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| Derivation / Formula | Reasoning |
|---|---|
| \[p_i = m v_i\] | Initial momentum is mass times initial velocity. |
| \[p_i = 0.10\times25 = 2.5\,\text{kg}\cdot\text{m/s}\] | Substitute \(m=0.10\,\text{kg}\) and \(v_i=25\,\text{m/s}\). |
| \[p_f = m(-v_x)\] | Velocity reverses direction after rebound, so it is negative. |
| \[p_f = -0.10\times19 = -1.9\,\text{kg}\cdot\text{m/s}\] | Insert \(m=0.10\,\text{kg}\) and \(v_x=19\,\text{m/s}\) (opposite direction). |
| \[\Delta p = p_f – p_i\] | Change in momentum is final minus initial momentum. |
| \[\Delta p = -1.9 – 2.5 = -4.4\,\text{kg}\cdot\text{m/s}\] | Compute numerical difference. |
| \[|\Delta p| = 4.4\,\text{kg}\cdot\text{m/s}\] | Take magnitude because question asks for magnitude of change. |
| \[\boxed{4.4\,\text{kg}\cdot\text{m/s}}\] | Final boxed answer. |
| Derivation / Formula | Reasoning for Each Choice |
|---|---|
| \[\text{(a)}\] | \(72\,\text{kg}\cdot\text{m/s}\) would require a speed change of \(720\,\text{m/s}\) for a \(0.10\,\text{kg}\) ball—unrealistic here. |
| \[\text{(b)}\] | \(1.8\,\text{kg}\cdot\text{m/s}\) results if one subtracts the magnitudes \(25-19\) but ignores direction in the momentum calculation. |
| \[\text{(c)}\] | \(1.2\,\text{kg}\cdot\text{m/s}\) arises from using \(19-25\) (still ignoring sign change) and multiplying by mass. |
| \[\text{(d)}\] | Correct; uses the full direction reversal: \(|(-19)-(25)|\times0.10=4.4\,\text{kg}\cdot\text{m/s}\). |
Just ask: "Help me solve this problem."
A 3800 kg open railroad car coasts along with a constant speed of 8.60 m/s along a level track. Snow begins to fall vertically and fills the car at rate of 3.50 kg/min. Ignoring friction with the tracks, what is the speed of the car after 90 min?
A 20 g piece of clay moving at a speed of 50 m/s strikes a 500 g pendulum bob at rest. The length of a string is 0.8 m. After the collision the clay-bob system starts to oscillate as a simple pendulum.
A pendulum consists of a mass \( M \) hanging at the bottom end of a massless rod of length \( \ell \) which has a frictionless pivot at its top end. A mass \( m \), moving with velocity \( v \), impacts \( M \) and becomes embedded. In terms of the given variables and constants, what is the smallest value of \( v \) sufficient to cause the pendulum (with embedded mass \( m \)) to swing clear over the top of its arc?
A block of mass [katex] m [/katex] is moving on a horizontal frictionless surface with a speed [katex] v_0 [/katex] as it approaches a block of mass [katex] 2m [/katex] which is at rest and has an ideal spring attached to one side.
When the two blocks collide, the spring is completely compressed and the two blocks momentarily move at the same speed, and then separate again, each continuing to move.
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?
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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] |
General Metric Conversion Chart
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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