# 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. |
Phy can also check your working. Just snap a picture!
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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
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} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
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