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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 block of mass [katex] m [/katex] is moving on a horizontal frictionless surface with a speed [katex] v_0 [/katex] as it approaches a block of mass [katex] 2m [/katex] which is at rest and has an ideal spring attached to one side.
When the two blocks collide, the spring is completely compressed and the two blocks momentarily move at the same speed, and then separate again, each continuing to move.
A small block of mass \( M \) is released from rest at the top of the curved frictionless ramp shown above. The block slides down the ramp and is moving with a speed \( 3.5v_0 \) when it collides with a larger block of mass \( 1.5M \) at rest at the bottom of the incline. The larger block moves to the right at a speed \( 2v_0 \) immediately after the collision.
Express your answers to the following questions in terms of the given quantities and fundamental constants.
A 0.5 kg cart, on a frictionless 2 m long table, is being pulled by a 0.1 kg mass connected by a string and hanging over a pulley. The system is released from rest. After the hanging mass falls 0.5 m, calculate the speed of the cart on the table. Use ONLY forces and energy.
A spring launches a 4 kg block across a frictionless horizontal surface. The block then ascends a 30° incline with a kinetic friction coefficient of 0.25, stopping after 55 m on the incline. If the spring constant is 800 N/m, find the initial compression of the spring. Disregard friction while in contact with the spring.
While traveling in its elliptical orbit around the Sun, Mars gains speed during the part of the orbit where it is getting closer to the Sun. Which of the following can be used to explain this gain in speed?
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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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