AP Physics

Unit 4 - Energy

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# (a) Direction of the resultant force acting on the marble at point C

The resultant force acting on the marble at point C is directed towards the center of the loop. This force is mainly comprised of the gravitational force pulling downwards and the normal force exerted by the track which also points towards the center of the loop during the circular motion.

# (b) Names of all the forces acting on the marble at point C

Force Description
Gravitational Force The force due to gravity acting downwards towards the center of the earth.
Normal Force The force exerted by the surface of the loop on the marble directed radially inward, toward the center of the loop.

# (c) Deduce the speed of the marble at point C. The working below uses two seperate conservation of energy equations. However, it can also be done in a single equation such that the postential energy at A transfroms into the potential energy at C and the kinetic energy at C. This is written as [katex] mgh_A = mgh_C + \frac{1}{2}mv^2 [/katex].

Step Derivation/Formula Reasoning
1 [katex] h_A = 0.8 \, \text{m} [/katex] Initial height from which the marble is released.
2 [katex] v_A = 0 \, \text{m/s} [/katex] Initial velocity (marble is released from rest).
3 [katex] v_B = \sqrt{2gh_A} [/katex] Re-arrange and solve for velocity at the bottom of the incline, using conservation of mechanical energy, where [katex] mgh_A = \frac{1}{2}m{v^2}_B [/katex].
4 [katex] v_B = \sqrt{2 \times 9.8 \times 0.8} [/katex] Calculating [katex] v_B [/katex].
5 [katex] v_B \approx 3.97 \, \text{m/s} [/katex] Approximate calculation of velocity at point B.
6 [katex] h_C = 0.35 \, \text{m} [/katex] The maximum height attained by the marble is at point C (top of loop).
7 [katex] v_C^2 = v_B^2 – 2gh_C [/katex] Using conservation of mechanical energy between points B and C.
8 [katex] v_C^2 = 3.97^2 – 2 \times 9.8 \times 0.35 [/katex] Calculating [katex] v_C [/katex] from [katex] v_B [/katex] and change in gravitational potential energy.
9 [katex] v_C \approx 3.0 \, \text{m/s} [/katex] Approximate calculation of velocity at point C.

# (d) Effect if the release height of the marble were to double

Step Derivation/Formula Reasoning
1 \( mgh_A = mgh_C + \frac{1}{2}mv^2 \) The conservation of energy between points A and C, as used in part B.
2 \( gh_A = gh_C + \frac{1}{2}v^2 \) Simplified formula, canceling out \( m \) on both sides.
3 \( g\Delta h = \frac{1}{2}v^2 \) Replace \( gh_A – gh_C \) with \( g\Delta h \), which represents the change in height.
4 \( \frac{\Delta h}{v^2} = \frac{1}{2g} \) Isolate \( \Delta h \) and \( v^2 \) to see their proportional relationship.
5 Proportional analysis In the initial setup: \( \Delta h = 0.8 – 0.35 = 0.45 \, \text{m} \). Doubling the release height, \( \Delta h \) becomes \( 1.6 – 0.35 = 1.25 \, \text{m} \). The ratio of \( \Delta h \) is: \[ \frac{1.25}{0.45} \approx 2.78 \] Since \( v^2 \propto \Delta h \), the velocity increases by \( \sqrt{2.78} \approx 1.67 \).
6 Conclusion The original velocity was \( 3 \, \text{m/s} \). With a velocity ratio of \( 1.67 \), the final velocity becomes: \[ 3 \times 1.67 \approx 5 \, \text{m/s} \]

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  1. The resultant force would point down (towards the center of the circle).
  2. Weight (a.k.a. force of gravity) and normal force.
  3. Using the law of conservation of energy, the potential energy at point A transforms into potential and kinetic energy at point C. So it can be deduced that:
    \( mgh_A = mgh_C + \frac{1}{2}mv^2 \)
    This would give a value of \( 3 \, \text{m/s} \).
  4. Using the same equation as above, we see that the velocity squared and the change in height are directly proportional. Therefore, if we double the initial height of \( 0.8 \, \text{m} \) to \( 1.6 \, \text{m} \), the velocity would have to increase proportionally by \( 1.67 \) times. So the original \( 3 \, \text{m/s} \) velocity would increase by \( 1.67 \) times to approximately \( 5 \, \text{m/s} \).

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\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}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

  1. 1. Some answers may vary by 1% due to rounding.
  2. Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
  3. Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
  4. Bookmark questions you can’t solve to revisit them later
  5. 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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