| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | v_{\text{initial}} = \frac{M_{\text{boat}} \cdot v_{\text{boat}}}{M_{\text{boat}}} | Initial velocity of the boat can be expressed in terms of its mass M_{\text{boat}} and its velocity v_{\text{boat}}. |
| 2 | v_{\text{final}} = \frac{(M_{\text{boat}} + m_{\text{sand}}) \cdot v_{\text{new}}}{M_{\text{boat}} + m_{\text{sand}}} | Final velocity of the boat after receiving the sack of sand of mass m_{\text{sand}} considers the total new mass and the new velocity v_{\text{new}} which conserves momentum. |
| 3 | M_{\text{boat}} \cdot v_{\text{boat}} = (M_{\text{boat}} + m_{\text{sand}}) \cdot v_{\text{new}} | Apply the law of conservation of momentum. The total momentum before the sack drops must equal the total momentum after, assuming no external forces. |
| 4 | v_{\text{new}} = \frac{M_{\text{boat}} \cdot v_{\text{boat}}}{M_{\text{boat}} + m_{\text{sand}}} | Solve the conservation of momentum equation for v_{\text{new}}, the new velocity of the system (boat plus sand). |
| 5 | v_{\text{new}} < v_{\text{boat}} | As m_{\text{sand}} is positive and added to M_{\text{boat}}, the denominator in the expression for v_{\text{new}} is larger than M_{\text{boat}}, hence v_{\text{new}} is less than v_{\text{boat}}. |
| 6 | Answer: (a) decrease | Since the final velocity v_{\text{new}} of the boat is less than its initial velocity v_{\text{boat}}, the boat will slow down, indicating that the correct answer is (a) decrease. |
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The figure shows a graph of the position x of two cars, C and D, as a function of time t. According to this graph, which statements about these cars must be true? (There could be more
than one correct choice.)
Which of the following graphs represent an object at rest?
Ball 1 is dropped from rest at time t = 0 from a tower of height h. At the same instant, ball 2 is launched upward from the ground with the initial speed v_0. If air resistance is negligible, at what time t will the two balls pass each other?
An object of mass 2 kg is thrown vertically downwards with an initial kinetic energy of 100 J. What is the distance fallen by the object at the instant when its kinetic energy has doubled?
An object is thrown downward at 23 m/s from the top of a 200 m tall building.
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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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