| Derivation/Formula | Reasoning |
|---|---|
| \[m_w v_{x,w} + m_s v_{x,s} = 0\] | Conservation of horizontal momentum; external horizontal forces are negligible during the push. |
| \[v_{x,s} = -\frac{m_w}{m_s} v_{x,w}\] | Algebraically solve for the son’s final velocity \(v_{x,s}\). |
| \[v_{x,s} = -\frac{70}{35}(0.55)\] | Substitute \(m_w = 70\,\text{kg}\), \(m_s = 35\,\text{kg}\), and \(v_{x,w} = 0.55\,\text{m/s}\). |
| \[\boxed{v_{x,s} = -1.1\,\text{m/s}}\] | The negative sign indicates motion opposite to the woman; speed is \(1.1\,\text{m/s}\). |
| Derivation/Formula | Reasoning |
|---|---|
| \[J = m_s (v_{x,s} – v_i)\] | Impulse–momentum theorem with \(v_i = 0\). |
| \[J = 35(-1.1 – 0)\] | Insert values for the son. |
| \[|J| = 38.5\,\text{Ns}\] | Magnitude of impulse. |
| \[F_{\text{avg}} = \frac{|J|}{\Delta t}\] | Average force equals impulse divided by the interaction time \(\Delta t\). |
| \[F_{\text{avg}} = \frac{38.5}{0.60} = 64\,\text{N}\] | Compute using \(\Delta t = 0.60\,\text{s}\). |
| \[\boxed{F_{\text{avg}} = 64\,\text{N}}\] | Magnitude of the force the mother exerts on the son. |
| Derivation/Formula | Reasoning |
|---|---|
| \[F_{s \rightarrow w} = -F_{w \rightarrow s}\] | Newton’s third law: forces between two bodies are equal in magnitude and opposite in direction. |
| \[\boxed{|F_{s \rightarrow w}| = |F_{w \rightarrow s}| = 64\,\text{N}}\] | The mother experiences a \(64\,\text{N}\) force directed opposite to the force on the son. |
| Derivation/Formula | Reasoning |
|---|---|
| \[a = \mu_k g\] | Kinetic-friction force \(\mu_k m g\) divided by mass gives a deceleration independent of mass. |
| \[\Delta x = \frac{v_i^2}{2a}\] | Stopping distance for constant deceleration. |
| \[\frac{\Delta x_s}{\Delta x_w} = \left(\frac{v_{x,s}}{v_{x,w}}\right)^2\] | Because both skaters have the same \(a = \mu_k g\), we can set the ratio of distances equal to the ratio of velocities (proportional analysis using the equation in step two). |
| \[\frac{\Delta x_s}{\Delta x_w} = \left(\frac{1.1}{0.55}\right)^2 = 4\] | Insert their initial speeds. The son’s stopping distance is \(4\) times greater than his mother’s stopping distance. |
| \[\boxed{\Delta x_s = 4(7.0) = 28\,\text{m}}\] | Multiply the woman’s \(7.0\,\text{m}\) by the ratio to find the son’s stopping distance. |
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A \(4 \, \text{kg}\) mass is traveling at \(10 \, \text{m/s}\) to the right when it collides inelastically with a stationary \(7 \, \text{kg}\) mass. The \(7 \, \text{kg}\) mass then travels at \(2 \, \text{m/s}\) at an angle of \(22^\circ\) below the horizontal. What are the velocity and the angle of the \(4 \, \text{kg}\) mass?
Two ice skaters suddenly push off against one another starting from a stationary position. The \(45 \, \text{kg}\) skater acquires a speed of \(0.375 \, \text{m/s}\) relative to the ice. What speed does the \(60 \, \text{kg}\) skater acquire relative to the ice?
Block 2 initially is at rest. Block 1 travels towards block 2 and collides with Block 2 as shown above. Find the final velocities of both blocks assuming the collision is elastic.
A bullet moving with an initial speed of \( v_o \) strikes and embeds itself in a block of wood which is suspended by a string, causing the bullet and block to rise to a maximum height \( h \). Which of the following statements is true of the collision.
A bowling ball moving with speed \(v\) collides head-on with a stationary tennis ball. The collision is elastic and there is no friction. The bowling ball barely slows down. What is the speed of the tennis ball after the collision?
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 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 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 \(15 \, \text{g}\) marble moves to the right at \(3.5 \, \text{m/s}\) and makes an elastic head-on collision with a \(22 \, \text{g}\) marble. The final velocity of the \(22 \, \text{g}\) marble is \(2.0 \, \text{m/s}\) to the right, and the final velocity of the \(15 \, \text{g}\) marble is \(5.4 \, \text{m/s}\) to the left. What was the initial velocity of the \(22 \, \text{g}\) marble?
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?
\(1.1\,\text{m/s}\)
\(64\,\text{N}\)
\(\text{equal magnitude, opposite direction}\)
\(28\,\text{m}\)
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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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