| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[p_i = m_1 v_{1i} + m_2 v_{2i}\] | Initial linear momentum is the sum of each ball’s momentum. Variables: \(m_1 = 0.7 \text{ kg}\), \(v_{1i} = 2 \text{ m/s}\) (chosen positive), \(m_2 = 0.5 \text{ kg}\), \(v_{2i} = -1.2 \text{ m/s}\) (opposite direction). |
| 2 | \[p_i = 0.7 \times 2 + 0.5 \times (-1.2) = 0.8\ \text{kg·m/s}\] | Substitute numeric values to find the total initial momentum. |
| 3 | \[p_f = m_1 v_{1x} + m_2 v_{2x}\] | Define the final linear momentum in terms of the unknown \(v_{2x}\). Given \(v_{1x} = -0.3 \text{ m/s}\) (cue ball rebounds). |
| 4 | \[m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1x} + m_2 v_{2x}\] | Apply conservation of momentum: total momentum before equals total momentum after because the collision is isolated. |
| 5 | \[0.8 = 0.7(-0.3) + 0.5 v_{2x}\] | Insert the known momenta into the conservation equation. |
| 6 | \[0.8 + 0.21 = 0.5 v_{2x}\] | Add \(0.21\) (which equals \(0.7 \times 0.3\)) to both sides to isolate the term with \(v_{2x}\). |
| 7 | \[1.01 = 0.5 v_{2x}\] | Simplify the left-hand side. |
| 8 | \[v_{2x} = 2.02\ \text{m/s}\] | Divide both sides by \(0.5\) to solve for \(v_{2x}\). |
| 9 | \[\boxed{v_{2x} = 2.02\ \text{m/s}}\] | The other ball moves in the positive (cueball’s original) direction at \(2.02 \text{ m/s}\). |
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An astronaut initially at rest in space throws a wrench, and recoils in the opposite direction. Select all that is true.
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 rubber ball bounces off of a wall with an initial speed \(v\) and reverses its direction so its speed is \(v\) right after the bounce. As a result of this bounce, which of the following quantities of the ball are conserved?
A rocket explodes into two fragments, one \(25\) times heavier than the other. The change in momentum of the lighter fragment is
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 \( m \).
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?
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?
Consider the following cases of inelastic collisions.
Case (1) – A car moving at \(75 \, \text{mph}\) collides with another car of equal mass moving at \(75 \, \text{mph}\) in the opposite direction and comes to a stop.
Case (2) A car moving at \(75 \, \text{mph}\) hits a stationary steel wall and rolls back.
The collision time is the same for both cases. In which of these cases would result in the greatest impact force?
A kickball is rolled by the pitcher at a speed of 10 m/s and it is kicked by another student. The kickball deforms a little during the kick, and then rebounds with a velocity of 15 m/s as its shape restores to a perfect sphere. Select all that must be true about the kickball and the kicking foot system.
A karate master is about to split a piece of wood with her hand. Select all she must do in order to deliver the maximum force to split the wood.
\(2.02\text{ m/s}\)
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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