{\Delta t}$.
- Newton’s Third Law: $\vec{F}_{AB} = -\vec{F}_{BA}$. Action and reaction act simultaneously on different bodies, belong to the same fundamental interaction, and never cancel each other. Internal forces sum to zero ($\sum \vec{F}_{\text{internal}} = \vec{0}$), ensuring conservation of total linear momentum when $\vec{F}_{\text{ext, net}} = \vec{0}$.
2. Connected Bodies & Pulley Constraints:
- Blocks in Contact: $a = \frac{F}{\sum m_i}, \quad N_{12} = \left(\frac{m_2 + m_3 + \dots}{\sum m_i}\right)F$.
- Blocks Connected by Strings: $a = \frac{F}{\sum m_i}, \quad T_1 = m_1 a, \quad T_2 = (m_1 + m_2)a$.
- Simple Atwood Machine:
$a = \left(\frac{m_1 – m_2}{m_1 + m_2}\right)g, \quad T = \frac{2m_1 m_2 g}{m_1 + m_2}, \quad F_{\text{clamp}} = 2T = \frac{4m_1 m_2 g}{m_1 + m_2} < (m_1 + m_2)g$.
Center of mass acceleration: $\vec{a}_{\text{cm}} = -\left(\frac{m_1 – m_2}{m_1 + m_2}\right)^2 g\hat{j}$ (always directed downwards). - Virtual Work Constraint Method: $\sum_{i} \vec{T}_i \cdot \vec{a}_i = 0$.
3. Friction Mechanics:
- Total Contact Force Resultant: $\vec{R} = \vec{N} + \vec{f} \implies R = \sqrt{N^2 + f^2}$.
- Static Friction: Self-adjusting $0 \le f_s \le f_L = \mu_s N$.
- Kinetic Friction: $f_k = \mu_k N$ ($\mu_k < \mu_s$).
- Angle of Friction & Angle of Repose: $\lambda = \theta_r = \arctan(\mu_s)$.
- Rough Incline Dynamics: $a_{\text{down}} = g(\sin\theta – \mu_k\cos\theta), \quad a_{\text{up}} = g(\sin\theta + \mu_k\cos\theta)$. Time ratio $\frac{t_{\text{down}}}{t_{\text{up}}} = \sqrt{\frac{\sin\theta + \mu_k\cos\theta}{\sin\theta – \mu_k\cos\theta}} > 1$. Rough vs. smooth incline time ratio: $\mu_k = \tan\theta\left(1 – \frac{1}{n^2}\right)$.
- Optimum Pulling Force Up an Incline: $F_{\text{min}} = Mg\sin(\theta + \lambda)$ at angle $\alpha = \lambda$ to the incline.
4. Circular Dynamics & Rotating Reference Frames:
- Centripetal Force (Inertial Frame): $F_c = \frac{mv^2}{r} = m\omega^2 r$. Work done in UCM is zero ($W_c = 0$).
- Level Turn Safe Speed: $v_{\text{max}} = \sqrt{\mu_s r g}$.
- Banked Roadway: Frictionless design speed $v_0 = \sqrt{rg\tan\theta}$. Safe speed limits: $v_{\text{max}} = \sqrt{rg\left(\frac{\tan\theta + \mu_s}{1 – \mu_s\tan\theta}\right)}$ and $v_{\text{min}} = \sqrt{rg\left(\frac{\tan\theta – \mu_s}{1 + \mu_s\tan\theta}\right)}$.
- Conical Pendulum: $T = \frac{mg}{\cos\theta} = m\omega^2 L, \quad \tau = 2\pi\sqrt{\frac{L\cos\theta}{g}} = 2\pi\sqrt{\frac{h}{g}}$.
- Rotating Non-Inertial Frame: Outward centrifugal force $\vec{F}_{\text{cf}} = m\omega^2\vec{r}$, Coriolis force $\vec{F}_{\text{Cor}} = -2m(\vec{\omega}\times\vec{v}_{\text{rel}})$.
Solved Examples
Example 1 (Impulse & Momentum in Oblique Wall Collision):
A ball of mass $m = 0.20\text{ kg}$ strikes a smooth vertical wall with speed $v = 25.0\text{ m/s}$ at an angle of incidence $\theta = 37^\circ$ to the normal ($\cos 37^\circ = 0.8, \sin 37^\circ = 0.6$). It rebounds elastically at the same angle with the same speed in contact time $\Delta t = 0.020\text{ s}$.
(a) Determine the impulse delivered to the ball by the wall.
(b) Find the average normal force exerted by the wall.
(c) Find the force parallel to the wall.
Solution:
(a) Choosing normal to the wall as $+x$ away from wall, and parallel along wall as $+y$:
$\vec{v}_i = -20.0\hat{i} + 15.0\hat{j}\text{ m/s}, \quad \vec{v}_f = +20.0\hat{i} + 15.0\hat{j}\text{ m/s}$.
$\Delta\vec{p} = m(\vec{v}_f – \vec{v}_i) = 0.20[40.0\hat{i}] = 8.0\hat{i}\text{ kg}\cdot\text{m/s}$.
$\vec{J} = \Delta\vec{p} = 8.0\hat{i}\text{ N}\cdot\text{s}$.
(b) Average normal force: $\vec{F}_{\text{avg}} = \frac{\vec{J}}{\Delta t} = \frac{8.0\hat{i}}{0.020} = 400.0\hat{i}\text{ N}$.
(c) Smooth wall implies zero shear friction $\implies F_y = 0$.
Example 2 (Block Stack with Wall Anchor & Friction):
Block $A$ ($m_1 = 3.0\text{ kg}$) rests on top of block $B$ ($m_2 = 6.0\text{ kg}$) on a frictionless horizontal floor. Block $A$ is tied to a rigid vertical wall by a horizontal cord. A horizontal pulling force $F = 45.0\text{ N}$ pulls block $B$ away from the wall. The coefficient of kinetic friction between $A$ and $B$ is $\mu_k = 0.40$. Taking $g = 10\text{ m/s}^2$:
(a) Find the tension $T$ in the cord tying block $A$.
(b) Find the acceleration $a_B$ of block $B$.
(c) If the cord tying block $A$ suddenly snaps, find the new common acceleration.
Solution:
(a) Normal force between $A$ and $B$: $N_A = m_1 g = 30.0\text{ N}$.
Kinetic friction: $f_k = \mu_k N_A = 0.40(30.0) = 12.0\text{ N}$.
For stationary block $A$: $T = f_k = 12.0\text{ N}$.
(b) On block $B$: $F – f_k = m_2 a_B \implies 45.0 – 12.0 = 6.0 a_B \implies 33.0 = 6.0 a_B \implies a_B = 5.5\text{ m/s}^2$.
(c) If cord snaps, both move together as single mass $M = 9.0\text{ kg}$: $a = \frac{45.0}{9.0} = 5.0\text{ m/s}^2$.
Example 3 (Movable Pulley on an Incline Constraint Synthesis):
A block $M = 4.0\text{ kg}$ on a smooth $30^\circ$ incline is connected via a cord over a fixed apex pulley to a light movable pulley $P$ supporting a hanging mass $m = 6.0\text{ kg}$. Taking $g = 10\text{ m/s}^2$:
(a) Determine the kinematic constraint between $a_M$ (up incline) and $a_m$ (downwards).
(b) Calculate the acceleration of each block and the cord tensions.
Solution:
(a) Virtual work: $T a_M – 2T a_m = 0 \implies a_M = 2 a_m$.
(b) On $M$: $T – M g \sin(30^\circ) = M a_M = 4.0(2 a_m) = 8.0 a_m \implies T = 20.0 + 8.0 a_m$.
On $m$: $m g – 2T = m a_m \implies 60.0 – 2(20.0 + 8.0 a_m) = 6.0 a_m \implies 20.0 = 22.0 a_m$.
$a_m = \frac{10}{11} \approx 0.909\text{ m/s}^2$.
$a_M = 2 a_m = \frac{20}{11} \approx 1.818\text{ m/s}^2$.
$T = 20.0 + 8.0\left(\frac{10}{11}\right) = \frac{300}{11} \approx 27.27\text{ N}$.
Example 4 (Rotating Rough Conical Funnel):
A conical funnel of semi-vertical angle $\alpha = 45^\circ$ rotates about its vertical axis with angular velocity $\omega$. A small block of mass $m$ rests on the inner rough surface ($\mu_s = 0.50$) at distance $R = 0.50\text{ m}$ from the axis. Taking $g = 10\text{ m/s}^2$:
(a) Find the optimum angular speed $\omega_0$ where no friction is required.
(b) Find the safe operating angular velocity range $[\omega_{\text{min}}, \omega_{\text{max}}]$.
Solution:
Incline angle to horizontal is $\theta = 90^\circ – 45^\circ = 45^\circ$.
(a) $\tan(45^\circ) = \frac{\omega_0^2 R}{g} \implies 1 = \frac{\omega_0^2(0.50)}{10} \implies \omega_0 = \sqrt{20} \approx 4.47\text{ rad/s}$.
(b) In rotating frame:
$\omega_{\text{max}} = \sqrt{\frac{g}{R}\left(\frac{1 + \mu_s}{1 – \mu_s}\right)} = \sqrt{20\left(\frac{1.5}{0.5}\right)} = \sqrt{60} \approx 7.75\text{ rad/s}$.
$\omega_{\text{min}} = \sqrt{\frac{g}{R}\left(\frac{1 – \mu_s}{1 + \mu_s}\right)} = \sqrt{20\left(\frac{0.5}{1.5}\right)} = \sqrt{\frac{20}{3}} \approx 2.58\text{ rad/s}$.
Safe window: $2.58\text{ rad/s} \le \omega \le 7.75\text{ rad/s}$.
Worksheet: 10 Practice Problems (Full Chapter Test)
Problem 1 (JEE Main – Single Correct):
A rocket of initial mass $M_0 = 1000\text{ kg}$ consumes fuel at rate $\frac{dm}{dt} = 20\text{ kg/s}$ with exhaust velocity $u_{\text{rel}} = 1000\text{ m/s}$. Taking $g = 10\text{ m/s}^2$, the initial upward acceleration of the rocket is:
(A) $10\text{ m/s}^2$
(B) $20\text{ m/s}^2$
(C) $30\text{ m/s}^2$
(D) $5\text{ m/s}^2$
Problem 2 (JEE Main – Single Correct):
A block of mass $m$ is placed on a rough horizontal floor with coefficient of static friction $\mu_s = 0.50$. A horizontal force $F = 0.30 mg$ is applied. The frictional force exerted by the floor on the block is:
(A) $0.50 mg$
(B) $0.30 mg$
(C) $0.20 mg$
(D) Zero
Problem 3 (JEE Main – Single Correct):
A block slides down a rough incline of angle $45^\circ$ in three times the time it takes to slide down a frictionless incline of the same length and slope. The coefficient of kinetic friction $\mu_k$ is:
(A) $\frac{1}{3}$
(B) $\frac{2}{3}$
(C) $\frac{8}{9}$
(D) $\frac{1}{9}$
Problem 4 (JEE Main – Single Correct):
A car is driven around a circular track of radius $r = 50.0\text{ m}$ on a horizontal road with $\mu_s = 0.50$. Taking $g = 10\text{ m/s}^2$, the maximum safe speed is:
(A) $25\text{ m/s}$
(B) $5\sqrt{10}\text{ m/s}$
(C) $10\text{ m/s}$
(D) $50\text{ m/s}$
Problem 5 (JEE Main – Single Correct):
Two masses $m_1 = 3\text{ kg}$ and $m_2 = 1\text{ kg}$ are connected by a light cord over a frictionless pulley. The tension in the cord during acceleration is ($g = 10\text{ m/s}^2$):
(A) $10\text{ N}$
(B) $15\text{ N}$
(C) $20\text{ N}$
(D) $30\text{ N}$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are correct regarding non-inertial reference frames and pseudo forces?
(A) Pseudo forces obey Newton’s Third Law and always possess an equal and opposite reaction force.
(B) The centrifugal force on a particle in a frame rotating with angular velocity $\vec{\omega}$ is $\vec{F}_{\text{cf}} = m\omega^2\vec{r}$, directed away from the axis of rotation.
(C) The Coriolis force on a particle moving with relative velocity $\vec{v}_{\text{rel}}$ in a rotating frame is $-2m(\vec{\omega}\times\vec{v}_{\text{rel}})$.
(D) The Coriolis force can never change the kinetic energy of a particle in the rotating frame.
Problem 7 (JEE Advanced – One or More Correct):
Block $A$ ($m_A = 2\text{ kg}$) is placed on block $B$ ($m_B = 4\text{ kg}$), which rests on a frictionless horizontal plane. The static and kinetic friction coefficients between $A$ and $B$ are $\mu_s = 0.40$ and $\mu_k = 0.30$. A horizontal force $F$ acts on $B$. Which of the following is/are correct? ($g = 10\text{ m/s}^2$):
(A) For $F \le 24\text{ N}$, both blocks accelerate together with common acceleration $a = \frac{F}{6}$.
(B) The maximum possible acceleration of block $A$ is $4.0\text{ m/s}^2$.
(C) When $F = 30\text{ N}$, slipping occurs between $A$ and $B$.
(D) When $F = 30\text{ N}$, the acceleration of block $A$ is $3.0\text{ m/s}^2$.
Problem 8 (JEE Advanced – One or More Correct):
For an ideal conical pendulum of string length $L$ and semi-vertical angle $\theta$:
(A) The tension in the string is strictly greater than the gravitational weight of the bob ($T > mg$).
(B) The period of revolution decreases as the semi-vertical angle $\theta$ increases.
(C) The net vertical force on the bob is zero.
(D) The work done by string tension over one complete circular orbit is zero.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
In an ideal Atwood machine, the two suspended masses are $m_1 = 9.0\text{ kg}$ and $m_2 = 1.0\text{ kg}$. Taking $g = 10\text{ m/s}^2$, find the magnitude of acceleration of the masses in $\text{m/s}^2$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A circular road of radius $r = 45.0\text{ m}$ is banked at an angle $\theta$ such that $\tan\theta = 0.50$. Taking $g = 10\text{ m/s}^2$, find the optimum design speed $v_0$ in $\text{m/s}$ for which no lateral friction is required.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (B) | 3. (C) | 4. (B) | 5. (B) | 6. (B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 8 | 10. 15
Solution 1:
$F_{\text{thrust}} = u_{\text{rel}}\left(\frac{dm}{dt}\right) = 1000 \times 20 = 20000\text{ N}$.
$F_{\text{net}} = F_{\text{thrust}} – M_0 g = 20000 – 10000 = 10000\text{ N} \implies a = \frac{10000}{1000} = 10\text{ m/s}^2$.
Correct Answer: (A)
Solution 2:
$f_L = \mu_s mg = 0.50 mg$. Since $F = 0.30 mg < f_L$, static friction balances applied force: $f_s = F = 0.30 mg$.
Correct Answer: (B)
Solution 3:
$\mu_k = \tan\theta\left(1 – \frac{1}{n^2}\right) = 1\left(1 – \frac{1}{9}\right) = \frac{8}{9}$.
Correct Answer: (C)
Solution 4:
$v_{\text{max}} = \sqrt{\mu_s r g} = \sqrt{0.50 \times 50.0 \times 10} = \sqrt{250} = 5\sqrt{10}\text{ m/s}$.
Correct Answer: (B)
Solution 5:
$T = \frac{2 m_1 m_2 g}{m_1 + m_2} = \frac{2(3)(1)(10)}{4} = 15\text{ N}$.
Correct Answer: (B)
Solution 6:
(A) False: pseudo forces have no physical agent. (B), (C), and (D) are verified properties of rotating reference frames.
Correct Answer: (B, C, D)
Solution 7:
$f_{s,\text{max}} = \mu_s m_A g = 8.0\text{ N} \implies a_{\text{max}} = 4.0\text{ m/s}^2$. Threshold force $F_{\text{max}} = 6 \times 4.0 = 24\text{ N}$. For $F = 30\text{ N}$, slipping occurs and $a_A = \frac{\mu_k m_A g}{m_A} = 3.0\text{ m/s}^2$. All options are correct.
Correct Answer: (A, B, C, D)
Solution 8:
All statements (A, B, C, D) are verified properties of the conical pendulum.
Correct Answer: (A, B, C, D)
Solution 9:
$a = \left(\frac{9.0 – 1.0}{9.0 + 1.0}\right)(10) = 8.0\text{ m/s}^2$.
Correct Answer: 8
Solution 10:
$v_0 = \sqrt{r g \tan\theta} = \sqrt{45.0 \times 10 \times 0.50} = \sqrt{225} = 15.0\text{ m/s}$.
Correct Answer: 15