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| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/katex] | Conservation of momentum |
| 2 | [katex]4,kg \cdot 10,m/s + 7,kg \cdot 0,m/s = 4,kg \cdot v_{1x} + 7,kg \cdot 2,m/s \cdot \cos(22^\circ)[/katex] | Plugging in given values and decomposing the 7 kg mass’s velocity into horizontal component. |
| 3 | [katex]40 = 4v_{1x} + 13.08[/katex] | Calculating the horizontal momentum contribution from the 7 kg mass post-collision. |
| 4 | [katex]4v_{1x} = 26.92[/katex] | Solve for the 4 kg mass’s horizontal velocity component. |
| 5 | [katex]v_{1x} = 6.73,m/s[/katex] | Calculating the horizontal velocity of the 4 kg mass. |
| 6 | [katex]m_1 u_{1y} + m_2 u_{2y} = m_1 v_{1y} + m_2 v_{2y}[/katex] | Conservation of momentum in the vertical direction. Since the initial vertical momentum is 0, the final combined vertical momentum must also be 0. |
| 7 | [katex]0 = 4,kg \cdot v_{1y} + 7,kg \cdot 2,m/s \cdot \sin(22^\circ)[/katex] | Recognizing that initial vertical velocities are 0 and calculating the vertical component for the 7 kg mass. |
| 8 | [katex]v_{1y} = -1.04,m/s[/katex] | Calculating the vertical velocity of the 4 kg mass (negative indicates opposite direction to the 7 kg mass’s vertical component). |
| 9 | [katex]v_1 = \sqrt{v_{1x}^2 + v_{1y}^2}[/katex] and [katex]\theta = \arctan\left(\frac{v_{1y}}{v_{1x}}\right)[/katex] | Combining horizontal and vertical components to find magnitude and direction of the 4 kg mass’s velocity. |
| 10 | [katex]v_1 \approx \sqrt{6.73^2 + (-1.04)^2}[/katex] | Plugging in horizontal and vertical components. |
| 11 | [katex]v_1 \approx 6.81,m/s[/katex] | Calculating the magnitude of velocity. |
| 12 | [katex]\theta \approx \arctan\left(\frac{-1.04}{6.73}\right)[/katex] | Calculating the direction of the velocity. |
| 13 | [katex]\theta \approx -8.8^\circ[/katex] | Determining the angle below the horizontal for the 4 kg mass’s velocity. |
Just ask: "Help me solve this problem."
A bullet moving with an initial speed of [katex] v_o [/katex] 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 [katex] h [/katex]. Which of the following statements is true of the collision.
A 0.0350 kg bullet moving horizontally at 425 m/s embeds itself into an initially stationary 0.550 kg block.
A block with mass m slides at speed [katex] v_0 [/katex] on a smooth surface and hits a stationary block with mass [katex] M [/katex] . They stick together and move at speed [katex] v_0/3 [/katex]. Find [katex] M [/katex] in terms of [katex] m [/katex] .
A 2,000 kg car collides with a stationary 1,000 kg car. Afterwards, they slide 6 m before coming to a stop. The coefficient of friction between the tires and the road is 0.7. Find the initial velocity of the 2,000 kg car before the collision?
A 4 kg mass is traveling at 10 m/s to the right when it collides inelastically with a stationary 7 kg mass. The 7 kg mass then travels at 2m/s at an angle of 22° below the horizontal. What are the velocity and the angle of the 4 kg mass?
6.81 m/s at 8.8° above the horizontal
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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] |
General Metric Conversion Chart
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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