# Part (a): Finding the final speed of the proton. Note you can also use conversation of energy to find the speed, where W_{\text{machine}} + KE_i = KE_f .

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

1 | v_f^2 = v_i^2 + 2ad | Use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance traveled, where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and d is the distance. |

2 | v_f^2 = (2.4 \times 10^7 \, \text{m/s})^2 + 2 \times (3.6 \times 10^{15} \, \text{m/s}^2) \times (0.035 \, \text{m}) | Substitute v_i = 2.4 \times 10^7 \, \text{m/s} , a = 3.6 \times 10^{15} \, \text{m/s}^2 , and d = 3.5 \, \text{cm} = 0.035 \, \text{m} . |

3 | v_f = \sqrt{ (2.4 \times 10^7)^2 + 2 \times 3.6 \times 10^{15} \times 0.035} | Simplify and solve for v_f . |

4 | v_f = \sqrt{5.76 \times 10^{14} + 2.52 \times 10^{14}} | Calculate inside the square root. |

5 | v_f = \sqrt{8.28 \times 10^{14}} | Sum the terms under the square root. |

6 | v_f = 2.88 \times 10^7 \, \text{m/s} |
Take the square root to find the final speed. |

# Part (b): Calculating the increase in kinetic energy

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

1 | \Delta KE =KE_f – KE_i | The change in kinetic energy is the difference between the initial and final kinetic energy. |

2 | \Delta KE = \frac{1}{2} m (v_f^2 – v_i^2) | Substitute in the formula for kinetic energy and factor out \frac{1}{2} m . |

3 | \Delta KE = \frac{1}{2} (1.67 \times 10^{-27} \, \text{kg}) [(2.88 \times 10^7 \, \text{m/s})^2 – (2.4 \times 10^7 \, \text{m/s})^2] | Substitute the values of m, v_f, v_i . |

4 | \Delta KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times 2.52 \times 10^{14} | Simplify the expression. |

5 | \Delta KE = 2.10 \times 10^{-13} \, \text{J} |
Calculate the final change in kinetic energy, which is the increase in kinetic energy of the proton. |

# Part (c): Effect of tripling the acceleration on the increase in kinetic energy

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

1 | W = \Delta KE | Use the work energy pricinple, which states the work applied to the proton is equal to the change in its kinetic energy. |

2 | Fd = \Delta KE | Substitute W with Fd since W = Fd . |

3 | mad = \Delta KE | Substitute F with ma since F = ma . |

4 | md = \frac{\Delta KE}{a} | Divide by acceleration on both sides. This equation clearly shows that \Delta KE is directly proportional to a . Hence tripling acceleration will also triple the the change in kinetic energy. |

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- Statistics

Advanced

Proportional Analysis

MCQ

A rocket of mass m is launched with kinetic energy *K _{0}*, from the surface of the Earth. How much less kinetic energy does the rocket have at an altitude of two Earth radii in terms of the gravitational constant,

- Energy

Advanced

Conceptual

MCQ

A lighter car and a heavier truck, each traveling to the right with the same speed v hit their brakes. The retarding frictional force F on both cars turns out to be constant and the same. After both vehicles travel a distance D (and both are still moving), which of the following statements is true?

- Energy

Intermediate

Conceptual

MCQ

You kick a ball straight up. Compare the sign of the work done by gravity on the ball while it goes up with the sign of the work done by gravity while it goes down.

- Energy

Intermediate

Conceptual

GQ

A satellite in circular orbit around the Earth moves at constant speed. This orbit is maintained by the force of gravity between the Earth and the satellite, yet no work is done on the satellite. How is this possible?

- Circular Motion, Energy, Gravitation

Intermediate

Mathematical

FRQ

A 0.035 kg bullet moving horizontally at 350 m/s embeds itself into an initially stationary 0.55 kg block. Air resistance is negligible.

- Energy, Linear Forces, Momentum

- v_f = 2.88 \times 10^7 \, \text{m/s}
- \Delta KE = 2.10 \times 10^{-13} \, \text{J}
- \Delta KE would also triple.

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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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