# Part (a): Solving Using Conservation of Energy

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

1 | U_i + K_i = U_f + K_f | The total mechanical energy (kinetic + potential) at the initial state equals the total mechanical energy at the final state due to conservation of energy. |

2 | U_i = m_g \cdot g \cdot h | Initial potential energy is due to m_g being at a height h from the ground. m_y is on the ground, so its potential energy is zero. |

3 | U_i = 38 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \cdot 2.5 \, \text{m} | Calculate the numerical value of U_i . g (acceleration due to gravity) is 9.8 m/s² and h is 2.5 m. |

4 | U_i = 931 \, \text{J} | Multiplying the values provides the initial potential energy. |

5 | K_i = 0 | Initially, the system is at rest so the initial kinetic energy is zero. |

6 | K_f = \frac{1}{2}(m_y + m_g + m_p) v^2 | Final kinetic energy calculated as the sum of the kinetic energies of m_g , m_y and the pulley. m_p is the mass of the pulley. The pulley contributes translational kinetic energy due to its mass. |

7 | U_f = 0 | At the final state, m_g is on the ground, so its potential energy is zero. |

8 | 931 \, \text{J} = \frac{1}{2} (32 \, \text{kg} + 38 \, \text{kg} + 3.1 \, \text{kg}) v^2 | Use U_i + K_i = U_f + K_f and solve for v . |

9 | v = \sqrt{\frac{2 \times 931 \, \text{J}}{73.1 \, \text{kg}}} | Simplify the equation to solve for v . |

10 | v = \sqrt{25.48 \, \text{m}^2/\text{s}^2} | Calculate the value under the square root. |

11 | v = 5.05 \, \text{m/s} |
The final velocity of m_g just before it strikes the ground. |

# Part (b): Solving Using Forces, Torque, and Kinematics

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

1 | F = m \cdot a | Newton’s second law, the net force applied on the system causes acceleration. |

2 | F_{my} – F_{mg} = (m_y + m_g + m_p) \cdot a | Forces due to weights of m_y and m_g , including the pulley’s inertia, where F_{my} = m_y \cdot g and F_{mg} = m_g \cdot g . |

3 | T = \frac{m_g g – m_y g}{1 + \frac{m_p}{m_y + m_g}} | Balance the forces taking into account the pulley rotation effect, where T is the net torque and m_p is moment of inertia derived as \frac{1}{2} m_p R^2 assuming a uniform cylinder. |

4 | a = \frac{m_g g – m_y g}{m_y + m_g + \frac{1}{2} m_p} | Solving the earlier equation for a , the acceleration of the system, where \frac{1}{2} m_p comes from dividing the pulley’s moment of inertia by R^2 . |

5 | v_f^2 = 2 \cdot a \cdot 2.5 \, \text{m} | Use kinematic equation for final velocity, incorporating the calculated acceleration and distance fallen, which is 2.5 m. |

6 | v_f = \sqrt{2 \cdot a \cdot 2.5 \, \text{m}} | Solve for final velocity using the value for a . |

7 | Approximate numerical calculation similar to Part (a) |
The velocity is essentially the same result as obtained from energy conservation, demonstrating the physics consistency. |

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

- Statistics

Intermediate

Mathematical

GQ

A wheel 31 cm in diameter accelerates uniformly from 240rpm to 360rpm in 6.8 s. How far will a point on the edge of the wheel have traveled in this time?

- Rotational Kinematics

Advanced

Mathematical

FRQ

A hoop with a mass m and unknown radius is rolling without slipping on a flat surface with an angular speed \omega*. *The hoop encounters a hill and continues to roll without slipping until it reaches a maximum height h*.*

- Rolling, Rotational Energy, Rotational Inertia, Rotational Motion

Advanced

Conceptual

MCQ

In both cases, a massless rod is supported by fulcrum, and a 200-kg hanging mass is suspended from the left end of the rod by a cable. A downward force *F* keeps the rod in rest. The rod in Case A is 50 cm long, and the rod in case B is 40 cm long (each rod is marked at 10-cm intervals). The magnitude of each vertical force F exerted on the rod will be

- Torque

Intermediate

Mathematical

MCQ

A rod is initially at rest on a rough horizontal surface. Three forces are exerted on the rod with the magnitudes and directions shown in the figure. The force exerted in the center of the rod is an equidistant 0.5 m from both ends of the rod. If friction between the rod and the table prevents the rod from rotating, what is the magnitude of the torque exerted on the rod about its center from frictional forces?

- Rotational Motion, Torque

Advanced

Mathematical

FRQ

A system consists of two small disks, of masses m and 2m, attached to ends of a rod of negligible mass of length 3x. The rod is free to turn about a vertical axis through point P. The first mass, m, is located x away from point P, and therefore the other mass, of 2m, is 2x from point P. The two disks rest on a rough horizontal surface; the coefficient of friction between the disks and the surface is . At time t = 0, the rod has an initial counterclockwise angular velocity ω_{i} about P. The system is gradually brought to rest by friction.

Derive an expressions for the following quantities in terms of µ, m, x, g, and ω_{i}.

- Angular Momentum, Rotational Kinematics, Rotational Motion, Torque

- v = 5.05 \, \text{m/s}
- v = 5.05 \, \text{m/s} . The velocity is essentially the same result as obtained from energy conservation, demonstrating the physics consistency.

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 |
---|---|

\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!

The most advanced version of Phy. Currently 50% off, for early supporters.

per month

Billed Monthly. Cancel Anytime.

Trial –> Phy Pro

- Unlimited Messages
- Unlimited Image Uploads
- Unlimited Smart Actions
- Unlimited UBQ Credits
- 30 --> 300 Word Input
- 3 --> 15 MB Image Size Limit
- 1 --> 3 Images per Message
- 200% Memory Boost
- 150% Better than GPT
- 75% More Accurate, 50% Faster
- 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