# (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 \( 0.0350 \) \( \text{kg} \) bullet moving horizontally at \( 425 \) \( \text{m/s} \) embeds itself into an initially stationary \( 0.550 \) \( \text{kg} \) block.
A block is released from rest and slides down a frictionless ramp inclined at \( 30^\circ \) from the horizontal. When the block reaches the bottom, the block-Earth system has mechanical energy \( \text{E}_i \). The experiment is repeated, but now horizontal and vertical forces of magnitude \( F \) are exerted on the block while it slides, as shown above. When the block reaches the bottom, the mechanical energy of the block-Earth system.
A spring with a spring constant of \( 600. \) \( \text{N/m} \) is used for a scale to weigh fish. What is the mass of a fish that would stretch the spring by \( 7.5 \) \( \text{cm} \) from its normal length?
A planet of constant mass orbits the sun in an elliptical orbit. Neglecting any friction effects, what happens to the planet’s rotational kinetic energy about the sun’s center?
Consider a neutron star with a mass equal to the sun, a radius of 10 km, and a rotation period of 1.0 s. What is the speed of a point on the equator of the star?
A car traveling to the right with a speed \( v \) brakes to a stop in a distance \( d \). What is the work done on the car by the frictional force \( F \)? (Assume that the frictional force is constant)
A child pushes horizontally on a box of mass m with constant speed v across a rough horizontal floor. The coefficient of friction between the box and the floor is µ. At what rate does the child do work on the box?
A ski lift carries skiers along a \(695 \, \text{m}\) slope inclined at \(34^\circ\). To lift a single rider, it is necessary to move \(72 \, \text{kg}\) of mass to the top of the lift. Under maximum load conditions, five riders per minute arrive at the top. If \(65\%\) of the energy supplied by the motor goes to overcoming friction, what average power must the motor supply?
A block of mass \( m \) is moving on a horizontal frictionless surface with a speed \( v_0 \) as it approaches a block of mass \( 2m \) which is at rest and has an ideal spring attached to one side.
When the two blocks collide, the spring is completely compressed and the two blocks momentarily move at the same speed, and then separate again, each continuing to move.
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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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