Physics 1 Exam Review: Dynamics, Circular Motion, and Projectile Motion

Physics 1 Exam Review

1. Banked Curves and Circular Motion

When a car travels around a banked curve, the banking angle helps provide the necessary centripetal force to keep the car moving in a circle without relying solely on friction.

• Banked Curve: A road or track that is tilted at an angle to help vehicles turn safely at a certain speed.

• Given: Speed of the car ($v$), radius of the curve ($r$).

• Find: The angle ($\theta$) at which the road should be banked so that no friction is required.

Key Equation:

• $\tan\theta = \frac{v^2}{rg}$

Example: For a car traveling at 20 m/s around a curve of radius 50 m, the required banking angle is $\tan\theta = \frac{(20)^2}{50 \times 9.8}$.

2. Ramps, Pulleys, and Acceleration

Systems involving ramps and pulleys are classic problems in Newtonian mechanics, often requiring the analysis of forces, masses, and acceleration.

• Scenario: An object hangs over a pulley at the top of a ramp; another object is on the ramp (with the pulley attached to it).

• Given: Angle of the ramp ($\theta$), masses of both objects ($m_1$, $m_2$), and no friction.

• Find: The acceleration of the system.

Key Equations:

• Forces on ramp: $m_1g\sin\theta$ (down the ramp)

• Forces on hanging mass: $m_2g$ (downward)

• Net force: $F_{net} = m_2g - m_1g\sin\theta$

• Total mass: $m_1 + m_2$

• Acceleration: $a = \frac{m_2g - m_1g\sin\theta}{m_1 + m_2}$

Example: If $m_1 = 2$ kg, $m_2 = 3$ kg, $\theta = 30^\circ$, $g = 9.8$ m/s2, then $a = \frac{3 \times 9.8 - 2 \times 9.8 \times \sin 30^\circ}{2 + 3}$.

3. Projectile Motion from a Height

Projectile motion problems often involve objects launched horizontally or at an angle from a certain height, requiring analysis of both horizontal and vertical motion.

• Scenario: An object is thrown horizontally from the top of a building.

• Given: Height of the building ($h$), horizontal distance to where the object lands ($d$).

• Find: The initial velocity ($v_0$) of the object.

Key Equations:

• Time to fall: $t = \sqrt{\frac{2h}{g}}$

• Horizontal velocity: $v_0 = \frac{d}{t}$

Example: If $h = 20$ m, $d = 30$ m, $g = 9.8$ m/s2, then $t = \sqrt{\frac{2 \times 20}{9.8}}$, $v_0 = \frac{30}{t}$.

4. Frictional Force in Unbanked Curves

When a car moves around an unbanked curve, friction provides the centripetal force needed to keep the car on the path.

• Given: Mass of the car ($m$), velocity ($v$), diameter of the curve ($d$).

• Find: The required frictional force ($F_f$).

Key Equation:

• Radius: $r = \frac{d}{2}$

• Centripetal force: $F_f = \frac{mv^2}{r}$

Example: For $m = 1000$ kg, $v = 15$ m/s, $d = 40$ m, $r = 20$ m, $F_f = \frac{1000 \times 15^2}{20}$.

5. General Projectile Motion

Projectile motion involves analyzing the path of an object launched at an angle with an initial velocity.

• Given: Initial velocity ($v_0$) and launch angle ($\theta$).

• Find: Range, maximum height, and time of flight.

Key Equations:

• Horizontal velocity: $v_{0x} = v_0 \cos\theta$

• Vertical velocity: $v_{0y} = v_0 \sin\theta$

• Time of flight: $t = \frac{2v_0\sin\theta}{g}$

• Range: $R = \frac{v_0^2 \sin 2\theta}{g}$

• Maximum height: $H = \frac{v_0^2 \sin^2\theta}{2g}$

Example: For $v_0 = 20$ m/s, $\theta = 45^\circ$, $g = 9.8$ m/s2, $R = \frac{20^2 \sin 90^\circ}{9.8}$.

6. Comparing Acceleration on Earth and the Moon

Objects experience different accelerations on Earth and the Moon due to differences in gravitational acceleration.

• Scenario: A block on a frictionless table on Earth is given a certain force, resulting in a specific acceleration.

• Given: Force ($F$), mass ($m$), gravity on Earth ($g_{Earth}$), gravity on Moon ($g_{Moon}$).

• Find: The acceleration on the Moon when the same force is applied.

Key Equation:

• Newton's Second Law: $a = \frac{F}{m}$ (acceleration is independent of gravity for a frictionless surface)

Additional info: If friction is present, the normal force (and thus friction) would depend on gravity, but for a frictionless table, only the applied force and mass matter.

7. Gravitational Pull Between Earth and Moon

Gravitational force acts between any two masses, such as the Earth and the Moon, or an object placed between them.

• Given: Mass of Earth ($M_{Earth}$), mass of Moon ($M_{Moon}$), distance between centers ($d$), and mass of the object ($m$).

• Find: The net gravitational force on the object at a point between Earth and Moon.

Key Equation:

• Newton's Law of Universal Gravitation: $F = G \frac{Mm}{r^2}$

• Net force: $F_{net} = F_{Earth} - F_{Moon}$ (direction depends on position)

Example: If the object is at a distance $x$ from Earth, $F_{Earth} = G \frac{M_{Earth} m}{x^2}$, $F_{Moon} = G \frac{M_{Moon} m}{(d-x)^2}$.

Additional info: The point where the net force is zero is called the "neutral point" or Lagrange point between Earth and Moon.