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

1 | [katex]\text{Displacement of Ball 1} = \frac{1}{2}gt^2[/katex] | Since Ball 1 is dropped from rest, its initial velocity [katex] v_i = 0 [/katex], and it accelerates downward due to gravity [katex] g [/katex]. |

2 | [katex]\Delta y_1 = \frac{1}{2}gt^2[/katex] | The displacement of Ball 1 after time [katex] t [/katex] is calculated using the equation [katex] \Delta y = v_i t + \frac{1}{2}gt^2 [/katex] with [katex] v_i = 0 [/katex]. |

3 | [katex]\text{Displacement of Ball 2} = v_0t – \frac{1}{2}gt^2[/katex] | Ball 2 is launched upward from the ground with an initial velocity [katex] v_0 [/katex]. The displacement is the initial velocity times time minus the effect of gravitational acceleration. |

4 | [katex]\Delta y_2 = v_0t – \frac{1}{2}gt^2[/katex] | The displacement of Ball 2 after time [katex] t [/katex] is calculated using the equation [katex] \Delta y = v_0t – \frac{1}{2}gt^2 [/katex]. |

5 | [katex]\Delta y_1 + \Delta y_2 = h[/katex] | The sum of the displacements of both balls will equal the height [katex]h[/katex]. This is the point where they pass each other because the combined distance traveled by both balls equals the initial height from which Ball 1 was dropped. |

6 | [katex]\frac{1}{2}gt^2 + (v_0t – \frac{1}{2}gt^2) = h[/katex] | Substitute the displacement formulas of both balls into the equation from step 5. |

7 | [katex]v_0t = h[/katex] | Combine like terms and simplify the equation. |

8 | [katex]t = \frac{h}{v_0}[/katex] | Solve for [katex] t [/katex]. |

9 | [katex]\boxed{t = \frac{h}{v_0}}[/katex] | The two balls will pass each other at time [katex] t = \frac{h}{v_0} [/katex]. |

Just ask: "Help me solve this problem."

- Statistics

Intermediate

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FRQ

Divers in Acapulco jump from a cliff that is \( 36 \, \text{m} \) above the water with an initial vertical velocity of \( 2 \, \text{m/s} \).

- 1D Kinematics

Beginner

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GQ

A car starts from rest and accelerates uniformly over a time of 5 seconds for a distance of 100 m. Determine the acceleration of the car.

- 1D Kinematics

Intermediate

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FRQ

A 10kg box is pushed to the right by an unknown force at an angle of 25° below the horizontal while a friction force of 50 N acts on the box as well. The box accelerates from rest and travels a distance of 4 m where it is moving at 3 m/s. Solve the following without the use of energy.

- 1D Kinematics, Linear Forces

Intermediate

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GQ

The International Space Station travels at \( 7660 \, \text{m/s} \). Find the average velocity of the space station if it takes \( 90 \, \text{minutes} \) to make one full orbit around Earth.

- 1D Kinematics

Beginner

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MCQ

A car increases its forward velocity uniformly from 40 m/s to 80 m/s while traveling a distance of 200 m. What is its acceleration during this time?

- 1D Kinematics

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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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