| Step | Reasoning |
|---|---|
| Identify the external forces acting on the ball-rod system during the collision to determine if linear momentum is conserved. \[ \vec{F}_{ext} = \dfrac{d\vec{p}}{dt} \] |
The question asks for a comparison of linear momentum before and after a collision. Linear momentum is conserved only if the net external force on the system is zero. |
| Use the conservation of angular momentum about the pivot to find the angular velocity \(\omega\) of the system after the collision. \[ L_i = L_f \implies mvL = (I_{rod} + I_{ball})\omega \\ mvL = \left(\dfrac{1}{3}ML^2 + mL^2\right)\omega \implies \omega = \dfrac{mv}{(1/3M + m)L} \] |
While the pivot exerts an external force (linear impulse), it exerts zero torque about its own axis. This allows us to find the final state of the system. |
| Calculate the linear momentum of the system after the collision using the velocity of the center of mass for the rod and the velocity of the ball. \[ p_f = m(\omega L) + M\left(\omega \dfrac{L}{2}\right) = \omega L \left(m + \dfrac{M}{2}\right) \] |
Linear momentum of a system is the sum of the momenta of its parts: \(p = \sum m_i v_i\). |
| Substitute the expression for \(\omega\) into the equation for \(p_f\) and compare it to \(p_i = mv\). \[ p_f = \left[ \dfrac{mv}{(1/3M + m)L} \right] L \left(m + \dfrac{M}{2}\right) = mv \left( \dfrac{m + 0.5M}{m + 0.33M} \right) \] |
This comparison allows us to determine if the momentum increased, decreased, or remained the same. |
| Evaluate the ratio to determine the relationship between \(p_f\) and \(p_i\). \[ \text{Since } \dfrac{m + 0.5M}{m + 0.33M} > 1, \text{ then } p_f > p_i \] |
Since the denominator has a smaller coefficient for \(M\) (0.33) than the numerator (0.5), the fraction is greater than 1. |
Why each choice is correct or incorrect:
(A) Linear momentum is not conserved because the pivot is fixed and must exert an external horizontal force to prevent the top of the rod from translating.
(B) Energy is lost in inelastic collisions, but momentum depends on the net external impulse, not the conservation of mechanical energy.
(C) This is the correct answer; the impulsive force from the pivot is directed forward when striking the rod below its center of percussion.
(D) When striking the very bottom of the rod, the rod’s tendency to rotate and translate would cause the top point to move backward if it were free; therefore, the pivot must pull the top point forward to keep it fixed.
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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 solid sphere of mass \(M\) and radius \(R\) is released from rest at the top of an incline of height \(H\) and rolls without slipping to the bottom. A block of mass \(M\) is released from rest at the top of an identical incline that is frictionless and slides to the bottom. If the translational speeds of the sphere and the block at the bottom of the incline are \(v_{sphere}\) and \(v_{block}\), respectively, what is the ratio \(\dfrac{v_{sphere}}{v_{block}}\)?

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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