# Part (a): Time Elapsed from Leaving the Table to Hitting the Floor

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

1 | t = \sqrt{\frac{2h}{g}} | Since the motion in the y-direction is a free fall, we use the kinematic equation y = \frac{1}{2}gt^2 for the vertical motion. Solving for t gives the time it takes to fall a distance h . |

# Part (b): Horizontal Component of the Velocity of the Block Just Before It Hits the Floor

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

1 | v_x = \frac{D}{t} | Using the result from part (a) and substituting t = \sqrt{\frac{2h}{g}} , we get v_x = \frac{D}{\sqrt{\frac{2h}{g}}} = \sqrt{\frac{D^2g}{2h}} . This is the velocity necessary to cover horizontal distance D in time t . |

2 | v_x = \frac{D}{\sqrt{\frac{2h}{g}}} | Replace t with the equation found in part a. |

# Part (c): Work Done on the Block by the Spring

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

1 | El = KE | The work done by the spring (elastic energy) is transforms into kinetic energy. Since we found the velocity in the previous part we can solve for the kinetic energy. |

2 | El = KE = \frac{1}{2}mv^2 | Formula for kinetic energy |

3 | KE = \frac{1}{2} m \left(\frac{D}{\sqrt{\frac{2h}{g}}}\right)^2 | Substitute in velocity found from previous step. |

4 | KE = \frac{mgD^2}{4h} | Simplify equation. Note that the Kinetic energy is the work done by the spring. |

# Part (d): Spring Constant

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

1 | \frac{1}{2}kx^2 = \frac{1}{2}mv^2 | The spring energy (EL) is equal to the kinetic energy as mentioned in part c. Hence we can set EL = KE and solve for k. |

2 | k = \frac{mv^2}{x^2} | Solve for k |

3 | k = \frac{m \left(\frac{D}{\sqrt{\frac{2h}{g}}}\right)^2}{x^2} | Substitute in the v , to get the final equation in terms of M, x, D, h, |

4 | k = \frac{mD^2 g}{2hx^2} | Simplify |

These steps address each part of the query based on principles of mechanics, conservation of energy, and kinematic equations.

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- Statistics

Intermediate

Mathematical

FRQ

A baseball is thrown at an angle of 25° relative to the ground at a speed of 23.0 m/s. The ball is caught 42.0 m from the thrower.

- Projectiles

Intermediate

Mathematical

MCQ

A block of mass 3.0 kg is hung from a spring, causing it to stretch 12 cm at equilibrium. The 3.0 kg block is then taken off and the spring returns to its original height. Now a 4.0 kg block is placed on the spring and released from rest. How far will the 4.0 kg block fall before its direction is reversed?

- Energy

Intermediate

Mathematical

GQ

A 25 g steel ball is attached to the top of a 24-cm-diameter vertical wheel. Starting from rest, the wheel accelerates at 470 \, \frac{rad}{s^2}. The ball is released after \frac{3}{4} of a revolution. How high does it go above the center of the wheel?

- 1D Kinematics, Energy, Rotational Motion

Advanced

Proportional Analysis

FRQ

On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 3.5 times as far as he would have on earth, given the same initial velocities on both planets. The ball is launched at a speed of 45 m/s at an angle of 29° above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet find:

- Projectiles

Advanced

Proportional Analysis

MCQ

Two blocks, m_2 > m_1 , having the same kinetic energy, move from a frictionless surface onto a surface having friction coefficient \mu_k . Which block will travel further before stopping.

- Energy

- t = \sqrt{\frac{2h}{g}}
- v_x = \frac{D}{\sqrt{\frac{2h}{g}}}
- KE = \frac{mgD^2}{4h}
- k = \frac{mD^2 g}{2hx^2}

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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_1m_2}{r^2} |

a = \frac{\Delta v}{\Delta t} | f = \mu N |

R = \frac{v_i^2 \sin(2\theta)}{g} |

Circular Motion | Energy |
---|---|

F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |

a_c = \frac{v^2}{r} | PE = mgh |

KE_i + PE_i = KE_f + PE_f |

Momentum | Torque and Rotations |
---|---|

p = m v | \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 |
---|

F = -k x |

T = 2\pi \sqrt{\frac{l}{g}} |

T = 2\pi \sqrt{\frac{m}{k}} |

Constant | Description |
---|---|

g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |

G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |

\mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |

k | Spring constant, in \text{N/m} |

M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |

M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |

M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |

Variable | SI Unit |
---|---|

s (Displacement) | \text{meters (m)} |

v (Velocity) | \text{meters per second (m/s)} |

a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |

t (Time) | \text{seconds (s)} |

m (Mass) | \text{kilograms (kg)} |

Variable | Derived SI Unit |
---|---|

F (Force) | \text{newtons (N)} |

E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |

P (Power) | \text{watts (W)} |

p (Momentum) | \text{kilogram meters per second (kgm/s)} |

\omega (Angular Velocity) | \text{radians per second (rad/s)} |

\tau (Torque) | \text{newton meters (Nm)} |

I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |

f (Frequency) | \text{hertz (Hz)} |

General Metric Conversion Chart

Conversion Example

Example of using unit analysis: Convert 5 kilometers to millimeters.

Start with the given measurement:

`\text{5 km}`

Use the conversion factors for kilometers to meters and meters to millimeters:

`\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}`

Perform the multiplication:

`\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}`

Simplify to get the final answer:

`\boxed{5 \times 10^6 \, \text{mm}}`

Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|

Pico- | p | 10^{-12} | 0.000000000001 |

Nano- | n | 10^{-9} | 0.000000001 |

Micro- | µ | 10^{-6} | 0.000001 |

Milli- | m | 10^{-3} | 0.001 |

Centi- | c | 10^{-2} | 0.01 |

Deci- | d | 10^{-1} | 0.1 |

(Base unit) | – | 10^{0} | 1 |

Deca- or Deka- | da | 10^{1} | 10 |

Hecto- | h | 10^{2} | 100 |

Kilo- | k | 10^{3} | 1,000 |

Mega- | M | 10^{6} | 1,000,000 |

Giga- | G | 10^{9} | 1,000,000,000 |

Tera- | T | 10^{12} | 1,000,000,000,000 |

- Some answers may be slightly off by 1% depending on rounding, etc.
- Answers will use different values of gravity. Some answers use 9.81 m/s
^{2}, and other 10 m/s^{2 }for calculations. - Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
- Bookmark questions that you can’t solve so you can come back to them later.
- Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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