0 attempts

0% avg

UBQ Credits

# Part (a): What is the difference in the time the balls spend in the air?

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

1 | y = v_0 t + \frac{1}{2}gt^2 | Use the kinematic equation for vertical motion to find the time of flight of each ball. |

2 | 19.6 = 0 + 14.7t + \frac{1}{2}(9.8)t^2 | For the ball thrown downward: [katex] v_0 = -14.7 \, \text{m/s} [/katex], [katex] g = 9.8 \, \text{m/s}^2 [/katex] |

3 | 19.6 = 14.7t + 4.9t^2 \Rightarrow 0 = 4.9t^2 + 14.7t – 19.6 | Solve the quadratic equation using the quadratic formula. |

4 | t = \frac{-14.7 \pm \sqrt{(14.7)^2 – 4(4.9)(-19.6)}}{2(4.9)} | The positive root is taken since time cannot be negative. |

5 | t = 1 \, \text{s} | Solving the quadratic equation gives [katex] t = 1 \, \text{s} [/katex]. |

6 | y = v_0 t + \frac{1}{2}gt^2 | Now for the ball thrown upward: initial speed [katex] v_0 = 14.7 \, \text{m/s} |

7 | 0 = 14.7t – \frac{1}{2}(9.8)t^2 \Rightarrow t(2.9 – 2.9t) = 0 \Rightarrow t = 3 \, \text{s} | Solving for \( t [/katex] when the ball reaches the top of its trajectory. |

8 | \text{Total time in air} = (2 \times 1 + 3)\ s = 5 \, \text{.} | The time the second ball spends in the air is double the ascent time plus the fall time. |

9 | Difference in time = 2 \, \text{s} – 1\,\text{s} = 1s |
The difference in the time the balls spend in the air. |

# Part (b): What is the velocity of each ball as it strikes the ground?

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

1 | v = v_0 + gt | Use the first kinematic equation to find the final velocity. |

2 | v = -14.7 + 9.8 \times 1 = -24.5 \, \text{m/s} | For the ball thrown downward: add gravitational acceleration effect. |

3 | v = 14.7 – 9.8 \times 3 = -14.7 \, \text{m/s} | For the ball thrown upward: subtract the effect of gravity over fall time. |

# Part (c): How far apart are the balls 0.800 s after they are thrown?

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

1 | d_{downward} = v_0 t + \frac{1}{2}gt^2 | Use kinematic equation for the ball thrown downward. |

2 | d_{downward} = 14.7 \times 0.800 + \frac{1}{2} \times 9.8 \times 0.800^2 = 15.68 \, \text{m} | Calculate distance traveled by the downward-thrown ball in 0.800s. |

3 | d_{upward} = v_0 t – \frac{1}{2}gt^2 | Use kinematic equation for the ball thrown upward. |

4 | d_{upward} = 14.7 \times 0.800 – \frac{1}{2} \times 9.8 \times 0.800^2 = 6.864 \, \text{m} | Calculate distance traveled by the upward-thrown ball in 0.800s. |

5 | 20 \, \text{m} | Distance between the balls after 0.800 s is [katex]15.68\, \text{m} + 6.864\, \text{m}[/katex] = 22.544 |

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

- Statistics

Beginner

Mathematical

MCQ

The graph above represents the motion of an object traveling in a straight line as a function of time. What is the average speed of the object during the first four seconds? Note the displacemnt from 0 to 4 seconds is 2 meters

- Motion Graphs

Intermediate

Conceptual

MCQ

Two balls are dropped off a cliff, 3 seconds apart. The first ball dropped is twice as heavy as the second ball dropped. Air resistance is negligible. While both balls are falling, the distance between the two balls is

- 1D Kinematics, Free Fall

Intermediate

Conceptual

MCQ

Two identical metal balls are being held side by side at the top of a ramp. Alex lets one ball, 4, start rolling down the hill. A few seconds later, Alex’ partner, Bob starts the second ball, B, down the hill by giving it a push. Ball B rolls down the hill along a line parallel to the path of the first ball and passes it. At the instant ball B passes ball A:

- 1D Kinematics

Advanced

Mathematical

GQ

An object of unknown mass is acted upon by multiple forces:

- 100 N to the right at 20°
- 400 N to the left at 40° below the horizontal
- 500 N to the right at 10° below horizontal.

The coefficients of friction are μ_{s}=0.6 and μ_{k}=0.2. Starting from rest, the object travels 10 meters in 4.5 seconds. What is the mass of the unknown object?

- 1D Kinematics, Friction, Linear Forces

Beginner

Mathematical

GQ

A car accelerates from rest with an acceleration of [katex]4.3 \text{ m/s}^2 [/katex] for a time of 6.8 s. The car then slows to a stop with an acceleration of [katex]5.1 \, \text{m/s}^2[/katex]. What is the total distance traveled by the car?

- 1D Kinematics

**Difference in Time Spent in the Air**: [katex]\boxed{\Delta t = 0.5 , \text{seconds}}[/katex]**Velocity of Each Ball as it Strikes the Ground**:- Ball Thrown Downward: [katex]\boxed{v_{\text{down}} = 53.9 , \text{m/s}}[/katex]
- Ball Thrown Upward: [katex]\boxed{v_{\text{up}} = 19.6 , \text{m/s}}[/katex]

**Distance Apart after 0.800 Seconds**: [katex]\boxed{\text{Distance Apart} = 2.352 , \text{meters}}[/katex]

By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

Kinematics | Forces |
---|---|

[katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |

[katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |

[katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |

[katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |

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

[katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |

[katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |

[katex]KE_i + PE_i = KE_f + PE_f[/katex] |

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

[katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |

[katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |

[katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |

Simple Harmonic Motion |
---|

[katex]F = -k x[/katex] |

[katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |

[katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |

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 |

- 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 [katex] v_i [/katex] is written as [katex] u [/katex]; sometimes [katex] \Delta x [/katex] is written as [katex] s [/katex].
- 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!

The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.

per month

Billed Monthly. Cancel Anytime.

Trial –> Phy Pro

- Unlimited Messages and Images
- Unlimited UBQ Credits
- 157% Better than GPT
- 30 --> 300 Word Input
- 3 --> 15 MB Image Size Limit
- 1 --> 3 Images per Message
- All Smart Actions
- Mobile Snaps
- Focus Mode
- No Ads

A quick explanation

UBQ credits are specifically used to grade your FRQs and GQs.

You can still view questions and see answers without credits.

Submitting an answer counts as 1 attempt.

Seeing answer or explanation counts as a failed attempt.

Lastly, check your average score, across every attempt, in the top left.

MCQs are 1 point each. GQs are 1 point. FRQs will state points for each part.

Phy can give partial credit for GQs & FRQs.

Phy sees everything.

It customizes responses, explanations, and feedback based on what you struggle with. Try your best on every question!

Understand you mistakes quicker.

For GQs and FRQs, Phy provides brief feedback as to how you can improve your answer.

Aim to increase your understadning and average score with every attempt!

10 Free Credits To Get You Started

*Phy Pro members get unlimited credits