AP Physics 1 Premium Prep Notes

Buoyant Force

• The buoyant force is calculated using Fb = pgVsub, where Fb is the buoyant force, p is the fluid's density, Vsub is the volume of the displaced fluid, and g is the acceleration due to gravity.

• $V_{sub}$ represents the volume of the object submerged in the fluid.

• For a balloon surrounded by air, $V_{sub}$ is the entire volume of the balloon.

• The volume of a sphere is given by $V = \frac{4}{3}πr^3$.

• If the radius of a sphere is doubled, its volume increases by a factor of 8.

Energy Conversion on a Ramp

• At the top of a ramp, a car possesses potential energy: $PE = mgh$.

• At the bottom, potential energy converts to kinetic energy: $KE = \frac{1}{2}mv^2 = mgh$.

• Multiplying both sides of $\frac{1}{2}mv^2 = mgh$ by 2 yields $mv^2 = 2mgh$.

• Dividing by $r$ gives $\frac{mv^2}{r} = \frac{2mgh}{r}$.

• The term $\frac{mv^2}{r}$ represents centripetal force, thus $\frac{2mgh}{r}$ is equivalent to centripetal force in this scenario.

Fluid Pressure

• Points at equal heights in fluid A have the same pressure.

• Gauge pressure from fluid B at the interface equals gauge pressure from fluid A at depth $y$: pBgh = pAgy

• Therefore, we can express this as ( pB = pA gy/h ) to show the relationship between the pressures at different depths in the two fluids.

• Therefore, the density of fluid B can be expressed as: pB = pA y/h

• This relationship indicates that the density of fluid B is proportional to the depth in fluid A and inversely proportional to the height ratio, which is essential for understanding pressure dynamics in different fluid systems. .

Projectile Motion

• Split initial velocity (vi) into horizontal (vx0) and vertical ( vy0) components:

  • Given: $v_i = 20 \text{ m/s}$, $\theta = 30^\circ$

  • $v_{x0} = (20 \text{ m/s})(\cos 30^\circ) = (20 \text{ m/s})(\frac{\sqrt{3}}{2}) = 10\sqrt{3} \text{ m/s}$

  • $v_{y0} = (20 \text{ m/s})(\sin 30^\circ) = (20 \text{ m/s})(\frac{1}{2}) = 10 \text{ m/s}$

• Vertical motion analysis:

  • $\Delta y = y - y_0 = 0$ (starts and ends on the ground)

  • $v_{y0} = 10 \text{ m/s}$

  • $v_y = \text{don't care}$

  • $a = -10 \text{ m/s}^2$

  • $t = ?$

• Using kinematic equation #3 to find the time:

  • Two solutions: $t = 0 \text{ s}$ (initial launch) and $t = 2 \text{ s}$ (landing).

• Alternative method: Calculate time to reach maximum height using kinematic equation #2, then double it.

Applying Kinematic equation

• Using the equation x = x0 + vit + 1/2at^2 to find displacement.

• Given: x0 = 0$,$vi = 10m/s, t = 2 s, a = -2.4m/s^2

• Calculate displacement (x) using the provided values:

  • x = 0 + (10 m/s)(2 s) + 1/2(-2.4 m/s²)(2 s)²

  • x = 20 m - 4.8 m

  • x = 15.2 m, indicating the maximum height achieved.

• $x = 0 + (10 \text{ m/s})(2 \text{ s}) + \frac{1}{2}(-2.4 \text{ m/s}^2)(2 \text{ s})^2 = 15.2 \text{ m}$

Net Force and Tension

• Net force is calculated using Newton's Second Law: $F_{net} = ma$.

• Tension (T) in a string connected to a hanging mass:

  • $F_{net} = mg - T$

  • $T = mg - ma$

• Kinetic friction acting on a block:

  • Fnet = T - Ff = T - F_N = T - (µ)mg

Forces on an Inclined Plane

• For an object at rest on an inclined plane, the net force is 0.

• Forces perpendicular to the plane: $F_N = mg\cos\theta$.

• Forces parallel to the plane: $F_{net} = T - mg\sin\theta - \mu mg\cos\theta = 0$.

• Tension equation: $T = mg\sin\theta + \mu mg\cos\theta$.

• For a hanging block connected to the block on the incline: $F_{net} = Mg - T = Mg - (mg\sin\theta + \mu mg\cos\theta) = 0$ .

• Solving for the coefficient of friction: $\mu = \frac{Mg - mg\sin\theta}{mg\cos\theta}$.

Bernoulli's Equation

• Bernoulli's equation: PA + ½ p air vA^2 + p air g yA = PB + ½ p air vB^2 + p air g yB

• Given: Point B is at rest (vB = 0$) and$yA = yB

• Simplified equation: PA + 1/2ho vA^2 = PB

• Pressure difference: PB - PA = p air gh

• Solving for vA$:$vA =

• This speed is determined by the liquid's density (ρ), the acceleration due to gravity (g), the height (h) of the liquid, and the density of the air (p air).

Uniform Accelerated Motion

• Using the equation: $d = v_0t + \frac{1}{2}at^2$

• Given: $d = 0.5 \text{ m}$, $v_0 = 0 \text{ m/s}$, $a = 5 \text{ m/s}^2$

• Solving for $t$: $0.5 = (0)t + \frac{1}{2}(5)t^2 \implies t = \sqrt{\frac{2 \cdot 0.5}{5}} = \frac{1}{\sqrt{5}} \approx 0.45 \text{ s}$.

Work-Energy Theorem

• Total work is the sum of work done by all forces: W{total} = W{person} + W{normal} + W{gravity} + W_{friction}.

• Normal force and gravity are perpendicular to displacement, so they do no work: W{normal} = W{gravity} = 0.

• Work-Energy Theorem: W{total} = \Delta KE = KEf - KE_i

• Work done by a person: W{person} = Δ KE - W{friction} = 1/2mv^2 - 0 - Fd cos(θ)

• Given: $m = 60 \text{ kg}$, $v = 3 \text{ m/s}$, $F = 300 \text{ N}$, $d = 5 \text{ m}$, $\theta = 20^\circ$

• $W_{person} = \frac{1}{2}(60 \text{ kg})(3 \text{ m/s})^2 - (300 \text{ N})(5 \text{ m})\cos 20^\circ = -1140 \text{ J}$

Torque

• Torque formula: $\tau = Fr\sin\theta$.

• If $\theta=90°$, then $\sin\theta = 1$, simplifying the formula to $\tau = Fr$

• Gravitational force: $F_g = mg$.

Projectile Motion - Horizontal

• Horizontal motion equations:

  • $x - x_0 = ?$

  • $v_x = 10\sqrt{3} \text{ m/s}$ (constant forward speed)

  • $t = 2 \text{ s}$

• Displacement: $d = v_xt = (10\sqrt{3} \text{ m/s})(2 \text{ s}) = 20\sqrt{3} \text{ m} \approx 34.6 \text{ m}$

Momentum Conservation

• Momentum conservation equation: m1v{1,0} + m2v{2,0} = m1v{1,f} + m2v{2,f}.

• Given: m1 = 10kg, v{1,0} = 10m/s, v{2,0} = 0m/s, v{1,f} = -2m/s, v_{2,f} = 8m/s.

• Solving for m2$:$(10)(10) + m2(0) = (10)(-2) + m2(8) \implies 100 = -20 + 8m2 \implies m_2 = 120/8 = 15 kg

• Therefore, the mass of m2 is 15 kg, confirming the conservation of momentum in this isolated system.

Collisions

• Perfectly Inelastic Collision: Objects stick together after the collision (not true if objects have different velocities post-collision)

• Perfectly Elastic Collision: Kinetic energy is conserved

• Checking for KE conservation:
  1/2m1v{1,0}^2 + 1/2m2v{2,0}^2 = 1/2m1v{1,f}^2 + 1/2m2v{2,f}^2

• Given: m1=4kg, v{1,0} = 10kg, m2=6kg, v{2,0} =0m/s, v{1,f} = -2 m/s,

  v{2,f} = 8m/s

• Calculated:

• $\frac{1}{2}(4)(10)^2 + \frac{1}{2}(6)(0)^2 = \frac{1}{2}(4)(-2)^2 + \frac{1}{2}(6)(8)^2$

• $200 + 0 = 8 + 192$
  Therefore, it is true that the energy is conserved in a perfectly elastic collision.

Mechanical Energy Conservation

• Mechanical energy is conserved in all cases; energy required is $mgh$.

Torque and Angular Acceleration

• $\tau = I\alpha$ (Torque equals moment of inertia times angular acceleration).

• For equal torques, the object with a smaller moment of inertia (I) will have greater angular acceleration ($\alpha$).

• Objects with mass concentrated closer to the center have smaller moment of inertia.

• Solid sphere has smaller I than hollow sphere -> solid sphere increases angular speed more quickly.

Friction

• Friction force: $F<em>f = \mu F</em>N$

• Vertical forces: gravity, normal force, and the vertical component of the applied force.

• Case 1: $F_N = mg - mg\sin\theta = mg - mg \sin 30^\circ = mg - \frac{mg}{2} = \frac{mg}{2}$

• Case 2: $F_N = mg + mg\sin\theta = mg + mg \sin 30^\circ = mg + \frac{mg}{2} = \frac{3mg}{2}$

• The force in the second case will be three times greater than in the first.

Newton's Second Law with Two Blocks

• Block 1: $F_{net} = ma = T - mg$

• Block 2: $F_{net} = M(-a) = T - Mg$

• Subtracting the second equation from the first: $ma + Ma = Mg - mg$

• Solving for acceleration: $a = \frac{g(M-m)}{m+M} = \frac{10(15-5)}{5+15} = \frac{100}{20} = 5 \text{ m/s}^2$

Summation of Forces

In AP Physics 1, summing forces involves applying Newton's Second Law, which states that the net force ([F]) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a), written as [F = ma]. To write a summation of forces:

1. Identify all forces: List all forces acting on the object (e.g., gravity (Fg$), tension ($T$), normal force ($FN$), friction ($Ff$), applied force ($Fa).

2. Choose a coordinate system: Select a coordinate system (usually x-y) to simplify the force analysis. Align one axis with the direction of acceleration if possible.

3. Resolve forces into components: Break each force into its x and y components using trigonometry (e.g., F{x} = F[Cos](θ)$,$F{y} = F[Sin](θ)).

4. Write equations for net force: Apply Newton's Second Law separately for each axis:

   • Σ Fx = ma_x

   • ΣFy = ma_y

   • (Σ: indicates that you are finding the sum of a series of terms)

5. Substitute and solve: Substitute known values into the equations and solve for unknowns.

Example:

Consider a block of mass (m) on an inclined plane with angle (θ), with an applied force ($T$) pulling it up the plane, and kinetic friction ($F_f$) opposing the motion.

1. Forces: Gravity (Fg = mg$), Tension ($ T$), Normal Force ($FN$), Friction ($Ff).

2. Coordinate System: x-axis along the inclined plane, y-axis perpendicular to the plane.

3. Components:

   • Gravity Components: Fgx= mg\sin(θ)$,$Fgy = mgcos(θ)

4. Net Force Equations:

   • ΣFx = T - mg\sin(θ) - Ff = ma_x

   • ΣFy = FN - mg\cos(θ) = 0

5. Solving:

   • From y-equation: $F_N = mg\cos(θ)$

   • Friction: Ff = \mu FN = μmgcos(θ)

   • Substitute into x-equation: $T - mg\sin(θ) - \mu mg\cos(θ) = ma_x$

   • Solve for $T$ or $a_x$ if other values are known.