| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | \(m_1 v_1 = m_2 v_2\) | Conservation of momentum. Before and after snow accumulation, the system’s momentum remains constant because there are no external forces. |
| 2 | \( \cdot (8.60,m/s) = (3800,kg + \Delta m) \cdot v_2\) | Initial momentum equals final momentum. \(\Delta m\) is the change in mass due to snow. |
| 3 | \(\Delta m = 3.50,kg/min \times 90,min\) | Calculating total mass of snow accumulated. |
| 4 | \(\Delta m = 315,kg\) | Total mass of snow accumulated. |
| 5 | \(32660 = (4115,kg) \cdot v_2\) | Plugging in the total mass after snow accumulation. |
| 6 | \(v_2 = \frac{32660}{4115}\) | Isolating \(v_2\) to solve for the final speed. |
| 7 | \(v_2 \approx 7.94,m/s\) | Calculating the final speed. |
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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 1.0-kg object is moving with a velocity of 6.0 m/s to the right. It collides and sticks to a 2.0-kg object moving with a velocity of 3.0 m/s in the same direction. How much kinetic energy was lost in the collision?
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 block of mass \( m \) is moving on a horizontal frictionless surface with a speed \( v_0 \) as it approaches a block of mass \( 2m \) which is at rest and has an ideal spring attached to one side.
When the two blocks collide, the spring is completely compressed and the two blocks momentarily move at the same speed, and then separate again, each continuing to move.
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 \( 1.0 \)\( \text{-kg} \) object is moving with a velocity of \( 6.0 \) \( \text{m/s} \) to the right. It collides and sticks to a \( 2.0 \)\( \text{-kg} \) object moving with a velocity of \( 3.0 \) \( \text{m/s} \) in the same direction. How much kinetic energy was lost in the collision?
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.
A child (\(m = 32 \, \text{kg}\)) in a boat (\(m = 71 \, \text{kg}\)) throws a \(7.1 \, \text{kg}\) package out horizontally with a speed of \(12.2 \, \text{m/s}\). Calculate the velocity of the boat immediately after, assuming it was initially at rest. Ignore water resistance.
| Experiment | Initial Velocity of Cart X \( (\text{m/s}) \) | Initial Velocity of Cart Y \( (\text{m/s}) \) | Final Velocity of Cart X \( (\text{m/s}) \) | Final Velocity of Cart Y \( (\text{m/s}) \) |
|---|---|---|---|---|
| \( 1 \) | \( 1 \) | \( 0 \) | \( 0 \) | \( 1 \) |
| \( 2 \) | \( 1 \) | \( -1 \) | \( -1 \) | \( 1 \) |
| \( 3 \) | \( 2 \) | \( 1 \) | \( 1 \) | \( 2 \) |
A student performs several experiments in which two carts collide as they travel along a horizontal surface. Cart X and Cart Y both have a mass of \( 1 \) \( \text{kg} \). Data collected from the three experiments are shown in the table above. During which experiment does the center of mass of the system of two carts have the greatest change in its momentum?
A rocket explodes into two fragments, one \(25\) times heavier than the other. The change in momentum of the lighter fragment is
7.94 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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