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

1 | a = \frac{F}{m} | Calculating horizontal acceleration (a) from thrust (F) and mass (m). |

2 | s = ut + \frac{1}{2}at^2 | Distance formula for motion with constant acceleration. |

3 | 2.10, \text{m} = \frac{1}{2}a t_{\text{table}}^2 | Distance covered on the table is 2.10 m, initial velocity (u) is 0. |

4 | t_{\text{table}} = \sqrt{\frac{2 \times 2.10, \text{m}}{a}} | Solving for the time (t_{\text{table}}) taken to travel across the table. |

5 | v_{x} = at_{\text{table}} | Calculating the horizontal velocity (v_{x}) at the edge of the table. |

6 | v_{x} = 4.2988, m/s | Plug in values and solve. |

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

1 | y = \frac{1}{2}gt_{\text{fall}}^2 | Distance formula for free fall, where y is the fall height and g is the acceleration due to gravity. |

2 | t_{\text{fall}} = \sqrt{\frac{2y}{g}} | Solving for the time (t_{\text{fall}}) to fall 2.10 m. |

3 | t_{\text{fall}} = .654, s | Plug in values and solve. |

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

1 | d_{\text{air}} = v_{x} t_{\text{fall}} | Horizontal distance (d_{\text{air}}) traveled while falling, using the horizontal velocity at the edge of the table and the fall time. |

2 | d_{\text{air}} =4.3 * .65 | Plug in values to get d_{\text{air}} = 2.81, m. Final Answer. |

Phy can also check your working. Just snap a picture!

- Statistics

Intermediate

Mathematical

GQ

A communications satellite orbits the Earth at an altitude of 35,000 km above the Earth’s surface. Take the mass of Earth to be 6 \times 10^{24} \text{ kg} the the radius of Earth to be 6.4 \times 10^6 \text{ m}. What is the satellite’s velocity?

- Centripetal Acceleration, Circular Motion, Gravitation, Linear Forces

Beginner

Mathematical

MCQ

The block is moving horizontally at a constant velocity. There are two applied forces on the object as shown in the image. In which direction is the friction force acting on the object?

- Linear Forces

Intermediate

Mathematical

GQ

A 2.0 kg wood box slides down a vertical wood wall while you push on it at a 45 ° angle. The coefficient of kinetic friction of wood µ_{k} = 0.200. What magnitude of force should you apply to cause the box to slide down at a constant speed?

- Linear Forces

Intermediate

Mathematical

GQ

A truck is traveling at 35 m/s when the driver realizes the truck as no breaks. He sees a ramp off the road, inclined at 20°, and decides to go up it to help the truck come to a stop. How far does the truck travel before coming to a stop (assume no friction).

- 1D Kinematics, Linear Forces

Intermediate

Conceptual

MCQ

A golfer hits her ball in a high arcing shot. Air resistance is negligible. When the ball is at its highest point, which of the following is true?

- Projectiles

Intermediate

Mathematical

GQ

A arrow is shot horizontally from a distance of 20 meters away. It lands .05 meters below the center of the target. If air resistance is negligible what was the initial speed of the arrow?

- Projectiles

Intermediate

Proportional Analysis

GQ

A child on Earth has a weight of 500N. Determine the weight of the child if the earth was to triple in both mass and radius (3M and 3r).

- Gravitation, Linear Forces

Advanced

Mathematical

GQ

A diver springs upward from a diving board. At the instant she contacts the water her speed is 8.90 m/s, and her body is extended at an angle of 75.0° with respect to the horizontal surface of the water. At this instant her vertical displacement is -3.00 m, where downward is the negative direction. Determine her initial velocity, both magnitude and direction.

- Projectiles

Intermediate

Mathematical

GQ

A 1509 g wood block is being pulled by the force meter at a constant velocity. Using the graph below find:

- The force applied to get the wood block moving
- The static coefficient of friction
- The kinetic coefficient of friction

- Linear Forces

Intermediate

Mathematical

FRQ

A 0.035 kg bullet moving horizontally at 350 m/s embeds itself into an initially stationary 0.55 kg block. Air resistance is negligible.

- Energy, Linear Forces, Momentum

2.81 meters from the base of the table

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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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