(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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| 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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