| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 | Conservation of momentum |
| 2 | 4,kg \cdot 10,m/s + 7,kg \cdot 0,m/s = 4,kg \cdot v_{1x} + 7,kg \cdot 2,m/s \cdot \cos(22^\circ) | Plugging in given values and decomposing the 7 kg mass’s velocity into horizontal component. |
| 3 | 40 = 4v_{1x} + 13.08 | Calculating the horizontal momentum contribution from the 7 kg mass post-collision. |
| 4 | 4v_{1x} = 26.92 | Solve for the 4 kg mass’s horizontal velocity component. |
| 5 | v_{1x} = 6.73,m/s | Calculating the horizontal velocity of the 4 kg mass. |
| 6 | m_1 u_{1y} + m_2 u_{2y} = m_1 v_{1y} + m_2 v_{2y} | Conservation of momentum in the vertical direction. Since the initial vertical momentum is 0, the final combined vertical momentum must also be 0. |
| 7 | 0 = 4,kg \cdot v_{1y} + 7,kg \cdot 2,m/s \cdot \sin(22^\circ) | Recognizing that initial vertical velocities are 0 and calculating the vertical component for the 7 kg mass. |
| 8 | v_{1y} = -1.04,m/s | Calculating the vertical velocity of the 4 kg mass (negative indicates opposite direction to the 7 kg mass’s vertical component). |
| 9 | v_1 = \sqrt{v_{1x}^2 + v_{1y}^2} and \theta = \arctan\left(\frac{v_{1y}}{v_{1x}}\right) | Combining horizontal and vertical components to find magnitude and direction of the 4 kg mass’s velocity. |
| 10 | v_1 \approx \sqrt{6.73^2 + (-1.04)^2} | Plugging in horizontal and vertical components. |
| 11 | v_1 \approx 6.81,m/s | Calculating the magnitude of velocity. |
| 12 | \theta \approx \arctan\left(\frac{-1.04}{6.73}\right) | Calculating the direction of the velocity. |
| 13 | \theta \approx -8.8^\circ | Determining the angle below the horizontal for the 4 kg mass’s velocity. |
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A rubber ball with a mass of 0.25 kg and a speed of 19.0 m/s collides perpendicularly with a wall and bounces off with a speed of 21 m/s in the opposite direction. What is the magnitude of the impulse acting on the rubber ball?
A pool cue ball, mass 0.7 kg, is traveling at 2 m/s when it collides head on with another ball, mass 0.5 kg, traveling in the opposite direction with a speed of 1.2 m/s. After the collision, the cue ball travels in the opposite direction at 0.3 m/s. What is the velocity of the other ball?
A truck going 15 km/h has a head-on collision with a small car going 30 km/h. Which statement best describes the situation?
A rubber ball and a lump of clay have equal mass. They are thrown with equal speed against a wall. The ball bounces back with nearly the same speed with which it hit. The clay sticks to the wall. Which one of these objects experiences the greater impulse?
An astronaut initially at rest in space throws a wrench, and recoils in the opposite direction. Select all that is true.
6.81 m/s at 8.8° above the horizontal
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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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