# (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. |
| 3 | g\Delta h = \frac{1}{2}v^2 | Further simplifying by replacing gh_A \,- \,gh_c with g\Delta h which is simply the change in height. |
| 4 | \frac{\Delta h}{v^2} = \frac{1}{2g} | Place \Delta h and v^2 on the same side of the equation to clearly see they are directly proportional. |
| 5 | Conclusion. | Since \Delta h and v^2 , if you doulbe height the velocity would increase proportionally by \boxed{\sqrt{2}} for a final speed of 3 \times \sqrt{2} \approx 4.24 \, \text{m/s}. |
Phy can also check your working. Just snap a picture!
A mechanic pushes a 2500 \, \text{kg} car from rest to a final speed v by doing 5.0 \times 10^3 \, \text{J} of work on the car. Frictional effect between the car and the ground are negligible. What is the final speed of the car?
Friction provides the force needed for a car to travel around a flat, circular race track. Answer the following:
A ball of mass m is fastened to a string. The ball swings at constant speed in a vertical circle of radius R with the other end of the string held fixed. Neglecting air resistance, what is the difference between the string’s tension at the bottom of the circle and at the top of the circle?
A bullet at speed v_0 trikes and embeds itself in a block of wood which is suspended by a string, causing the bullet and block to rise to a maximum height h. Which of the following statements is true?
A ball is thrown straight up. At what point does the ball have the most energy?
By continuing you (1) agree to our Terms of Sale and Terms of Use 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_1m_2}{r^2} |
| a = \frac{\Delta v}{\Delta t} | f = \mu N |
| R = \frac{v_i^2 \sin(2\theta)}{g} |
| Circular Motion | Energy |
|---|---|
| F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
| a_c = \frac{v^2}{r} | PE = mgh |
| KE_i + PE_i = KE_f + PE_f |
| Momentum | Torque and Rotations |
|---|---|
| p = m v | \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 |
|---|
| F = -k x |
| T = 2\pi \sqrt{\frac{l}{g}} |
| T = 2\pi \sqrt{\frac{m}{k}} |
| Constant | Description |
|---|---|
| g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
| G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
| \mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
| k | Spring constant, in \text{N/m} |
| M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
| M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
| M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| s (Displacement) | \text{meters (m)} |
| v (Velocity) | \text{meters per second (m/s)} |
| a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
| t (Time) | \text{seconds (s)} |
| m (Mass) | \text{kilograms (kg)} |
| Variable | Derived SI Unit |
|---|---|
| F (Force) | \text{newtons (N)} |
| E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
| P (Power) | \text{watts (W)} |
| p (Momentum) | \text{kilogram meters per second (kgm/s)} |
| \omega (Angular Velocity) | \text{radians per second (rad/s)} |
| \tau (Torque) | \text{newton meters (Nm)} |
| I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |
| f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \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}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | 10^{-12} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
The most advanced version of Phy. Currently 50% off, for early supporters.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Understand you mistakes quicker.
For GQs and FRQs, Phy provides brief feedback as to how you can improve your answer.
Aim to increase your understadning and average score with every attempt!
10 Free Credits To Get You Started
*Phy Pro members get unlimited credits
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
