Comprehensive Physics Notes: Motion in a Straight Line, Motion in a Plane, and Laws of Motion

Motion in a Straight Line

• Rectilinear Motion

  • The study of the motion of objects along a straight line is known as rectilinear motion.

• Average Velocity and Average Speed

  • Average Velocity: Defined as the ratio of total displacement to the total time interval.     $\text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time interval}}$$\bar{v} = \frac{x_2 - x_1}{t_2 - t_1} = \frac{\Delta x}{\Delta t}$     where $x_1$ and $x_2$ are the positions of the object at time $t_1$ and $t_2$.
  • Average Speed: Defined as the ratio of total path length (distance travelled) to the total time interval.     $\text{Average speed} = \frac{\text{Total path length}}{\text{Total time interval}}$

• Uniform Motion

  • If an object moving along a straight line covers equal distances in equal intervals of time, it is said to be in uniform motion along a straight line.
  • In uniform motion, the velocity of the object remains constant.

• Instantaneous Velocity and Speed

  • Instantaneous Velocity: The average velocity tells us how fast an object moves over a given time interval, but not at specific instants. The velocity at a specific instant $t$ is called instantaneous velocity and is defined as the limit of the average velocity as the time interval $\Delta t$ becomes infinitesimally small.     $v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$     where $\frac{dx}{dt}$ is the differential coefficient of $x$ with respect to $t$, representing the rate of change of position with respect to time.
  • Determining Velocity from Position-Time Graph: Velocity at any instant (e.g., $t = 4.0\,\text{s}$) is equal to the slope of the tangent to the position-time graph at that instant.     
  • Limiting Value Analysis at $t = 4.0\,\text{s}$:     
  • For $\Delta t = 2.0\,\text{s}$ ($t_1 = 3.0\,\text{s}$, $t_2 = 5.0\,\text{s}$): $x(t_1) = 2.16\,\text{m}$, $x(t_2) = 10.0\,\text{m}$, $\Delta x = 7.84\,\text{m}$, $\frac{\Delta x}{\Delta t} = 3.92\,\text{m\,s}^{-1}$.
  • For $\Delta t = 1.0\,\text{s}$ ($t_1 = 3.5\,\text{s}$, $t_2 = 4.5\,\text{s}$): $x(t_1) = 3.43\,\text{m}$, $x(t_2) = 7.29\,\text{m}$, $\Delta x = 3.86\,\text{m}$, $\frac{\Delta x}{\Delta t} = 3.86\,\text{m\,s}^{-1}$.
  • For $\Delta t = 0.5\,\text{s}$ ($t_1 = 3.75\,\text{s}$, $t_2 = 4.25\,\text{s}$): $x(t_1) = 4.21875\,\text{m}$, $x(t_2) = 6.14125\,\text{m}$, $\Delta x = 1.9225\,\text{m}$, $\frac{\Delta x}{\Delta t} = 3.845\,\text{m\,s}^{-1}$.
  • For $\Delta t = 0.1\,\text{s}$ ($t_1 = 3.95\,\text{s}$, $t_2 = 4.05\,\text{s}$): $x(t_1) = 4.93039\,\text{m}$, $x(t_2) = 5.31441\,\text{m}$, $\Delta x = 0.38402\,\text{m}$, $\frac{\Delta x}{\Delta t} = 3.8402\,\text{m\,s}^{-1}$.
  • For $\Delta t = 0.01\,\text{s}$ ($t_1 = 3.995\,\text{s}$, $t_2 = 4.005\,\text{s}$): $x(t_1) = 5.100824\,\text{m}$, $x(t_2) = 5.139224\,\text{m}$, $\Delta x = 0.0384\,\text{m}$, $\frac{\Delta x}{\Delta t} = 3.8400\,\text{m\,s}^{-1}$.
  • As $\Delta t$ approaches $0$, the average velocity approaches the limiting value $3.84\,\text{m\,s}^{-1}$, which is $\frac{dx}{dt}$ at $t = 4.0\,\text{s}$.
  • Instantaneous Speed: Instantaneous speed is the magnitude of instantaneous velocity. For example, velocities of $24\,\text{m\,s}^{-1}$ and $-24\,\text{m\,s}^{-1}$ both correspond to a speed of $24.0\,\text{m\,s}^{-1}$.

• Problem: Velocity and Average Velocity from Position Equation

  • Given position equation: $x = a + b t^2$, where $a = 8.5\,\text{m}$, $b = 2.5\,\text{m\,s}^{-2}$, and $t$ is in seconds.
  • Part (a): Instantaneous velocity at $t = 0\,\text{s}$ and $t = 2.0\,\text{s}$:     $v = \frac{dx}{dt} = \frac{d}{dt}(a + b t^2) = 2 b t$
  • At $t = 0\,\text{s}$: $v = 0\,\text{m\,s}^{-1}$.
  • At $t = 2.0\,\text{s}$: $v = 2 \times 2.5 \times 2.0 = 10\,\text{m\,s}^{-1}$.
  • Part (b): Average velocity between $t = 2.0\,\text{s}$ and $t = 4.0\,\text{s}$:     $\bar{v} = \frac{x_4 - x_2}{4 - 2} = \frac{(a + 16b) - (a + 4b)}{2} = \frac{12b}{2} = 6b = 6 \times 2.5 = 15\,\text{m\,s}^{-1}$

• Acceleration

  • Acceleration describes how velocity changes with time. It is the rate of change of velocity.
  • Average Acceleration: Ratio of change of velocity to the time interval.     $\bar{a} = \frac{v_2 - v_1}{t_2 - t_1} = \frac{\Delta v}{\Delta t}$
  • Unit: $\text{m\,s}^{-2}$.
  • Dimensions: $[\text{a}] = [\text{L T}^{-2}]$.
  • Vector quantity.
  • If velocity increases with time, acceleration is positive ($+ve$).
  • If velocity decreases with time, acceleration is negative ($-ve$), termed retardation or deceleration.
  • Uniform Acceleration: Velocity changes by equal amounts in equal intervals of time.
  • Instantaneous Acceleration: The limit of average acceleration as $\Delta t \to 0$:     $a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} = \frac{d^2 x}{dt^2}$
  • Position-Time Graphs for Acceleration:     
  • Curves upward for positive acceleration ($a > 0$).
  • Curves downward/flattens for negative acceleration ($a < 0$).
  • Straight inclined line for zero acceleration ($a = 0$).
  • Velocity-Time Graphs for Motion with Constant Acceleration:
  • Positive direction with positive acceleration.
  • Positive direction with negative acceleration.
  • Negative direction with negative acceleration.
  • Motion with negative acceleration changing direction at time $t_1$ (moves in $+x$ direction from $0$ to $t_1$, then in $-x$ direction from $t_1$ to $t_2$).

• Displacement from Velocity-Time Graph

  • The area under the velocity-time graph equals the displacement of the particle.
  • Proof: In uniform motion, velocity $u$ is constant (parallel to time axis). Over time $T$, $\text{Area} = u T = \text{Displacement}$.

• Kinematic Equations for Uniformly Accelerated Motion

  • Velocity-Time Relation:     $\text{Slope} = a = \frac{BC}{AC} = \frac{v - u}{t}$$v - u = a t \implies v = u + a t \quad \text{or} \quad (v = v_0 + a t)$
  • Position-Time Relation:     $\text{Displacement } s = \text{Area of rectangle} + \text{Area of triangle} = u t + \frac{1}{2}(v - u)t$     Substituting $v - u = a t$ gives:     $s = u t + \frac{1}{2} a t^2 \quad \text{or} \quad (s = v_0 t + \frac{1}{2} a t^2)$
  • Position-Velocity Relation:     $\text{Displacement } s = \text{Average velocity} \times \text{time} = \left(\frac{v + u}{2}\right) \left(\frac{v - u}{a}\right) = \frac{v^2 - u^2}{2a}$$v^2 - u^2 = 2 a s \implies v^2 = u^2 + 2 a s \quad \text{or} \quad (v^2 = v_0^2 + 2 a s)$

• Stopping Distance of Vehicles

  • The distance a vehicle travels before stopping after applying brakes.     $v^2 = u^2 + 2 a s$$0 = u^2 + 2 a s \implies -u^2 = 2 a s \implies s = -\frac{u^2}{2a}$

• Motion under Free Fall

  • Free fall is motion with uniform acceleration due to gravity directed downwards: $a = -g = -9.8\,\text{m\,s}^{-2}$.
  • Released from rest ($u = 0$) at $y = 0$:     $v = -g t = -9.8 t$$y = -\frac{1}{2} g t^2 = -4.9 t^2$$v^2 = -2 g y = -19.6 y$

• Problem: Motion of a Ball Thrown Vertically Upwards

  • A ball is thrown vertically upwards at $20\,\text{m\,s}^{-1}$ from the top of a $25.0\,\text{m}$ tall building ($g = 10\,\text{m\,s}^{-2}$).
  • Part (a): Maximum height risen above the launch point:     $v^2 - u^2 = 2 a s \implies 0 - 20^2 = 2(-10)s \implies -400 = -20s \implies s = 20\,\text{m}$$\text{Total height above ground} = 20\,\text{m} + 25\,\text{m} = 45\,\text{m}$
  • Part (b): Total time before hitting the ground:
  • Upward time ($t_1$): $v = u + a t_1 \implies 0 = 20 - 10 t_1 \implies t_1 = 2\,\text{s}$.
  • Downward time ($t_2$) from height of $45\,\text{m}$:       $s = u t_2 + \frac{1}{2} a t_2^2 \implies -45 = 0 - \frac{1}{2}(10) t_2^2 \implies -45 = -5 t_2^2 \implies t_2^2 = 9 \implies t_2 = 3\,\text{s}$$\text{Total time} = 2\,\text{s} + 3\,\text{s} = 5\,\text{s}$

• Galileo's Law of Odd Numbers

  • Statement: "The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [1 : 3 : 5 : 7…]."
  • Proof:
  • Dividing time into intervals $\tau$ with $u = 0$ and $y = -\frac{1}{2} g t^2$:
  • At $t = 0$: $y = 0$, $y$ in terms of $y_0 \left[= -\frac{1}{2} g \tau^2\right] = 0$, interval distance = $-$
  • At $t = \tau$: $y = -\frac{1}{2} g \tau^2$, $y_0$, interval distance = $y_0$, ratio = $1$
  • At $t = 2\tau$: $y = -4\left(\frac{1}{2} g \tau^2\right)$, $4 y_0$, interval distance = $3 y_0$, ratio = $3$
  • At $t = 3\tau$: $y = -9\left(\frac{1}{2} g \tau^2\right)$, $9 y_0$, interval distance = $5 y_0$, ratio = $5$
  • At $t = 4\tau$: $y = -16\left(\frac{1}{2} g \tau^2\right)$, $16 y_0$, interval distance = $7 y_0$, ratio = $7$
  • At $t = 5\tau$: $y = -25\left(\frac{1}{2} g \tau^2\right)$, $25 y_0$, interval distance = $9 y_0$, ratio = $9$
  • At $t = 6\tau$: $y = -36\left(\frac{1}{2} g \tau^2\right)$, $36 y_0$, interval distance = $11 y_0$, ratio = $11$

Motion in a Plane

• Scalars and Vectors

  • Scalar Quantity: Has magnitude only and no direction. Specified completely by a single number with appropriate units. Examples: distance, mass, temperature, time.
  • Vector Quantity: Has both magnitude and direction and obeys the triangle law or parallelogram law of addition. Examples: displacement, velocity, acceleration, force.

• Vector Representation

  • Represented using bold letters (e.g., $\mathbf{A}$) or an arrow over a letter (e.g., $\vec{A}$).
  • Absolute magnitude is denoted by $|\vec{A}| = A$
  • Graphically, a vector is represented by a directed line segment. Point $P$ is the tail and point $Q$ is the head. Line length indicates magnitude; arrowhead indicates direction.

• Position and Displacement Vectors

  • Position vectors of points $P$ and $P'$ at times $t$ and $t'$ are $\mathbf{r} = OP$ and $\mathbf{r}' = OP'$.
  • Displacement vector $\vec{PP'}$ is the straight line joining initial position $P$ to final position $P'$: $\Delta \mathbf{r} = \mathbf{r}' - \mathbf{r}$.
  • Displacement is independent of the actual path taken.

• Equality of Vectors

  • Two vectors $\mathbf{A}$ and $\mathbf{B}$ are equal if and only if they have identical magnitudes and directions.

• Multiplication of Vectors by Real Numbers

  • Multiplying vector $\mathbf{A}$ by a scalar $\lambda > 0$ yields vector $\lambda \mathbf{A}$ with same direction and magnitude $\lambda |\mathbf{A}|$.
  • Multiplying vector $\mathbf{A}$ by a scalar $\lambda < 0$ flips direction and yields magnitude $-\lambda |\mathbf{A}|$.

• Graphical Method of Vector Addition and Subtraction

  • Triangle Law of Vector Addition (Head-to-Tail Method): If two vectors are represented in magnitude and direction by two sides of a triangle taken in order, their resultant is given by the third side taken in reverse order.
  • Vector addition is commutative: $\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}$
  • Vector addition is associative: $(\mathbf{A} + \mathbf{B}) + \mathbf{C} = \mathbf{A} + (\mathbf{B} + \mathbf{C})$
  • Subtraction of Vectors: Defined via addition: $\mathbf{A} - \mathbf{B} = \mathbf{A} + (-\mathbf{B})$.
  • Parallelogram Law of Vector Addition: If two vectors are represented in magnitude and direction by adjacent sides of a parallelogram, their resultant is given by the diagonal passing through their common point.     

• Problem: Direction of Umbrella in Rain and Wind

  • Rain falls vertically at $v_r = 35\,\text{m\,s}^{-1}$. Wind blows East to West at $v_w = 12\,\text{m\,s}^{-1}$.
  • Resultant speed $R$:     $R = \sqrt{v_r^2 + v_w^2} = \sqrt{35^2 + 12^2} = \sqrt{1225 + 144} = \sqrt{1369} = 37\,\text{m\,s}^{-1}$
  • Angle $\theta$ with vertical:     $\tan(\theta) = \frac{v_w}{v_r} = \frac{12}{35} = 0.343 \implies \theta = \tan^{-1}(0.343) \approx 19^\circ$
  • The boy should hold his umbrella in the vertical plane at an angle of $19^\circ$ with the vertical towards the East.

• Unit Vectors and Vector Resolution

  • Unit Vector: A vector of unit magnitude pointing in a specific direction. Dimensionless and unitless. $\hat{A} = \frac{\mathbf{A}}{|\mathbf{A}|} \implies \mathbf{A} = |\mathbf{A}| \hat{A}$.
  • Orthogonal Unit Vectors: $\hat{i}$, $\hat{j}$, $\hat{k}$ along $x$, $y$, $z$ axes. $|\hat{i}| = |\hat{j}| = |\hat{k}| = 1$
  • Resolution in a Plane: $\mathbf{A} = A_x \hat{i} + A_y \hat{j}$, where $A_x = A \cos(\theta)$ and $A_y = A \sin(\theta)$.
  • Magnitude $A = \sqrt{A_x^2 + A_y^2}$, angle $\theta = \tan^{-1}\left(\frac{A_y}{A_x}\right)$.
  • In 3D: $\mathbf{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ with magnitude $A = \sqrt{A_x^2 + A_y^2 + A_z^2}$.

• Analytical Method of Vector Addition

  • From geometry:     $OS^2 = ON^2 + SN^2$$ON = OP + PN = A + B \cos(\theta), \quad SN = B \sin(\theta)$$R^2 = (A + B \cos(\theta))^2 + (B \sin(\theta))^2 = A^2 + 2 A B \cos(\theta) + B^2 \cos^2(\theta) + B^2 \sin^2(\theta)$$R = \sqrt{A^2 + B^2 + 2 A B \cos(\theta)}$
  • Direction $\alpha$ of resultant with vector $\mathbf{A}$:     $\tan(\alpha) = \frac{SN}{OP + PN} = \frac{B \sin(\theta)}{A + B \cos(\theta)}$
  • Law of Sines: $\frac{R}{\sin(\theta)} = \frac{B}{\sin(\alpha)} \implies \sin(\alpha) = \frac{B}{R} \sin(\theta)$.

• Problem: Resultant Velocity of a Motorboat

  • Motorboat heading North at $v_b = 25\,\text{km/h}$, current $v_c = 10\,\text{km/h}$ at $60^\circ$ East of South.
  • Angle between vectors $\theta = 180^\circ - 60^\circ = 120^\circ$
  • Resultant velocity magnitude $R$:     $R = \sqrt{v_b^2 + v_c^2 + 2 v_b v_c \cos(120^\circ)} = \sqrt{25^2 + 10^2 + 2(25)(10)(-1/2)} = \sqrt{625 + 100 - 250} = \sqrt{475} \approx 22\,\text{km/h}$
  • Direction $\phi$ East of North:     $\sin(\phi) = \frac{v_c}{R} \sin(120^\circ) = \frac{10 \times \frac{\sqrt{3}}{2}}{21.8} = \frac{10 \sqrt{3}}{2 \times 21.8} \approx 0.397 \implies \phi \approx 23.4^\circ$

• Motion in Two Dimensions

  • Position vector: $\mathbf{r} = x \hat{i} + y \hat{j}$
  • Displacement vector: $\Delta \mathbf{r} = \Delta x \hat{i} + \Delta y \hat{j}$
  • Velocity vector: $\mathbf{v} = \frac{d\mathbf{r}}{dt} = v_x \hat{i} + v_y \hat{j}$, where $v_x = \frac{dx}{dt}$, $v_y = \frac{dy}{dt}$.
  • Magnitude $v = \sqrt{v_x^2 + v_y^2}$, angle $\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)$.
  • Acceleration vector: $\mathbf{a} = \frac{d\mathbf{v}}{dt} = a_x \hat{i} + a_y \hat{j}$, where $a_x = \frac{dv_x}{dt}$, $a_y = \frac{dv_y}{dt}$.

• Problem: Particle Derivatives from Position Function

  • Position function: $\mathbf{r} = 3.0 t \hat{i} + 2.0 t^2 \hat{j} + 5.0 \hat{k}$
  • Part (a): Velocity and acceleration functions:     $\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = 3.0 \hat{i} + 4.0 t \hat{j}$$\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = 4.0 \hat{j} \quad (a = 4.0\,\text{m\,s}^{-2} \text{ along } y\text{-direction})$
  • Part (b): Magnitude and direction of $\mathbf{v}(t)$ at $t = 1.0\,\text{s}$:     $\mathbf{v}(1.0) = 3.0 \hat{i} + 4.0 \hat{j}$$\text{Magnitude } v = \sqrt{3^2 + 4^2} = 5.0\,\text{m\,s}^{-1}$$\theta = \tan^{-1}\left(\frac{4.0}{3.0}\right) \approx 53^\circ \text{ with } x\text{-axis}$

• Projectile Motion

  • An object in flight after being thrown or projected is called a projectile. Its path (trajectory) is parabolic.
  • Initial velocity components: $u_x = u \cos(\theta)$ along horizontal; $u_y = u \sin(\theta)$ along vertical.
  • Acceleration components: $a_x = 0$ (horizontal velocity $u \cos(\theta)$ is constant); $a_y = -g$. At peak, v_y = 0$.\n - **Equation of Path**:\n    x = u \cos(\theta) t \implies t = \frac{x}{u \cos(\theta)}\n    y = u \sin(\theta) t - \frac{1}{2} g t^2 = u \sin(\theta) \left(\frac{x}{u \cos(\theta)}\right) - \frac{1}{2} g \left(\frac{x}{u \cos(\theta)}\right)^2\n    y = \tan(\theta) x - \frac{g}{2 u^2 \cos^2(\theta)} x^2\n    This takes the parabolic form y = a x + b x^2.\n - **Time of Flight (T)**:\n    s_y = u_y T + \frac{1}{2} a_y T^2 \implies 0 = u \sin(\theta) T - \frac{1}{2} g T^2 \implies T = \frac{2 u \sin(\theta)}{g}\n - **Horizontal Range (R)**:\n    R = u \cos(\theta) \times T = u \cos(\theta) \times \frac{2 u \sin(\theta)}{g} = \frac{u^2 \sin(2\theta)}{g}\n - Maximum range occurs at \theta = 45^\circ$:$R_{\max} = \frac{u^2}{g}.\n - Range is identical for complementary angles \theta$and$90^\circ - \theta:\n      \text{For } (90^\circ - \theta), \quad R = \frac{u^2 \sin(2(90^\circ - \theta))}{g} = \frac{u^2 \sin(180^\circ - 2\theta)}{g} = \frac{u^2 \sin(2\theta)}{g}\n - **Maximum Height (H)**:\n    v_y^2 - u_y^2 = 2 a_y H \implies 0 - u^2 \sin^2(\theta) = -2 g H \implies H = \frac{u^2 \sin^2(\theta)}{2g}\n\n- **Problem: Cricket Ball Projectile**\n - u = 28\,\text{m\,s}^{-1}$,$\theta = 30^\circ$,$g = 9.8\,\text{m\,s}^{-2}.\n - **(a) Maximum Height**: H = \frac{28^2 \sin^2(30^\circ)}{2 \times 9.8} = \frac{784 \times 0.25}{19.6} = 10\,\text{m}.\n - **(b) Time of Flight**: T = \frac{2 \times 28 \times \sin(30^\circ)}{9.8} = \frac{28}{9.8} = 2.9\,\text{s}.\n - **(c) Horizontal Range**: R = \frac{28^2 \sin(60^\circ)}{9.8} = \frac{784 \times 0.866}{9.8} = 69\,\text{m}.\n\n- **Uniform Circular Motion**\n - Motion of an object along a circular path at constant speed.\n - **Period (T)**: Time for one full revolution.\n - **Frequency (\nu$)**: Number of revolutions per second ($\nu = \frac{1}{T}$, unit: Hertz$\text{Hz}).\n - **Angular Velocity (\omega$)**: Rate of change of angular displacement$\Delta \theta:\n    \omega = \lim_{\Delta t \to 0} \frac{\Delta \theta}{\Delta t} = \frac{d\theta}{dt} = \frac{2\pi}{T} = 2\pi \nu \quad (\text{unit: rad/s})\n - **Relation between Linear and Angular Velocity**: v = r \omega.\n - **Angular Acceleration (\alpha$)**:$\alpha = \frac{d\omega}{dt} = \frac{d^2 \theta}{dt^2}.\n - **Centripetal Acceleration (a_c)**: Acceleration directed towards the centre along the radius:\n    a_c = \frac{v^2}{R} = \omega^2 R = v \omega\n\n- **Problem: Insect Trapped in Circular Groove**\n - Radius R = 12\,\text{cm} = 0.12\,\text{m}$, completes 7 revolutions in$100\,\text{s}$($T = \frac{100}{7}\,\text{s}).\n - **(a) Angular and Linear Speed**:\n    \omega = \frac{2\pi}{T} = \frac{2\pi}{100/7} = \frac{14\pi}{100} = 0.44\,\text{rad/s}\n    v = \omega R = 0.44 \times 0.12 = 5.3 \times 10^{-2}\,\text{m\,s}^{-1}\n - **(b) Acceleration Vector**: Direction changes continuously towards the centre (not constant). Magnitude:\n    a = \omega^2 R = (0.44)^2 \times 0.12 = 2.3 \times 10^{-2}\,\text{m\,s}^{-2}\n\n# Laws of Motion\n\n- **Galileo's Law of Inertia**\n - If net external force is zero, a body at rest continues to remain at rest, and a body in motion continues to move with uniform velocity. Inertia means resistance to change.\n - Example: A standing passenger jerks backward when a bus starts suddenly (inertia of rest) and thrown forward when a moving bus stops suddenly (inertia of motion).\n\n- **Newton's First Law of Motion**\n - Every body continues in its state of rest or of uniform motion in a straight line unless compelled by an external force. Net external force equals zero implies zero acceleration.\n\n- **Momentum**\n - Momentum \mathbf{p}$is the product of mass$m$and velocity$\mathbf{v}$:$\mathbf{p} = m \mathbf{v}.\n - Vector quantity; unit: \text{kg\,m\,s}^{-1}$; dimensions:$[\text{M L T}^{-1}].\n - Heavier/faster objects require greater force to stop in a given time interval.\n\n- **Newton's Second Law of Motion**\n - The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force:\n    F \propto \frac{\Delta \mathbf{p}}{\Delta t} \implies F = \frac{d\mathbf{p}}{dt}\n - Catching a ball: Drawing hands backward increases time interval \Delta t$, reducing impact force$F = \frac{\Delta p}{\Delta t}.\n\n- **Derivation of Force Equation (F = m a)**\n - For constant mass m:\n    F = \frac{d}{dt}(m \mathbf{v}) = m \frac{d\mathbf{v}}{dt} = m a\n - Unit: \text{kg\,m\,s}^{-2}$or Newton ($\text{N}$). 1 N is the force causing$1\,\text{m\,s}^{-2}$acceleration in a$1\,\text{kg} mass.\n - **Vector Form**: F_x = m a_x$,$F_y = m a_y$,$F_z = m a_z.\n - Important features: Consistent with 1st law (F = 0 \implies a = 0$); force component affects only the parallel velocity component;$F_{\text{net}} = m a_{\text{system}}; instantaneous acceleration depends on instantaneous force.\n\n- **Problem: Bullet Fired into Wooden Block**\n - Bullet mass m = 0.04\,\text{kg}$, initial speed$u = 90\,\text{m/s}$, stopped ($v = 0$) after distance$s = 60\,\text{cm} = 0.6\,\text{m}.\n - Constant retardation a:\n    v^2 - u^2 = 2 a s \implies 0 - 90^2 = 2 \times a \times 0.6 \implies a = -\frac{8100}{1.2} = -6750\,\text{m\,s}^{-2}\n - Retarding force F = m a = 0.04 \times (-6750) = -270\,\text{N}.\n\n- **Problem: Force from Vertical Motion Equation**\n - Motion given by y = u t + \frac{1}{2} g t^2.\n - Velocity v = \frac{dy}{dt} = u + g t.\n - Acceleration a = \frac{dv}{dt} = g.\n - Force F = m a = m g\n\n- **Impulse and Impulsive Force**\n - **Impulse (I)**: Product of large force acting for a short duration and time duration; equals change in momentum.\n    I = F \times t = \Delta p\n - Unit: \text{kg\,m\,s}^{-1}$or$\text{N\,s}$; dimensions:$[\text{M L T}^{-1}].\n - **Impulse-Momentum Principle**: F = \frac{dp}{dt} \implies F dt = dp \implies I = \Delta p.\n\n- **Problem: Impulse Imparted to a Baseball/Cricketball**\n - Batsman hits back ball (m = 0.15\,\text{kg}$) straight at initial speed$12\,\text{m\,s}^{-1} without changing speed.\n - \text{Initial momentum} = 0.15 \times (-12) = -1.8\,\text{kg\,m\,s}^{-1}.\n - \text{Final momentum} = 0.15 \times 12 = 1.8\,\text{kg\,m\,s}^{-1}.\n - \text{Impulse} = \Delta p = 1.8 - (-1.8) = 3.6\,\text{N\,s}.\n\n- **Newton's Third Law of Motion**\n - To every action, there is an equal and opposite reaction: F_{AB} = -F_{BA}.\n - Forces occur in pairs simultaneously on different bodies (do not cancel out).\n\n- **Apparent Weight of a Body in an Elevator**\n  \n - Normal reaction R gives apparent weight.\n - **Case 1 (Rest or Uniform Velocity, a = 0$)**:$R - m g = 0 \implies R = m g (No change in weight).\n - **Case 2 (Accelerating Upwards at a$)**:$R - m g = m a \implies R = m(g + a) (Weight increases).\n - **Case 3 (Accelerating Downwards at a$)**:$m g - R = m a \implies R = m(g - a) (Weight decreases).\n - **Case 4 (Free Fall, a = g$)**:$R = m(g - g) = 0 (Weightlessness).\n\n- **Problem: Passenger Scale Reading in Elevator**\n - Man of mass m = 70\,\text{kg}$,$g = 10\,\text{m\,s}^{-2}.\n - **(a) Upward uniform speed (10\,\text{m\,s}^{-1}$,$a = 0$)**:$R = m g = 700\,\text{N}$, scale reading$= \frac{700}{10} = 70\,\text{kg}.\n - **(b) Downward acceleration (a = 5\,\text{m\,s}^{-2}$)**:$R = m(g - a) = 70(10 - 5) = 350\,\text{N}$, scale reading$= 35\,\text{kg}.\n - **(c) Upward acceleration (a = 5\,\text{m\,s}^{-2}$)**:$R = m(g + a) = 70(10 + 5) = 1050\,\text{N}$, scale reading$= 105\,\text{kg}.\n - **(d) Free fall (a = g$)**:$R = 0\,\text{N}$, scale reading$= 0\,\text{kg}.\n\n- **Law of Conservation of Momentum**\n - Total momentum of an isolated system (F_{\text{ext}} = 0) is conserved.\n - **Recoil of Gun**: Initial momentum = 0$. Final momentum$\mathbf{p}b + \mathbf{p}_g = 0 \implies m \mathbf{v} + M \mathbf{V} = 0.\n    \text{Recoil velocity } V = -\frac{m v}{M}, \quad \text{Muzzle velocity } v = -\frac{M V}{m}\n - **Collision of Two Bodies**: F{AB} \Delta t = p_A' - p_A$and$F_{BA} \Delta t = p_B' - p_B$. Since$F_{AB} = -F_{BA}$,$p_A' - p_A = -(p_B' - p_B) \implies p_A' + p_B' = p_A + p_B.\n\n- **Equilibrium of a Particle**\n - Net external force is zero (\sum \mathbf{F}i = 0).\n - Two forces: \mathbf{F}_1 + \mathbf{F}_2 = 0 \implies \mathbf{F}_1 = -\mathbf{F}_2\n - Three forces: \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = 0 \implies \mathbf{F}_1 + \mathbf{F}_2 = -\mathbf{F}_3\n\n- **Problem: Suspended Mass in Equilibrium**\n - Mass m = 6\,\text{kg}$($W = m g = 60\,\text{N}$) suspended by$2\,\text{m}$rope. Horizontal force$50\,\text{N}$applied at midpoint$P.\n - At midpoint P$, upper rope tension$T_1$at angle$\theta$to vertical; lower rope tension$T_2 = 60\,\text{N}.\n - Vertical equilibrium: T_1 \cos(\theta) = 60\n - Horizontal equilibrium: T_1 \sin(\theta) = 50\n - Dividing gives: \tan(\theta) = \frac{50}{60} = \frac{5}{6} \implies \theta = \tan^{-1}\left(\frac{5}{6}\right) \approx 40^\circ\n\n- **Friction**\n - Force opposing relative motion between surfaces in contact.\n - **Static Friction (f_s$)**: Opposes impending motion. Maximum value$f{s,\max} = \mu_s N$, where$\mu_s is coefficient of static friction.\n    \text{Law of Static Friction: } f_s \le \mu_s N\n - **Kinetic Friction (f_k$)**: Opposes actual sliding motion.$f_k = \mu_k N$, where$\mu_k < \mu_s. Independent of contact area and velocity.\n\n- **Body on an Inclined Plane and Angle of Friction**\n  \n - Forces on block at point of sliding: Weight m g$, Normal force$N$, Static friction$f_{s,\max}.\n - m g \sin(\theta) = f_{s,\max} = \mu_s N\n - m g \cos(\theta) = N\n - Dividing equations: \mu_s = \tan(\theta) \implies \theta = \tan^{-1}(\mu_s)$.$\theta is the angle of friction.\n - Example: If \theta = 15^\circ$,$\mu_s = \tan(15^\circ) = 0.27.\n\n- **Problem: Train Box Static Acceleration**\n - Maximum acceleration of train for box (\mu_s = 0.15) to remain stationary:\n    m a = f_s \le \mu_s m g \implies a \le \mu_s g \implies a_{\max} = 0.15 \times 10 = 1.5\,\text{m\,s}^{-2}\n\n- **Problem: Block and Trolley System Acceleration and Tension**\n  \n - Trolley mass 20\,\text{kg}$, hanging block mass$3\,\text{kg}$,$\mu_k = 0.04$,$g = 10\,\text{m\,s}^{-2}.\n - Hanging block motion: 30 - T = 3 a\n - Trolley motion: T - f_k = 20 a$, where$f_k = \mu_k N = 0.04 \times (20 \times 10) = 8\,\text{N}.\n - System equations: T - 8 = 20 a\n - Adding equations: 22 = 23 a \implies a = \frac{22}{23} \approx 0.96\,\text{m\,s}^{-2}.\n - Tension: T = 30 - 3(0.96) = 27.1\,\text{N}.\n\n- **Rolling Friction, Disadvantages, Advantages, and Reduction**\n - Rolling friction is significantly smaller than static or kinetic friction.\n - Disadvantage: Dissipates power as heat in moving machine parts.\n - Advantage: Essential for walking, automobile braking, tyre traction.\n - Methods to reduce friction: Lubricants, ball bearings, cushion of air between surfaces.\n\n- **Circular Motion Dynamics**\n - Centripetal force directed towards centre: f_c = \frac{m v^2}{R}.\n - **Car on Level Curved Road**: Frictional force provides centripetal acceleration:\n    f_s = \frac{m v^2}{R} \le \mu_s N = \mu_s m g \implies v_{\max} = \sqrt{\mu_s R g}\n - **Problem: Cyclist on Turn** (v = 18\,\text{km/h} = 5\,\text{m\,s}^{-1}$,$R = 3\,\text{m}$,$\mu_s = 0.1$,$g = 9.8\,\text{m\,s}^{-2}):\n    v^2 = 25\,\text{m}^2\text{s}^{-2}, \quad \mu_s R g = 0.1 \times 3 \times 9.8 = 2.94\,\text{m}^2\text{s}^{-2}\n    Since v^2 > \mu_s R g, the cyclist will slip.\n\n- **Car on a Banked Curved Road**\n  \n - Vertical force balance: N \cos(\theta) - f \sin(\theta) = m g\n - Horizontal centripetal force: N \sin(\theta) + f \cos(\theta) = \frac{m v^2}{R}\n - **Maximum Safe Speed**: v_{\max} = \sqrt{R g \frac{\mu_s + \tan(\theta)}{1 - \mu_s \tan(\theta)}}\n - **Optimum Speed** (without friction, \mu_s = 0$):$v_{\text{optimum}} = \sqrt{R g \tan(\theta)}.\n\n- **Problem: Racetrack Speeds**\n - R = 300\,\text{m}$,$\theta = 15^\circ$,$\mu_s = 0.2$,$g = 9.8\,\text{m\,s}^{-2}.\n - **(a) Optimum Speed**:\n    v_o = \sqrt{R g \tan(\theta)} = \sqrt{300 \times 9.8 \times \tan(15^\circ)} = \sqrt{2940 \times 0.2679} = 28.1\,\text{m\,s}^{-1}\n - **(b) Maximum Permissible Speed**:\n    v_{\max} = \sqrt{\frac{300 \times 9.8 \times (\tan(15^\circ) + 0.2)}{1 - 0.2 \times \tan(15^\circ)}} = \sqrt{\frac{2940 \times 0.4679}{1 - 0.05358}} = 38.1\,\text{m\,s}^{-1}$$