| 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 girl throws a stone from a bridge. Consider the following ways she might throw the stone. The speed of the stone as it leaves her hand is the same in each case.
Case A: Thrown straight up.
Case B: Thrown straight down.
Case C: Thrown out at an angle of 45° above horizontal.
Case D: Thrown straight out horizontally.
In which case will the speed of the stone be greatest when it hits the water below if there is no significant air resistance, assuming equal initial speeds?
A super dart of mass \(20 \, \text{g}\), traveling at \(350 \, \text{m/s}\), strikes a steel plate at an angle of \(30^\circ\) with the plane of the plate, as shown in the figure. It bounces off the plate at the same angle but at a speed of \(320 \, \text{m/s}\). What is the magnitude of the impulse that the plate gives to the bullet?
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.
In a controlled experiment, engineers test a firecracker. The firecracker has mass \( m \) and is placed at rest on a horizontal surface. When the firecracker is lit, it explodes and breaks apart into two pieces. In the first trial, one piece with mass \( \frac{m}{2} \) moves to the left with speed \( v_L \) and the other piece moves to the right with speed \( v_R \). A second trial is performed with an identical firecracker, and one piece with mass \( \frac{3m}{4} \) moves to the left, again with speed \( v_L \). What will the speed of the other piece be in this second trial?
A ball of radius \( r \) rolls on the inside of a circular track of radius \( R \). If the ball starts from rest at the left vertical edge of the track, what will be its speed when it reaches the lowest point of the track, rolling without slipping? For a solid spherical ball, the moment of inertia is \(\frac{2}{5} m r^2\).
A crate is pulled 2.5 m at constant velocity along a 25° incline. The coefficient of kinetic friction between the crate and the plane is 0.250. What is the efficiency of this procedure?
A horizontal spring with spring constant 162 N/m is compressed 50 cm and used to launch a 3 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the rough surface is 0.2. Find the total distance the box travels before stopping.
A baseball is thrown vertically into the air with a velocity \( v \), and reaches a maximum height \( h \). At what height was the baseball moving with one-half its original velocity? Assume air resistance is negligible.
In which one of the following circumstances does the principle of conservation of mechanical energy apply, even though a nonconservative force acts on the moving object?
A child (\(m = 32 \, \text{kg}\)) in a boat (\(m = 71 \, \text{kg}\)) throws a \(7.1 \, \text{kg}\) package out horizontally with a speed of \(12.2 \, \text{m/s}\). Calculate the velocity of the boat immediately after, assuming it was initially at rest. Ignore water resistance.
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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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