(a) Calculate the maximum speed of the car at the lowest point of the track.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ PE_A = mgh_A \] | Calculate the potential energy at point \( A \) where \( h_A = 90 \, \text{m} \). |
| 2 | \[ KE_P = \frac{1}{2}mv_{\text{max}}^2 \] | Calculate the kinetic energy at point \( P \). The speed is at maximum here, so \(\Delta h = 0\). |
| 3 | \[ PE_A = KE_P \] | Use the conservation of energy: all potential energy at \( A \) converts into kinetic energy at \( P \). |
| 4 | \[ mgh_A = \frac{1}{2}mv_{\text{max}}^2 \] | Set the potential energy equal to the kinetic energy. The mass \( m \) cancels out. |
| 5 | \[ v_{\text{max}} = \sqrt{2gh_A} \] | Solve for \( v_{\text{max}} \) by rearranging and simplifying the equation. |
| 6 | \[ v_{\text{max}} = \sqrt{2 \times 9.8 \times 90} \] | Substitute \( g = 9.8 \, \text{m/s}^2 \) and \( h_A = 90 \, \text{m} \) into the equation. |
| 7 | \[\boxed{v_{\text{max}} = 42 \, \text{m/s}}\] | Calculate the final value of the maximum velocity \( v_{\text{max}} \). |
(b) Calculate the speed \( v_B \) at the top of the loop.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ PE_B = mgh_B \] | Calculate the potential energy at point \( B \) where \( h_B = 50 \, \text{m} \). |
| 2 | \[ KE_B = \frac{1}{2}mv_B^2 \] | Calculate the kinetic energy at point \( B \). |
| 3 | \[ PE_A = PE_B + KE_B \] | Apply conservation of energy between points \( A \) and \( B \). |
| 4 | \[ mgh_A = mgh_B + \frac{1}{2}mv_B^2 \] | Set the potential and kinetic energy sum at \( B \) equal to the potential energy at \( A \). Mass \( m \) cancels out. |
| 5 | \[ gh_A = gh_B + \frac{1}{2}v_B^2 \] | Mass cancels out; simplify the equation for \( v_B \). |
| 6 | \[ v_B^2 = 2g(h_A – h_B) \] | Rearrange to solve for the speed \( v_B \). |
| 7 | \[ v_B = \sqrt{2 \times 9.8 \times (90 – 50)} \] | Substitute values \( g = 9.8 \, \text{m/s}^2 \), \( h_A = 90 \, \text{m} \), and \( h_B = 50 \, \text{m} \) into the equation. |
| 8 | \[\boxed{v_B = 28 \, \text{m/s}}\] | Calculate the speed \( v_B \) at the top of the loop. |
(c) Forces at Point B:
– Normal Force (\( F_N \)): Points towards the center of the loop.
– Gravitational Force (\( F_g \)): Points downwards (also towards the center when car is at the top of the loop).
(d) Calculate the magnitude of all forces and explain loop modification.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ F_N + F_g = \frac{mv_B^2}{r} \] | The centripetal force required at point \( B \). Both forces point towards the center. |
| 2 | \[ F_g = mg \] | Substitute gravitational force where \( m = 700 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). |
| 3 | \[ F_N + mg = \frac{mv_B^2}{r} \] | Substitute the gravitational force into the centripetal force equation. |
| 4 | \[ F_N = \frac{mv_B^2}{r} – mg \] | Rearrange to solve for the normal force \( F_N \). |
| 5 | \[ F_N = \frac{700 \times 28^2}{20} – 700 \times 9.8 \] | Substitute known values: \( m = 700 \, \text{kg} \), \( v_B = 28 \, \text{m/s} \), \( r = 20 \, \text{m} \). |
| 6 | \[\boxed{F_N = 20580 \, \text{N}}\] | Calculate the normal force at the top of the loop. |
Loop Modification:
If the loop’s shape or radius is modified, the drop height should be adjusted accordingly to ensure energy conservation and keep \( v_B \) unchanged. This is supported by the equation derived in part B: \(v_B = \sqrt{2g (h_A – h_B)}\) which shows that \( v_B \) depends only on the height difference of the rollercoaster.
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A \(2 \, \text{kg}\) model rocket is launched with a thrust force of \(275 \, \text{N}\) and reaches a height of \(90 \, \text{m}\), at which point the thrust cuts out, but the rocket continues moving at \(150 \, \text{m/s}\). What is the average air resistance force acting on the rocket during its ascent?
Two wires are tied to the \(500 \, \text{g}\) sphere as shown above. The sphere revolves in a horizontal circle at a constant speed of \(7.2 \, \text{m/s}\). What is the tension in the upper wire? What is the tension in the lower wire?
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?
Two blocks, \( m_2 > m_1 \), having the same kinetic energy, move from a frictionless surface onto a surface having friction coefficient \( \mu_k \). Which block will travel further before stopping.
Using only work and energy, find the velocity of the masses after they have traveled \(0.8 \, \text{m}\). Refer to the image above.
A centripetal force of \( 5.0 \) newtons is applied to a rubber stopper moving at a constant speed in a horizontal circle. If the same force is applied, but the radius is made smaller, what happens to the speed, \( v \), and the frequency, \( f \), of the stopper?
A 0.4 kg object is attached to a horizontal spring undergoes SHM with the total energy of 0.2 J. The potential energy as a function of position presented by the graph.
A compressed spring mounted on a disk can project a small ball. When the disk is not rotating, as shown in the top view above, the ball moves radially outward. The disk then rotates in a counterclockwise direction as seen from above, and the ball is projected outward at the instant the disk is in the position shown above. Which of the following best shows the subsequent path of the ball relative to the ground?
A \( 0.0350 \) \( \text{kg} \) bullet moving horizontally at \( 425 \) \( \text{m/s} \) embeds itself into an initially stationary \( 0.550 \) \( \text{kg} \) block.
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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] |
Metric Prefixes
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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