| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v_{\text{initial}} = \frac{M_{\text{boat}} \cdot v_{\text{boat}}}{M_{\text{boat}}}[/katex] | Initial velocity of the boat can be expressed in terms of its mass [katex]M_{\text{boat}}[/katex] and its velocity [katex]v_{\text{boat}}[/katex]. |
| 2 | [katex]v_{\text{final}} = \frac{(M_{\text{boat}} + m_{\text{sand}}) \cdot v_{\text{new}}}{M_{\text{boat}} + m_{\text{sand}}}[/katex] | Final velocity of the boat after receiving the sack of sand of mass [katex]m_{\text{sand}}[/katex] considers the total new mass and the new velocity [katex]v_{\text{new}}[/katex] which conserves momentum. |
| 3 | [katex]M_{\text{boat}} \cdot v_{\text{boat}} = (M_{\text{boat}} + m_{\text{sand}}) \cdot v_{\text{new}}[/katex] | Apply the law of conservation of momentum. The total momentum before the sack drops must equal the total momentum after, assuming no external forces. |
| 4 | [katex]v_{\text{new}} = \frac{M_{\text{boat}} \cdot v_{\text{boat}}}{M_{\text{boat}} + m_{\text{sand}}}[/katex] | Solve the conservation of momentum equation for [katex]v_{\text{new}}[/katex], the new velocity of the system (boat plus sand). |
| 5 | [katex]v_{\text{new}} < v_{\text{boat}}[/katex] | As [katex]m_{\text{sand}}[/katex] is positive and added to [katex]M_{\text{boat}}[/katex], the denominator in the expression for [katex]v_{\text{new}}[/katex] is larger than [katex]M_{\text{boat}}[/katex], hence [katex]v_{\text{new}}[/katex] is less than [katex]v_{\text{boat}}[/katex]. |
| 6 | Answer: (a) decrease | Since the final velocity [katex]v_{\text{new}}[/katex] of the boat is less than its initial velocity [katex]v_{\text{boat}}[/katex], the boat will slow down, indicating that the correct answer is (a) decrease. |
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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 \( \mu_k \).
A block of mass \(m\) is initially sliding to the right with speed \(v_0\) on a horizontal, frictionless track. At point A, the block enters a rough section of track of length \(L\) where the coefficient of kinetic friction between the block and the track is \(\mu_k\). After passing point B, the track becomes frictionless again and transitions into a ramp inclined at an angle \(\theta\) to the horizontal. The block travels up the ramp and reaches a maximum vertical height \(h\) above the horizontal track before momentarily coming to rest. Assume the block-Earth system has zero gravitational potential energy when the block is on the horizontal track.
A uniform solid cylinder of mass \(M\) and radius \(R\) is placed on a rough horizontal surface. A light string is attached to a frictionless axle that passes through the center of mass of the cylinder. The string is pulled horizontally to the right, passes over a light, frictionless pulley at the edge of the surface, and is attached to a hanging block of mass \(m\), as shown in Figure 1. The rotational inertia of a uniform solid cylinder about its center is \(I = \dfrac{1}{2}MR^2\). The system is released from rest, and the cylinder rolls to the right without slipping.
A uniform solid cylinder of mass \(M\) and radius \(R\) is held a negligible distance above a rough horizontal surface. The cylinder is given an initial clockwise angular velocity \(\omega_0\) about its central axis and is then dropped so that it lands gently on the surface without bouncing. The coefficient of kinetic friction between the cylinder and the surface is \(\mu_k\). The cylinder initially slips on the surface, but after a time interval \(\Delta t\), it begins to roll without slipping. The rotational inertia of a uniform solid cylinder is \(I = \dfrac{1}{2}MR^2\).
A spool of mass \(M\) and rotational inertia \(I\) is mounted on a fixed, frictionless, horizontal axle. The spool consists of an inner hub of radius \(r\) and an outer disk of radius \(R\). A light string is wrapped around the inner hub. A block of mass \(m\) is attached to the free end of the string. The system is released from rest, and the block descends as the string unwinds without slipping on the inner hub.
A block of mass \(m\) is placed on a horizontal track and pushed against an ideal spring of spring constant \(k\), compressing it a distance \(D\). The block is released from rest. The track is frictionless except for a rough section of length \(L\) with a coefficient of kinetic friction \(\mu_k\). After passing the rough section, the block slides up a frictionless incline of angle \(\theta\). The block remains on the track at all times, and you may assume the block’s size is negligible. The block completely passes through the rough section and reaches a maximum vertical height \(H\) on the incline.
A stepped pulley consists of two joined solid cylinders: an outer cylinder of radius \(R\) and an inner cylinder of radius \(r\). The pulley has a total rotational inertia \(I\) and mass \(M_p\), and it is mounted on a frictionless horizontal axle. A block of mass \(m\) is suspended from a lightweight string wrapped around the right side of the outer cylinder. A student holds a second lightweight string wrapped around the left side of the inner cylinder. The student pulls downward on the second string with a constant force \(F_A\). Neither string slips on the pulley.
A 2.0 kg block is sliding along a rough horizontal surface. A sensor measures the kinetic energy K of the block as a function of its position x. The data are plotted in the graph shown below.
Based on the graph, what is the magnitude of the net force acting on the block?
A student performs an experiment to determine the acceleration due to gravity, \(g\), by dropping a small metal sphere from rest. The sphere is released from various heights, \(h\), and the time, \(t\), it takes to fall to the floor is measured using an electronic timer. The student intends to create a linear graph of the data such that the slope of the best-fit line is exactly equal to \(g\). Which of the following correctly identifies the quantities that should be plotted on the vertical and horizontal axes?
A student performs an experiment to determine the acceleration due to gravity \(g\) by launching a small marble horizontally from the edge of a table with a constant initial speed \(v_0\). The student varies the height \(H\) of the table for several trials and measures the horizontal range \(D\) from the base of the table to the landing point on the floor. Which of the following pairs of quantities should be plotted to produce a linear graph, and what is the physical meaning of the slope of that graph?
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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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