0 attempts
0% avg
UBQ Credits
# 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 runner is moving at \( 4 \) \( \text{m/s} \). She is opposed by magic in the form of air resistance, which exerts a constant \( 20 \) \( \text{Newtons} \) force in the direction opposite her velocity. At what rate is she using energy to remain at constant velocity?
A ball is thrown straight up. At what point does the ball have the most energy?
If you want to double the momentum of a gas molecule, by what factor must you increase its kinetic energy?
A \( 50 \) \( \text{g} \) ice cube can slide up and down a frictionless \( 30^{\circ}\) slope. At the bottom, a spring with spring constant \( 25 \) \( \text{N/m} \) is compressed \( 10 \) \( \text{cm} \) and is used to launch the ice cube up the slope. How high does it go above its starting point? Express your answer with the appropriate units.
A \( 25.0 \) \( \text{kg} \) block is placed at the top of an inclined plane set at an angle of \( 35 \) degrees to the horizontal. The block slides down the \( 1.5 \) \( \text{m} \) slope at a constant rate. How much work did friction do on the block?
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 |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
