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First Light AcademyHELIOS PROVING GROUNDS / FLIGHT SCHOOL

FLIGHT ANALYSIS / PHYSICS CLUSTERS

Three cases. One justified decision at a time.

Three independent cases from Empires of a Shattered Sky

Use the evidence in each case to make a justified decision. Show an equation before substitution, include units, and explain your reasoning. Use a calculator in degree mode. Keep unrounded values until the final answer. Each cluster is worth 20 points.

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CLUSTER 01 / 20 POINTS

Across the Breach

Scene and supplied evidence

A launcher at (0, 0) releases a capsule at 20.0 m/s. Available angles are 30, 45 and 60 degrees. A 7.00 m beam stands at x = 18.0 m. The horizontal net extends from x = 34.0 to 37.0 m at y = 0. Gravity acts down at 9.81 m/s².
Scene 1 · Questions 1.1–1.8. Open the diagram for a larger view.

A damaged service bridge has isolated a maintenance annex near the Antarctic Arcology. A Codex Naviga repair crew must send a protected communications module across the gap. Their delivery drone is disabled, so they use a cargo launcher with three angle settings. The capsule accelerates along a straight rail and then moves freely through the air. A raised support beam obstructs the route, and a horizontal receiving net covers a narrow landing zone on the far side. Your task is to choose a launch setting that clears the beam and reaches the net.

Model: Treat the capsule as a particle. Ignore air resistance, wind, and Earth curvature. After release, ax = 0 and ay = -9.81 m/s². The release point is (0, 0); right and up are positive. The receiving net is at y = 0. All angles are measured above horizontal.

Site or launcher measurementValue
Available launch angles30.0°, 45.0°, 60.0°
Speed at release for every setting20.0 m/s
Initial speed on rail0 m/s
Receiving net horizontal extentx = 34.0 m to 37.0 m
Beam location and top heightx = 18.0 m; y = 7.00 m
Graph 1: Rail speed rises linearly from 0 m/s at 0 s to 20.0 m/s at 2.00 s.
Graph 1: Rail speed rises linearly from 0 m/s at 0 s to 20.0 m/s at 2.00 s.

Graph 1. The line is straight. Its time origin is the start of rail acceleration; projectile time begins again at zero at release.

Cluster 1 Motion evidence

1.1. Use Graph 1 to determine the capsule’s acceleration along the rail. Show how the graph supports your value. [2 points]View cluster 1 diagram
1.2. Determine the distance traveled along the rail before release using a kinematics equation. Explain how the area under Graph 1 checks your result. [2 points]View cluster 1 diagram
1.3. For the 60.0° setting, calculate the initial horizontal and vertical velocity components. [2 points]View cluster 1 diagram
1.4. On the axes below, sketch horizontal velocity and vertical velocity versus time for the 60.0° flight. For this item, use vx0 = 10.0 m/s, vy0 = 17.3 m/s, and a total flight time of 3.53 s. Mark when the capsule reaches its highest point. [3 points]View cluster 1 diagram
Blank horizontal-velocity and vertical-velocity axes for item 1.4; time runs from 0 to 4 seconds.
Blank horizontal-velocity and vertical-velocity axes for item 1.4; time runs from 0 to 4 seconds.

Cluster 1 The delivery decision

1.5. Calculate the time to return to y = 0 and the horizontal range for the 60.0° setting. Use vx0 = 10.0 m/s and vy0 = 17.3 m/s if needed. [3 points]View cluster 1 diagram
1.6. Calculate the capsule’s height when it reaches the beam at x = 18.0 m using the 60.0° setting. State whether it clears the 7.00 m beam. [2 points]View cluster 1 diagram

The crew’s model gives these results for the other settings. The predicted range assumes no obstruction.

Launch anglePredicted rangeHeight at x = 18.0 m
30.0°35.3 m5.10 m
45.0°40.8 m10.1 m
1.7. Recommend one of the three launch settings. Use numerical evidence to address both the beam and the receiving net. Explain why each other setting fails. [3 points]View cluster 1 diagram
1.8. A technician says, “At the highest point of the 60° flight, the capsule’s speed and acceleration are both zero.” Evaluate both parts of this claim. State the velocity components and acceleration there. [3 points]View cluster 1 diagram

CLUSTER 02 / 20 POINTS

Catch Without Breaking

Scene and supplied evidence

A 200 kg pod moving right at 6.00 m/s approaches an 800 kg tug initially at rest. Thrusters are off during capture. After capture they move together; a 200 N braking force acts left. Clearance is 4.00 m when braking begins.
Scene 2 · Questions 2.1–2.8. Open the diagram for a larger view.

A cargo pod has broken loose near a Codex Naviga orbital transfer station. A recovery tug waits ahead of it with its capture system aligned through both centers of mass. The pilot can use a rapid locking mechanism or a damped mechanism that gradually brings the tug and pod to a common velocity. Thrusters remain off during capture. After capture, the pilot uses a braking thruster to stop both objects relative to the station. Choose a capture method and evaluate whether the combined system can stop safely.

Model: Use a local frame moving with the station over this short interaction. Motion is one-dimensional; right is positive. Neglect external impulse and differences in gravitational acceleration during capture. Ignore rotation. After capture the tug and pod remain connected. Thruster propellant loss is negligible.

Measured or specified quantityValue
Pod mass and initial velocity200 kg; +6.00 m/s
Tug mass and initial velocity800 kg; 0 m/s
Maximum allowable coupling force1,000 N
Net braking force after capture-200 N
Clear stopping distance at start of braking4.00 m
Graph 2: Rapid-lock triangular force pulse lasts 0.500 s and peaks at 3840 N. Damped pulse lasts 4.00 s and peaks at 480 N. Dashed coupling limit is 1000 N. Force on the pod acts left; the panels have different scales.
Graph 2: Rapid-lock triangular force pulse lasts 0.500 s and peaks at 3840 N. Damped pulse lasts 4.00 s and peaks at 480 N. Dashed coupling limit is 1000 N. Force on the pod acts left; the panels have different scales.

Graph 2. Each pulse is a straight-sided triangle. Both graphs show the magnitude of the force on the pod; its direction is left. The dashed line is the coupling limit. The two panels use different time and force scales.

Cluster 2 Capture analysis

2.1. Before capture, the pod has no propulsion and negligible net force in the local model. Explain why its velocity does not fall to zero. Identify the relevant Newton’s law. [2 points]View cluster 2 diagram
2.2. During capture, compare the force exerted by the tug on the pod with the force exerted by the pod on the tug. Compare their acceleration magnitudes and support the comparison using their masses. [2 points]View cluster 2 diagram
2.3. Use conservation of momentum to calculate the common velocity immediately after capture. State the condition that allows you to conserve the tug-and-pod system’s momentum during this interaction. [3 points]View cluster 2 diagram
2.4. Use the area under either force-time pulse to determine the impulse on the pod. Include its sign in the stated coordinate system. [2 points]View cluster 2 diagram

Cluster 2 Equipment and braking decisions

2.5. Which original capture mechanism stays within the coupling’s force limit? Support your answer with both peak forces and the specified limit. [2 points]View cluster 2 diagram
2.6. Calculate the total kinetic energy before and immediately after capture, then account for the change. Use a common final velocity of +1.20 m/s for this item. [3 points]View cluster 2 diagram
2.7. Starting at +1.20 m/s after capture, calculate the acceleration and stopping distance under the -200 N net braking force. Decide whether the available 4.00 m is sufficient. Show your equations. [3 points]View cluster 2 diagram
2.8. An engineer proposes a symmetric triangular capture pulse lasting 2.00 s while keeping the impulse magnitude at 960 N·s. Calculate the required peak force, check the coupling limit, and explain why the common final velocity would remain the same as for the original captures. [3 points]View cluster 2 diagram

CLUSTER 03 / 20 POINTS

Shutdown Is Not Enough

Scene and supplied evidence

After shutdown, residual heat passes from the reactor assembly through a heat exchanger to a cooler sink. Heat capacity is 12.0 MJ/°C and baseline removal is 0.400 MW. A 500 V source limited to 60.0 A supplies parallel essential services of 8.00 kW and one pump A, B or C. The main generator is unavailable. The fission inset leaves the released-neutron count blank.
Scene 3 · Questions 3.1–3.8. Open the diagram for a larger view.

An equipment fault at an auxiliary fission plant serving the Antarctic Arcology causes an unintended power increase and disrupts normal cooling. The protective shutdown stops the sustained chain reaction. The main generator is unavailable, but the reactor assembly continues to warm. An engineering crew has a backup electrical supply and three auxiliary pumps. You must explain the continuing heat production and choose a pump that removes enough heat without overloading the backup supply.

Science briefing: A critical reactor has a self-sustaining chain reaction; the term does not itself mean overheating. Shutdown makes the chain reaction subcritical. Unstable fission products continue to decay and release energy. The remaining heat must still be transferred to cooler surroundings.

QuantityScenario value
Normal thermal power before the fault20.0 MW
Normal electrical output before the fault6.00 MW
Mean assembly temperature at monitoring time t = 0305 °C
Effective heat capacity C of modeled assembly12.0 MJ/°C
Maximum permitted mean temperature in this scenario320 °C
Graph 3: Heat generation is 1.20 MW from 0 to 120 s, 1.00 MW from 120 to 240 s, and 0.800 MW from 240 to 360 s. Baseline heat removal is 0.400 MW throughout.
Graph 3: Heat generation is 1.20 MW from 0 to 120 s, 1.00 MW from 120 to 240 s, and 0.800 MW from 240 to 360 s. Baseline heat removal is 0.400 MW throughout.

Graph 3. Monitoring begins shortly after shutdown. Use constant heat generation of 1.20 MW for 0–120 s, 1.00 MW for 120–240 s, and 0.800 MW for 240–360 s. Baseline heat removal is 0.400 MW throughout. These are simplified interval averages.

Cluster 3 Nuclear and cooling evidence

Thermal model: Treat the reactor assembly as one uniform-temperature system with constant heat capacity. There is no phase change. The listed heat-removal rates include all modeled cooling and are net of pump heating. Use ΔE = (Pgenerated - Premoved)Δt and ΔT = ΔE/C. Here 1 MW = 1 MJ/s. The heat sink remains cooler than the assembly.

At t = 120 s, exactly one auxiliary pump can start and remain on until t = 360 s. Its additional heat removal adds to the 0.400 MW baseline. Rates are constant over this short interval. The backup source maintains 500 V and can supply at most 60.0 A. Essential controls and services require 8.00 kW continuously. Treat the listed currents as steady operating currents; ignore start-up transients.

PumpSupply voltageCurrentAdditional net thermal removal
A500 V20.0 A0.300 MW
B500 V40.0 A0.700 MW
C500 V70.0 A1.10 MW

Nuclear model data: A uranium-235 nucleus has 92 protons; barium has 56 protons and krypton has 36. A neutron has mass number 1 and charge 0. One possible fission channel is represented below; mass numbers appear after the element names.

neutron + uranium-235 → barium-141 + krypton-92 + ____ neutrons + released energy

3.1. Complete the neutron count in the model and show conservation of total nucleon number. Describe the nuclear change and explain how the released neutrons can sustain a chain reaction. [3 points]View cluster 3 diagram
3.2. Explain why the statement “The reactor is no longer critical, so it cannot continue producing heat” is incorrect. Identify the source of heat after shutdown. [2 points]View cluster 3 diagram

Cluster 3 Energy and temperature predictions

3.3. Use Graph 3 to calculate the net thermal energy added to the assembly from 0 to 120 s with baseline cooling only. Explain what the area between the two lines represents. [2 points]View cluster 3 diagram
3.4. With baseline cooling only, calculate the temperature at each time below. Show your work and determine whether the 320 °C limit is exceeded during the modeled 360 s. [3 points]View cluster 3 diagram
Monitoring timeMean assembly temperature
0 s305 °C
120 s
240 s
360 s
3.5. Before the fault, what percentage of the reactor’s thermal power became electrical output? At steady operation, how much thermal power was not converted to electricity and had to be rejected to the surroundings? Use the normal-operation values, not the post-shutdown graph. [2 points]View cluster 3 diagram

Cluster 3 The cooling decision

3.6. Use P = IV to complete the electrical-power table. Include the essential 8.00 kW load. Show the backup source’s maximum power and identify every option that exceeds it. [3 points]View cluster 3 diagram
PumpPump electrical powerTotal with essential loads
A
B
C
3.7. Choose the pump that meets both requirements from 120 to 360 s: total electrical demand stays within the backup supply’s capacity, and total heat removal is at least the heat-generation rate throughout that interval. Cite numerical evidence for both requirements. [2 points]View cluster 3 diagram
3.8. For a separate check, assume Pump B starts at 120 s when the assembly is at 313 °C. Build an interval-by-interval energy calculation to predict the temperature at 240 s and 360 s. Explain how the result follows from the balance of heat generation and heat removal. [3 points]View cluster 3 diagram

All measurements and equipment specifications are fictional instructional data. Use the scene diagrams, graphs, and stated model together.