| 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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What force is necessary to stretch an ideal spring with a spring constant of \( 120 \) \( \text{N/m} \) by \( 30 \) \( \text{cm} \)?
A cart with a mass of \( 20 \) \( \text{kg} \) is pressed against a wall by a horizontal spring with spring constant \( k = 244 \) \( \text{N/m} \) placed between the cart and the wall. The spring is compressed by \( 0.1 \) \( \text{m} \). While the spring is compressed, an additional constant horizontal force of \( 20 \) \( \text{N} \) continues to push the cart toward the wall. What is the resulting acceleration of the cart?
The two blocks of masses \( M \) and \( 2M \) travel at the same speed \( v \) but in opposite directions. They collide and stick together. How much mechanical energy is lost to other forms of energy during the collision?
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 box of mass \( 20 \) \( \text{kg} \) moves to the right on a horizontal frictionless surface with a speed of \( 4.0 \) \( \text{m/s} \). The box collides with and remains attached to one end of a spring of negligible mass whose other end is fixed to a wall. After the collision, the spring compresses a maximum distance of \( 0.50 \) \( \text{m} \), and the box then oscillates back and forth.
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 big bird has a mass of about 0.021 kg. Suppose it does 0.36 J of work against gravity, so that it ascends straight up with a net acceleration of 0.625 m/s2. How far up does it move?
Two ice skaters suddenly push off against one another starting from a stationary position. The \(45 \, \text{kg}\) skater acquires a speed of \(0.375 \, \text{m/s}\) relative to the ice. What speed does the \(60 \, \text{kg}\) skater acquire relative to the ice?
A \(0.50 \, \text{kg}\) mass is attached to a spring constant \(20 \, \text{N/m}\) along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of \(1.5 \, \text{m/s}\) at the equilibrium position. What is the total energy of the system?
The maximum energy a bone can absorb without breaking is surprisingly small. Experimental data show that a leg bone of a healthy, \( 80 \) \( \text{kg} \) human can absorb about \( 240 \) \( \text{J} \). From what maximum height could a \( 80 \) \( \text{kg} \) person jump and land rigidly upright on both feet without breaking their legs? Assume that all energy is absorbed by the leg bones in a rigid landing. Express your answer with the appropriate units.
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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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