# (a) Direction of the resultant force acting on the marble at point C

The resultant force acting on the marble at point C is directed towards the center of the loop. This force is mainly comprised of the gravitational force pulling downwards and the normal force exerted by the track which also points towards the center of the loop during the circular motion.

# (b) Names of all the forces acting on the marble at point C

Force | Description |
---|---|

Gravitational Force | The force due to gravity acting downwards towards the center of the earth. |

Normal Force | The force exerted by the surface of the loop on the marble directed radially inward, toward the center of the loop. |

# (c) Deduce the speed of the marble at point C. The working below uses two seperate conservation of energy equations. However, it can also be done in a single equation such that the postential energy at A transfroms into the potential energy at C and the kinetic energy at C. This is written as mgh_A = mgh_C + \frac{1}{2}mv^2 .

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

1 | h_A = 0.8 \, \text{m} | Initial height from which the marble is released. |

2 | v_A = 0 \, \text{m/s} | Initial velocity (marble is released from rest). |

3 | v_B = \sqrt{2gh_A} | Re-arrange and solve for velocity at the bottom of the incline, using conservation of mechanical energy, where mgh_A = \frac{1}{2}m{v^2}_B . |

4 | v_B = \sqrt{2 \times 9.8 \times 0.8} | Calculating v_B . |

5 | v_B \approx 3.97 \, \text{m/s} | Approximate calculation of velocity at point B. |

6 | h_C = 0.35 \, \text{m} | The maximum height attained by the marble is at point C (top of loop). |

7 | v_C^2 = v_B^2 – 2gh_C | Using conservation of mechanical energy between points B and C. |

8 | v_C^2 = 3.97^2 – 2 \times 9.8 \times 0.35 | Calculating v_C from v_B and change in gravitational potential energy. |

9 | v_C \approx 3.0 \, \text{m/s} | Approximate calculation of velocity at point C. |

# (d) Effect if the release height of the marble were to double

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

1 | mgh_A = mgh_C + \frac{1}{2}mv^2 | The conservation of energy between points A and C, as used in part B. |

2 | gh_A = gh_C + \frac{1}{2}v^2 | Simplified formula. |

3 | g\Delta h = \frac{1}{2}v^2 | Further simplifying by replacing gh_A \,- \,gh_c with g\Delta h which is simply the change in height. |

4 | \frac{\Delta h}{v^2} = \frac{1}{2g} | Place \Delta h and v^2 on the same side of the equation to clearly see they are directly proportional. |

5 | Conclusion. | Since \Delta h and v^2 , if you doulbe height the velocity would increase proportionally by \boxed{\sqrt{2}} for a final speed of 3 \times \sqrt{2} \approx 4.24 \, \text{m/s}. |

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An object of mass 2 kg is thrown vertically downwards with an initial kinetic energy of 100 J. What is the distance fallen by the object at the instant when its kinetic energy has doubled?

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A rubber ball bounces off of a wall with an initial speed v and reverses its direction so its speed is v right after the bounce. As a result of this bounce, which of the following quantities of the ball are conserved?

- Energy, Momentum

Intermediate

Conceptual

MCQ

How does the speed *v _{1}* of a block

- Energy

Advanced

Mathematical

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Riders in a carnival ride stand with their backs against the wall of a circular room of diameter 8.0 m. The room is spinning horizontally about an axis through its center at a rate of 45 rev/min when the floor drops so that it no longer provides any support for the riders. What is the *minimum* coefficient of static friction between the wall and the rider required so that the rider does not slide down the wall?

- Circular Motion

- part (a) The resultant force would point down (towards the center of the circle).
- part (b) Weight (a.k.a force of gravity) and normal force
- part (c) Using law of conservation of energy the potential energy at point A transforms into potential and kinetic energy at point C. So it can be deduced that: mgh_A = mgh_C + \frac{1}{2}mv^2 . This would give a value of 3 m/s.
- part (d) The speed would increase by \sqrt{2} , therefore the new speed would be 3 \times \sqrt{2} \approx 4.24 \, \text{m/s}.

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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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