# (a) Direction of the resultant force acting on the marble at point C
The resultant force acting on the marble at point C is directed towards the center of the loop. This force is mainly comprised of the gravitational force pulling downwards and the normal force exerted by the track which also points towards the center of the loop during the circular motion.
# (b) Names of all the forces acting on the marble at point C
| Force | Description |
|---|---|
| Gravitational Force | The force due to gravity acting downwards towards the center of the earth. |
| Normal Force | The force exerted by the surface of the loop on the marble directed radially inward, toward the center of the loop. |
# (c) Deduce the speed of the marble at point C. The working below uses two seperate conservation of energy equations. However, it can also be done in a single equation such that the postential energy at A transfroms into the potential energy at C and the kinetic energy at C. This is written as \( mgh_A = mgh_C + \frac{1}{2}mv^2 \).
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( h_A = 0.8 \, \text{m} \) | Initial height from which the marble is released. |
| 2 | \( v_A = 0 \, \text{m/s} \) | Initial velocity (marble is released from rest). |
| 3 | \( v_B = \sqrt{2gh_A} \) | Re-arrange and solve for velocity at the bottom of the incline, using conservation of mechanical energy, where \( mgh_A = \frac{1}{2}m{v^2}_B \). |
| 4 | \( v_B = \sqrt{2 \times 9.8 \times 0.8} \) | Calculating \( v_B \). |
| 5 | \( v_B \approx 3.97 \, \text{m/s} \) | Approximate calculation of velocity at point B. |
| 6 | \( h_C = 0.35 \, \text{m} \) | The maximum height attained by the marble is at point C (top of loop). |
| 7 | \( v_C^2 = v_B^2 – 2gh_C \) | Using conservation of mechanical energy between points B and C. |
| 8 | \( v_C^2 = 3.97^2 – 2 \times 9.8 \times 0.35 \) | Calculating \( v_C \) from \( v_B \) and change in gravitational potential energy. |
| 9 | \( v_C \approx 3.0 \, \text{m/s} \) | Approximate calculation of velocity at point C. |
# (d) Effect if the release height of the marble were to double
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( mgh_A = mgh_C + \frac{1}{2}mv^2 \) | The conservation of energy between points A and C, as used in part B. |
| 2 | \( gh_A = gh_C + \frac{1}{2}v^2 \) | Simplified formula, canceling out \( m \) on both sides. |
| 3 | \( g\Delta h = \frac{1}{2}v^2 \) | Replace \( gh_A – gh_C \) with \( g\Delta h \), which represents the change in height. |
| 4 | \( \frac{\Delta h}{v^2} = \frac{1}{2g} \) | Isolate \( \Delta h \) and \( v^2 \) to see their proportional relationship. |
| 5 | Proportional analysis | In the initial setup: \( \Delta h = 0.8 – 0.35 = 0.45 \, \text{m} \). Doubling the release height, \( \Delta h \) becomes \( 1.6 – 0.35 = 1.25 \, \text{m} \). The ratio of \( \Delta h \) is: \[ \frac{1.25}{0.45} \approx 2.78 \] Since \( v^2 \propto \Delta h \), the velocity increases by \( \sqrt{2.78} \approx 1.67 \). |
| 6 | Conclusion | The original velocity was \( 3 \, \text{m/s} \). With a velocity ratio of \( 1.67 \), the final velocity becomes: \[ 3 \times 1.67 \approx 5 \, \text{m/s} \] |
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A block of mass 3.0 kg is hung from a spring, causing it to stretch 12 cm at equilibrium. The 3.0 kg block is then taken off and the spring returns to its original height. Now a 4.0 kg block is placed on the spring and released from rest. How far will the 4.0 kg block fall before its direction is reversed?
An object undergoing simple harmonic motion has a maximum displacement of \(6.2\) \(\text{m}\) at \(t = 0.0\) \(\text{s}\). If the angular frequency of oscillation is \(1.6\) \(\text{rad/s}\), what is the object’s displacement when \(t = 3.5\) \(\text{s}\)?
A car is going over the top of a hill whose curvature approximates a circle of radius \( 350 \) \( \text{m} \). At what velocity will the occupants of the car appear to weigh \( 10\% \) less than their normal weight?
A \(90 \, \text{kg}\) individual is cycling up a hill inclined at \(30^\circ\) on a \(12 \, \text{kg}\) bicycle. The hill is quite steep, and the coefficient of static friction is \(0.85\). The cyclist ascends \(12 \, \text{m}\) up the hill and then pauses at the summit. They then start descending from rest and travel \(9 \, \text{m}\) before firmly applying the brakes, causing the wheels to lock.
A spring launches a \(4 \, \text{kg}\) block across a frictionless horizontal surface. The block then ascends a \(30^\circ\) incline with a kinetic friction coefficient of \(\mu_k = 0.25\), stopping after \(55 \, \text{m}\) on the incline. If the spring constant is \(800 \, \text{N/m}\), find the initial compression of the spring. Disregard friction while in contact with the spring.
A bullet of mass \(0.0500 \, \text{kg}\) traveling at \(50.0 \, \text{m/s}\) is fired horizontally into a wooden block suspended from a long rope. The mass of the wooden block is \(0.300 \, \text{kg}\) and it is initially at rest. The collision is completely inelastic and after impact the bullet + wooden block move together until the center of mass of the system rises a vertical distance \(h\) above its initial position.
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A mass is attached to the end of a spring and set into simple harmonic motion with an amplitude \( A \) on a horizontal frictionless surface. Determine the following in terms of only the variable \( A \).
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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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