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| 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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Two identical arrows, one with \( 2 \) times the speed of the other, are fired into a bale of hay. Assuming the hay exerts a constant “frictional” force on the arrows, the faster arrow will penetrate how much farther than the slower arrow?
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A 0.4 kg object is attached to a horizontal spring undergoes SHM with the total energy of 0.2 J. The potential energy as a function of position presented by the graph.
A rescue helicopter lifts a 79 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s2 and is lifted through a distance of 11 m.
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.
For an object oscillating in SHM, what is the relationship between its displacement, velocity, and acceleration graphs as a function of time?
Two blocks are on a horizontal, frictionless surface. Block \( A \) is moving with an initial velocity of \( v_0 \) toward block \( B \), which is stationary. The two blocks collide, stick together, and move off with a velocity of \( \frac{v_0}{3} \). Which block, if either, has the greater mass?
A spring with a spring constant of \( 600. \) \( \text{N/m} \) is used for a scale to weigh fish. What is the mass of a fish that would stretch the spring by \( 7.5 \) \( \text{cm} \) from its normal length?
A pendulum has a period of \(2.0 \, \text{s}\) on Earth. What is its length?
A rubber ball with a mass of \(0.25 \, \text{kg}\) and a speed of \(19.0 \, \text{m/s}\) collides perpendicularly with a wall and bounces off with a speed of \(21 \, \text{m/s}\) in the opposite direction. What is the magnitude of the impulse acting on the rubber ball?
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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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