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# Part A: Calculate velocity of the bullet + wooden block just after impact

Step | Derivation/Formula | Reasoning |
---|---|---|

1 | [katex]m_1 = 0.0500 \, \text{kg}[/katex] | Mass of the bullet. |

2 | [katex]u_1 = 50.0 \, \text{m/s}[/katex] | Initial velocity of the bullet. |

3 | [katex]m_2 = 0.300 \, \text{kg}[/katex] | Mass of the wooden block. |

4 | [katex]u_2 = 0 \, \text{m/s}[/katex] | Initial velocity of the wooden block (at rest). |

5 | [katex]v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}[/katex] | Formula for the final velocity after a completely inelastic collision, where [katex] m_1 [/katex] and [katex] m_2 [/katex] are masses of the bullet and block, [katex] u_1 [/katex] and [katex] u_2 [/katex] are their initial velocities respectively. |

6 | [katex]v = \frac{(0.0500 \, \text{kg}) (50.0 \, \text{m/s}) + (0.300 \, \text{kg}) (0 \, \text{m/s})}{0.0500 \, \text{kg} + 0.300 \, \text{kg}}[/katex] | Substitute the values of [katex] m_1 [/katex], [katex] m_2 [/katex], [katex] u_1 [/katex], and [katex] u_2 [/katex] into the formula. |

7 | [katex]\mathbf{v = 7.14 \, \text{m/s}}[/katex] | Compute to find the final velocity of the bullet + wooden block system just after the collision. |

# Part B: Calculate the vertical distance [katex] h [/katex] reached by the bullet and wooden block

Step | Derivation/Formula | Reasoning |
---|---|---|

1 | [katex]v = 7.14 \, \text{m/s}[/katex] | Initial velocity of the system after impact (from part A). |

2 | [katex]v^2 = u^2 + 2gh[/katex] | Use the kinematic equation relating initial velocity ([katex]u[/katex]), final velocity ([katex]v[/katex]), acceleration due to gravity ([katex]g[/katex]), and height ([katex]h[/katex]). Since the system comes to a stop at the highest point, final velocity [katex]v[/katex] is 0. |

3 | [katex]0 = (7.14 \, \text{m/s})^2 + 2(-9.81 \, \text{m/s}^2)h[/katex] | At the highest point the final velocity is 0, and acceleration due to gravity (a negative value, since it acts downwards) impacts the rising object. |

4 | [katex]h = \frac{(7.14 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2}[/katex] | Rearrange to solve for [katex]h[/katex]. |

5 | [katex]\mathbf{h = 2.59 \, \text{m}}[/katex] | Compute the vertical height. |

# Part C: Compare effects of doubling the bullet’s mass vs. velocity on [katex] h [/katex]

Comparison | Analysis | Conclusion |
---|---|---|

Doubling bullet’s velocity | [katex] h_f \propto v^2 [/katex], doubling [katex] v [/katex] would quadruple [katex] h [/katex] (since [katex] h_f = \frac{4v^2}{2g} [/katex]) | Increase in velocity has a significant effect, leading to an increase by a factor of four for [katex] h [/katex]. |

Doubling bullet’s mass | [katex] h_f [/katex] determined more by [katex] v [/katex] (system velocity) than by mass directly. Increasing mass lowers system velocity, hence slightly lowering [katex] h [/katex]. | Little to no increase in [katex] h [/katex]; could actually decrease due to lower velocity after collision. |

None | Analysis shows that velocity change impacts height significantly compared to mass change. | (ii) Doubling the bullet’s velocity has a greater effect on height than doubling the bullet’s mass. |

Just ask: "Help me solve this problem."

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Block 2 initially is at rest. Block 1 travels towards block 2 and collides with Block 2 as shown above. Find the final velocities of both blocks assuming the collision is elastic.

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A satellite in circular orbit around the Earth moves at constant speed. This orbit is maintained by the force of gravity between the Earth and the satellite, yet no work is done on the satellite. How is this possible?

- Circular Motion, Energy, Gravitation

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A horizontal force of 110 N is applied to a 12 kg object, moving it 6 m on a horizontal surface where the kinetic friction coefficient is 0.25. The object then slides up a 17° inclined plane. Assuming the 110 N force is no longer acting on the incline, and the coefficient of kinetic friction there is 0.45, calculate the distance the object will slide on the incline.

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A 84.4 kg climber is scaling the vertical wall. His safety rope is made of a material that behaves like a spring that has a spring constant of 1.34 x 10^{3} N/m. He accidentally slips and falls 0.627 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

- Energy, Springs

(a) 7.1 m/s

(b) 2.6 m

(c) ii

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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

Conversion Example

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 |

- 1. Some answers may vary by 1% due to rounding.
- Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
- Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
- Bookmark questions you can’t solve to revisit them later
- 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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