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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( PE = KE \) | Use conservation of mechanical energy from the top to the bottom of the ramp, where potential energy (PE) is converted to kinetic energy (KE). |
| 2 | \( Mgh = \frac{1}{2}M(3.5v_0)^2 \) | Initial potential energy \( Mgh \) at the top is equal to the kinetic energy \( \frac{1}{2}M(3.5v_0)^2 \) at the bottom. |
| 3 | \( gh = \frac{1}{2}(3.5v_0)^2 \) | Cancel the mass \( M \) from both sides (as it does not affect the energy balance). |
| 4 | \( h = \frac{(3.5v_0)^2}{2g} \) | Solve for the height \( h \). |
| 5 | \( h = \frac{12.25v_0^2}{2g} \) | Calculate \( (3.5)^2 = 12.25 \) to obtain the final expression. |
| 6 | \( h = \frac{12.25v_0^2}{2g} \) | Final simplified expression for the height \( h \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( p_i = p_f \) | Apply conservation of momentum before and after the collision. |
| 2 | \( M(3.5v_0) = 1.5M(2v_0) + Mv_x \) | Initial momentum of the system equals the final momentum. Find the speed \( v_x \) of the small block immediately after the collision. |
| 3 | \( 3.5v_0 = 3v_0 + v_x \) | Cancel \( M \) from both sides and solve for \( v_x \). |
| 4 | \( v_x = 0.5v_0 \) | Calculate \( v_x \). |
| 5 | \( \text{Work done by friction} = \Delta KE \) | The work done by friction is equal to the change in kinetic energy of the larger block. |
| 6 | \( \mu \cdot 1.5MgD = \frac{1}{2}1.5M(2v_0)^2 \) | The friction work (\( \mu \times \text{force} \times \text{distance} \)) equals the initial kinetic energy of the larger block. |
| 7 | \( \mu \cdot gD = 2v_0^2 \) | Simplify the equation by canceling \( 1.5M \) from both sides. |
| 8 | \( \mu = \frac{2v_0^2}{gD} \) | Solve for \( \mu \). |
| 9 | \( \mu = \frac{2v_0^2}{gD} \) | Final expression for the coefficient of friction \(\mu\). |
Just ask: "Help me solve this problem."
A \( 240 \) \( \text{kg} \) block is dropped from \( 3.0 \) meters onto a spring, compresses the spring and comes to rest.
A small block moving with a constant speed v collides inelastically with a block M attached to one end of a spring k. The other end of the spring is connected to a stationary wall. Ignore friction between the blocks and the surface.
A 75.0kg log floats downstream with a speed of 1.80 m/s. Eight frogs hop onto the log in a series of perfectly inelastic collisions. If each frog has a mass of 0.30 kg and an upstream speed of 1.3 m/s, what is the change in kinetic energy for this system?
A car accelerates uniformly from rest to [katex] 29.4 [/katex] m/s in [katex] 6.93 [/katex] s along a level stretch of road. Ignoring friction, determine the average power in both watts and horsepower ([katex] 1 \text{ horsepower} = 745.7 \text{ Watts} [/katex]) required to accelerate the car if:
A rubber ball bounces off of a wall with an initial speed v and reverses its direction so its speed is v right after the bounce. As a result of this bounce, which of the following quantities of the ball are conserved?
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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] |
General Metric Conversion Chart
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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