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 mgh_A = mgh_C + \frac{1}{2}mv^2 .

Step Derivation/Formula Reasoning
1 h_A = 0.8 \, \text{m} Initial height from which the marble is released.
2 v_A = 0 \, \text{m/s} Initial velocity (marble is released from rest).
3 v_B = \sqrt{2gh_A} Re-arrange and solve for velocity at the bottom of the incline, using conservation of mechanical energy, where mgh_A = \frac{1}{2}m{v^2}_B .
4 v_B = \sqrt{2 \times 9.8 \times 0.8} Calculating v_B .
5 v_B \approx 3.97 \, \text{m/s} Approximate calculation of velocity at point B.
6 h_C = 0.35 \, \text{m} The maximum height attained by the marble is at point C (top of loop).
7 v_C^2 = v_B^2 – 2gh_C Using conservation of mechanical energy between points B and C.
8 v_C^2 = 3.97^2 – 2 \times 9.8 \times 0.35 Calculating v_C from v_B and change in gravitational potential energy.
9 v_C \approx 3.0 \, \text{m/s} 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.
3 g\Delta h = \frac{1}{2}v^2 Further simplifying by replacing gh_A \,- \,gh_c with g\Delta h which is simply the change in height.
4 \frac{\Delta h}{v^2} = \frac{1}{2g} Place \Delta h and v^2 on the same side of the equation to clearly see they are directly proportional.
5 Conclusion. Since \Delta h and v^2 , if you doulbe height the velocity would increase proportionally by \boxed{\sqrt{2}} for a final speed of 3 \times \sqrt{2} \approx 4.24 \, \text{m/s}.

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  1. part (a) The resultant force would point down (towards the center of the circle).
  2. part (b) Weight (a.k.a force of gravity) and normal force
  3. part (c) Using 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 m/s.
  4. part (d) The speed would increase by \sqrt{2} , therefore the new speed would be 3 \times \sqrt{2} \approx 4.24 \, \text{m/s}.


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\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_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 MotionEnergy
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
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients 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.
kSpring 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
VariableSI 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)}
VariableDerived 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. 

  1. Start with the given measurement: \text{5 km}

  2. 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}}

  3. 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}

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



Power of Ten




















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  1. Some answers may be slightly off by 1% depending on rounding, etc.
  2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
  3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
  4. Bookmark questions that you can’t solve so you can come back to them later. 
  5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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