| Step | Reasoning |
|---|---|
| Identify the relationship between impulse and the change in the momentum vector. \[ \vec{J} = \Delta \vec{p} = M(\vec{v}_f – \vec{v}_i) \] |
The question asks for the impulse vector, which is defined by the impulse-momentum theorem as the change in the linear momentum of the object. |
| Decompose the initial velocity into x and y components using the given geometry. \[ \vec{v}_i = \langle -v \sin 45^\circ, v \cos 45^\circ \rangle = \langle -\dfrac{v\sqrt{2}}{2}, \dfrac{v\sqrt{2}}{2} \rangle \] |
To perform vector subtraction, we must resolve the velocities into the specified coordinate system. With the angle \(45^\circ\) relative to the wall surface, the x-component (normal) uses sine and the y-component (parallel) uses cosine. Since positive x is away from the wall and positive y is the direction of initial travel, the ball moves in the negative x-direction initially. |
| Decompose the final velocity into x and y components. \[ \vec{v}_f = \langle \dfrac{v}{2} \sin 45^\circ, \dfrac{v}{2} \cos 45^\circ \rangle = \langle \dfrac{v\sqrt{2}}{4}, \dfrac{v\sqrt{2}}{4} \rangle \] |
The final velocity has a magnitude of \(\dfrac{v}{2}\) and moves away from the wall (positive x) while continuing in the same parallel direction (positive y). |
| Calculate the components of the impulse vector by subtracting initial momentum from final momentum. \[ \begin{align*} J_x &= M \left( \dfrac{v\sqrt{2}}{4} – \left( -\dfrac{v\sqrt{2}}{2} \right) \right) = M \left( \dfrac{v\sqrt{2}}{4} + \dfrac{2v\sqrt{2}}{4} \right) = \dfrac{3Mv\sqrt{2}}{4} \\ J_y &= M \left( \dfrac{v\sqrt{2}}{4} – \dfrac{v\sqrt{2}}{2} \right) = -\dfrac{Mv\sqrt{2}}{4} \end{align*} \] |
Subtracting the components yields the change in momentum in each direction. |
| Combine the components into the final vector expression. \[ \vec{J} = \dfrac{3Mv\sqrt{2}}{4} \hat{i} – \dfrac{Mv\sqrt{2}}{4} \hat{j} = \dfrac{Mv\sqrt{2}}{4} (3\hat{i} – \hat{j}) \] |
The vector is the sum of its scaled unit vectors. |
Why each choice is correct or incorrect:
(A) This is the correct answer.
(B) Incorrectly adds the y-components of velocity instead of finding the difference, resulting in a positive j-term.
(C) Incorrectly calculates the change in the x-component as \( v_{fx} – v_{ix} \) without accounting for the fact that the initial x-velocity was in the negative direction.
(D) Only accounts for the impulse in the x-direction, ignoring the fact that the loss of speed also changed the momentum component parallel to the wall.
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A car travels along a straight road between two cities separated by a total distance of \(2D\). The car travels the first distance \(D\) at a constant speed \(v_0\) and the remaining distance \(D\) at a constant speed \(3v_0\). Which of the following correctly identifies the average speed \(v_{avg}\) of the car for the entire trip and provides a valid justification?
A student conducts a laboratory experiment where a cart is moved \(2.0 \text{ m}\) to the right and then \(1.0 \text{ m}\) to the left along a straight, horizontal track. The student calculates the total distance traveled and the final displacement of the cart. Which of the following correctly classifies these quantities and provides a valid justification?
An experimental automated cart is tested on a linear track. A computer-controlled sensor measures the cart’s velocity \(v\) as a function of time \(t\), as shown in the graph. What is the displacement of the cart during the time interval from \(t = 0 \text{ s}\) to \(t = 10 \text{ s}\)?
A hiker starts at a trailhead and walks \(3.0 \text{ km}\) due North. The hiker then turns and walks \(5.0 \text{ km}\) in a direction \(37^\circ\) South of East to reach a campsite. (Note: \(\sin 37^\circ \approx 0.60\); \(\cos 37^\circ \approx 0.80\)). What is the magnitude of the hiker’s total displacement from the trailhead to the campsite?
A laboratory cart is restricted to motion along a horizontal track. A motion sensor records the direction of the cart’s velocity and the direction of its acceleration at three different times, as shown in the table below.
| Time | Direction of Velocity | Direction of Acceleration |
| :— | :— | :— |
| \(t_1\) | Right | Left |
| \(t_2\) | Left | Left |
| \(t_3\) | Left | Right |
Which of the following correctly describes the motion of the cart at each time?
A test rocket moves along a straight, horizontal track. A sensor records the rocket’s acceleration as a function of time, as shown in the graph below.
What is the average acceleration of the rocket during the time interval from \(t = 0 \text{ s}\) to \(t = 5 \text{ s}\)?
A sprinter crosses the finish line of a race moving with a velocity of \(12 \text{ m/s}\). The sprinter continues to run at this constant velocity for a reaction time of \(1.0 \text{ s}\) before beginning to slow down with a constant acceleration. If the sprinter comes to a complete stop exactly \(4.0 \text{ s}\) after crossing the finish line, what is the magnitude of the sprinter’s acceleration during the braking phase?
A surveyor starts at point \(P\) and walks a distance \(d\) due north. The surveyor then turns and walks an equal distance \(d\) in a direction \(60^{\circ}\) west of north to reach point \(Q\). What is the magnitude of the surveyor’s total displacement from point \(P\) to point \(Q\)?
A train is traveling at a constant speed \( v_0 \) when the engineer applies the brakes, resulting in a constant deceleration of magnitude \( a_0 \) that brings the train to a stop in a distance \( d \). In a second trial, the train is traveling at a speed \( 2v_0 \) when the brakes are applied and is brought to a stop with a constant deceleration of magnitude \( 2a_0 \). Which of the following is the stopping distance for the train in the second trial?
A small experimental vehicle starts from rest and accelerates with a constant acceleration \(a\) over a horizontal distance \(d\). After this distance, the vehicle’s engine is adjusted such that it continues to accelerate at a constant rate of \(\dfrac{a}{2}\) for an additional horizontal distance \(d\). Which of the following expressions represents the speed of the vehicle after it has traveled the total distance \(2d\)?
A
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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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