| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] E_{\text{total}} = E_{\text{kinetic}} + E_{\text{potential}} [/katex] | Total mechanical energy is the sum of kinetic and potential energy. |
| 2 | [katex] E_{\text{kinetic}} = \frac{1}{2}mv^2 [/katex] | Kinetic energy formula, where [katex] E_{\text{kinetic}} [/katex] is kinetic energy, [katex] m [/katex] is mass, and [katex] v [/katex] is velocity. |
| 3 | [katex] E_{\text{potential}} = mgh [/katex] | Potential energy formula, where [katex] E_{\text{potential}} [/katex] is potential energy, [katex] h [/katex] is height, and [katex] g [/katex] is gravitational acceleration. |
| 4 | At the top, [katex] E_{\text{total}} = E_{\text{potential}} [/katex] | Initially, all energy is potential energy since velocity is zero. |
| 5 | At the bottom, [katex] E_{\text{total}} = E_{\text{kinetic}} [/katex] | At the bottom, all energy is converted to kinetic energy, assuming negligible air resistance. |
| 6 | [katex] mgh = \frac{1}{2}mv^2 [/katex] | Equating potential energy at the top with kinetic energy at the bottom. |
| 7 | [katex] 2gh = v^2 [/katex] | Cancel [katex] m [/katex] and rearrange the equation. |
| 8 | [katex] v = \sqrt{2gh} [/katex] | Take the square root to find [katex] v [/katex]. |
| 9 | [katex] v_A = \sqrt{2gH} [/katex] | Apply the formula to ball A, dropped from height [katex] H [/katex]. |
| 10 | [katex] v_B = \sqrt{2g \cdot 3.5H} [/katex] | Apply the formula to ball B, dropped from height [katex] 3.5H [/katex]. |
| 11 | [katex] \frac{v_A}{v_B} = \frac{\sqrt{2gH}}{\sqrt{7gH}} [/katex] | Compare the velocities of the two balls. |
| 12 | [katex] \boxed{\frac{v_A}{v_B} = \sqrt{\frac{2}{7}}} [/katex] | Simplify to find the ratio. |
The derivation uses energy principles to arrive at the final velocity formula, and the ratio of velocities of ball A to ball B is [katex] \sqrt{\frac{2}{7}} [/katex].
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A ski tow carries people to the top of a nearby mountain. It operates on a slope of angle \( 15.7^\circ \) of length \( 260 \) \( \text{m} \). The rope moves at a speed of \( 13.0 \) \( \text{km/h} \) and provides power for \( 54 \) riders at one time, with an average mass per rider of \( 67.0 \) \( \text{kg} \).
A block of mass 3.0 kg is hung from a spring, causing it to stretch 12 cm at equilibrium. The 3.0 kg block is then taken off and the spring returns to its original height. Now a 4.0 kg block is placed on the spring and released from rest. How far will the 4.0 kg block fall before its direction is reversed?
An apple is released from rest \(500 \, \text{m}\) above the ground. Due to the combined forces of air resistance and gravity, it has a speed of \(40 \, \text{m/s}\) when it reaches the ground. What percentage of the initial mechanical energy of the apple–Earth system was dissipated due to air resistance? Take the potential energy of the apple–Earth system to be zero when the apple reaches the ground.
An object at rest suddenly explodes into two fragments (\(m_1\) and \(m_2\)) by an explosion. Fragment \(m_1\) acquires \(3\) times the kinetic energy of the other. What is the ratio of \(m_1\) to \(m_2\)?
A particle of mass \(m\) slides down a fixed, frictionless sphere of radius \(R\), starting from rest at the top.
In terms of \(m\), \(g\), \(R\), and \(\theta\), determine each of the following for the particle while it is sliding on the sphere.
A 0.2 kg object is attached to a horizontal spring undergoes SHM with the total energy of 0.4 J. The kinetic energy as a function of position presented by the graph.
A comet of mass \( m_c = 3.2 \times 10^{14} \) \( \text{kg} \) is orbiting a star with mass \( m_s = 1.8 \times 10^{30} \) \( \text{kg} \). The comet’s orbit is elliptical. At its closest point, the comet is a distance \( r_1 = 8.3 \times 10^{10} \) \( \text{m} \) from the star, and at its farthest point, the comet is a distance \( r_2 = 4.9 \times 10^{11} \) \( \text{m} \) from the star. What is the change in the kinetic energy of the comet as it moves along its orbit from distance \( r_2 \) to distance \( r_1 \) from the star?
A \( 1.0 \)\( \text{-kg} \) object is moving with a velocity of \( 6.0 \) \( \text{m/s} \) to the right. It collides and sticks to a \( 2.0 \)\( \text{-kg} \) object moving with a velocity of \( 3.0 \) \( \text{m/s} \) in the same direction. How much kinetic energy was lost in the collision?
If the coefficient of static friction is \( \mu_s = 0.5 \), how much force must be applied to a spring (spring constant of \( 0.8 \) \( \text{N/m} \)) which is attached to a block of wood (mass \( 4.0 \) \( \text{kg} \)) in order to just begin to move the block?
A person holds a book at rest a few feet above a table. The person then lowers the book at a slow constant speed and places it on the table. Which of the following accurately describes the change in the total mechanical energy of the Earth–book system?
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque 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 Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [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 |
| Variable | SI 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] |
| Variable | Derived 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] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
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]
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]
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] |
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