0 attempts
0% avg
(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.
Just ask: "Help me solve this problem."
We'll help clarify entire units in one hour or less — guaranteed.
Riders in a carnival ride stand with their backs against the wall of a circular room of diameter \(8.0 \, \text{m}\). The room is spinning horizontally about an axis through its center at a rate of \(45 \, \text{rev/min}\) when the floor drops so that it no longer provides any support for the riders. What is the minimum coefficient of static friction between the wall and the rider required so that the rider does not slide down the wall?
A person is making homemade ice cream. She exerts a force of magnitude \(23 \, \text{N}\) on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius \(0.25 \, \text{m}\). The force is always applied parallel to the motion of the handle. If the handle is turned once every \(1.7 \, \text{s}\), what is the average power being expended?
The ultracentrifuge is an important tool for separating and analyzing proteins. Because of the enormous centripetal accelerations, the centrifuge must be carefully balanced, with each sample matched by a sample of identical mass on the opposite side. Any difference in the masses of opposing samples creates a net force on the shaft of the rotor, potentially leading to a catastrophic failure of the apparatus. Suppose a scientist makes a slight error in sample preparation and one sample has a mass \( 10 \) \( \text{mg} \) larger than the opposing sample.
If the samples are \( 12 \) \( \text{cm} \) from the axis of the rotor and the ultracentrifuge spins at \( 60000 \) \( \text{rpm} \), what is the magnitude of the net force on the rotor due to the unbalanced samples?
From the figure above, determine which characteristic fits this collision best.
A small block moving with a constant speed \(v\) collides inelastically with a block \(M\) attached to one end of a spring \(k\). The other end of the spring is connected to a stationary wall. Ignore friction between the blocks and the surface.
A speed skater goes around a turn that has a radius of \(31 \, \text{m}\). The skater has a speed of \(14 \, \text{m/s}\) and experiences a centripetal force of \(460 \, \text{N}\). What is the mass of the skater?
Why do pilots sometimes black out while pulling out at the bottom of a dive?
A theme park ride consists of a large vertical wheel of radius \( R \) that rotates counterclockwise on a horizontal axle through its center. The cars on the wheel move at a constant speed \( v \). Points \( A \) and \( D \) represent the position of a car at the highest and lowest point of the ride, respectively. While passing point \( A \), a student releases a small rock of mass \( m \), which falls to the ground without hitting anything. Which of the following best represents the kinetic energy of the rock when it is at the same height as point \( D \)?
A \( 0.20 \) \( \text{kg} \) object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown in the graph above. The object starts from rest at displacement \( x = 0 \) and time \( t = 0 \) and is displaced a distance of \( 20 \) \( \text{m} \). Determine each of the following.
Which of the following do not affect the maximum speed that a car can drive in a circle? Choose both correct answers.
By continuing you (1) agree to our Terms of Use and Terms of Sale 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] |
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] |
One price to unlock most advanced version of Phy across all our tools.
per month
Billed Monthly. Cancel Anytime.
We crafted THE Ultimate A.P Physics 1 Program so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the 2026 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.
Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed
