| Step | Reasoning |
|---|---|
| Identify the expression for the block’s final translational speed using conservation of energy. \[ M g H = \dfrac{1}{2} M v_{block}^2 \implies v_{block} = \sqrt{2 g H} \] |
The block slides without friction, meaning all initial gravitational potential energy is converted into translational kinetic energy. |
| Identify the expression for the sphere’s final translational speed using conservation of energy for a rolling object. \[ M g H = \dfrac{1}{2} M v_{sphere}^2 + \dfrac{1}{2} I \omega^2 \] |
The sphere rolls without slipping, so the initial gravitational potential energy is shared between translational and rotational kinetic energy. |
| Substitute the moment of inertia and the rolling condition for the solid sphere. \[ M g H = \dfrac{1}{2} M v_{sphere}^2 + \dfrac{1}{2} \left( \dfrac{2}{5} M R^2 \right) \left( \dfrac{v_{sphere}}{R} \right)^2 = \dfrac{1}{2} M v_{sphere}^2 + \dfrac{1}{5} M v_{sphere}^2 = \dfrac{7}{10} M v_{sphere}^2 \] |
To relate everything to translational speed, we use the moment of inertia for a solid sphere \(I = \dfrac{2}{5} M R^2\) and the rolling-without-slipping condition \(\omega = \dfrac{v_{sphere}}{R}\). |
| Solve for the sphere’s translational speed and calculate the ratio. \[ v_{sphere} = \sqrt{\dfrac{10}{7} g H} \implies \dfrac{v_{sphere}}{v_{block}} = \dfrac{\sqrt{\frac{10}{7} g H}}{\sqrt{2 g H}} = \sqrt{\dfrac{10}{14}} = \sqrt{\dfrac{5}{7}} \] |
Combining the results for both objects allows for the determination of the final ratio requested. |
Why each choice is correct or incorrect:
(A) Calculates the ratio of the kinetic energies or the squares of the speeds but forgets to take the square root.
(B) Uses the incorrect rotational inertia (hollow sphere instead of solid sphere).
(C) This is the correct answer.
(D) Fails to account for the energy required to make the sphere rotate, which reduces the energy available for translational motion compared to the sliding block.
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An engineer is testing two different designs for a rotating flywheel. Design A consists of a uniform solid disk with mass \(M\) and radius \(R\) mounted on a low-friction axle. When a motor applies a constant net torque \(\tau\) to Design A, it experiences an angular acceleration \(\alpha_0\). Design B consists of a uniform solid disk with mass \(2M\) and radius \(2R\). If the motor is adjusted to apply a constant net torque of \(4\tau\) to Design B, what is the resulting angular acceleration of Design B in terms of \(\alpha_0\)?

A spring of ideal spring constant \(k\) hangs vertically from a ceiling. When the spring is unextended, its bottom end is at position \(y = 0\). The positive \(y\)-direction is defined as downward. A block of mass \(m\) is attached to the spring and gently lowered until it hangs at rest at its equilibrium position \(y_{eq}\). The block is then pulled down an additional distance \(A\) to a maximum position \(y_{max} = y_{eq} + A\) and released from rest.
A block of mass \(M\) is attached to an ideal spring and undergoes simple harmonic motion on a frictionless horizontal surface. The equilibrium position of the block is at \(x = 0\). Which of the following graphs best represents the acceleration \(a\) of the block as a function of its displacement \(x\)?

A block of mass \(M\) is attached to the lower end of a vertical spring with spring constant \(k\), while the upper end of the spring is fixed to a ceiling. A second block of mass \(m\) is placed on top of the first block. The two-block system is set into vertical simple harmonic motion with amplitude \(A\). At the instant the blocks pass through the equilibrium position while moving downward, which of the following correctly describes the magnitude of the force exerted by the spring on the bottom block, \(F_{spring}\), and the magnitude of the normal force exerted by the bottom block on the top block, \(F_{normal}\)?

A block of mass \(m\) is attached to an ideal horizontal spring with spring constant \(k\). The system oscillates on a frictionless surface with amplitude \(A\). Which of the following expressions represents the kinetic energy of the block when its displacement from the equilibrium position is \(x = \dfrac{1}{3}A\)?

Two uniform, thin rods, Rod A and Rod B, both have the same mass \(M\). Rod A has length \(L\) and Rod B has length \(2L\). Both rods are pivoted at one end so they can rotate in a vertical plane. The rods are held in a horizontal position and released from rest at the same time. What is the ratio \(\dfrac{\alpha_B}{\alpha_A}\) of the magnitude of the initial angular acceleration of Rod B to that of Rod A?

A drone of mass \( m \) is flying in a straight horizontal line at a constant speed \( v \) and a constant altitude \( H \) above the ground. A stationary observer is located at point \( P \) on the ground. At a certain moment, the horizontal distance between the drone and the observer is \( x \). Which of the following is a correct expression for the magnitude of the angular momentum of the drone relative to the observer at this moment?

A uniform rigid rod of mass \(M\) and length \(D\) is suspended vertically from a horizontal, frictionless pivot at its top end. The rotational inertia of the rod about the pivot is \(I_{rod} = \dfrac{1}{3}MD^2\). A small lump of clay of mass \(m\) is thrown horizontally with speed \(v_0\) toward the rod. The clay strikes the rod at a distance \(L\) from the pivot and sticks to it. The collision occurs almost instantaneously, and the rod-clay system swings upward after the collision.

A small ball of mass \(m\) travels horizontally with speed \(v\) on a frictionless surface. It strikes the bottom end of a uniform rod of mass \(M\) and length \(L\) that is suspended vertically from a frictionless pivot at its top end. The ball sticks to the rod, and the ball-rod system begins to rotate about the pivot. Which of the following correctly compares the linear momentum of the ball-rod system immediately before the collision, \(p_i\), to the linear momentum of the system immediately after the collision, \(p_f\), and provides the correct physical justification?

A uniform solid cylinder of mass \(M\) and radius \(R\) is released from rest at the top of an incline of height \(h\). The cylinder rolls down the incline without slipping. Which of the following statements correctly characterizes the work done by the static frictional force and its effect on the mechanical energy of the cylinder-Earth system?
Consider the following possible justifications:
I. The force acts through a displacement.
II. The point of contact is instantaneously at rest.
III. Mechanical energy is converted to thermal energy.
C
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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