6.1 Kinematics, Dynamics, Newton's Laws & Vector Addition

6.1 Kinematics, Dynamics, Newton's Laws & Vector Addition

Welcome to Chapter 6 of the Pakistan Army Medical Cadet (AMC) Initial Test preparation guide. Physics forms a central pillar of the academic assessment for candidates seeking entry into the Army Medical College. This section covers fundamental physical mechanics, vector algebra, kinematics, dynamics, Newton's laws, friction, linear impulse, and projectile motion.

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1. Vector Quantities & Rectangular Component Method

Physical quantities are divided into scalars (possessing magnitude only, such as mass, temperature, and energy) and vectors (possessing both magnitude and specific direction, complying with vector addition rules, such as displacement, velocity, force, and acceleration).

Cartesian Resolution of Vectors

In a two-dimensional Cartesian coordinate system, a vector $\mathbf{A}$ making an angle $\theta$ with the positive x-axis can be resolved into two mutually perpendicular (rectangular) components:

• Horizontal component: $A_x = A \cos\theta$
• Vertical component: $A_y = A \sin\theta$
• Vector expression: $\mathbf{A} = A_x \hat{i} + A_y \hat{j}$
• Magnitude: $A = |\mathbf{A}| = \sqrt{A_x^2 + A_y^2}$

Vector Addition by Rectangular Components

When adding multiple coplanar vectors $\mathbf{A}, \mathbf{B}, \mathbf{C},\dots$:

1. Resolve each vector into horizontal and vertical components.
2. Sum all horizontal components: $R_x = \sum A_x = A_x + B_x + C_x + \dots$
3. Sum all vertical components: $R_y = \sum A_y = A_y + B_y + C_y + \dots$
4. Calculate the resultant magnitude: $R = \sqrt{R_x^2 + R_y^2}$
5. Find the reference angle $\phi = \tan^{-1}\left(\frac{|R_y|}{|R_x|}\right)$. The actual directional angle $\theta$ depends on the quadrant:

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2. Vector Multiplication: Dot & Cross Products

Vectors can be multiplied in two fundamental ways depending on whether the product yield is a scalar or a vector.

Scalar (Dot) Product

The scalar product of two vectors $\mathbf{A}$ and $\mathbf{B}$ is defined as: $\mathbf{A} \cdot \mathbf{B} = A B \cos\theta$ where $\theta$ is the angle between the vectors when placed tail-to-tail ($0^\circ \le \theta \le 180^\circ$).

• Commutative property: $\mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A}$
• Unit vector dot products: $\hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1$, and $\hat{i}\cdot\hat{j} = \hat{j}\cdot\hat{k} = \hat{k}\cdot\hat{i} = 0$.
• Component form: $\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z$
• Orthogonality test: If $\mathbf{A} \cdot \mathbf{B} = 0$ for non-zero vectors, the vectors are perpendicular ($\theta = 90^\circ$).

Vector (Cross) Product

The vector product of two vectors $\mathbf{A}$ and $\mathbf{B}$ yields a vector $\mathbf{C} = \mathbf{A} \times \mathbf{B}$ defined by: $\mathbf{A} \times \mathbf{B} = (A B \sin\theta) \hat{n}$ where $\hat{n}$ is a unit vector perpendicular to the plane containing $\mathbf{A}$ and $\mathbf{B}$, determined by the Right-Hand Rule.

• Non-commutative (Anti-commutative): $\mathbf{A} \times \mathbf{B} = -(\mathbf{B} \times \mathbf{A})$
• Unit vector cross products: $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0$; $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$.
• Geometrical significance: The magnitude $|\mathbf{A} \times \mathbf{B}| = A B \sin\theta$ equals the area of a parallelogram with adjacent sides formed by $\mathbf{A}$ and $\mathbf{B}$.
• Parallelism test: If $\mathbf{A} \times \mathbf{B} = 0$ for non-zero vectors, the vectors are parallel ($\theta = 0^\circ$) or anti-parallel ($\theta = 180^\circ$).

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3. Equilibrium of Rigid Bodies

A rigid body is in complete mechanical equilibrium when both its translational and rotational accelerations are zero.

Conditions of Equilibrium

1. First Condition (Translational Equilibrium): The vector sum of all external forces acting on the body must be zero: $\sum \mathbf{F} = 0 \implies \sum F_x = 0, \quad \sum F_y = 0, \quad \sum F_z = 0$
2. Second Condition (Rotational Equilibrium): The vector sum of all external torques acting on the body about any arbitrary axis must be zero: $\sum \boldsymbol{\tau} = 0 \implies \tau = \mathbf{r} \times \mathbf{F} = r F \sin\theta = 0$

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4. Rectilinear Kinematics & Newton's Laws of Motion

Equations of Motion under Constant Acceleration

For rectilinear motion under uniform acceleration $a$:

1. $v_f = v_i + a t$
2. $S = v_i t + \frac{1}{2} a t^2$
3. $2 a S = v_f^2 - v_i^2$
4. $S_n = v_i + \frac{a}{2}(2n - 1)$ (Distance traveled in the $n$-th second)

Newton's Laws of Motion

• First Law (Law of Inertia): A body continues in its state of rest or uniform motion in a straight line unless acted upon by a net external force. Inertia is quantified by mass.
• Second Law (Momentum Formulation): The time rate of change of linear momentum of a body is directly proportional to the applied force and takes place in the direction of the force: $\mathbf{F}_{net} = \frac{d\mathbf{p}}{dt} = \frac{d(m\mathbf{v})}{dt} = m \mathbf{a}$
• Third Law: To every action, there is an equal and opposite reaction. Action and reaction forces act on different bodies simultaneously.

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5. Friction Mechanics & Linear Impulse

Friction

Friction is the retarding force opposing relative motion between two contacting surfaces.

• Static Friction ($f_s$): Self-adjusting force that prevents relative motion. Its maximum value (limiting friction) is: $f_{s,\text{max}} = \mu_s N$ where $\mu_s$ is the coefficient of static friction and $N$ is the normal force ($N = mg$ on horizontal plane).
• Kinetic Friction ($f_k$): Retarding force operating during active sliding motion: $f_k = \mu_k N$ Note: $\mu_k < \mu_s$ always, meaning less force is needed to maintain sliding than to initiate it.

Impulse ($\mathbf{J}$)

Impulse is defined as the product of a large force acting over a short time interval, equal to the net change in linear momentum: $\mathbf{J} = \mathbf{F}_{avg} \Delta t = \Delta \mathbf{p} = m \mathbf{v}_f - m \mathbf{v}_i$

• SI Unit: $\text{N}\cdot\text{s} = \text{kg}\cdot\text{m/s}$
• Graphical representation: The area under a force-time ($F-t$) graph equals total impulse.

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6. Projectile Motion Trajectory & Key Formulae

Projectile motion is two-dimensional motion under uniform gravitational acceleration $g$, ignoring air resistance. The horizontal component of velocity remains constant ($a_x = 0$), while the vertical component experiences downward gravitational acceleration ($a_y = -g$).

v_x &= v_i \cos\theta & \quad (\text{constant}) \\ v_y &= v_i \sin\theta - g t & \quad (\text{varies with time}) \end{aligned}$$ ``` Projectile Motion Trajectory y ^ | (H_max at peak: v_y = 0) | . * . | * * | * * | * * | (v_i) * * | * \theta * +------*---------------------*-----> x (Launch) (Landing) |------ Range (R) -----| ``` ### Essential Projectile Formulae for AMC Test - **Time to reach maximum height ($t_h$):** $t_h = \frac{v_i \sin\theta}{g}$ - **Total Time of Flight ($T$):** $T = 2 t_h = \frac{2 v_i \sin\theta}{g}$ - **Maximum Height ($H_{max}$):** $H_{max} = \frac{v_i^2 \sin^2\theta}{2g}$ - **Horizontal Range ($R$):** $R = v_x T = \frac{v_i^2 \sin 2\theta}{g}$ - **Maximum Range Angle:** Range is maximum when $\sin 2\theta = 1 \implies \theta = 45^\circ$, giving $R_{max} = \frac{v_i^2}{g}$. - **Complementary Angles:** Launch angles $\theta$ and $(90^\circ - \theta)$ yield identical horizontal ranges $R$ for the same initial velocity. --- ## 7. Worked AMC Numerical Examples ### Example 1: Vector Cross Product Calculation **Question:** Vector $\mathbf{A}$ has magnitude $6\text{ units}$ and vector $\mathbf{B}$ has magnitude $8\text{ units}$. If the angle between them is $30^\circ$, calculate the magnitude of their vector cross product $|\mathbf{A} \times \mathbf{B}|$. **Solution:** 1. Use cross product magnitude formula: $|\mathbf{A} \times \mathbf{B}| = A B \sin\theta$ 2. Substitute values: $A = 6$, $B = 8$, $\sin(30^\circ) = 0.5$ 3. Calculate: $|\mathbf{A} \times \mathbf{B}| = 6 \times 8 \times 0.5 = 24\text{ units}$. ### Example 2: Projectile Kinematics **Question:** A artillery shell is fired with initial velocity $v_i = 98\text{ m/s}$ at an angle of $30^\circ$ above the horizontal. Taking $g = 9.8\text{ m/s}^2$, find (a) time of flight, (b) maximum height, and (c) horizontal range. **Solution:** 1. Vertical velocity component: $v_{iy} = v_i \sin(30^\circ) = 98 \times 0.5 = 49\text{ m/s}$. 2. Horizontal component: $v_{ix} = v_i \cos(30^\circ) = 98 \times 0.866 = 84.87\text{ m/s}$. 3. **Time of flight ($T$):** $T = \frac{2 v_{iy}}{g} = \frac{2 \times 49}{9.8} = 10\text{ seconds}$. 4. **Maximum height ($H$):** $H = \frac{v_{iy}^2}{2g} = \frac{49^2}{2 \times 9.8} = \frac{2401}{19.6} = 122.5\text{ meters}$. 5. **Horizontal Range ($R$):** $R = v_{ix} \times T = 84.87 \times 10 = 848.7\text{ meters}$. ### Example 3: Braking Dynamics & Linear Impulse **Question:** A military jeep of mass $1000\text{ kg}$ traveling at $20\text{ m/s}$ comes to rest in $4\text{ seconds}$ under constant braking. Determine the average retarding force and total impulse delivered. **Solution:** 1. Acceleration: $a = \frac{v_f - v_i}{t} = \frac{0 - 20}{4} = -5\text{ m/s}^2$. 2. Retarding force magnitude: $F = m |a| = 1000 \times 5 = 5000\text{ N}$. 3. Impulse magnitude: $J = F \Delta t = 5000 \times 4 = 20000\text{ N}\cdot\text{s}$ (or $\Delta p = m(v_f - v_i) = 1000 \times 20 = 20000\text{ kg}\cdot\text{m/s}$).