Physics for the IB Diploma: Workbook

Exercise 2.2 Newton's Laws of Motion

1. Newton's First Law and Equilibrium

• a. Newton's First Law: A body remains in a state of rest or uniform motion in a straight line unless acted upon by an external unbalanced force.
• b. Equilibrium: A state where the net force acting on a body is zero, resulting in no acceleration. \sum F = 0
• c. Equilibrium and Motion: Yes, a body can be in equilibrium while moving if its velocity is constant (i.e., moving at a constant speed in a straight line).

2. Situations of Equilibrium

• a. Helicopter Hovering: In equilibrium (net force is zero).
• b. Car at Constant Speed on Straight Road: In equilibrium (constant velocity).
• c. Cyclist Around a Bend: Not in equilibrium (velocity changing direction, hence accelerating).
• d. Skydiver at Terminal Velocity: In equilibrium (weight balanced by air resistance).

3. Newton's Second Law

• a. Newton's Second Law: The net force acting on a body is equal to the rate of change of its momentum, or \sum F = ma, where F is the net force, m is the mass, and a is the acceleration.
• b. Definition of Newton: 1 Newton is the force required to accelerate a 1 kg mass at 1 ms^{-2}. 1 N = 1 kg \cdot ms^{-2}
• c. Table Completion (using F = ma):
  • 120 N, Mass = ?, Acceleration = 50 ms^{-2} => Mass = \frac{120}{50} = 2.4 kg
  • Net Force = 900 N, Mass = 4.5 kg, Acceleration = ? => Acceleration = \frac{900}{4.5} = 200 ms^{-2}
  • Net Force = ?, Mass = 6 kg, Acceleration = 0.25 ms^{-2} => Net Force = 6 \cdot 0.25 = 1.5 N

4. Bullet Acceleration

• Calculations:
  • Initial speed, v_i = 0 ms-1
  • Final speed, v_f = 1500 ms-1
  • Time, t = 0.1 s
  • Mass, m = 0.05 kg
  • Acceleration, a = \frac{vf - vi}{t} = \frac{1500 - 0}{0.1} = 15000 ms^{-2}
  • Average force, F = ma = 0.05 \cdot 15000 = 750 N

5. Paper Cone

• Given: Force F = 0.05 N, mass m = 12 g = 0.012 kg.
• a. Acceleration:
  • a = \frac{F}{m} = \frac{0.05}{0.012} \approx 4.17 ms^{-2}
• b. Graph: A sketch of acceleration vs. time would show an initial high acceleration that gradually decreases as the cone approaches terminal velocity. Eventually, the acceleration becomes zero when the air resistance equals the weight of the cone.
• c. Explanation: Initially, the net force is high, leading to a large acceleration. As the cone gains speed, air resistance increases, reducing the net force and thus the acceleration. At terminal velocity, the weight of the cone is balanced by the air resistance, resulting in zero net force and zero acceleration.

6. Newton's Third Law

• a. Newton's Third Law: For every action, there is an equal and opposite reaction. Forces always occur in pairs, acting on different objects.
• b. Examples:
  • i. Moon and Earth Gravitational Forces: This is a Newton's Third Law pair. The force of the Moon on the Earth is equal and opposite to the force of the Earth on the Moon.
  • ii. Book on Table: This is not a Newton's Third Law pair. The weight of the book is the gravitational force of the Earth on the book. The Newton's Third Law pair would be the gravitational force of the book on the Earth. The normal contact force is the reaction to the force the book exerts on the table.
  • iii. Proton and Electron: This is a Newton's Third Law pair. The electrical force exerted by a proton on an electron is equal and opposite to the electrical force exerted by the electron on the proton.
  • iv. Raindrop and Ground: This is a Newton's Third Law pair. The force exerted by the raindrop on the ground is equal and opposite to the force exerted by the ground on the raindrop.

7. Elevator Problem

• Given: Mass m = 60 kg, g = 9.81 ms^{-2}.
• a. Constant Speed:
  • The reading on the weighing scales is equal to the weight of the person: W = mg = 60 \cdot 9.81 = 588.6 N.
• b. Downward Acceleration:
  • Acceleration a = 0.25g = 0.25 \cdot 9.81 = 2.4525 ms^{-2}
  • The net force is F_{net} = mg - R = ma, where R is the reading on the scales.
  • R = m(g - a) = 60 \cdot (9.81 - 2.4525) = 60 \cdot 7.3575 = 441.45 N
  • Feeling: The person feels lighter.
• c. Upward Acceleration:
  • Acceleration a = 0.2g = 0.2 \cdot 9.81 = 1.962 ms^{-2}
  • The net force is F_{net} = R - mg = ma
  • R = m(g + a) = 60 \cdot (9.81 + 1.962) = 60 \cdot 11.772 = 706.32 N
  • Feeling: The person feels heavier.

8. Laws in Natural Sciences

• a. How We Know Things: Through observation, experimentation, and formulating theories that can be tested and potentially falsified.
• b. Definition of Law: A law is a descriptive statement about how nature behaves, based on repeated observations and experiments.
• c. Proving a Law: No, it is not possible to definitively prove a law. Laws are always subject to revision or replacement if new evidence contradicts them.
• d. Impact of Relativity: Einstein's theory of relativity showed that Newton's laws are approximations that work well at low speeds and weak gravitational fields. At very high speeds or in strong gravitational fields, relativistic effects become significant, and Newton's laws are no longer accurate.

Exercise 2.3 Circular Motion

1. Angular Speed

• a. Equations:
  • i. \omega = \frac{2\pi}{T}, where \omega is angular speed and T is the period.
  • ii. \omega = 2\pi f, where f is the frequency.
• b. Calculations:
  • i. Earth around the Sun (1 year = 365.25 \cdot 24 \cdot 60 \cdot 60 seconds):
    • \omega = \frac{2\pi}{365.25 \cdot 24 \cdot 60 \cdot 60} \approx 1.99 \times 10^{-7} rad/s
  • ii. Children's Carousel (1 rotation per minute):
    • \omega = \frac{2\pi}{60} \approx 0.105 rad/s
  • iii. Moon around the Earth (13 rotations in 1 year):
    • \omega = \frac{13 \cdot 2\pi}{365.25 \cdot 24 \cdot 60 \cdot 60} \approx 2.59 \times 10^{-6} rad/s
• c. Singapore Flyer:
  • Radius r = 75 m, Period T = 30 minutes = 1800 seconds
    • i. Frequency: f = \frac{1}{T} = \frac{1}{1800} \approx 0.000556 Hz
    • ii. Angular Speed: \omega = 2\pi f = \frac{2\pi}{1800} \approx 0.00349 rad/s
    • iii. Linear Speed: v = r\omega = 75 \cdot 0.00349 \approx 0.262 m/s

2. Circular Motion and Force

• a. Acceleration: A body in circular motion is always accelerating because its velocity (direction) is constantly changing.
• b. Unbalanced Force Direction: The unbalanced force must be directed towards the center of the circle.
• c. Name: Centripetal force.
• d. Equation: F = \frac{mv^2}{r}, where F is the centripetal force, m is the mass, v is the linear speed, and r is the radius of the circle.

3. Graphs of Acceleration and Force

• a. Acceleration:
  • i. Acceleration vs. mass (constant v, r): Linear relationship (a ∝ m).
  • ii. Acceleration vs. linear speed (constant m, r): Quadratic relationship (a ∝ v^2).
  • iii. Acceleration vs. radius (constant m, v): Inverse relationship (a ∝ 1/r).
• b. Force:
  • i. Force vs. mass (constant v, r): Linear relationship (F ∝ m).
  • ii. Force vs. linear speed (constant m, r): Quadratic relationship (F ∝ v^2).
  • iii. Force vs. radius (constant m, v): Inverse relationship (F ∝ 1/r).

4. Real Forces Providing Centripetal Force

• a. Moon around Earth: Gravitational force.
• b. Electron around Nucleus: Electrostatic force.
• c. Proton in Large Hadron Collider: Magnetic force.
• d. Car on a Bend: Friction between tyres and the road.
• e. Ball on a String: Tension in the string.

5. Hydrogen Atom Model

• Given: Radius r = 5.29 \times 10^{-11} m, Electrostatic force F = 8.23 \times 10^{-8} N, Mass of electron m = 9.1 \times 10^{-31} kg.
• F = \frac{mv^2}{r} => v = \sqrt{\frac{Fr}{m}} = \sqrt{\frac{8.23 \times 10^{-8} \cdot 5.29 \times 10^{-11}}{9.1 \times 10^{-31}}} \approx 2.19 \times 10^6 m/s

6. Car on Circular Road

• a.
  • i. Friction Force: The friction force must be greater for the more massive car, since F = \frac{mv^2}{r}. The needed centripetal force grows with mass.
  • ii. Maximum Speed: The maximum friction force is F_{max} = \mu mg, where \mu is the coefficient of friction. Therefore, \frac{mv^2}{r} = \mu mg. v = \sqrt{\mu gr}; which is independent of mass.
  • iii. Wet/Icy Road: A wet or icy road reduces the coefficient of friction, \mu, hence the maximum speed is lower because the maximum available centripetal force is reduced.
• b.
  • i. Speed Limit: The speed limit is lower because the radius of curvature on the exit road is smaller than on the motorway. Since v = \sqrt{\mu gr}, a smaller radius implies a lower maximum speed.
  • ii. Calculation: Given \mu = 0.75, r = 80 m, g = 9.81 ms^{-2}
    • v = \sqrt{0.75 \cdot 9.81 \cdot 80} \approx 24.26 m/s
• c. Worn Tyres: Halving the coefficient of friction halves the value of \mu. v = \sqrt{\mu gr}, the maximum safe speed becomes v_{new} = v / \sqrt{2}, which is about 0.707v. So the max speed around the bend would be reduced by a factor of \sqrt{2}.

7. Ball on a String

• Given: Mass m = 100 g = 0.1 kg, Length l = 60 cm = 0.6 m, Frequency f = 2.5 Hz.
• a.
  • i. Linear Speed: v = 2\pi r f = 2 \pi (0.6)(2.5) \approx 9.42 m/s
  • ii. Tension: T = \frac{mv^2}{r} = \frac{0.1 \cdot (9.42)^2}{0.6} \approx 14.8 N
• b.
  • i. Not Perfectly Horizontal: Gravity acts on the ball, so the string must have a vertical component of tension to balance the weight of the ball.
  • ii. Free-Body Diagram: The diagram shows the weight (mg) acting downward, and the tension (T) acting along the string at an angle to the horizontal.
• c. Actual Tension: The actual tension must be larger because it has to support the weight of the ball in addition to providing the centripetal force.
• d. Centripetal Force Component: The horizontal component of the tension provides the centripetal force.
• e. Other Component: The vertical component of the tension balances the weight of the ball.

8. Cyclist on Banked Track

• a. Free-Body Diagram: Shows the weight (mg) acting downwards, and the normal reaction force (N) acting perpendicular to the track surface.
• b. Derivation:
  • The horizontal component of the normal force provides the centripetal force: N \sin\theta = \frac{mv^2}{r}.
  • The vertical component of the normal force balances the weight: N \cos\theta = mg.
  • Dividing the two equations: \tan\theta = \frac{v^2}{gr}, so \theta = \arctan(\frac{v^2}{gr}).
• c. Angle of Banking: Olympic cyclists have higher speeds (v), so the angle of banking \theta must be larger to provide the necessary centripetal force without relying on friction.

9. Mass on Vertical Circle

• Description: A mass is swung in a vertical circle of radius r.
• Note: The net unbalanced force acts as the centripetal force.