| 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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A space probe far from the Earth is travelling at \( 14.8 \) \( \text{km s}^{-1} \). It has mass \( 1\,312 \) \( \text{kg} \). The probe fires its rockets to give a constant thrust of \( 156 \) \( \text{kN} \) for \( 220. \) \( \text{s} \). It accelerates in the same direction as its initial velocity. In this time it burns \( 150. \) \( \text{kg} \) of fuel.
Calculate the final speed of the space probe in \( \text{km s}^{-1} \).
A \(70 \, \text{kg}\) woman and her \(35 \, \text{kg}\) son are standing at rest on an ice rink. They push against each other for a time of \(0.60 \, \text{s}\), causing them to glide apart. The speed of the woman immediately after they separate is \(0.55 \, \text{m/s}\). Assume that during the push, friction is negligible compared with the forces the people exert on each other.
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?
A \(0.025 \, \text{kg}\) golf ball moving at \(18.0 \, \text{m/s}\) crashes through the window of a house in \(5.0 \times 10^{-4} \, \text{s}\). After the crash, the ball continues in the same direction with a speed of \(10.0 \, \text{m/s}\).
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.
Two identical blocks are connected to the opposite ends of a compressed spring. The blocks initially slide together on a frictionless surface with velocity \( v \) to the right. The spring is then released by remote control. At some later instant, the left block is moving at \( \frac{v}{2} \) to the left, and the other block is moving to the right. What is the speed of the center of mass of the system at that instant?
Two people, one of mass \( 88 \) \( \text{kg} \) and the other of mass \( 55 \) \( \text{kg} \), sit in a rowboat of mass \( 70 \) \( \text{kg} \). With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat \( 3.1 \) \( \text{m} \) apart from each other, now exchange seats.
A \(2 \, \text{kg}\) object slides east at \(4 \, \text{m/s}\) and collides with a stationary \(3 \, \text{kg}\) object. After the collision, the \(2 \, \text{kg}\) object is traveling at an unknown velocity at \(15^\circ\) north of east and the \(3 \, \text{kg}\) object is traveling at \(38^\circ\) south of east. What is each object’s final velocity?
A boy of mass \( m \) and a girl of mass \( 2m \) are initially at rest at the center of a frozen pond. They push each other so that she slides to the left at speed \( v \) across the frictionless ice surface and he slides to the right. What is the total work done by the children?
A “doomsday” asteroid with a mass of \( 1010 \, \text{kg} \) is hurtling through space. Unless the asteroid’s speed is changed by about \( 0.20 \, \text{cm/s} \), it will collide with Earth and cause tremendous damage. Researchers suggest that a small “space tug” sent to the asteroid’s surface could exert a gentle constant force of \( 2.5 \, \text{N} \). For how long must this force act?
\(2.02\text{ m/s}\)
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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] | |
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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