Rocket Science Physics — Part 2: Extended Examples, Practice Problems & Projects
6th Grade Physics · Supplementary Volume

Rocket Science Physics
Part 2

Extended worked examples (6–8 per topic), 80 practice problems with answer key, hands-on lab projects, and a complete reference glossary.

60+New Examples
80Practice Problems
3Lab Projects
30Glossary Terms
Extended Examples Unit 1 — Forces & Newton's Laws
1.1Newton's First Law — Additional Examples
RecallIf ΣF = 0 → a = 0 (object stays at rest or constant velocity)
Examples 6 – 8
EXAMPLE 6Parachute deployed — rocket reaches terminal velocity
Given
  • m: 0.3 kg (rocket descending)
  • F_gravity: 2.94 N downward
  • F_drag (chute): 2.94 N upward
Analysis
  1. Gravity pulls down: 2.94 N
  2. Parachute drag pushes up: 2.94 N
  3. ΣF = 2.94 − 2.94 = 0 N
  4. First Law: ΣF = 0 → a = 0 → constant velocity (terminal velocity)
AnswerΣF = 0 → rocket descends at constant terminal velocity — parachute works perfectly.
EXAMPLE 7Rocket drifting sideways in orbit
Given
  • v_sideways: 2 m/s (bump from docking)
  • Location: ISS interior (free-fall)
  • ΣF lateral: 0 N
Analysis
  1. No lateral force acts on the astronaut.
  2. ΣF = 0 → First Law applies.
  3. Astronaut drifts sideways at exactly 2 m/s indefinitely.
  4. Must grab a handhold to stop — applies a new force to break inertia.
AnswerInertia keeps astronaut moving sideways at 2 m/s until a force acts.
EXAMPLE 8Comparing inertia of two rockets — which is harder to accelerate?
Given
  • Rocket A: 0.2 kg
  • Rocket B: 1.5 kg
  • Same engine: 10 N
Analysis
  1. Greater mass = greater inertia = more resistance to change.
  2. Rocket A: a = 10 / 0.2 = 50 m/s²
  3. Rocket B: a = 10 / 1.5 = 6.7 m/s²
  4. Rocket B's inertia is 7.5× greater — takes far more effort to change its motion.
AnswerRocket B is 7.5× harder to accelerate — mass IS inertia (First Law).
1.2Newton's Second Law — Additional Examples
RecallF = m × a | a = F/m | m = F/a
Examples 7 – 8
EXAMPLE 7Rocket burns fuel — mass decreases, a increases
Given
  • F (constant): 500 N
  • m at t=0: 100 kg
  • m at t=30s: 60 kg (fuel burned)
Solution
  1. At launch: a = 500/100 = 5 m/s²
  2. At t=30s: a = 500/60 = 8.33 m/s²
  3. Acceleration grows as fuel burns — explains why rockets accelerate faster over time!
Answera grows from 5 → 8.33 m/s² as fuel burns — same engine, less mass = more acceleration.
EXAMPLE 8Four-engine cluster vs. single engine
Given
  • Single engine: 30 N
  • Cluster (4 engines): 4 × 30 = 120 N
  • m: 0.8 kg
Solution
  1. Single: a = 30/0.8 = 37.5 m/s²
  2. Cluster: a = 120/0.8 = 150 m/s²
  3. 4× the force → 4× the acceleration (direct proportion, F=ma).
AnswerCluster gives 150 m/s² vs. 37.5 m/s² — 4× engines = 4× acceleration.
1.3Newton's Third Law — Additional Examples
RecallF_action = −F_reaction (equal magnitude, opposite direction)
Examples 6 – 8
EXAMPLE 6Fire hose reaction force
Given
  • Water pushed forward: 800 N (action)
  • Firefighter mass: 90 kg
Solution
  1. By 3rd Law: Hose pushes firefighter back with 800 N.
  2. a = 800/90 = 8.89 m/s² backward!
  3. Same physics as a rocket — water out the front, firefighter pushed back.
AnswerFirefighter experiences 800 N backward — must brace or be knocked over.
EXAMPLE 7Jet engine — exhaust vs. aircraft thrust
Given
  • Exhaust mass flow: 300 kg/s
  • Exhaust speed: 500 m/s
Solution
  1. Exhaust momentum per second = 300 × 500 = 150,000 kg·m/s per second.
  2. Force = momentum/time = 150,000 N backward (action).
  3. By 3rd Law: aircraft thrust = 150,000 N forward (reaction).
AnswerThrust = 150,000 N forward — direct application of 3rd Law.
EXAMPLE 8Unequal masses pushing apart — cannon and cannonball
Given
  • Cannonball mass: 5 kg
  • Cannon mass: 500 kg
  • Push force: 10,000 N each
Solution
  1. Forces are equal: both experience 10,000 N (3rd Law).
  2. Cannonball: a = 10,000/5 = 2,000 m/s² (flies fast!)
  3. Cannon: a = 10,000/500 = 20 m/s² (recoils slowly)
  4. Equal force, very different accelerations — mass determines response.
AnswerSame force: ball at 2,000 m/s² vs cannon at 20 m/s² — mass controls acceleration.
1.4Net Force — Additional Examples
RecallF_net = F_thrust − F_gravity − F_drag
Examples 6 – 8
EXAMPLE 6Net force changes as rocket burns fuel (t = 0 and t = 3 s)
Given
  • F_thrust: 40 N (constant)
  • m at t=0: 0.8 kg
  • m at t=3s: 0.5 kg (fuel burned)
  • F_drag: 1 N (approx. constant)
Solution
  1. t=0: F_grav = 0.8×9.8 = 7.84 N; F_net = 40 − 7.84 − 1 = 31.16 N; a = 31.16/0.8 = 38.95 m/s²
  2. t=3s: F_grav = 0.5×9.8 = 4.9 N; F_net = 40 − 4.9 − 1 = 34.1 N; a = 34.1/0.5 = 68.2 m/s²
  3. Acceleration nearly doubled as fuel burned! This is why rockets accelerate more over time.
Answera grows from 38.95 → 68.2 m/s² as fuel burns off — lighter = faster.
EXAMPLE 7Rocket descending under parachute — find terminal velocity condition
Given
  • m: 0.3 kg
  • F_drag (chute): 2.94 N upward
  • F_thrust: 0 N
Solution
  1. F_gravity = 0.3 × 9.8 = 2.94 N (downward)
  2. F_net = 0 − 2.94 + 2.94 = 0 N
  3. a = 0 → terminal velocity reached — no acceleration, constant descent.
AnswerF_net = 0 → terminal velocity. Drag perfectly balances gravity.
EXAMPLE 8Which engine is "just right" for a 1.2 kg rocket?
Given
  • m: 1.2 kg
  • Engine X: F = 10 N
  • Engine Y: F = 20 N
  • Engine Z: F = 35 N
  • F_drag: 2 N (all cases)
Solution
  1. F_grav = 1.2 × 9.8 = 11.76 N
  2. X: F_net = 10 − 11.76 − 2 = −3.76 N ✗ Won't lift off
  3. Y: F_net = 20 − 11.76 − 2 = 6.24 N; a = 5.2 m/s² ✓ Gentle liftoff
  4. Z: F_net = 35 − 11.76 − 2 = 21.24 N; a = 17.7 m/s² ✓✓ Fast liftoff
AnswerX: won't fly | Y: good liftoff at 5.2 m/s² | Z: aggressive at 17.7 m/s²
Extended Examples Unit 2 — Thrust, Propulsion & Momentum
2.1Momentum — Additional Examples
Recallp = m × v
Examples 6 – 8
EXAMPLE 6Collision — rocket recovery parachute snaps taut
Given
  • Rocket m: 0.3 kg at 50 m/s downward
  • Parachute: stops rocket in 0.5 s
Solution
  1. Initial momentum: p = 0.3 × 50 = 15 kg·m/s
  2. Final momentum = 0 kg·m/s (at rest)
  3. Δp = 15 kg·m/s changed in 0.5 s
  4. Force on rocket: F = Δp/Δt = 15/0.5 = 30 N — quite a jerk!
AnswerΔp = 15 kg·m/s → parachute exerts 30 N stopping force over 0.5 s.
EXAMPLE 7Two-stage separation — momentum split
Given
  • Combined mass: 10 kg at 300 m/s
  • Stage 1 mass: 7 kg (drops off)
  • Stage 2 mass: 3 kg (continues)
Solution
  1. Total initial momentum: p = 10 × 300 = 3,000 kg·m/s
  2. Conservation: p is shared between both pieces.
  3. If Stage 1 separates at same speed 300 m/s: p₁ = 7×300 = 2,100, p₂ = 3×300 = 900 kg·m/s
  4. Stage 2 engine then fires, adding more momentum to reach orbit.
AnswerTotal p = 3,000 kg·m/s splits: Stage 1 takes 2,100, Stage 2 takes 900 kg·m/s.
EXAMPLE 8Momentum of exhaust — single fuel pulse
Given
  • Exhaust mass: 0.01 kg per second
  • Exhaust speed: 3,000 m/s
  • Burn time: 2 s
Solution
  1. Total exhaust mass = 0.01 × 2 = 0.02 kg
  2. Exhaust momentum = 0.02 × 3,000 = 60 kg·m/s backward
  3. By conservation, rocket gains 60 kg·m/s forward.
  4. If rocket mass = 1 kg: Δv = 60/1 = 60 m/s speed gain.
AnswerRocket gains Δp = 60 kg·m/s → Δv = 60 m/s from 2-second burn.
2.2Impulse — Additional Examples
RecallJ = F × Δt = Δp = m × Δv
Examples 6 – 8
EXAMPLE 6Engine class comparison — A through D
Engine Class Chart (Total Impulse ranges)
  1. Class A: 1.26–2.5 N·s → Δv = 2.5/0.3 = 8.3 m/s for 0.3 kg rocket
  2. Class B: 2.51–5.0 N·s → Δv = 5.0/0.3 = 16.7 m/s (2× A)
  3. Class C: 5.01–10.0 N·s → Δv = 10/0.3 = 33.3 m/s (2× B)
  4. Class D: 10.01–20.0 N·s → Δv = 20/0.3 = 66.7 m/s (2× C)
  5. Each letter class doubles the total impulse of the previous.
PatternEach class doubles impulse: A=8.3, B=16.7, C=33.3, D=66.7 m/s for a 0.3 kg rocket.
EXAMPLE 7Soft landing — reducing impulse over longer time
Given
  • m: 500 kg spacecraft
  • v_approach: 10 m/s
  • Δp needed: 5,000 kg·m/s
Solution
  1. Hard landing (0.1 s): F = 5000/0.1 = 50,000 N — catastrophic!
  2. Soft landing (10 s): F = 5000/10 = 500 N — gentle!
  3. Same total impulse (Δp = 5,000 N·s), but spreading it over time reduces peak force.
AnswerSame Δp, but longer time (10s) reduces peak force from 50,000 N to just 500 N.
EXAMPLE 8Vernier thruster on a spacecraft — small course correction
Given
  • F_thruster: 22 N
  • Δt: 0.5 s (short burst)
  • m spacecraft: 2,200 kg
Solution
  1. J = F × Δt = 22 × 0.5 = 11 N·s
  2. Δv = J/m = 11/2,200 = 0.005 m/s = 5 mm/s
  3. Tiny velocity change — but in space this precisely adjusts the orbit!
AnswerΔv = 5 mm/s — tiny but precise. Orbital mechanics amplifies small corrections.
2.3Tsiolkovsky Rocket Equation — Additional Examples
RecallΔv = v_e × ln(m₀ / m_f) | ln(2)≈0.693 | ln(3)≈1.099 | ln(5)≈1.609 | ln(10)≈2.303
Examples 6 – 8
EXAMPLE 6Comparing two propellants for same rocket
Given
  • m₀/m_f: 3 (same for both)
  • Propellant A v_e: 2,500 m/s (kerosene)
  • Propellant B v_e: 4,400 m/s (liquid H₂)
Solution
  1. ln(3) = 1.099
  2. Propellant A: Δv = 2500 × 1.099 = 2,748 m/s
  3. Propellant B: Δv = 4400 × 1.099 = 4,836 m/s
  4. Better propellant gives 76% more speed from the same mass ratio!
AnswerA: 2,748 m/s | B: 4,836 m/s — propellant quality is as important as mass ratio.
EXAMPLE 7Ion thruster — tiny exhaust mass, incredibly fast
Given
  • v_e (ion): 30,000 m/s
  • m₀/m_f: 1.1 (only 10% fuel!)
  • ln(1.1): ≈ 0.095
Solution
  1. Δv = 30,000 × 0.095 = 2,850 m/s
  2. With only 10% fuel mass by weight!
  3. Compared to chemical: needs 75% fuel for same Δv.
  4. Ion drives are 8× more fuel-efficient — but very low thrust (slow to build up).
AnswerIon drive: Δv = 2,850 m/s with only 10% fuel — incredible efficiency for deep space.
EXAMPLE 8Why 100% fuel still can't reach orbital speed in one stage
Given
  • v_e: 3,000 m/s
  • Scenario: 90% of mass is fuel (m₀/m_f = 10)
Solution
  1. ln(10) = 2.303
  2. Δv = 3000 × 2.303 = 6,909 m/s
  3. Still below the 9,400 m/s needed for orbit!
  4. Even 99% fuel (m₀/m_f=100): Δv = 3000 × ln(100) = 3000 × 4.605 = 13,815 m/s — but then you have no room for payload or structure!
Answer90% fuel → only 6,909 m/s. Staging solves this by ditching empty mass mid-flight.
2.4Thrust-to-Weight Ratio — Additional Examples
RecallTWR = F_thrust / (m × g)
Examples 6 – 8
EXAMPLE 6TWR improves as fuel burns — track over time
Given
  • F_thrust: 200 N (constant)
  • m at t=0: 10 kg
  • m at t=10s: 7 kg
  • m at t=20s: 4 kg
Solution
  1. t=0: TWR = 200/(10×9.8) = 200/98 = 2.04
  2. t=10s: TWR = 200/(7×9.8) = 200/68.6 = 2.92
  3. t=20s: TWR = 200/(4×9.8) = 200/39.2 = 5.10
  4. TWR more than doubles as fuel burns — rocket accelerates harder and harder!
AnswerTWR grows: 2.04 → 2.92 → 5.10 as fuel burns off — acceleration keeps increasing!
EXAMPLE 7Design challenge — maximum payload for given TWR
Given
  • F_thrust: 50 N
  • Engine + body mass: 0.4 kg
  • Min TWR needed: 2.0
Solution
  1. Min TWR = 2.0 means max weight = F/TWR = 50/2 = 25 N
  2. Max total mass = 25/9.8 = 2.55 kg
  3. Payload mass = 2.55 − 0.4 = 2.15 kg maximum
AnswerMax payload = 2.15 kg to maintain TWR ≥ 2.0 with this engine.
EXAMPLE 8Space Shuttle comparison — TWR at liftoff vs. SRB separation
Given
  • Total thrust (liftoff): 30,160,000 N
  • Mass at liftoff: 2,041,000 kg
  • Mass at SRB separation: 1,050,000 kg
  • Thrust at SRB sep (3 SSMEs only): 5,925,000 N
Solution
  1. Liftoff TWR: 30,160,000/(2,041,000×9.8) = 30,160,000/20,001,800 = 1.51
  2. At SRB sep: 5,925,000/(1,050,000×9.8) = 5,925,000/10,290,000 = 0.58
  3. TWR drops below 1 at SRB separation — but now rocket is almost in orbit! Orbital velocity provides most of the "lift".
AnswerLiftoff TWR 1.51 → SRB sep TWR 0.58 — but by then it's nearly in orbit.
Extended Examples Units 3–5 — Selected Additional Examples
3.2Kinetic Energy — Example 6: Energy to reach Mach 1
EXAMPLE 6How much KE does a rocket need to reach the speed of sound?
Given
  • m: 0.5 kg
  • v (Mach 1): 343 m/s
Solution
  1. KE = ½ × 0.5 × (343)²
  2. KE = 0.25 × 117,649 = 29,412 J
  3. Nearly 30,000 J — equivalent to lifting 300 kg by 10 meters!
AnswerKE = 29,412 J needed to reach Mach 1 — huge energy required.
3.3Potential Energy — Example 6: Energy stored at geostationary orbit
EXAMPLE 6PE gained by a 1,000 kg satellite reaching GEO (35,786 km)
Given
  • m: 1,000 kg
  • h: 35,786,000 m
  • g: 9.8 m/s² (approx.)
Solution
  1. PE = 1000 × 9.8 × 35,786,000
  2. PE = 3.507 × 10¹¹ J
  3. That's 350 billion Joules! Equivalent to ~84 tons of TNT.
AnswerPE = 3.51 × 10¹¹ J — this is the minimum energy to lift 1,000 kg to GEO.
4.1Drag Force — Example 6: Maximum speed rocket (drag = net force)
EXAMPLE 6Rocket reaches top speed when F_net = 0
Given
  • F_thrust: 25 N
  • m: 0.4 kg at high altitude
  • ρ (at altitude): 0.7 kg/m³
  • C_d: 0.4, A: 0.001 m²
Solution
  1. At max speed: F_thrust = F_gravity + F_drag
  2. F_gravity = 0.4 × 9.8 = 3.92 N
  3. F_drag max = 25 − 3.92 = 21.08 N
  4. 21.08 = ½ × 0.7 × v² × 0.4 × 0.001 = 0.00014 × v²
  5. v = √(21.08/0.00014) = √150,571 = 388 m/s
AnswerMaximum speed ≈ 388 m/s (> Mach 1!) when all thrust balances gravity + drag.
4.3Altitude Formula — Example 6: Pre-launch altitude prediction worksheet
EXAMPLE 6Full flight altitude calculation — burn + coast combined
Complete Flight Profile — Step by Step
  1. Given: m=0.3 kg, F_thrust=25 N, F_drag=1 N, burn time=1.5 s
  2. Net a during burn: a = (25 − 2.94 − 1)/0.3 = 21.06/0.3 = 70.2 m/s²
  3. Speed at end of burn: v = 0 + 70.2 × 1.5 = 105.3 m/s
  4. Altitude during burn: d₁ = ½ × 70.2 × 1.5² = 78.98 m
  5. Coasting: a = −9.8 m/s² (gravity only, ignoring drag for simplicity)
  6. Coast time to apex: t = 105.3/9.8 = 10.74 s
  7. Coast altitude gain: d₂ = 105.3×10.74 − ½×9.8×10.74² = 1131 − 565 = 566 m
  8. Total altitude: h = 78.98 + 566 = 645 m
Predicted Altitude≈ 645 m — full 8-step calculation matching real launch prediction methods!
5.1Orbital Velocity — Example 6: Earth's orbital speed around the Sun
EXAMPLE 6How fast does Earth orbit the Sun?
Given
  • GM_sun: 1.327 × 10²⁰ m³/s²
  • r (Earth-Sun): 1.496 × 10¹¹ m
Solution
  1. v = √(GM/r) = √(1.327×10²⁰ / 1.496×10¹¹)
  2. v = √(886,898,396) = 29,780 m/s
  3. Earth travels at ~29.8 km/s around the Sun — 107,000 km/h!
AnswerEarth's orbital speed = 29,780 m/s ≈ 107,000 km/h around the Sun.
5.4Multi-Stage Rockets — Example 6: Student team staging competition
EXAMPLE 6Team A (single stage) vs Team B (two stage) — same total fuel
Given
  • Total fuel: 200 g = 0.2 kg each team
  • Dry mass (A): 0.3 kg single stage
  • Dry mass (B): 0.25 kg (two stages, lighter per stage)
  • v_e (both): 2,000 m/s
Solution
  1. Team A: m₀=0.5 kg, m_f=0.3 kg; ratio=1.667; Δv = 2000×ln(1.667) = 2000×0.511 = 1,022 m/s
  2. Team B Stage 1: m₀=0.35, m_f=0.225; ratio=1.556; Δv₁ = 2000×0.443 = 886 m/s
  3. Team B Stage 2: m₀=0.225, m_f=0.125; ratio=1.8; Δv₂ = 2000×0.588 = 1,176 m/s
  4. Team B total: Δv = 886 + 1176 = 2,062 m/s
AnswerTeam A: 1,022 m/s | Team B: 2,062 m/s — staging delivers 2× the speed with the same fuel!
Student Exercises 80 Practice Problems

Try to solve each problem without looking at examples first. Show all work in the space provided. Answers are in the Answer Key at the end of this section. Use g = 9.8 m/s² and GM_Earth = 3.986 × 10¹⁴ m³/s² unless told otherwise.

Unit 1 — Forces & Newton's Laws Problems 1–20
1

A rocket of mass 0.5 kg is pushed with a force of 15 N. What is its acceleration?

F = 15 N | m = 0.5 kg | Find: a
WORK SPACE
2

What force is needed to accelerate a 3 kg rocket at 12 m/s²?

m = 3 kg | a = 12 m/s² | Find: F
WORK SPACE
3

A force of 90 N produces an acceleration of 6 m/s². What is the rocket's mass?

F = 90 N | a = 6 m/s² | Find: m
WORK SPACE
4

A rocket has thrust 60 N, gravity force 8 N, and drag 3 N. Find the net force and acceleration if m = 0.5 kg.

F_thrust=60N | F_grav=8N | F_drag=3N | m=0.5kg
WORK SPACE
5

A model rocket sits on the pad. Its weight is 4.9 N and the pad provides 4.9 N upward. What is ΣF and what does the First Law say about motion?

ΣF = ? | What happens to the rocket?
WORK SPACE
6

Rocket engine fires and pushes exhaust backward with 35 N. By Newton's Third Law, what is the thrust force on the rocket, and in which direction?

F_exhaust = 35 N backward | Find: F_rocket
WORK SPACE
7

Two identical rockets (m = 1 kg each) push apart with equal forces of 20 N. What is each rocket's acceleration?

F = 20 N each | m = 1 kg each | Find: a
WORK SPACE
8

What net force is required to give a 0.25 kg rocket an acceleration of 100 m/s²?

m = 0.25 kg | a = 100 m/s² | Find: F_net
WORK SPACE
9

A rocket with thrust 20 N and weight 22 N and no drag. Will it lift off? What is F_net?

F_thrust=20N | F_grav=22N | F_drag=0
WORK SPACE
10

In the vacuum of deep space, a 200 kg rocket traveling at 500 m/s has its engine off. What happens to its velocity over the next hour?

ΣF = 0 | v = 500 m/s | Engine off
WORK SPACE
11

Find the thrust needed for a 2 kg rocket to accelerate upward at 25 m/s² with 4 N of drag.

m=2kg | a=25m/s² | F_drag=4N
WORK SPACE
12

A rocket engine produces 500 N of action force on the exhaust. What is the reaction force? Does the direction matter — explain.

F_action = 500 N | Find: F_reaction and direction
WORK SPACE
13

Engine A produces 15 N for 0.4 kg. Engine B produces 15 N for 0.8 kg. Compare their accelerations.

F same | Different masses | Find both accelerations
WORK SPACE
14

A 0.6 kg rocket has thrust 18 N, drag 2 N. At what point does the engine need to shut off to keep acceleration below 20 m/s²?

m=0.6kg | Target: a ≤ 20 m/s²
WORK SPACE
15

Write all three forms of Newton's Second Law and state what each one is used to find.

F=? | a=? | m=? | State the formula and its use
WORK SPACE
16

An astronaut (m = 80 kg) pushes off a wall at 1.5 m/s. How much force did the wall exert on the astronaut if the push lasted 0.3 s?

m=80kg | Δv=1.5m/s | Δt=0.3s
WORK SPACE
17

A rocket on Mars (g = 3.7 m/s²) has mass 1 kg and thrust 5 N. Find F_net and acceleration.

g_Mars=3.7m/s² | m=1kg | F_thrust=5N | drag≈0
WORK SPACE
18

Rocket A (m = 0.5 kg) has thrust 40 N. Rocket B (m = 2 kg) has thrust 80 N. Which has greater acceleration?

A: F=40N, m=0.5kg | B: F=80N, m=2kg
WORK SPACE
19

A free body diagram shows: thrust 30 N up, gravity 9.8 N down, drag 1.2 N down. Find F_net and state the direction.

F_up=30N | F_down=9.8+1.2=11N
WORK SPACE
20

A spacecraft coasting in deep space at 12,000 m/s fires a braking thruster producing 500 N for 10 s. Mass = 2,500 kg. Find new velocity.

v₀=12,000m/s | F=−500N | Δt=10s | m=2,500kg
WORK SPACE
Unit 2 — Momentum & Propulsion Problems 21–40
21

Calculate the momentum of a 0.4 kg rocket moving at 55 m/s.

m=0.4kg | v=55m/s | Find: p
WORK SPACE
22

A rocket has momentum 180 kg·m/s and mass 3 kg. Find its velocity.

p=180 kg·m/s | m=3kg | Find: v
WORK SPACE
23

A Class B engine (total impulse = 5 N·s) burns for 0.8 s. What is the average thrust?

J=5N·s | Δt=0.8s | Find: F_avg
WORK SPACE
24

A 0.35 kg rocket starts from rest and a Class C engine (impulse = 10 N·s) fires. Find the final velocity.

m=0.35kg | J=10N·s | v₀=0 | Find: v_final
WORK SPACE
25

Use the Rocket Equation: v_e = 2,500 m/s, m₀ = 8 kg, m_f = 4 kg. Find Δv. (ln 2 = 0.693)

v_e=2500 | m₀/m_f=2 | ln(2)=0.693
WORK SPACE
26

A rocket with TWR = 1.8 and weight 50 N. What is its thrust force?

TWR=1.8 | Weight=50N | Find: F_thrust
WORK SPACE
27

Find the TWR of a rocket: thrust = 120 N, mass = 5 kg.

F=120N | m=5kg | g=9.8 | Find: TWR
WORK SPACE
28

A 1,000 kg spacecraft changes velocity by 300 m/s. Find the impulse needed.

m=1000kg | Δv=300m/s | Find: J
WORK SPACE
29

Using the Rocket Equation with v_e = 3,500 m/s and mass ratio m₀/m_f = 5, find Δv. (ln 5 = 1.609)

v_e=3500 | m₀/m_f=5 | ln(5)=1.609
WORK SPACE
30

Two rockets have the same momentum (p = 600 kg·m/s). Rocket A has m = 30 kg, Rocket B has m = 150 kg. Find each velocity.

p=600 for both | m_A=30kg | m_B=150kg
WORK SPACE
31

What is the minimum thrust needed for a 4 kg rocket to achieve TWR = 1.2?

m=4kg | TWR=1.2 | Find: F_thrust
WORK SPACE
32

An engine fires with 80 N for 3 seconds. How much does it change the momentum of a 4 kg rocket?

F=80N | Δt=3s | m=4kg
WORK SPACE
33

A water rocket expels 0.1 kg of water at 15 m/s. If the rocket mass after water expulsion is 0.3 kg and started from rest, what is its velocity? (Use conservation of momentum.)

m_water=0.1kg | v_water=15m/s | m_rocket=0.3kg | v₀=0
WORK SPACE
34

Rocket Equation: v_e = 4,000 m/s, m₀/m_f = 3. Find Δv. (ln 3 = 1.099)

v_e=4000 | ratio=3 | ln(3)=1.099
WORK SPACE
35

A 2 kg rocket starts at rest and ends at 40 m/s. What impulse did the engine deliver?

m=2kg | v₀=0 | v_f=40m/s
WORK SPACE
36

Calculate the TWR of the Apollo Lunar Module on the Moon: thrust = 45,040 N, mass = 15,103 kg, g_moon = 1.6 m/s².

F=45040N | m=15103kg | g=1.6m/s²
WORK SPACE
37

A 5 N thruster fires for 20 seconds. What is the total impulse, and how much does it change the velocity of a 250 kg spacecraft?

F=5N | Δt=20s | m=250kg
WORK SPACE
38

Why does a rocket work in the vacuum of space where there is no air to push against? Answer using momentum conservation.

Conceptual — use p = mv and conservation of momentum
WORK SPACE
39

A rocket (m = 0.6 kg) has v_e = 1,800 m/s and m₀/m_f = 2. Find Δv. Is TWR > 1 if thrust = 30 N? (ln 2 = 0.693)

v_e=1800 | ratio=2 | m=0.6kg | F=30N
WORK SPACE
40

A two-stage rocket: Stage 1 Δv = 3,200 m/s, Stage 2 Δv = 4,100 m/s. Does it have enough Δv for orbit (need 9,400 m/s)? By how much does it fall short or exceed?

Δv₁=3200 | Δv₂=4100 | Needed=9400
WORK SPACE
Unit 3 — Gravity, Weight & Energy Problems 41–56
41

What is the weight of a 2.5 kg rocket on Earth?

m=2.5kg | g=9.8m/s²
WORK SPACE
42

A spacecraft weighs 600 N on Earth. What is its mass? What does it weigh on Mars (g = 3.7 m/s²)?

W_Earth=600N | g_Mars=3.7m/s²
WORK SPACE
43

Find the kinetic energy of a 0.4 kg rocket at 80 m/s.

m=0.4kg | v=80m/s
WORK SPACE
44

What speed does a 0.5 kg rocket need to have kinetic energy of 5,000 J?

KE=5000J | m=0.5kg | Find: v
WORK SPACE
45

Find the potential energy of a 0.35 kg rocket at 120 m altitude.

m=0.35kg | h=120m | g=9.8m/s²
WORK SPACE
46

A rocket has PE = 2,940 J at its peak. If m = 0.6 kg, what is the maximum altitude?

PE=2940J | m=0.6kg | Find: h
WORK SPACE
47

Using energy conservation: a 0.3 kg rocket reaches 80 m/s at engine cutoff. What maximum altitude does it coast to? (Ignore drag.)

v=80m/s | m=0.3kg | KE→PE | Find: h
WORK SPACE
48

How much more KE does a rocket have at 100 m/s vs 50 m/s? (m = 0.4 kg both cases) — express as a ratio.

v₁=50m/s | v₂=100m/s | m=0.4kg
WORK SPACE
49

Find gravitational force between Earth (5.97×10²⁴ kg) and a 100 kg satellite at altitude 300 km (r = 6.671×10⁶ m). G = 6.674×10⁻¹¹.

m₂=100kg | r=6.671×10⁶m
WORK SPACE
50

If a rocket has KE = 12,800 J and m = 0.8 kg, what is its speed?

KE=12800J | m=0.8kg | Find: v
WORK SPACE
51

What is the weight of a 70 kg astronaut on the Moon (g = 1.6 m/s²) and on Mars (g = 3.7 m/s²)?

m=70kg | g_Moon=1.6 | g_Mars=3.7
WORK SPACE
52

A rocket falls from 200 m with no parachute. What is its impact speed? (Start from rest, ignore drag.)

h=200m | v₀=0 | Use KE=PE
WORK SPACE
53

At 4× Earth's radius from Earth's center, gravity is reduced by what factor compared to the surface? What is the new g?

r = 4 × r_surface | g_surface=9.8m/s²
WORK SPACE
54

A 5 kg payload is lifted to 500 m. Calculate the work done against gravity (= PE gained).

m=5kg | h=500m | g=9.8m/s²
WORK SPACE
55

A rocket (m = 0.5 kg) is launched straight up at 60 m/s. Find: (a) KE at launch, (b) PE at peak altitude, (c) peak altitude.

m=0.5kg | v=60m/s | Energy conservation
WORK SPACE
56

Using F = G×m₁×m₂/r², if the distance between two objects triples, what happens to the gravitational force between them?

Conceptual: r → 3r | What happens to F?
WORK SPACE
Unit 4 — Aerodynamics & Flight Mechanics Problems 57–70
57

Find the drag force on a rocket: ρ = 1.2 kg/m³, v = 40 m/s, C_d = 0.45, A = 0.0008 m².

F_drag = ½ρv²C_dA
WORK SPACE
58

A rocket starts from rest and accelerates at 35 m/s² for 3 seconds. Find final velocity.

v₀=0 | a=35m/s² | t=3s
WORK SPACE
59

A rocket reaches 105 m/s then engine cuts off. It decelerates at 9.8 m/s² (gravity). How long until it stops at peak altitude?

v₀=105m/s | a=−9.8m/s² | v_f=0
WORK SPACE
60

Using d = v₀t + ½at², find how far a rocket travels from rest with a = 50 m/s² in 2.5 s.

v₀=0 | a=50m/s² | t=2.5s
WORK SPACE
61

Drag doubles when speed doubles — True or False? Justify with the drag formula.

F_drag = ½ρv²C_dA | Consider v vs 2v
WORK SPACE
62

A rocket has CG at 28 cm from nose, CP at 36 cm from nose, diameter = 4 cm. Is it stable? Calculate the stability margin in calibers.

CG=28cm | CP=36cm | d=4cm
WORK SPACE
63

A rocket is launched and accelerates at 60 m/s² for 1 s. It then coasts (a = −9.8 m/s²). Find altitude at engine cutoff and maximum altitude.

a_burn=60m/s² | t_burn=1s | a_coast=−9.8m/s²
WORK SPACE
64

How do you experimentally find the Center of Gravity of a model rocket? What tool do you need?

Conceptual / lab procedure question
WORK SPACE
65

At altitude where ρ = 0.5 kg/m³, what is drag on the same rocket from problem 57 at the same speed?

ρ=0.5 | v=40m/s | C_d=0.45 | A=0.0008m²
WORK SPACE
66

A rocket (m = 0.4 kg) falls freely from peak for 4 s. Find: (a) speed at end of fall, (b) distance fallen.

v₀=0 | a=9.8m/s² | t=4s
WORK SPACE
67

CG is at 20 cm from nose, CP is at 17 cm from nose, diameter = 5 cm. Is this rocket stable or unstable? What should be done to fix it?

CG=20cm | CP=17cm | d=5cm
WORK SPACE
68

Find the acceleration of a rocket given v₀ = 10 m/s, v_f = 70 m/s, and t = 4 s.

v₀=10 | v_f=70 | t=4s | Find: a
WORK SPACE
69

Why does drag decrease as a rocket climbs into higher altitudes? Which term in the drag equation explains this?

F_drag = ½ρv²C_dA — which variable changes with altitude?
WORK SPACE
70

A rocket (m = 0.3 kg) is dropped from 300 m with a parachute that opens immediately, giving terminal velocity of 5 m/s. How long does it take to reach the ground?

h=300m | v_terminal=5m/s (constant) | Find: t
WORK SPACE
Unit 5 — Orbital Mechanics & Space Flight Problems 71–80
71

Find the orbital speed of a satellite at 600 km altitude (r = 6.971 × 10⁶ m).

GM=3.986×10¹⁴ | r=6.971×10⁶m
WORK SPACE
72

Find the escape velocity from the Moon's surface: GM_moon = 4.90 × 10¹², r_moon = 1.737 × 10⁶ m.

v_esc = √(2GM/r)
WORK SPACE
73

A satellite has orbital period T = 5,400 s (90 minutes). Using Kepler's Third Law, find its orbital radius r. (4π²/GM = 9.901×10⁻¹⁴)

T=5400s | r³ = GM(T/2π)²
WORK SPACE
74

If orbital speed at 400 km = 7,673 m/s, what is escape velocity from the same altitude? (Use √2 × v_orbit.)

v_orbit=7673m/s | √2=1.4142
WORK SPACE
75

A two-stage rocket has Stage 1 Δv = 4,500 m/s and Stage 2 Δv = 5,200 m/s. What is total Δv? Can it reach orbit AND have fuel reserve?

Δv₁=4500 | Δv₂=5200 | Orbital need=9400
WORK SPACE
76

Why do satellites at higher orbits move slower than satellites at lower orbits? Use the orbital velocity formula to explain.

v_orbit = √(GM/r) — what happens as r increases?
WORK SPACE
77

Find the period of a satellite at r = 8.0 × 10⁶ m from Earth's center. T = 2π√(r³/GM).

r=8.0×10⁶m | GM=3.986×10¹⁴
WORK SPACE
78

Earth's orbital speed around the Sun is 29,780 m/s. What is the escape velocity from Earth's orbit around the Sun? (Use √2 × v_orbit.)

v_orbit=29780m/s | √2=1.4142
WORK SPACE
79

A three-stage rocket: Δv₁ = 2,800 m/s, Δv₂ = 3,100 m/s, Δv₃ = 3,500 m/s. Does it reach orbital speed? By how much does it exceed or fall short?

Sum the stages | Orbital need = 9,400 m/s
WORK SPACE
80

In your own words: explain why astronauts on the ISS feel "weightless" even though gravity at 400 km altitude is still 8.67 m/s².

Conceptual — relate to orbital free-fall and Newton's First Law
WORK SPACE
Answer Key Problems 1–80

Compare your answers below. If you got a different answer, go back and check each step. Remember: always include units in your final answer!

1
a = 30 m/s²
a = F/m = 15/0.5
2
F = 36 N
F = ma = 3×12
3
m = 15 kg
m = F/a = 90/6
4
F_net = 49 N; a = 98 m/s²
F_net = 60−8−3; a = 49/0.5
5
ΣF = 0; rocket stays at rest (1st Law)
4.9 − 4.9 = 0 N
6
35 N forward (upward thrust)
3rd Law: equal and opposite
7
a = 20 m/s² each (opposite directions)
a = F/m = 20/1
8
F_net = 25 N
F = 0.25×100
9
F_net = −2 N; will NOT lift off
20−22 = −2 N downward
10
Velocity stays at 500 m/s (1st Law)
ΣF=0 in space → no change
11
F_thrust = 73.6 N
F = ma+mg+drag = 50+19.6+4
12
500 N forward; YES direction is opposite
3rd Law: equal magnitude, opposite direction
13
A: 37.5 m/s² | B: 18.75 m/s²
a = F/m both cases
14
Engine already gives a = (18−5.88−2)/0.6 = 16.87 m/s² — already under 20 m/s²
F_net = 18−5.88−2 = 10.12 N
15
F=ma (find force); a=F/m (find accel.); m=F/a (find mass)
Three algebraic rearrangements
16
F = 400 N
F = mΔv/Δt = 80×1.5/0.3
17
F_net = 1.3 N; a = 1.3 m/s²
F_net = 5 − (1×3.7) = 1.3 N
18
A: 80 m/s² | B: 40 m/s² — A wins!
A=40/0.5; B=80/2
19
F_net = 19 N upward
30 − 9.8 − 1.2 = 19 N
20
v = 12,000 − 2 = 11,998 m/s
Δv=FΔt/m = 500×10/2500=2 m/s decel.
21
p = 22 kg·m/s
p = 0.4×55
22
v = 60 m/s
v = p/m = 180/3
23
F_avg = 6.25 N
F = J/Δt = 5/0.8
24
v = 28.57 m/s
Δv = J/m = 10/0.35
25
Δv = 1,732.5 m/s
2500×0.693
26
F_thrust = 90 N
F = TWR × Weight = 1.8×50
27
TWR = 2.45
120/(5×9.8) = 120/49
28
J = 300,000 N·s
J = mΔv = 1000×300
29
Δv = 5,631.5 m/s
3500×1.609
30
v_A = 20 m/s | v_B = 4 m/s
v = p/m each
31
F = 47.04 N
F = TWR × mg = 1.2×4×9.8
32
Δp = 240 kg·m/s; Δv = 60 m/s
J = 80×3; Δv = J/m
33
v_rocket = 5 m/s forward
p_total=0; 0.1×15 = 0.3×v; v=5 m/s
34
Δv = 4,396 m/s
4000×1.099
35
J = 80 N·s
J = mΔv = 2×40
36
TWR ≈ 1.87
45040/(15103×1.6) = 45040/24165
37
J=100N·s; Δv=0.4 m/s
J=5×20; Δv=100/250
38
Exhaust momentum backward = rocket momentum forward. No air needed — total momentum stays zero.
Conservation of momentum
39
Δv=1,247 m/s; TWR=30/(0.6×9.8)=5.10 ✓
1800×0.693; TWR>1
40
Δv_total=7,300 m/s — short by 2,100 m/s
3200+4100=7300 < 9400
41
W = 24.5 N
2.5×9.8
42
m = 61.2 kg; W_Mars = 226 N
m=600/9.8; W=61.2×3.7
43
KE = 1,280 J
½×0.4×6400
44
v = 141.4 m/s
v=√(2×5000/0.5)=√20000
45
PE = 411.6 J
0.35×9.8×120
46
h = 500 m
h = 2940/(0.6×9.8)
47
h = 326.5 m
KE=½×0.3×6400=960J; h=960/(0.3×9.8)
48
KE₁=400J; KE₂=1600J; Ratio = 4× (speed doubled → 4× KE)
v² effect; ½×0.4×2500 vs ½×0.4×10000
49
F ≈ 897 N
GM/r² × m = (3.986×10¹⁴/4.45×10¹³)×100
50
v = 179 m/s
v=√(2×12800/0.8)=√32000
51
Moon: 112 N | Mars: 259 N
70×1.6; 70×3.7
52
v = 62.6 m/s
v=√(2×9.8×200)=√3920
53
g is 1/16 of surface = 0.6125 m/s²
r×4 → r²×16 → g÷16
54
PE = 24,500 J
5×9.8×500
55
KE=900J; PE=900J; h=183.7m
½×0.5×3600; h=900/(0.5×9.8)
56
F reduces to 1/9 of original
F∝1/r²; r×3 → F÷9
57
F_drag = 0.3456 N
½×1.2×1600×0.45×0.0008
58
v = 105 m/s
v = 0+35×3
59
t = 10.71 s
t = 105/9.8
60
d = 156.25 m
½×50×6.25
61
FALSE — drag quadruples when speed doubles (v² in formula)
F∝v²; (2v)²=4v²
62
Stable; margin = 2.0 calibers
(36−28)/4 = 8/4 = 2.0 ≥ 1.0 ✓
63
Altitude at cutoff: 30m; max altitude: 214m
d₁=½×60×1; v_cutoff=60m/s; coast up via v²/2g
64
Balance rocket horizontally on fingertip — the balance point IS the CG.
No tools needed beyond a finger!
65
F_drag = 0.144 N (58% less than sea level)
½×0.5×1600×0.45×0.0008; ρ halved → drag halved
66
v=39.2m/s; d=78.4m
v=9.8×4; d=½×9.8×16
67
UNSTABLE (CP ahead of CG). Add nose weight or larger fins.
CG=20 > CP=17 from nose → flip!
68
a = 15 m/s²
a = (70−10)/4 = 60/4
69
Air density ρ decreases with altitude — drag drops proportionally.
F_drag = ½ρv²C_dA → ρ is the key variable
70
t = 60 s (1 minute)
d = v×t → t = 300/5 = 60s
71
v = 7,558 m/s
√(3.986×10¹⁴/6.971×10⁶)
72
v_esc = 2,375 m/s
√(2×4.90×10¹²/1.737×10⁶)
73
r ≈ 6.62 × 10⁶ m (≈ 251 km altitude)
r³=GM(T/2π)²; cube root
74
v_esc = 10,852 m/s
1.4142×7673
75
Δv=9,700 m/s; exceeds orbital need by 300 m/s ✓
4500+5200=9700; 9700−9400=300 spare
76
v ∝ 1/√r — as r increases, v decreases. Larger orbit → slower speed needed.
v=√(GM/r): r larger → smaller denominator under sqrt → smaller v
77
T ≈ 7,116 s ≈ 118.6 min
r³=5.12×10²⁰; r³/GM=1.284×10⁶; T=2π×√1.284×10⁶
78
v_esc_sun = 42,096 m/s ≈ 42.1 km/s
1.4142×29,780
79
Δv=9,400 m/s — exactly reaches orbital speed!
2800+3100+3500=9400
80
ISS and astronauts are in continuous free-fall (orbit). Everything falls together, so no contact forces — apparent weightlessness despite real gravity of 8.67 m/s².
Free-fall = apparent weightlessness; 1st Law in orbit
Hands-On Labs 3 Rocket Science Projects

Project 1 — Water Bottle Rocket Build & Launch

Topics: Newton's Laws · TWR · Altitude · Stability · Flight Data Recording

Build a pressurized water rocket from a 2-liter soda bottle. This project applies Newton's Second and Third Laws directly — water expelled downward launches the rocket upward. You will predict altitude before launch, record real flight data, and compare predicted vs. actual results.

Materials Needed

2-liter plastic soda bottle (clean, empty)
Cork or rubber stopper (to fit bottle neck)
Bicycle pump with pressure gauge
PVC launch stand or commercial water rocket launcher
Cardstock or balsa wood (for fins)
Modeling clay or metal fishing weights (nose ballast)
Tape, scissors, ruler
Scale (to measure rocket mass)
Clinometer or altitude tracker app (optional)
Water (fill bottle 30–40%)

Pre-Launch Calculations (complete before building)

  1. Measure empty bottle mass (m_dry). Fill 35% with water and measure filled mass (m₀). Calculate fuel mass = m₀ − m_dry.
  2. Estimate average thrust F from fill pressure using: F ≈ (gauge pressure in Pa) × (bottle cross-section area). At 60 PSI (414,000 Pa) with A = 0.003 m²: F ≈ 1,242 N!
  3. Calculate weight: W = m₀ × 9.8. Then find TWR = F / W. Is it greater than 1?
  4. Estimate burn time: the water empties in roughly 0.2–0.5 seconds at high pressure. Use Δt = 0.3 s as an estimate.
  5. Estimate burnout speed: v = F_net/m × Δt. (Simplify: use average mass during burn = (m₀ + m_dry)/2.)
  6. Predict peak altitude: h = v²/(2×9.8). Record your prediction on the data table below.

Build Procedure

  1. Cut 3–4 fins from cardstock (trapezoidal shape, about 10 cm tall). Tape or glue fins evenly around the bottle bottom (neck = nose in this rocket).
  2. Create a nose cone from a cone of cardstock and tape to the bottle bottom (which becomes the "top" in flight).
  3. Check stability: balance rocket on fingertip to find CG. The CG should be above (closer to nose than) the geometric center.
  4. If CG is too low, add clay weight inside the nose cone and re-check balance.
  5. Fill rocket 35% with water, insert stopper/valve, place on launcher.

Launch & Data Collection

  1. Set launch angle to vertical (90°). Clear all people from a 30 m radius. Launch area must be an open field.
  2. Pump to target pressure (40–60 PSI). Record pressure used.
  3. Launch! Observer at distance records time from launch to apex (can use phone slow-motion video).
  4. Measure or estimate actual altitude using the time-of-flight: h ≈ ½ × g × (t_apex)². Compare to your prediction!
  5. Repeat 3 launches varying fill amount (25%, 35%, 50%) and record results.
Launch #Fill %Pressure (PSI)Predicted Height (m)Measured Height (m)% Error
135%____________
225%____________
350%____________
Physics Connections in This Lab

Newton's 3rd Law (water out = rocket up) · F=ma (net force during burn) · KE and PE (energy conversion during flight) · Drag (why actual height < predicted) · Stability (CG vs CP fin design).

⚠ Safety Rules

Never exceed 80 PSI. Never stand in front of the nozzle. Always use goggles. Maintain 10 m minimum distance from launch site during pressurization. Adult supervision required at all times.

Project 2 — Balloon Rocket Momentum Lab

Topics: Momentum · Impulse · Newton's Third Law · Mass vs. Velocity

Use a balloon on a string guide to explore momentum, impulse, and Newton's Third Law in a controlled classroom setting. This experiment lets you vary mass and measure speed directly.

Materials Needed

Long balloons (at least 5)
Fishing line or thin string (5 meters)
Plastic drinking straw
Tape (masking or scotch)
Ruler or measuring tape
Stopwatch or phone timer
Small paper clips (to add mass to straw)
Scale (to weigh balloon + straw)

Setup

  1. Thread fishing line through a straw. Tie line between two chairs at the same height, pulled taut (this is the "track").
  2. Inflate balloon, hold closed without tying. Tape balloon to straw. This is your "rocket."
  3. Start balloon at one end of line. Release and time how long it takes to travel the measured track length.
  4. Calculate speed: v = distance / time.
  5. Weigh the balloon + straw system with a scale and calculate momentum: p = m × v.

Experiments to Run

  1. Vary inflation: Inflate balloon to three different sizes (small, medium, large). Record speed for each. How does more "propellant" affect speed?
  2. Vary mass: Add 1, 2, 3 paper clips to the straw to increase mass without changing propellant. Does more mass mean slower speed? Calculate p = mv — does momentum stay approximately constant?
  3. Direction test: Aim balloon at 45° off horizontal. Measure horizontal and vertical distance. Which direction did the air go? Which direction does the balloon move? (Newton's Third Law illustration.)
TrialInflation SizeAdded Mass (g)Distance (m)Time (s)Speed (m/s)Momentum (kg·m/s)
1Small0____________
2Medium0____________
3Large0____________
4Medium2g____________
5Medium5g____________
Physics Connections

Conservation of momentum · Impulse-momentum theorem · Newton's Third Law · Effect of mass on acceleration (F=ma) · Qualitative Tsiolkovsky: more "propellant" (air) → more Δv.

Project 3 — Stability Testing & Nose Cone Design Challenge

Topics: CG vs. CP · Drag Coefficients · Aerodynamic Shape · F=ma

Design and compare three paper rocket nose cones — flat, conical, and ogive (rounded) — and test which design flies straightest and highest. This directly applies the drag formula and stability rules.

Materials Needed

Paper tubes (toilet paper rolls or rolled cardstock)
Cardstock for nose cones and fins
Modeling clay (nose ballast)
Tape, scissors, ruler, compass
Model rocket engine (Class A or B) + launcher
Scale and meter stick
String and finger (for CG check)

Design & Build Three Rockets

  1. Rocket A — Flat nose: Cap the tube with a flat circular piece of cardstock. Add 3 small fins at base. Measure and record CG. Predict this will have highest drag (C_d ≈ 0.8).
  2. Rocket B — Cone nose: Roll cardstock into a 45° half-angle cone. Attach to tube. Same fins as A. Measure CG. Predict moderate drag (C_d ≈ 0.4).
  3. Rocket C — Ogive nose (rounded): Cut a half-circle of cardstock, curl into a smooth dome shape. Attach. Same fins. Measure CG. Predict lowest drag (C_d ≈ 0.25).
  4. For each rocket: find CG by balancing on fingertip. Find approximate CP at about 65% of body length from nose. Verify CG is ahead of (closer to nose than) CP.
  5. If unstable: add clay to nose until CG moves forward of CP by at least 1 diameter (1 caliber).

Pre-Flight Calculations for Each Rocket

  1. Weigh each rocket (m). Calculate weight W = m × 9.8.
  2. Select same engine class for all three (use Class A: average F ≈ 5 N, burn ≈ 0.5 s).
  3. Calculate net force: F_net = F_thrust − W − F_drag (estimate drag at v = 20 m/s using each rocket's C_d and A = 0.0005 m²).
  4. Predict acceleration and burnout speed for each.
  5. Predict peak altitude and fill in the table.
RocketNose TypeC_dMass (kg)Stability (CG−CP)/dPred. Alt (m)Actual Alt (m)Flew Straight?
AFlat0.80____________Y/N
BConical0.40____________Y/N
COgive0.25____________Y/N

Analysis Questions (answer after flights)

  1. Which rocket flew the highest? Does this match your drag coefficient prediction?
  2. Did any rocket fly in a curved path or tumble? Was its stability margin below 1 caliber?
  3. Calculate the percentage improvement in altitude from Rocket A to Rocket C. Was the difference larger or smaller than you predicted?
  4. Using the drag formula, calculate what the drag force would be on each rocket at 30 m/s. How does this compare to the weight of each rocket?
  5. If you had to design a 4th rocket that flew even higher than Rocket C, what would you change and why?
Physics Connections

Drag formula F_drag = ½ρv²C_dA · Net force and acceleration (F=ma) · Altitude prediction using kinematics · CG vs CP stability analysis · Shape-drag relationship in aerodynamics.

⚠ Safety Rules

All model rocket launches require adult supervision and must follow NAR (National Association of Rocketry) safety code. Use only certified model rocket engines. Launch only in open fields away from people, trees, and power lines. Always use launch controllers with safety keys.

Reference Glossary of Key Terms & Constants
Acceleration a
Rate of change of velocity. Speeding up, slowing down, or changing direction all count as acceleration.
Unit: m/s²
Center of Gravity CG
The point where a rocket's mass is balanced. Found by balancing the rocket on one finger horizontally.
Unit: distance from nose (cm or m)
Center of Pressure CP
The point where aerodynamic forces act on the rocket. Must be behind (below) CG for stable flight.
Unit: distance from nose (cm or m)
Delta-v Δv
The total change in velocity a rocket can achieve by burning all its propellant. The "budget" for reaching orbit or another planet.
Unit: m/s
Drag Coefficient C_d
A dimensionless number representing how aerodynamic a shape is. Pointed rockets have C_d ≈ 0.3; blunt shapes ≈ 0.8+.
Unit: dimensionless (no units)
Drag Force F_drag
Air resistance opposing a rocket's motion. Grows with the square of speed. Decreases with altitude as air thins.
Unit: Newtons (N)
Escape Velocity v_esc
Minimum speed to escape a planet's gravity permanently without further thrust. On Earth ≈ 11.2 km/s.
Unit: m/s
Exhaust Velocity v_e
Speed at which combustion gases exit the rocket nozzle. Higher v_e means more efficient propellant. Chemical rockets: 2,500–4,400 m/s.
Unit: m/s
Force F
A push or pull that can change an object's motion. Forces cause acceleration when they are unbalanced (net force ≠ 0).
Unit: Newtons (N) = kg·m/s²
Free Fall
Motion under gravity alone, with no other forces. Orbiting astronauts are in continuous free fall — they fall around Earth without hitting it.
Condition: only gravity acts
Gravitational Constant G
Universal constant in Newton's gravity law. Same everywhere in the universe.
G = 6.674 × 10⁻¹¹ N·m²/kg²
Impulse J
Force multiplied by the time it acts. Equals the change in momentum. Model rocket engines are rated in Newton-seconds (N·s).
Unit: N·s = kg·m/s
Inertia
The tendency of an object to resist changes in its motion. More mass = more inertia. Described by Newton's First Law.
Related to mass (kg)
Kinetic Energy KE
Energy of motion. KE = ½mv². Quadruples when speed doubles. The energy needed to reach orbital speed is enormous.
Unit: Joules (J)
Mass m
The amount of matter in an object. Does NOT change with location (unlike weight). More mass = harder to accelerate.
Unit: kilograms (kg)
Momentum p
Mass × velocity. A measure of how hard it is to stop a moving object. Conserved in closed systems — the basis of how rockets work.
Unit: kg·m/s
Net Force F_net
The vector sum of all forces on an object. If F_net = 0, motion doesn't change. If F_net ≠ 0, acceleration occurs.
Unit: Newtons (N)
Orbital Period T
Time for one complete orbit. The ISS takes 92 minutes; the Moon takes 27.5 days; GPS satellites take 12 hours.
Unit: seconds (s) or hours
Orbital Velocity v_orbit
Speed needed to orbit at a given altitude. At ISS altitude: 7,673 m/s. Higher orbit = slower speed needed.
Unit: m/s
Potential Energy PE
Stored gravitational energy based on height. PE = mgh. Converts to kinetic energy on descent.
Unit: Joules (J)
Propellant
The fuel and oxidizer burned in a rocket engine. The mass of propellant ejected creates thrust via Newton's Third Law.
Measured in kg or % of total mass
Staging
Dropping empty rocket stages during flight to reduce mass. Allows higher total delta-v from the same amount of propellant.
Each stage adds its Δv to total
Terminal Velocity
Constant falling speed when drag = gravity (F_net = 0). A rocket under a good parachute falls at low terminal velocity.
Unit: m/s (constant speed)
Thrust F_thrust
The force produced by a rocket engine pushing exhaust backward. Reaction to exhaust momentum by Newton's Third Law.
Unit: Newtons (N)
Thrust-to-Weight Ratio TWR
Thrust divided by rocket weight. Must be greater than 1.0 to lift off. Increases as fuel burns off during flight.
Dimensionless ratio (no units)
Tsiolkovsky Equation
Δv = v_e × ln(m₀/m_f). The fundamental equation of rocketry relating speed change to propellant mass and exhaust speed.
Δv in m/s
Velocity v
Speed with direction included. Changing direction (like turning) is a change in velocity even if speed is constant.
Unit: m/s
Weight W
Gravitational force on a mass. W = mg. Changes depending on location (less on Moon, Mars). Different from mass!
Unit: Newtons (N)
g (Earth surface)
Gravitational acceleration at Earth's surface. Every second of free fall, speed increases by 9.8 m/s.
g = 9.8 m/s²
GM (Earth)
Earth's gravitational parameter — used in orbital calculations. Combines G and Earth's mass into one convenient constant.
GM = 3.986 × 10¹⁴ m³/s²
Quick Formula Reference Card — All 20 Topics
1.1 First Law
If ΣF = 0 → a = 0
No net force = no acceleration
1.2 Second Law
F = m × a
Also: a=F/m · m=F/a
1.3 Third Law
F_action = −F_reaction
Equal magnitude, opposite direction
1.4 Net Force
F_net = F_T − F_g − F_d
Thrust minus gravity minus drag
2.1 Momentum
p = m × v
Unit: kg·m/s
2.2 Impulse
J = F × Δt = Δp
Unit: N·s
2.3 Rocket Equation
Δv = v_e × ln(m₀/m_f)
ln(2)=0.693, ln(3)=1.099, ln(5)=1.609
2.4 TWR
TWR = F / (m × g)
Must be >1.0 to lift off
3.1 Weight
W = m × g
g=9.8 Earth, 1.6 Moon, 3.7 Mars
3.2 Kinetic Energy
KE = ½ × m × v²
Unit: Joules (J)
3.3 Potential Energy
PE = m × g × h
Unit: Joules (J)
3.4 Gravitation
F = G × m₁m₂ / r²
G = 6.674×10⁻¹¹
4.1 Drag
F_d = ½ρv²C_dA
ρ_sea level = 1.2 kg/m³
4.2 Velocity
v = v₀ + a × t
Final speed from initial + accel × time
4.3 Displacement
d = v₀t + ½at²
Distance during accelerated flight
4.4 Stability
(CG − CP) / d ≥ 1.0
CG must be closer to nose than CP
5.1 Orbital Velocity
v_orb = √(GM/r)
GM_Earth = 3.986×10¹⁴
5.2 Escape Velocity
v_esc = √(2GM/r)
= √2 × v_orbit at same r
5.3 Kepler's 3rd
T = 2π√(r³/GM)
Orbital period from radius
5.4 Multi-Stage
Δv_total = Σ Δv_i
Sum each stage's rocket equation
Rocket Science Physics · Part 2 · 60+ Examples · 80 Practice Problems · 3 Lab Projects · 30-Term Glossary