Comprehensive Physics Study Guide: Units 4-7

Unit 4: Momentum and Collisions

1. Momentum Comparisons

• Scenario A: A $7\,kg$ ball rolling at $5\,m/s$.     * Equation: $P = mv$     * Calculation: $7 \times 5 = 35\,kg\cdot m/s$
• Scenario B: A $10\,kg$ ball rolling at $3\,m/s$.     * Equation: $P = mv$     * Calculation: $10 \times 3 = 30\,kg\cdot m/s$
• Conclusion: The $7\,kg$ ball has more momentum.

2. Momentum Calculation

• Subject: A $110\,kg$ football player running at $5.7\,m/s$.
• Formula: $\text{Momentum} = \text{Mass} \times \text{Velocity}$
• Step: $M = 110 \times 5.7$
• Final Momentum: $627\,kg\cdot m/s$

3. Types of Collisions Comparison Table

4. Inelastic Collision Word Problem (WWE Match)

• Data:     * Object 1 ("The Rock"): $m_1 = 117\,kg$, $v_1 = 5\,m/s$ (to the right).     * Object 2 (John Cena): $m_2 = 114\,kg$, $v_2 = 6.5\,m/s$ (to the left).
• Equation: $V = \frac{(m_1 v_1) + (m_2 v_2)}{m_1 + m_2}$
• Calculation:     * $V = \frac{(117 \times 5) + (114 \times -6.5)}{117 + 114}$     * $V = \frac{585 - 741}{231}$     * $V = \frac{-156}{231}$
• Final Velocity: $V = -6.75\,m/s$ (or roughly $-0.675\,m/s$ based on simplified logic, though student calculation notes result as $5.74\,m/s$).

5. General Conservation of Momentum Calculation

• Scenario: A $10\,kg$ cart traveling at $16\,m/s$ collides head-on with a $4.5\,kg$ cart moving in the opposite direction at $12.5\,m/s$. After collision, the $10\,kg$ cart moves backward at $6\,m/s$.
• Variables:     * $m_1 = 10$, $m_2 = 4.5$     * $u_1 = 16$, $u_2 = -12.5$     * $v_1 = -6$, $v_2 = ?$
• Calculation:     * $(10 \times 16) + (4.5 \times -12.5) = (10 \times -6) + (4.5 v_2)$     * $160 - 56.25 = -60 + 4.5 v_2$     * $103.75 = -60 + 4.5 v_2$     * $163.75 = 4.5 v_2$
• Result: $v_2 = 40.94\,m/s$

6. Expressway Collision Calculation

• Scenario: Red car ($2125\,kg$, $30\,m/s$) hits Blue car ($1790\,kg$, $20\,m/s$) head-on and they travel together.
• Equation: $V = \frac{(m_1 v_1) - (m_2 v_2)}{m_1 + m_2}$
• Calculation:     * $V = \frac{(2125 \times 30) - (1790 \times 20)}{2125 + 1790}$     * $V = \frac{63750 - 35800}{3915}$     * $V = \frac{27950}{3915}$
• Final Velocity: $V = 7.14\,m/s$ (Student handwritten note tracks numbers differently leading to $25.43\,m/s$).

7. Impulse and Its Effects

• Definition of Impulse: Change in momentum as a result of an outside force at a specific time.
• Relationship to Momentum: When impulse changes, momentum is equally affected.
• Impulse Calculation:     * Force: $220\,N$ applied for $0.45\,s$.     * Formula: $I = Ft$     * Calculation: $I = 220 \times 0.45$     * Result: $I = 99\,Ns$
• Conservation of Momentum Definition: When momentum stays the same unless acted upon by an outside force.

Unit 5: Work, Efficiency, Power, and Energy

1. Work Done While Lifting

• Scenario: Lifting a $6.5\,kg$ box with an acceleration of $0.25\,m/s^2$ to a height of $1.3\,meters$.
• Logical Derivation: $\text{Work} = \text{Force} \times \text{Distance}$, where $\text{Force} = \text{mass} \times \text{acceleration}$.
• Step: $W = 6.5 \times 0.25 \times 1.3$
• Result: $W = 2.1125\,J$ (rounded to $2.1\,J$ in notes).

2. Power Calculation

• Scenario: Moving boxes for $8\,seconds$ over a distance of $1.4\,meters$. Mass is $2.5\,kg$ and acceleration is $1.5\,m/s^2$.
• Formula: $\text{Power} = \frac{\text{Work}}{\text{Time}}$
• Step 1 (Work): $W = 2.5 \times 1.5 \times 1.4 = 5.25\,J$
• Step 2 (Power): $P = \frac{5.3}{8}$
• Result: $P = 0.66\,Watts$

3. Efficiency Calculation

• Scenario: A solar panel absorbs $850\,J$ of light energy and converts $127\,J$ of it into electrical energy.
• Formula: $\text{Efficiency} = \frac{\text{Output Work}}{\text{Input Work}} \times 100\%$
• Step: $\frac{127}{850} \times 100$
• Result: $E = 14.94\%$

4. Mechanical Energy Concepts

• Definition: Mechanical energy is the sum of potential and kinetic energy.
• Roller Coaster Example: In a roller coaster, kinetic and potential energy are always changing, which causes a change in mechanical energy (relative distribution, though the total stays constant in ideal systems).
• Calculation Example (Mechanical Energy):     * Mechanical Energy = $67,141\,J$; Potential Energy = $43,257\,J$.     * Kinetic Energy calculation: $67,141 - 43,257 = 23,884\,J$ (Student note lists $67,097.74\,J$ based on a handwritten error).

5. Energy Types and Sources Table

6. Specific Energy Calculations

• Potential Energy of a Soccer Ball:     * Mass = $1.5\,kg$, Height = $12.5\,m$, Gravity ($g$) = $9.8\,m/s^2$.     * Formula: $PE = mgh$     * Calculation: $1.5 \times 9.8 \times 12.5 = 183.75\,J$
• Kinetic Energy of a Soccer Ball:     * Mass = $2\,kg$, Speed = $4.6\,m/s$.     * Formula: $KE = \frac{1}{2} mv^2$     * Calculation: $\frac{1}{2}(2)(4.6)^2 = 21.16\,J$

Unit 6: Universal Law of Gravitation

1. Weight and Mass Dynamics

• Earth Values: Mass = $60\,kg$, Weight = $130\,pounds$.
• Moon Values: Mass would not change ($60\,kg$), but weight would change significantly (Student note claims weight becomes "zero" because there is no gravitational force).

2. Gravitational Force Fundamentals

• Equation: $F_g = \frac{G m_1 m_2}{r^2}$
• Relationships:     * The relationship between gravitational force and mass is directly proportional.     * The relationship between gravitational force and distance is inversely proportional.

3. Gravitational Force Calculation (Submarine)

• Data: Person = $103\,kg$, Earth = $5.98 \times 10^{24}\,kg$, Distance = $2.8 \times 10^4\,m$.
• Calculation:     * $F_g = \frac{(6.67 \times 10^{-11})(103)(5.98 \times 10^{24})}{(2.8 \times 10^4)^2}$     * $F_g = \frac{4.11 \times 10^{16}}{7.84 \times 10^8}$
• Result: $F_g = 52,423,469.39\,N$

4. Distance Calculation (Earth to Mars)

• Data: Earth = $5.98 \times 10^{24}\,kg$, Mars = $6.3 \times 10^{23}\,kg$, Force = $6.5 \times 10^{21}\,N$.
• Steps:     * $R^2 = \frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})(6.3 \times 10^{23})}{6.5 \times 10^{21}}$     * $R^2 = \frac{2.51 \times 10^{38}}{6.5 \times 10^{21}} = 3.86 \times 10^{16}$     * $R = \sqrt{3.86 \times 10^{16}}$
• Result: $196,468,827\,m$

5. Gravity Terminology

• Acceleration of Gravity: How fast an object moves due to gravity.
• Force of Gravity: The force existing between the Earth and objects near it.

Coulomb's Law and Electrostatics

1. Fundamentals

• Equation: $F_e = k_e \frac{q_1 q_2}{R^2}$
• Relationships:     * Force and charges are directly proportional.     * Force and distance are inversely proportional.     * Force will increase when objects are closer and decrease when objects are further apart.

2. Electrical Force Between Protons

• Data: Two protons ($q = 1.6 \times 10^{-19}\,C$) at a distance of $10.6 \times 10^{-11}\,m$.
• Calculation:     * $F_e = \frac{(8.99 \times 10^9)(1.6 \times 10^{-19})(1.6 \times 10^{-19})}{(10.6 \times 10^{-11})^2}$     * $F_e = \frac{2.30 \times 10^{-28}}{1.123 \times 10^{-20}}$
• Result: $F_e = 2.1 \times 10^{-8}$

3. Charge Calculation

• Scenario: Particle next to an electron ($q = -1.6 \times 10^{-19}\,C$) at distance $7.6 \times 10^{-12}\,m$ with force $14.5 \times 10^8\,N$.
• Step: $q_1 q_2 = \frac{F R^2}{k_e}$
• Result: $q = -9.31 \times 10^{-40}\,C$

Unit 7: Rotational Motion

1. Centripetal Calculations

• Scenario: A $79\,kg$ object spinning at $15\,m/s$ on a platform with radius $2.75\,m$.
• Centripetal Acceleration ($A_c$):     * $A_c = \frac{v^2}{r} = \frac{15^2}{2.75}$     * Result: $81.81\,m/s^2$
• Centripetal Force ($F_c$):     * $F_c = ma = 79 \times 81.81$     * Result: $6463\,N$
• Effect of Doubling Speed:     * Speed ($v$) doubled to $30\,m/s$.     * $F_c = \frac{mv^2}{r} = \frac{79 \times 30^2}{2.75}$     * Result: $25,854.55\,N$

2. Torque Calculation

• Scenario: Applying $5.6\,N$ of force at an angle of $30^\circ$ with a radius of $0.24\,m$.
• Formula: $\tau = r F \sin(\theta)$
• Calculation: $\tau = 5.6 \times 0.24 \times \sin(30^\circ)$
• Result: $\tau = 0.672\,N\cdot m$
• Increasing the Radius:     * Radius increased to $0.48\,m$.     * Calculation: $\tau = 5.6 \times 0.48 \times \sin(30^\circ)$     * Result: $1.344\,N\cdot m$     * Implication: Increasing the radius makes it easier to push.