Unit 2 · Unit Review

06Unit 2 Review

Consolidate your understanding of circular motion, gravity, momentum, energy, and special relativity with mixed conceptual questions, calculation problems, and a student self-check.

Consolidating Unit 2 prepares you for thermal physics and waves in Unit 3, where energy concepts reappear in new contexts.

How do the laws of motion, gravity, energy, and relativity work together to describe the behavior of objects — from a spinning satellite to a particle traveling near the speed of light?

Unit 2 Summary

Ch 1 · Circular Motion

Objects moving in a circle require a net inward (centripetal) force. Centripetal acceleration equals v²/r, and the centripetal force equals mv²/r. Period, frequency, and angular velocity describe how fast an object completes each revolution.

Ch 2 · Gravity and Orbital Mechanics

Newton's law of universal gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between them (Fg = Gm₁m₂/r²). Orbital speed and period follow directly from setting gravitational force equal to centripetal force.

Ch 3 · Momentum and Collisions

Momentum (p = mv) is conserved in all closed systems. Impulse (J = FΔt) equals the change in momentum. Elastic collisions conserve both momentum and kinetic energy; inelastic collisions conserve only momentum.

Ch 4 · Work, Energy, and Simple Machines

Work equals force times displacement times the cosine of the angle between them (W = Fd cosθ). The work-energy theorem states that net work equals the change in kinetic energy. Mechanical energy (KE + PE) is conserved when only conservative forces act. Simple machines trade force for distance while conserving work.

Ch 5 · Special Relativity

At speeds approaching c, time dilates (t′ = γt), lengths contract (L′ = L/γ), and mass-energy equivalence holds (E = mc²). The Lorentz factor γ = 1/√(1 − v²/c²) quantifies relativistic effects.

Key Equations

Fc = mv²/r

Centripetal force

ac = v²/r

Centripetal acceleration

Fg = Gm₁m₂/r²

Newton's gravitation

p = mv

Momentum

J = FΔt = Δp

Impulse

W = Fd cosθ

Work

KE = ½mv²

Kinetic energy

PE = mgh

Gravitational PE

ME = KE + PE

Mechanical energy

t' = γt

Time dilation

L' = L/γ

Length contraction

E = mc²

Mass-energy

Worked Examples

Example 1

A 1,200 kg car rounds a flat circular curve of radius 80 m at 20 m/s. What centripetal force does the road exert on the car?

Identify: m = 1,200 kg, v = 20 m/s, r = 80 m

Formula: Fc = mv²/r

Fc = (1,200)(20²) / 80

Fc = (1,200)(400) / 80

Fc = 480,000 / 80

Answer:Fc = 6,000 N (directed toward the center of the curve)
Example 2

Calculate the gravitational force between Earth (m₁ = 5.97 × 10²⁴ kg) and a 70 kg person standing on its surface (r = 6.37 × 10⁶ m). G = 6.674 × 10⁻¹¹ N·m²/kg².

Formula: Fg = Gm₁m₂/r²

Fg = (6.674 × 10⁻¹¹)(5.97 × 10²⁴)(70) / (6.37 × 10⁶)²

Numerator: 6.674 × 10⁻¹¹ × 4.179 × 10²⁶ ≈ 2.789 × 10¹⁶

Denominator: (6.37 × 10⁶)² = 4.058 × 10¹³

Fg ≈ 2.789 × 10¹⁶ / 4.058 × 10¹³

Answer:Fg ≈ 687 N (consistent with weight = mg = 70 × 9.8 = 686 N ✓)
Example 3

A 3 kg ball moving at 6 m/s east collides with a stationary 5 kg ball. After the collision the 3 kg ball moves at 1 m/s east. Find the velocity of the 5 kg ball after the collision.

Conservation of momentum: p_before = p_after

p_before = (3)(6) + (5)(0) = 18 kg·m/s

p_after = (3)(1) + (5)v₂

18 = 3 + 5v₂

5v₂ = 15

Answer:v₂ = 3 m/s east
Example 4

A 2 kg block starts from rest and slides down a frictionless ramp of height 5 m. What is its speed at the bottom? (g = 9.8 m/s²)

Conservation of mechanical energy: PE_top = KE_bottom

mgh = ½mv²

gh = ½v² (mass cancels)

v² = 2gh = 2(9.8)(5) = 98

v = √98

Answer:v ≈ 9.9 m/s
Example 5

A muon created in the upper atmosphere has a proper lifetime of 2.2 μs. It travels at v = 0.98c relative to Earth. How long does the muon's lifetime appear to an observer on Earth? (γ = 1/√(1 − 0.98²) ≈ 5.03)

Time dilation: t′ = γt₀

t₀ = 2.2 × 10⁻⁶ s (proper lifetime in muon frame)

γ ≈ 5.03

t' = 5.03 × 2.2 × 10⁻⁶

Answer:t′ ≈ 11.1 μs (the muon appears to live ~5× longer from Earth's frame)

Guided Practice

Guided Problem 1

A 0.5 kg ball on a 1.2 m string is swung in a horizontal circle at 4 m/s. What is the tension in the string?

Hint: Tension provides the centripetal force. Use Fc = mv²/r with r = 1.2 m.

Guided Problem 2

Two asteroids have masses 4.0 × 10¹² kg and 6.0 × 10¹² kg and are separated by 500 m. Find the gravitational force between them. (G = 6.674 × 10⁻¹¹ N·m²/kg²)

Hint: Apply Fg = Gm₁m₂/r². Square the distance first: r² = (500)² = 250,000 m².

Guided Problem 3

A 0.2 kg hockey puck slides at 8 m/s and is brought to rest by a 4 N friction force. How long does the friction act? Use the impulse-momentum theorem.

Hint: J = FΔt = Δp. The change in momentum is 0 − (0.2)(8) = −1.6 kg·m/s. Solve for Δt.

Guided Problem 4

A 10 kg crate is pushed 6 m along a floor by a 50 N force applied at 30° below the horizontal. How much work is done by the applied force?

Hint: W = Fd cosθ. The angle between force and displacement is 30°. cos 30° ≈ 0.866.

Guided Problem 5

A spaceship travels at v = 0.60c. Its proper length is 200 m. What length does a stationary observer measure? (γ = 1.25)

Hint: Length contraction: L′ = L/γ. Divide the proper length by γ.

Key Vocabulary

Centripetal force

The net inward force required to keep an object moving in a circular path. It always points toward the center of the circle.

Example: Fc = mv²/r — for a 2 kg ball at 3 m/s on a 1 m string: Fc = 18 N

Gravitational field

A region of space in which a mass experiences a gravitational force. Field strength g = Fg/m (units: N/kg or m/s²).

Example: At Earth's surface, g ≈ 9.8 N/kg directed toward Earth's center.

Momentum

The product of an object's mass and velocity (p = mv). A vector quantity conserved in closed systems.

Example: A 5 kg object at 10 m/s has p = 50 kg·m/s.

Impulse

The product of a net force and the time interval over which it acts (J = FΔt). Equal to the change in momentum.

Example: A 20 N force acting for 3 s delivers J = 60 N·s of impulse.

Work

Energy transferred by a force acting through a displacement. W = Fd cosθ, where θ is the angle between force and displacement.

Example: Pushing a box 5 m with a 10 N force parallel to the floor: W = 50 J.

Kinetic energy

Energy an object possesses due to its motion. KE = ½mv². Doubles when speed increases by √2; quadruples when speed doubles.

Example: A 4 kg ball at 6 m/s: KE = ½(4)(36) = 72 J.

Potential energy

Stored energy due to an object's position in a force field. Gravitational PE = mgh, measured relative to a chosen reference level.

Example: A 3 kg book on a 2 m shelf: PE = (3)(9.8)(2) ≈ 58.8 J.

Time dilation

The relativistic effect by which a moving clock ticks more slowly than a stationary one. Described by t′ = γt₀, where γ ≥ 1.

Example: At v = 0.87c, γ ≈ 2, so 1 s of proper time appears as 2 s to a stationary observer.

Workbook Check

Interactive Practice — 5 Questions

1

A 0.8 kg ball moves in a circle of radius 0.5 m at 4 m/s. What is the centripetal force?

2

If the distance between two masses is doubled, the gravitational force between them becomes:

3

A 4 kg object moving at 5 m/s collides with and sticks to a stationary 6 kg object. What is their combined speed after the collision?

4

A 5 kg object is lifted 3 m. How much gravitational potential energy does it gain? (g = 10 m/s²)

5

A rocket travels at 0.6c. Its proper length is 100 m. What length does a stationary observer measure? (γ = 1.25)