(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 75.0kg log floats downstream with a speed of 1.80 m/s. Eight frogs hop onto the log in a series of perfectly inelastic collisions. If each frog has a mass of 0.30 kg and an upstream speed of 1.3 m/s, what is the change in kinetic energy for this system?
Two satellites of equal mass, \( S_1 \) and \( S_2 \), orbit the Earth. \( S_1 \) is orbiting at a distance \( r \) from the Earth’s center at speed \( v \). \( S_2 \) orbits at a distance \( 2r \) from the Earth’s centre at speed \( \dfrac{v}{\sqrt{2}} \). The ratio of the centripetal force on \( S_1 \) to the centripetal force on \( S_2 \) is
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A car with speed \( v \) and an identical car with speed \( 3v \) both travel the same circular section of an unbanked (flat) road. If the frictional force required to keep the faster car on the road without skidding is \( F \), then the frictional force required to keep the slower car on the road without skidding is
Which of the following do not affect the maximum speed that a car can drive in a circle? Choose both correct answers.
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
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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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