AP Physics

Unit 4 - Energy




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# Part (a): Finding the final speed of the proton. Note you can also use conversation of energy to find the speed, where W_{\text{machine}} + KE_i = KE_f .

Step Derivation/Formula Reasoning
1 v_f^2 = v_i^2 + 2ad Use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance traveled, where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and d is the distance.
2 v_f^2 = (2.4 \times 10^7 \, \text{m/s})^2 + 2 \times (3.6 \times 10^{15} \, \text{m/s}^2) \times (0.035 \, \text{m}) Substitute v_i = 2.4 \times 10^7 \, \text{m/s} , a = 3.6 \times 10^{15} \, \text{m/s}^2 , and d = 3.5 \, \text{cm} = 0.035 \, \text{m} .
3 v_f = \sqrt{ (2.4 \times 10^7)^2 + 2 \times 3.6 \times 10^{15} \times 0.035} Simplify and solve for v_f .
4 v_f = \sqrt{5.76 \times 10^{14} + 2.52 \times 10^{14}} Calculate inside the square root.
5 v_f = \sqrt{8.28 \times 10^{14}} Sum the terms under the square root.
6 v_f = 2.88 \times 10^7 \, \text{m/s} Take the square root to find the final speed.

# Part (b): Calculating the increase in kinetic energy

Step Derivation/Formula Reasoning
1 \Delta KE =KE_f – KE_i The change in kinetic energy is the difference between the initial and final kinetic energy.
2 \Delta KE = \frac{1}{2} m (v_f^2 – v_i^2) Substitute in the formula for kinetic energy and factor out \frac{1}{2} m .
3 \Delta KE = \frac{1}{2} (1.67 \times 10^{-27} \, \text{kg}) [(2.88 \times 10^7 \, \text{m/s})^2 – (2.4 \times 10^7 \, \text{m/s})^2] Substitute the values of m, v_f, v_i .
4 \Delta KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times 2.52 \times 10^{14} Simplify the expression.
5 \Delta KE = 2.10 \times 10^{-13} \, \text{J} Calculate the final change in kinetic energy, which is the increase in kinetic energy of the proton.

# Part (c): Effect of tripling the acceleration on the increase in kinetic energy

Step Derivation/Formula Reasoning
1 W = \Delta KE Use the work energy pricinple, which states the work applied to the proton is equal to the change in its kinetic energy.
2 Fd = \Delta KE Substitute W with Fd since W = Fd .
3 mad = \Delta KE Substitute F with ma since F = ma .
4 md = \frac{\Delta KE}{a} Divide by acceleration on both sides. This equation clearly shows that \Delta KE is directly proportional to a . Hence tripling acceleration will also triple the the change in kinetic energy.

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  1. v_f = 2.88 \times 10^7 \, \text{m/s}
  2. \Delta KE = 2.10 \times 10^{-13} \, \text{J}
  3. \Delta KE would also triple.

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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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