| 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 block with mass \( m \) slides at speed \( v_0 \) on a smooth surface and hits a stationary block with mass \( M \). They stick together and move at speed \( \frac{v_0}{3} \). Find \( M \) in terms of \( m \).
An object at rest suddenly explodes into two fragments (\(m_1\) and \(m_2\)) by an explosion. Fragment \(m_1\) acquires \(3\) times the kinetic energy of the other. What is the ratio of \(m_1\) to \(m_2\)?
In a controlled experiment, engineers test a firecracker. The firecracker has mass \( m \) and is placed at rest on a horizontal surface. When the firecracker is lit, it explodes and breaks apart into two pieces. In the first trial, one piece with mass \( \frac{m}{2} \) moves to the left with speed \( v_L \) and the other piece moves to the right with speed \( v_R \). A second trial is performed with an identical firecracker, and one piece with mass \( \frac{3m}{4} \) moves to the left, again with speed \( v_L \). What will the speed of the other piece be in this second trial?
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?
An egg dropped on the road usually beaks, while one dropped on the grass usually does not break because for the egg dropped on the grass:
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.
A rubber ball and a lump of clay have equal mass. They are thrown with equal speed against a wall. The ball bounces back with nearly the same speed with which it hit. The clay sticks to the wall. Which one of these objects experiences the greater impulse?
A rubber ball and a piece of clay have equal masses. They are dropped from the same height on horizontal steel platform. The ball bounces back with nearly the same speed with which it hit. The clay sticks to the platform. Choose all that is true about the ball/platform and the clay/platform systems.
On a frictionless horizontal air table, puck A (with mass \( 0.249 \) \( \text{kg} \)) is moving toward puck B (with mass \( 0.375 \) \( \text{kg} \)), which is initially at rest. After the collision, puck A has velocity \( 0.115 \) \( \text{m/s} \) to the left, and puck B has velocity \( 0.645 \) \( \text{m/s} \) to the right.
In which of the following is the rate of change of the particle’s momentum zero?
\(1.1\,\text{m/s}\)
\(64\,\text{N}\)
\(\text{equal magnitude, opposite direction}\)
\(28\,\text{m}\)
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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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