| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ m_b \, v_i = \Bigl(m_b + m_{\text{block}}\Bigr) \, v_x \] | Apply conservation of momentum for the inelastic collision where the bullet embeds in the block. |
| 2 | \[ 0.05 \times 200 = (0.05 + 1.3) \, v_x \] | Substitute the given values: bullet mass \(m_b=0.05\,\text{kg}\), bullet initial speed \(v_i=200\,\text{m/s}\), and block mass \(1.3\,\text{kg}\). |
| 3 | \[ 10 = 1.35 \, v_x \] | Simplify the multiplication and sum of masses. |
| 4 | \[ v_x = \frac{10}{1.35} \] | Solve for the block’s speed immediately after the collision. |
| 5 | \[ \boxed{ v_x \approx 7.41\,\text{m/s} } \] | This is the final speed of the block (with bullet embedded) immediately after impact. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ \frac{1}{2} (m_b+m_{\text{block}}) \, v_x^2 = \frac{1}{2} k \, (\Delta x)^2 \] | At maximum compression of the spring the block’s kinetic energy is completely converted into spring potential energy. |
| 2 | \[ (m_b+m_{\text{block}}) \, v_x^2 = k \, (\Delta x)^2 \] | Simplify by canceling the common factor \( \frac{1}{2} \) on both sides. |
| 3 | \[ (\Delta x)^2 = \frac{(m_b+m_{\text{block}}) \, v_x^2}{k} \] | Rearrange to solve for the square of the displacement (amplitude) \( \Delta x \). |
| 4 | \[ \Delta x = \sqrt{\frac{1.35 \times (7.41)^2}{2500}} \] | Substitute \(m_b+m_{\text{block}}=1.35\,\text{kg}\), \(v_x\approx7.41\,\text{m/s}\), and \(k=2500\,\text{N/m}\). |
| 5 | \[ \Delta x \approx \sqrt{\frac{1.35 \times 54.93}{2500}} \] | Since \((7.41)^2 \approx 54.93\), the numerator calculates to approximately \(74.15\). |
| 6 | \[ \Delta x \approx \sqrt{0.02966} \] | Divide the numerator \(74.15\) by \(2500\) to obtain the value inside the square root. |
| 7 | \[ \boxed{ \Delta x \approx 0.172\,\text{m} } \] | This is the amplitude of the resulting oscillation of the block-spring system. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ \omega = \sqrt{\frac{k}{m_b+m_{\text{block}}}} \] | For a mass-spring system executing simple harmonic motion, the angular frequency \( \omega \) is determined by this formula. |
| 2 | \[ \omega = \sqrt{\frac{2500}{1.35}} \] | Substitute \(k=2500\,\text{N/m}\) and \(m_b+m_{\text{block}}=1.35\,\text{kg}\) into the formula. |
| 3 | \[ \omega \approx 43.03\,\text{rad/s} \] | Calculate the square root to approximate the angular frequency. |
| 4 | \[ f = \frac{\omega}{2\pi} \] | The relationship between angular frequency \( \omega \) and frequency \( f \) is given by \( f = \omega/(2\pi) \). |
| 5 | \[ f \approx \frac{43.03}{2\pi} \] | Substitute the computed value of \( \omega \) into the frequency expression. |
| 6 | \[ \boxed{ f \approx 6.85\,\text{Hz} } \] | This is the frequency of the oscillatory motion of the block. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ x(t) = \Delta x \, \sin(\omega t) \] | The standard equation for simple harmonic motion where the displacement is zero at \(t=0\) and the velocity is maximum. |
| 2 | \[ x(t) = 0.172 \, \sin(43.03\,t) \] | Substitute the amplitude \(\Delta x \approx 0.172\,\text{m}\) and angular frequency \(\omega \approx 43.03\,\text{rad/s}\) into the general equation. |
| 3 | \[ \boxed{ x(t) = 0.172 \, \sin(43.03\,t) } \] | This is the complete equation of motion for the block on the spring, with \(x(0)=0\). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ T = \frac{2\pi}{\omega} \] | The period \(T\) of a simple harmonic oscillator is given by this formula. |
| 2 | \[ T = \frac{2\pi}{43.03} \] | Substitute the angular frequency \(\omega \approx 43.03\,\text{rad/s}\) into the period formula. |
| 3 | \[ \boxed{ T \approx 0.146\,\text{s} } \] | This is the period of the oscillation of the block-spring system. |
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A simple pendulum consists of a bob of mass 1.8 kg attached to a string of length 2.3 m. The pendulum is held at an angle of 30° from the vertical by a light horizontal string attached to a wall, as shown above.
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A pendulum consists of a ball of mass \( m \) suspended at the end of a massless cord of length \( L \). The pendulum is drawn aside through an angle of \( 60^\circ \) with the vertical and released. At the low point of its swing, the speed of the pendulum ball is
Two blocks of ice, one five times as heavy as the other, are at rest on a frozen lake. A person then pushes each block the same distance \(d\). Ignore friction and assume that an equal force \(F\) is exerted on each block. Which of the following statements is true about the kinetic energy of the heavier block after the push?
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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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