# Part (a): Solving Using Conservation of Energy
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | U_i + K_i = U_f + K_f | The total mechanical energy (kinetic + potential) at the initial state equals the total mechanical energy at the final state due to conservation of energy. |
| 2 | U_i = m_g \cdot g \cdot h | Initial potential energy is due to m_g being at a height h from the ground. m_y is on the ground, so its potential energy is zero. |
| 3 | U_i = 38 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \cdot 2.5 \, \text{m} | Calculate the numerical value of U_i . g (acceleration due to gravity) is 9.8 m/s² and h is 2.5 m. |
| 4 | U_i = 931 \, \text{J} | Multiplying the values provides the initial potential energy. |
| 5 | K_i = 0 | Initially, the system is at rest so the initial kinetic energy is zero. |
| 6 | K_f = \frac{1}{2}(m_y + m_g + m_p) v^2 | Final kinetic energy calculated as the sum of the kinetic energies of m_g , m_y and the pulley. m_p is the mass of the pulley. The pulley contributes translational kinetic energy due to its mass. |
| 7 | U_f = 0 | At the final state, m_g is on the ground, so its potential energy is zero. |
| 8 | 931 \, \text{J} = \frac{1}{2} (32 \, \text{kg} + 38 \, \text{kg} + 3.1 \, \text{kg}) v^2 | Use U_i + K_i = U_f + K_f and solve for v . |
| 9 | v = \sqrt{\frac{2 \times 931 \, \text{J}}{73.1 \, \text{kg}}} | Simplify the equation to solve for v . |
| 10 | v = \sqrt{25.48 \, \text{m}^2/\text{s}^2} | Calculate the value under the square root. |
| 11 | v = 5.05 \, \text{m/s} | The final velocity of m_g just before it strikes the ground. |
# Part (b): Solving Using Forces, Torque, and Kinematics
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | F = m \cdot a | Newton’s second law, the net force applied on the system causes acceleration. |
| 2 | F_{my} – F_{mg} = (m_y + m_g + m_p) \cdot a | Forces due to weights of m_y and m_g , including the pulley’s inertia, where F_{my} = m_y \cdot g and F_{mg} = m_g \cdot g . |
| 3 | T = \frac{m_g g – m_y g}{1 + \frac{m_p}{m_y + m_g}} | Balance the forces taking into account the pulley rotation effect, where T is the net torque and m_p is moment of inertia derived as \frac{1}{2} m_p R^2 assuming a uniform cylinder. |
| 4 | a = \frac{m_g g – m_y g}{m_y + m_g + \frac{1}{2} m_p} | Solving the earlier equation for a , the acceleration of the system, where \frac{1}{2} m_p comes from dividing the pulley’s moment of inertia by R^2 . |
| 5 | v_f^2 = 2 \cdot a \cdot 2.5 \, \text{m} | Use kinematic equation for final velocity, incorporating the calculated acceleration and distance fallen, which is 2.5 m. |
| 6 | v_f = \sqrt{2 \cdot a \cdot 2.5 \, \text{m}} | Solve for final velocity using the value for a . |
| 7 | Approximate numerical calculation similar to Part (a) | The velocity is essentially the same result as obtained from energy conservation, demonstrating the physics consistency. |
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| Kinematics | Forces |
|---|---|
| \Delta x = v_i t + \frac{1}{2} at^2 | F = ma |
| v = v_i + at | F_g = \frac{G m_1m_2}{r^2} |
| a = \frac{\Delta v}{\Delta t} | f = \mu N |
| R = \frac{v_i^2 \sin(2\theta)}{g} |
| Circular Motion | Energy |
|---|---|
| F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
| a_c = \frac{v^2}{r} | PE = mgh |
| KE_i + PE_i = KE_f + PE_f |
| Momentum | Torque and Rotations |
|---|---|
| p = m v | \tau = r \cdot F \cdot \sin(\theta) |
| J = \Delta p | I = \sum mr^2 |
| p_i = p_f | L = I \cdot \omega |
| Simple Harmonic Motion |
|---|
| F = -k x |
| T = 2\pi \sqrt{\frac{l}{g}} |
| T = 2\pi \sqrt{\frac{m}{k}} |
| Constant | Description |
|---|---|
| g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
| G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
| \mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
| k | Spring constant, in \text{N/m} |
| M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
| M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
| M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| s (Displacement) | \text{meters (m)} |
| v (Velocity) | \text{meters per second (m/s)} |
| a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
| t (Time) | \text{seconds (s)} |
| m (Mass) | \text{kilograms (kg)} |
| Variable | Derived SI Unit |
|---|---|
| F (Force) | \text{newtons (N)} |
| E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
| P (Power) | \text{watts (W)} |
| p (Momentum) | \text{kilogram meters per second (kgm/s)} |
| \omega (Angular Velocity) | \text{radians per second (rad/s)} |
| \tau (Torque) | \text{newton meters (Nm)} |
| I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |
| f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | 10^{-12} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
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