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# Part (a): Finding the final speed of the proton. Note you can also use conversation of energy to find the speed, where [katex] W_{\text{machine}} + KE_i = KE_f [/katex].
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] v_f^2 = v_i^2 + 2ad [/katex] | Use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance traveled, where [katex] v_f [/katex] is the final velocity, [katex] v_i [/katex] is the initial velocity, [katex] a [/katex] is the acceleration, and [katex] d [/katex] is the distance. |
| 2 | [katex] 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}) [/katex] | Substitute [katex] v_i = 2.4 \times 10^7 \, \text{m/s} [/katex], [katex] a = 3.6 \times 10^{15} \, \text{m/s}^2 [/katex], and [katex] d = 3.5 \, \text{cm} = 0.035 \, \text{m} [/katex]. |
| 3 | [katex] v_f = \sqrt{ (2.4 \times 10^7)^2 + 2 \times 3.6 \times 10^{15} \times 0.035} [/katex] | Simplify and solve for [katex] v_f [/katex]. |
| 4 | [katex] v_f = \sqrt{5.76 \times 10^{14} + 2.52 \times 10^{14}} [/katex] | Calculate inside the square root. |
| 5 | [katex] v_f = \sqrt{8.28 \times 10^{14}} [/katex] | Sum the terms under the square root. |
| 6 | [katex] v_f = 2.88 \times 10^7 \, \text{m/s} [/katex] | Take the square root to find the final speed. |
# Part (b): Calculating the increase in kinetic energy
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] \Delta KE =KE_f – KE_i [/katex] | The change in kinetic energy is the difference between the initial and final kinetic energy. |
| 2 | [katex] \Delta KE = \frac{1}{2} m (v_f^2 – v_i^2) [/katex] | Substitute in the formula for kinetic energy and factor out [katex] \frac{1}{2} m [/katex]. |
| 3 | [katex] \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] [/katex] | Substitute the values of [katex] m, v_f, v_i [/katex]. |
| 4 | [katex] \Delta KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times 2.52 \times 10^{14} [/katex] | Simplify the expression. |
| 5 | [katex] \Delta KE = 2.10 \times 10^{-13} \, \text{J} [/katex] | 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 | [katex] W = \Delta KE [/katex] | Use the work energy pricinple, which states the work applied to the proton is equal to the change in its kinetic energy. |
| 2 | [katex] Fd = \Delta KE [/katex] | Substitute [katex] W [/katex] with [katex] Fd [/katex] since [katex] W = Fd [/katex]. |
| 3 | [katex] mad = \Delta KE [/katex] | Substitute [katex] F [/katex] with [katex] ma [/katex] since [katex] F = ma [/katex]. |
| 4 | [katex] md = \frac{\Delta KE}{a} [/katex] | Divide by acceleration on both sides. This equation clearly shows that [katex] \Delta KE [/katex] is directly proportional to [katex] a [/katex]. Hence tripling acceleration will also triple the the change in kinetic energy. |
Just ask: "Help me solve this problem."
A typical \( 68 \)\(-\text{kg} \) person can maintain a steady energy expenditure of \( 480 \) \( \text{W} \) on a bicycle. Approximately how many Calories are “burned” when the person rides a bicycle for \( 15 \) minutes? A typical energy efficiency for the human body is \( 25\% \), which takes into account the release of thermal energy.
A linear spring of force constant \( k \) is used in a physics lab experiment. A block of mass \( m \) is attached to the spring and the resulting frequency, \( f \), of the simple harmonic oscillations is measured. Blocks of various masses are used in different trials, and in each case, the corresponding frequency is measured and recorded. If \( f^{2} \) is plotted versus \( \frac{1}{m} \), the graph will be a straight line with slope
A person is making homemade ice cream. She exerts a force of magnitude 23 N on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius 0.25 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 1.7 s, what is the average power being expended?
A 90 kg individual is cycling up a hill inclined at 30 degrees on a 12 kg bicycle. The hill is quite steep, and the coefficient of static friction is 0.85. The cyclist ascends 12 meters up the hill and then pauses at the summit. If they then start descending from the peak at rest and travel 9 meters before firmly applying the brakes, causing the wheels to lock.
How does the time t1 of a block m reaching the bottom of slide 1 compare with t2, the time taken block of mass 2m to reach the end of slide 2 that’s curved? The blocks are released from the same height.
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| 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
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] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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