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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ KE_{\text{initial}} = \frac{1}{2} m_b v_0^2 \] | Initial kinetic energy of the bullet of mass \( m_b \) moving with speed \( v_0 \). |
| 2 | \[ p_{\text{initial}} = m_b v_0 \] | Initial linear momentum of the bullet (the block is at rest). |
| 3 | \[ p_{\text{after}} = (m_b + m_w) \, v \] | After impact the bullet and wooden block (mass \( m_w \)) move together with speed \( v \); their momenta add because they form a single body. |
| 4 | \[ m_b v_0 = (m_b + m_w) \, v \] | During the short collision interval external impulses (tension, gravity) are negligible, so linear momentum is conserved. |
| 5 | \[ v = \frac{m_b}{m_b + m_w} \, v_0 \] | Solving the conservation-of-momentum equation for the common speed immediately after the collision. |
| 6 | \[ KE_{\text{after}} = \frac{1}{2}(m_b + m_w) v^2 = \frac{1}{2}\, \frac{m_b^2}{m_b + m_w} \, v_0^2 \] | Kinetic energy of the bullet–block system just after the perfectly inelastic collision (note \( KE_{\text{after}} < KE_{\text{initial}} \)). |
| 7 | \[ PE_{\text{max}} = (m_b + m_w) g h \] | Gravitational potential energy when the system rises to its maximum height \( h \) above the collision point; \( g \) is the acceleration due to gravity. |
| 8 | \[ \frac{1}{2}(m_b + m_w) v^2 = (m_b + m_w) g h \] | After the collision, while the system swings upward, mechanical energy is conserved (tension does no work and air resistance is neglected). Thus all post-collision kinetic energy converts to gravitational potential energy. |
| 9 | \[ \frac{KE_{\text{after}}}{KE_{\text{initial}}} = \frac{m_b}{m_b + m_w} \lt 1 \] | Only a fraction \( \tfrac{m_b}{m_b + m_w} \) of the bullet’s original kinetic energy reappears as potential energy; the rest is dissipated as heat, sound, and deformation during the inelastic impact. |
| 10 | Evaluation of statements | (a) False – kinetic energy is not conserved in an inelastic collision. (b) True – \( KE_{\text{after}} = PE_{\text{max}} \) (Step 8). (c) False – bullet alone does not conserve momentum; only the system does. (d) False – at the highest point velocity is zero, so momentum is zero. (e) False – Step 9 shows initial KE exceeds final PE. Therefore the correct statement is (b). |
Just ask: "Help me solve this problem."
A rocket of mass \( m \) is launched with kinetic energy \( K_0 \), from the surface of the Earth. How much less kinetic energy does the rocket have at an altitude of two Earth radii? Give your answer in terms of the gravitational constant \( G \), the mass of the Earth \( m_E \), the radius of the Earth \( R_E \), and the mass of the rocket?
List at least 2 everyday forces that are not conservative, and explain why they aren’t.
A spring is connected to a wall and a horizontal force of \( 80.0 \) \( \text{N} \) is applied. It stretches \( 25 \) \( \text{cm} \); what is its spring constant?
A constant force of strength \( 20 \) \( \text{N} \) acts on an object of mass \( 3 \) \( \text{kg} \) as it moves a distance of \( 4 \) \( \text{m} \). If this force is applied perpendicular to the \( 4 \) \( \text{m} \) displacement, the work done by the force is equal to:
A \(3800 \, \text{kg}\) open railroad car coasts along with a constant speed of \(8.60 \, \text{m/s}\) along a level track. Snow begins to fall vertically and fills the car at a rate of \(3.50 \, \text{kg/min}\). Ignoring friction with the tracks, what is the speed of the car after \(90 \, \text{min}\)?
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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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