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

1 | h = L – L \cos(\theta) | Calculate the vertical height the pendulum rises. This uses the initial angle of 60 degrees and the length of the pendulum. |

2 | PE_{\text{top}} = mgh | Calculate the potential energy at the highest point. Potential energy is defined as the product of mass, gravitational acceleration, and height. |

3 | KE_{\text{bottom}} = \frac{1}{2}mv^2 | Calculate the kinetic energy at the bottom. Kinetic energy is defined as half the product of mass and the square of the velocity. |

4 | PE_{\text{top}} = KE_{\text{bottom}} | Apply the conservation of mechanical energy. The potential energy at the highest point is converted to kinetic energy at the lowest point. |

5 | mgh = \frac{1}{2}mv^2 | Set the potential energy equal to the kinetic energy to find the relationship between the height and the speed at the lowest point. |

6 | v = \sqrt{2gh} | Solve for the velocity. Here, mass cancels out, showing that the velocity does not depend on the mass of the pendulum. |

7 | v = \sqrt{2g(L – L\cos(\theta))} | Substitute the height equation into the velocity equation. This gives the velocity in terms of the pendulum length and the angle. |

8 | v = \sqrt{2gL(1 – \cos(\theta))} | Final simplification to find the expression for the velocity at the lowest point of the swing. |

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

- Statistics

Advanced

Mathematical

GQ

- Energy

Advanced

Proportional Analysis

MCQ

A rock is whirled on the end of a string in a horizontal circle of radius R with a constant period T. If the radius of the circle is reduced to R/3, while the period remains T, what happens to the centripetal acceleration (a_{c}) of the rock?

- Circular Motion

Intermediate

Conceptual

MCQ

A bullet at speed v_0 trikes 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 h. Which of the following statements is true?

- Energy, Momentum

Advanced

Proportional Analysis

MCQ

A block starts at rest on a frictionless inclined track which then turns into a circular loop of radius *R* and is vertical. In terms of *R *and constants, find the minimum height *h *above the bottom of the loop the block must start from so it makes it around the loop.

- Circular Motion

Advanced

Mathematical

GQ

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.

- Energy

Intermediate

Conceptual

MCQ

A girl throws a stone from a bridge. Consider the following ways she might throw the stone. The speed of the stone as it leaves her hand is the same in each case.

Case A: Thrown straight up.

Case B: Thrown straight down.

Case C: Thrown out at an angle of 45° above horizontal.

Case D: Thrown straight out horizontally.

In which case will the speed of the stone be greatest when it hits the water below if there is no significant air resistance, assuming equal initial speeds?

- Energy

Advanced

Mathematical

FRQ

A conical pendulum is formed by attaching a ball of mass m to a string of length L, then allowing the ball to move in a horizontal circle of radius R.

- Circular Motion

Intermediate

Mathematical

FRQ

Friction provides the force needed for a car to travel around a flat, circular race track. Answer the following:

- Centripetal Acceleration, Circular Motion, Friction

Advanced

Proportional Analysis

MCQ

A car is safely negotiating an unbanked circular turn at a speed of 17 m/s on dry road. However, a long wet patch in the road appears and decreases the maximum static frictional force to one-fifth of its dry-road value. If the car is to continue safely around the curve, by what factor would the it need to change the original velocity?

- Circular Motion

Advanced

Mathematical

MCQ

A crate is pulled 2.5 m at constant velocity along a 25° incline. The coefficient of kinetic friction between the crate and the plane is 0.250. What is the efficiency of this procedure?

- Energy

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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