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| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[p_i = m v_i\] | Linear momentum is the product of mass and velocity. We take motion toward the wall as positive. |
| 2 | \[p_i = (0.10\,\text{kg})(+25\,\text{m/s}) = +2.5\,\text{kg}\cdot\text{m/s}\] | Substitute the given mass and initial speed. |
| 3 | \[v_f = -19\,\text{m/s}\] | After rebounding, the ball moves in the opposite direction, so the velocity is negative in our sign convention. |
| 4 | \[p_f = m v_f = (0.10\,\text{kg})(-19\,\text{m/s}) = -1.9\,\text{kg}\cdot\text{m/s}\] | Compute the final momentum with the reversed velocity. |
| 5 | \[\Delta p = p_f – p_i \] | Change in momentum equals final minus initial momentum (vector subtraction). |
| 6 | \[\Delta p = -1.9\,\text{kg}\cdot\text{m/s} – (+2.5\,\text{kg}\cdot\text{m/s}) = -4.4\,\text{kg}\cdot\text{m/s}\] | Carry out the subtraction, keeping the signs. |
| 7 | \[|\Delta p| = 4.4\,\text{kg}\cdot\text{m/s}\] | The problem asks for the magnitude of the change, so we take the absolute value. |
| 8 | \[\boxed{4.4\,\text{kg}\cdot\text{m/s}}\] | Numerical result corresponds to choice (d). |
| Choice | Reasoning for Each Choice |
|---|---|
| (a) \(72\,\text{kg}\cdot\text{m/s}\) | This value might come from confusing momentum with kinetic–energy terms, e.g. writing \(m(v_i^2+v_f^2)\) (forgetting the factor 1/2 and the unit mismatch): \(0.10(25^2+19^2)\approx72\). It ignores both proper units and the vector nature of momentum. |
| (b) \(1.8\,\text{kg}\cdot\text{m/s}\) | A student could mis-read the rebound speed as \(18\,\text{m/s}\) and then report “final momentum only,” i.e. \(p_f \approx m(18)=1.8\,\text{kg}\cdot\text{m/s}\), forgetting to compute the change and omitting the sign. |
| (c) \(1.2\,\text{kg}\cdot\text{m/s}\) | This can result from ignoring direction but doubling the mass by mistake: using \(m=0.20\,\text{kg}\) and \(|v_i-v_f|=6\,\text{m/s}\): \(0.20\times6=1.2\,\text{kg}\cdot\text{m/s}\). |
| (d) \(4.4\,\text{kg}\cdot\text{m/s}\) | Correct: \(|\Delta p| = m|v_f – v_i| = 0.10\times| -19 – 25 | = 4.4\,\text{kg}\cdot\text{m/s}\). |
Just ask: "Help me solve this problem."
A super dart of mass \(20 \, \text{g}\), traveling at \(350 \, \text{m/s}\), strikes a steel plate at an angle of \(30^\circ\) with the plane of the plate, as shown in the figure. It bounces off the plate at the same angle but at a speed of \(320 \, \text{m/s}\). What is the magnitude of the impulse that the plate gives to the bullet?
A baseball, mass \(0.5 \, \text{kg}\), is traveling to the right at \(32.2 \, \text{m/s}\) when it is hit by a bat and travels the opposite direction at \(72.2 \, \text{m/s}\). The bat hits the ball with a force of \(1,222 \, \text{N}\). What is the ball’s change in momentum and how long was the ball in contact with the bat?
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 rubber ball with a mass of \(0.25 \, \text{kg}\) and a speed of \(19.0 \, \text{m/s}\) collides perpendicularly with a wall and bounces off with a speed of \(21 \, \text{m/s}\) in the opposite direction. What is the magnitude of the impulse acting on the rubber ball?
A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular to each other, at \( 18 \) \( \text{m/s} \). The third piece has \( 2.5 \) times the mass as the other two.
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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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