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

1 | x_{M_1} = \frac{L}{2} | The boy is sitting at one end of the seesaw, which places him at a distance of half the length of the plank (L) from the fulcrum. |

2 | x_{M_2} = \frac{L}{2} | The girl is sitting at the other end of the seesaw opposite to the boy, also at a distance of half the length of the plank from the fulcrum. |

3 | x_M = \frac{L}{2} | The mass of the plank (M) is uniformly distributed, thus its center of mass is at the midpoint of the plank, which coincides with the fulcrum. |

4 | \tau_{M_1} = M_1 \cdot g \cdot \frac{L}{2} | Calculate the torque due to the boy’s mass at one end of the seesaw. Torque is given by \tau = r \times F where r is the distance from the pivot point and F is the force due to weight, which is M_1 \cdot g . |

5 | \tau_{M_2} = M_2 \cdot g \cdot \frac{L}{2} | Calculate the torque due to the girl’s mass at the other end. Similar to step 4, using the girl’s mass. |

6 | \tau_M = M \cdot g \cdot 0 | Calculate the torque due to the seesaw’s own mass. Since the seesaw’s center of mass is exactly at the fulcrum, the distance r is zero, thus the torque is zero. |

7 | \tau_{total} = \tau_{M_1} – \tau_{M_2} | Sum the torques. Torque due to the boy is assumed counterclockwise and positive, while that due to the girl is clockwise and negative (or vice versa depending on assignment). |

8 | M_1 \cdot g \cdot \frac{L}{2} = M_2 \cdot g \cdot \frac{L}{2} | For the seesaw to be balanced, the total torque must be 0. Setting the torques equal gives this balance condition. |

9 | M_1 = M_2 | Solve for the relationship between M_1 and M_2 . Since all other factors are equal and cancel out, the masses must be equal for balance. |

10 | M_1 = M_2 |
This shows that for the seesaw to remain balanced with a plank mass placed uniformly, the masses of the boy and girl must be equal. This is the condition for mechanical equilibrium. |

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

- Statistics

Advanced

Mathematical

GQ

A sphere of mass *M* and radius *r*, and rotational inertia *I* is released from the top of a inclined plane of height *h*. The surface has considerable friction. Using only the variable mentioned, derive an expression for the sphere’s center of mass velocity.

- Rotational Motion

Advanced

Mathematical

FRQ

A hungry bear weighing 700 N walks out on a beam in an attempt to retrieve a basket of goodies hanging at the end of the beam. The beam is uniform, weighs 200 N, and is 6.00 m long. The goodies weigh 80 N.

- Linear Forces, Rotational Motion, Torque

Beginner

Mathematical

MCQ

A boy and a girl are balanced on a massless seesaw. The boy has a mass of 60 kg and the girl’s mass is 50 kg. If the boy sits 1.5 m from the pivot point on one side of the seesaw, where must the girl sit on the other side for equilibrium?

- Rotational Motion, Torque

Advanced

Conceptual

MCQ

While traveling in its elliptical orbit around the Sun, Mars gains speed during the part of the orbit where it is getting closer to the Sun. Which of the following can be used to explain this gain in speed?

- Angular Momentum, Energy, Gravitation, Rotational Motion, Torque

Intermediate

Mathematical

MCQ

An object’s angular momentum changes by 10\,\text{kg-m}^2\text{/s} in 2.0 s. What magnitude average torque acted on this object?

- Angular Momentum, Rotational Inertia, Rotational Motion, Torque

Yes. Both children have identical masses. See working in explanation.

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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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