| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[p_{\text{before}} = m v_i \] | The clay’s initial momentum equals its mass times its speed. |
| 2 | \[p_{\text{after}} = (m+M) v_x \] | After sticking, the clay + bob move together with speed \(v_x\). |
| 3 | \[m v_i = (m+M) v_x \] | Linear momentum is conserved because the collision is perfectly inelastic and external horizontal forces are negligible. |
| 4 | \[v_x = \frac{m}{m+M} v_i \] | Solve algebraically for the common speed. |
| 5 | \[v_x = \frac{0.020}{0.020+0.500}\,(50) \;=\;1.92\;\text{m/s}\] | Insert \(m = 0.020\,\text{kg}\), \(M = 0.500\,\text{kg}\), and \(v_i = 50\,\text{m/s}\). |
| 6 | \[\boxed{v_x \approx 1.92\,\text{m/s}}\] | Speed of the combined mass just after impact. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[\tfrac12 (m+M) v_x^2 = (m+M) g h \] | Kinetic energy right after collision changes into gravitational potential energy at the highest point. |
| 2 | \[h = \frac{v_x^2}{2g} \] | Cancel \((m+M)\) and solve for the height rise \(h\). |
| 3 | \[h = \frac{(1.92)^2}{2(9.81)} = 0.189\,\text{m}\] | Substitute \(v_x\) and \(g = 9.81\,\text{m/s}^2\). |
| 4 | \[h = L(1-\cos\theta)\] | The vertical rise of a pendulum bob is related to its length \(L\) and angular displacement \(\theta\). |
| 5 | \[\cos\theta = 1-\frac{h}{L} = 1-\frac{0.189}{0.800} = 0.764\] | Insert \(h\) and the string length \(L = 0.800\,\text{m}\). |
| 6 | \[\theta = \arccos(0.764) \approx 40.1^{\circ}\] | Take the inverse cosine to find the maximum angle. |
| 7 | \[\boxed{\theta_{\max} \approx 40.1^{\circ}}\] | Maximum angular displacement. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[T = 2\pi \sqrt{\frac{L}{g}}\] | For small oscillations, the period of a simple pendulum depends only on its length and \(g\). |
| 2 | \[T = 2\pi \sqrt{\frac{0.800}{9.81}} = 1.79\,\text{s}\] | Insert \(L = 0.800\,\text{m}\) and \(g = 9.81\,\text{m/s}^2\). |
| 3 | \[\boxed{T \approx 1.79\,\text{s}}\] | Period of the clay-bob system. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[E = \tfrac12 (m+M) v_x^2\] | Total mechanical energy is the kinetic energy right after the collision (identical to potential energy at the highest point). |
| 2 | \[E = \tfrac12 (0.520) (1.92)^2 = 0.961\,\text{J}\] | Use \(m+M = 0.520\,\text{kg}\) and \(v_x = 1.92\,\text{m/s}\). |
| 3 | \[\boxed{E \approx 0.961\,\text{J}}\] | Total energy of the oscillating system. |
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A box of mass \(m\) is initially at rest at the top of a ramp that is at an angle \(\theta\) with the horizontal. The block is at a height \(h\) and length \(L\) from the bottom of the ramp. The coefficient of kinetic friction between the block and the ramp is \(\mu\). What is the kinetic energy of the box at the bottom of the ramp?
A space probe far from the Earth is travelling at \( 14.8 \) \( \text{km s}^{-1} \). It has mass \( 1\,312 \) \( \text{kg} \). The probe fires its rockets to give a constant thrust of \( 156 \) \( \text{kN} \) for \( 220. \) \( \text{s} \). It accelerates in the same direction as its initial velocity. In this time it burns \( 150. \) \( \text{kg} \) of fuel.
Calculate the final speed of the space probe in \( \text{km s}^{-1} \).
An object with a mass \(m = 80 \, \text{g}\) is attached to a spring with a force constant \(k = 25 \, \text{N/m}\). The spring is stretched \(52.0 \, \text{cm}\) and released from rest. If it is oscillating on a horizontal frictionless surface, determine the velocity of the mass when it is halfway to the equilibrium position.
A \(90 \, \text{kg}\) individual is cycling up a hill inclined at \(30^\circ\) on a \(12 \, \text{kg}\) bicycle. The hill is quite steep, and the coefficient of static friction is \(0.85\). The cyclist ascends \(12 \, \text{m}\) up the hill and then pauses at the summit. They then start descending from rest and travel \(9 \, \text{m}\) before firmly applying the brakes, causing the wheels to lock.
A runner is moving at \( 4 \) \( \text{m/s} \). She is opposed by magic in the form of air resistance, which exerts a constant \( 20 \) \( \text{Newtons} \) force in the direction opposite her velocity. At what rate is she using energy to remain at constant velocity?
A block starts at rest on a frictionless inclined track which then turns into a circular loop of radius \( R \) and is vertical. In terms of \( R \) and constants, find the minimum height \( h \) above the bottom of the loop the block must start from so it makes it around the loop.
A block is attached to a horizontal spring. The block is held so the spring is stretched and the block is released from rest, undergoing simple harmonic motion with a frequency of \( 2 \) \( \text{Hz} \). How long after release will the block first reach a point where it is momentarily at rest?
A \(10 \, \text{meter}\) long pendulum on the earth, is set into motion by releasing it from a maximum angle of less than \(10^\circ\) relative to the vertical. At what time \(t\) will the pendulum have fallen to a perfectly vertical orientation?
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?
A spring with spring constant \( k = 2.3 \) \( \text{N/m} \) is attached to an object of mass \( 10 \) \( \text{kg} \). If the object is hung from the ceiling by this spring, how much would the spring be stretched?
\(1.92\,\text{m/s}\)
\(40.1^{\circ}\)
\(1.79\,\text{s}\)
\(0.961\,\text{J}\)
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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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