| Derivation / Formula | Reasoning |
|---|---|
| \[m = 20\,\text{g} = 0.020\,\text{kg}\] | Convert the dart’s mass to kilograms so that impulse \(J\) comes out in \(\text{N}\cdot\text{s}\). |
| \[v_{\perp}=v\sin30^{\circ},\; v_{\text{tan}}=v\cos30^{\circ}\] | Resolve any velocity \(v\) into components perpendicular and parallel (tangential) to the plate; the dart makes \(30^{\circ}\) with the plate. |
| \[v_{\perp i}=350\,(0.500)=175\,\text{m/s},\;v_{\text{tan} i}=350\,(0.866) =303.1\,\text{m/s}\] | Compute the magnitudes of the incoming perpendicular and tangential components. |
| \[v_{\perp f}=-320\,(0.500)=-160\,\text{m/s},\;v_{\text{tan} f}=320\,(0.866)=277.9\,\text{m/s}\] | After reflection the perpendicular component reverses sign, while the tangential component keeps the same upwards direction but is reduced because the speed drops to \(320\,\text{m/s}\). Note: Upwards is set to be negative. |
| \[\Delta v_{\perp}=v_{\perp f}-v_{\perp i}=-160-175=-335\,\text{m/s}\] | Change in the perpendicular velocity component. |
| \[\Delta v_{\text{tan}}=v_{\text{tan} f}-v_{\text{tan} i}=277.9-303.1=-25.2\,\text{m/s}\] | Change in the tangential velocity component (small but not zero—must be included). |
| \[J_{\perp}=m\,\Delta v_{\perp}=0.020(-335)=-6.70\,\text{N}\cdot\text{s}\] | Impulse component from the perpendicular momentum change. |
| \[J_{\text{tan}}=m\,\Delta v_{\text{tan}}=0.020(-25.2)=-0.504\,\text{N}\cdot\text{s}\] | Impulse component from the tangential momentum change. |
| \[|\mathbf J|=\sqrt{J_{\perp}^{2}+J_{\text{tan}}^{2}}=\sqrt{6.70^{2}+0.504^{2}}\approx 6.7\,\text{N}\cdot\text{s}\] | Combine the components vectorially to obtain the magnitude of the total impulse delivered by the plate. |
| \[\boxed{|\mathbf J|\;\approx\;6.7\,\text{N}\cdot\text{s}}\] | Matches option (b). |
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A moderate force will break an egg. However, an egg dropped on the road usually breaks, while one dropped on the grass usually does not break because for the egg dropped on the grass:
A 0.035 kg bullet moving horizontally at 350 m/s embeds itself into an initially stationary 0.55 kg block. Air resistance is negligible.
A \( 1.0 \)\( \text{-kg} \) object is moving with a velocity of \( 6.0 \) \( \text{m/s} \) to the right. It collides and sticks to a \( 2.0 \)\( \text{-kg} \) object moving with a velocity of \( 3.0 \) \( \text{m/s} \) in the same direction. How much kinetic energy was lost in the collision?
A \(3800 \, \text{kg}\) open railroad car coasts along with a constant speed of \(8.60 \, \text{m/s}\) along a level track. Snow begins to fall vertically and fills the car at a rate of \(3.50 \, \text{kg/min}\). Ignoring friction with the tracks, what is the speed of the car after \(90 \, \text{min}\)?
A cardinal (Richmondena cardinalis) of mass \( 3.80 \times 10^{-2} \) \( \text{kg} \) and a baseball of mass \( 0.150 \) \( \text{kg} \) have the same kinetic energy. What is the ratio of the cardinal’s magnitude \( p_c \) of momentum to the magnitude \( p_b \) of the baseball’s momentum?
A small block of mass \( M \) is released from rest at the top of the curved frictionless ramp shown above. The block slides down the ramp and is moving with a speed \( 3.5v_0 \) when it collides with a larger block of mass \( 1.5M \) at rest at the bottom of the incline. The larger block moves to the right at a speed \( 2v_0 \) immediately after the collision.
Express your answers to the following questions in terms of the given quantities and fundamental constants.
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 platform is initially rotating on smooth ice with negligible friction, as shown above. A stationary disk is dropped directly onto the center of the platform. A short time later, the disk and platform rotate together at the same angular velocity, as shown at right in the figure. How does the angular momentum of only the platform change, if at all, after the disk drops? And what is the best justification.
A golf club exerts an average horizontal force of \(1000 \, \text{N}\) on a \(0.045 \, \text{kg}\) golf ball that is initially at rest on the tee. The club is in contact with the ball for \(1.8 \, \text{milliseconds}\). What is the speed of the golf ball just as it leaves the tee?
Car A, mass 1000 kg, is traveling at 40 m/s when it collides with a stationary car B. They stick together and travel at 7 m/s. What is the mass of car B?
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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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