Mechanics | HSC - Wyatt's Notes

HSC physics study notes - Mechanics

flowchart TD
    A[Mechanics] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Equations of motion (constant acceleration): $v = u + at$ $s = ut + \frac{1}{2}at^2$ $v^2 = u^2 + 2as$

Average velocity: $\bar{v} = \frac{u + v}{2}$

Newton’s Second Law: $F = ma$

Weight: $W = mg$ (where $g \approx 9.8\,\text{m/s}^2$)

Friction: $f = \mu N$ where $\mu$ is the coefficient of friction

Linear momentum: $p = mv$

Impulse: $J = F\Delta t = \Delta p$

Conservation of momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ (when no external forces)

Kinetic energy: $E_k = \frac{1}{2}mv^2$

Gravitational potential energy: $E_p = mgh$

Work done: $W = Fs\cos\theta$

Work-energy theorem: $W_{\text{net}} = \Delta E_k$

Problem: A ball is thrown horizontally at $20\,\text{m/s}$ from a height of $45\,\text{m}$. Find the horizontal range ($g = 10\,\text{m/s}^2$).

Solution:

Step 1: Find time of flight using vertical motion: $s = \frac{1}{2}gt^2 \Rightarrow 45 = \frac{1}{2}(10)t^2 \Rightarrow t = 3\,\text{s}$

Step 2: Find horizontal range: $\text{Range} = u_x \cdot t = 20 \times 3 = 60\,\text{m}$

Answer: The horizontal range is $60\,\text{m}$

Problem: A $5\,\text{kg}$ block is pushed across a rough surface with a $20\,\text{N}$ horizontal force. If the coefficient of kinetic friction is $0.3$, find the acceleration ($g = 10\,\text{m/s}^2$).

Solution:

Step 1: Calculate the normal force: $N = mg = 5 \times 10 = 50\,\text{N}$

Step 2: Calculate friction: $f = \mu N = 0.3 \times 50 = 15\,\text{N}$

Step 3: Find net force: $F_{\text{net}} = 20 - 15 = 5\,\text{N}$

Step 4: Apply Newton’s second law: $a = \frac{F_{\text{net}}}{m} = \frac{5}{5} = 1\,\text{m/s}^2$

Answer: The acceleration is $1\,\text{m/s}^2$

Problem: A $2\,\text{kg}$ ball moving at $3\,\text{m/s}$ collides with a stationary $3\,\text{kg}$ ball. After the collision, the $2\,\text{kg}$ ball moves at $1\,\text{m/s}$. Find the velocity of the $3\,\text{kg}$ ball.

Solution:

Step 1: Apply conservation of momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ $2 \times 3 + 3 \times 0 = 2 \times 1 + 3 \times v_2$

Step 2: Solve for $v_2$: $6 = 2 + 3v_2 \Rightarrow v_2 = \frac{4}{3} \approx 1.33\,\text{m/s}$

Answer: The velocity of the $3\,\text{kg}$ ball is $\frac{4}{3}\,\text{m/s}$

1. Always draw a free-body diagram before applying Newton’s laws
2. For projectile motion, separate horizontal and vertical components
3. Momentum is conserved only when there are no external forces
4. Check units and significant figures in your final answer

1. A car accelerates from rest at $2\,\text{m/s}^2$ for $5\,\text{s}$. Find the distance covered.
2. A $10\,\text{kg}$ object is pulled up a $30°$ incline with a force of $80\,\text{N}$. Find the acceleration (frictionless).
3. Two objects of mass $4\,\text{kg}$ and $6\,\text{kg}$ collide head-on. The $4\,\text{kg}$ object was moving at $5\,\text{m/s}$ and the $6\,\text{kg}$ object at $-3\,\text{m/s}$. After the collision, the $4\,\text{kg}$ object moves at $-1\,\text{m/s}$. Find the final velocity of the $6\,\text{kg}$ object.

Problem: A $20\,\text{kg}$ block is pushed $10\,\text{m}$ along a horizontal surface by a $50\,\text{N}$ force at $30°$ above the horizontal. The coefficient of kinetic friction is $0.2$. Find the work done by each force and the net work.

Solution:

Step 1: Work done by applied force: $W_F = Fd\cos\theta = 50 \times 10 \times \cos 30° = 50 \times 10 \times 0.866 = 433\,\text{J}$

Step 2: Normal force: $N = mg - F\sin\theta = 20 \times 10 - 50 \times \sin 30° = 200 - 25 = 175\,\text{N}$

Step 3: Friction force: $f = \mu N = 0.2 \times 175 = 35\,\text{N}$

Step 4: Work done by friction: $W_f = -f \times d = -35 \times 10 = -350\,\text{J}$

Step 5: Work done by gravity and normal force is zero (perpendicular to displacement).

Step 6: Net work: $W_{\text{net}} = 433 - 350 = 83\,\text{J}$

Answer: Net work is $83\,\text{J}$

Problem: A ball is thrown from a $20\,\text{m}$ cliff at $25\,\text{m/s}$ at $45°$ above the horizontal. Find the horizontal range ($g = 10\,\text{m/s}^2$).

Solution:

Step 1: Components: $v_{0x} = 25\cos 45° = 17.68\,\text{m/s}$, $v_{0y} = 25\sin 45° = 17.68\,\text{m/s}$

Step 2: Vertical displacement: $s = -20\,\text{m}$ (below starting point)

Step 3: Using $s = v_{0y}t - \frac{1}{2}gt^2$: $-20 = 17.68t - 5t^2$ $5t^2 - 17.68t - 20 = 0$

Step 4: Solving: $t = \frac{17.68 + \sqrt{17.68^2 + 400}}{10} = \frac{17.68 + 26.83}{10} = 4.45\,\text{s}$

Step 5: Range: $R = v_{0x} \times t = 17.68 \times 4.45 = 78.7\,\text{m}$

Answer: The horizontal range is approximately $78.7\,\text{m}$

Problem: A $3\,\text{kg}$ ball moving at $4\,\text{m/s}$ collides elastically with a $1\,\text{kg}$ ball at rest. Find the velocities after collision.

Solution:

For elastic collisions: $v_1' = \frac{m_1 - m_2}{m_1 + m_2}v_1 = \frac{3-1}{3+1} \times 4 = \frac{2}{4} \times 4 = 2\,\text{m/s}$

$v_2' = \frac{2m_1}{m_1 + m_2}v_1 = \frac{2 \times 3}{3+1} \times 4 = \frac{6}{4} \times 4 = 6\,\text{m/s}$

Answer: The $3\,\text{kg}$ ball continues at $2\,\text{m/s}$ and the $1\,\text{kg}$ ball moves at $6\,\text{m/s}$

Mechanics is the foundation of physics and engineering. From designing bridges and vehicles to understanding planetary motion, the principles of Newton’s laws, energy conservation, and momentum are universally applicable.

1. For problems involving height, always define a coordinate system and be consistent with signs
2. Kinetic energy is always positive; work can be positive or negative
3. In elastic collisions, both momentum and kinetic energy are conserved
4. Use energy methods when speed and height are involved — they are often simpler than force methods

Problem: A $5\,\text{kg}$ block is placed on a $30°$ incline with coefficient of kinetic friction $\mu = 0.2$. The block is given an initial velocity of $4\,\text{m/s}$ up the incline. Find how far up the incline the block travels before stopping ($g = 10\,\text{m/s}^2$).

Solution:

Step 1: Forces along the incline (taking up as positive): $F_{\text{gravity along incline}} = -mg\sin\theta = -5 \times 10 \times \sin 30° = -25\,\text{N}$

Step 2: Normal force: $N = mg\cos\theta = 5 \times 10 \times \cos 30° = 43.3\,\text{N}$

Step 3: Friction (opposes motion, so acts down the incline): $f = \mu N = 0.2 \times 43.3 = 8.66\,\text{N}$

Step 4: Net force: $F_{\text{net}} = -25 - 8.66 = -33.66\,\text{N}$

Step 5: Acceleration: $a = \frac{F_{\text{net}}}{m} = \frac{-33.66}{5} = -6.73\,\text{m/s}^2$

Step 6: Using $v^2 = u^2 + 2as$: $0 = 16 + 2(-6.73)s \implies s = \frac{16}{13.46} = 1.19\,\text{m}$

Answer: The block travels $1.19\,\text{m}$ up the incline before stopping

Common mistake: Forgetting that friction acts down the incline when the block is moving up. Friction always opposes the direction of motion.

Problem: A $2\,\text{kg}$ ball is thrown vertically upward at $15\,\text{m/s}$. Find the height at which its speed is $5\,\text{m/s}$ ($g = 10\,\text{m/s}^2$).

Solution:

Step 1: Apply the work-energy theorem: $W_{\text{net}} = \Delta E_k$

Step 2: The only force doing work is gravity: $W = -mgh$

Step 3: Change in kinetic energy: $\Delta E_k = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2 = \frac{1}{2}(2)(25) - \frac{1}{2}(2)(225) = 25 - 225 = -200\,\text{J}$

Step 4: Set equal: $-mgh = -200$ $-2 \times 10 \times h = -200$ $h = 10\,\text{m}$

Answer: The height is $10\,\text{m}$

Common mistake: Forgetting the negative sign for work done by gravity. Gravity does negative work when the object moves upward.

Problem: A $0.5\,\text{kg}$ ball is tied to a string of length $1\,\text{m}$ and swung in a horizontal circle at $3\,\text{rev/s}$. Find the tension in the string.

Solution:

Step 1: Find the angular velocity: $\omega = 2\pi f = 2\pi \times 3 = 6\pi\,\text{rad/s}$

Step 2: The tension provides the centripetal force: $T = m\omega^2 r = 0.5 \times (6\pi)^2 \times 1 = 0.5 \times 36\pi^2 = 18\pi^2 \approx 177.7\,\text{N}$

Answer: The tension is approximately $178\,\text{N}$

Common mistake: Using $v = \omega r$ incorrectly. For uniform circular motion, $v = \omega r$ and $T = mv^2/r = m\omega^2 r$.

Mechanics describes how objects move under forces: Think of forces as pushes and pulls. When you push a shopping cart, you apply a force. Friction pushes back. The net force determines how the cart accelerates. This simple principle — force equals mass times acceleration — explains everything from falling apples to orbiting planets.

Why it matters: Mechanics is the foundation of physics and engineering. Every bridge, building, vehicle, and aircraft is designed using these principles. Understanding kinematics, forces, momentum, and energy is essential for solving real-world problems.

The key insight: Energy methods are often simpler than force methods — instead of tracking every force at every moment, just compare the beginning and end states.

Confusing mass with weight in force calculations. Mass (kg) is the amount of matter, while weight (N) is the gravitational force W = mg. When applying Newton’s second law F = ma, the mass in kg is used, not the weight in Newtons. Students often substitute weight where mass is required, giving acceleration values that are too large.

• Calculus — Differentiation and integration are used to derive kinematic equations and analyse variable acceleration.
• Algebra — Solving systems of equations and quadratic formulas are essential for force and projectile problems.
• Waves — Oscillatory motion and wave mechanics extend the principles of simple harmonic motion from mechanics.
• Inorganic — Energy conservation in thermochemistry parallels the work-energy theorem in mechanics.

Forgetting that friction opposes motion, not force direction. Friction acts opposite to the direction of relative motion or attempted motion, not necessarily opposite to the applied force. When pushing a block up a ramp, friction acts down the ramp. When the block slides down, friction acts up the ramp. The direction changes with the motion.

Using displacement instead of distance in kinematics. Displacement is a vector (can be negative), while distance is a scalar (always positive). When an object changes direction, the total distance travelled is the sum of the magnitudes of each segment, while the net displacement may be small or zero. Confusing the two gives incorrect speed calculations.