| 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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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?
The efficiency of a pulley system is 55%. The
pulleys are used to raise a mass of 90.0 kg to a height of
5.60 m. What force is exerted on the rope of the pulley
system if the rope is pulled for 22 m in order to raise
the mass to the required height?
Two people, one of mass \( 88 \) \( \text{kg} \) and the other of mass \( 55 \) \( \text{kg} \), sit in a rowboat of mass \( 70 \) \( \text{kg} \). With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat \( 3.1 \) \( \text{m} \) apart from each other, now exchange seats.
A \( 1.0 \, \text{kg} \) lump of clay is sliding to the right on a frictionless surface with a speed of \( 2 \, \text{m/s} \). It collides head-on and sticks to a \( 0.5 \, \text{kg} \) metal sphere that is sliding to the left with a speed of \( 4 \, \text{m/s} \). What is the kinetic energy of the combined objects after the collision?
Two masses \(m_1\) and \(4m_1\) are on an incline. Both surfaces have the same coefficient of kinetic friction. Both objects start from rest at the same height. Which mass has the largest speed at the bottom?
A block of mass 3.0 kg is hung from a spring, causing it to stretch 12 cm at equilibrium. The 3.0 kg block is then taken off and the spring returns to its original height. Now a 4.0 kg block is placed on the spring and released from rest. How far will the 4.0 kg block fall before its direction is reversed?
From the figure above, determine which characteristic fits this collision best.
Ball \(A\) of mass \(m\) is dropped from a building of height \(H\). Ball \(B\) of mass \(1.7 \, \text{m}\) is dropped from a building of height \(1.7H\). Using energy, what the ratio of \(v_A\) to \(v_B\) (final velocity of ball \(A\) to final velocity of ball \(B\)). Air resistance is negligible.
A block is initially at rest on top of an inclined ramp that makes an angle \( \theta_0 \) with the horizontal. The distance measured along the base of the ramp is \( D \). After the block is released from rest, it slides down the frictionless ramp and then continues onto a rough horizontal surface until it finally comes to rest at the position \( x = 4D \) measured from the base of the ramp. The coefficient of kinetic friction between the block and the rough horizontal surface is \( \mu_k \).
An object is projected vertically upward from ground level. It rises to a maximum height \( H \). If air resistance is negligible, which of the following must be true for the object when it is at a height \( H/2 \) ?
\(1.92\,\text{m/s}\)
\(40.1^{\circ}\)
\(1.79\,\text{s}\)
\(0.961\,\text{J}\)
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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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