Docking Clearance: teacher answer key
Public sample teacher key. Includes model answers; open the student worksheet when assigning the activity.
HS-PS2-1: Analyze data to support the claim that Newton's second law of motion describes the mathematical relationship among the net force on a macroscopic object, its mass, and its acceleration.
Using the worksheet
- Worksheet opens the questions. New questions follow each launch; misses add a reflection prompt.
- Recorded automatically: the prediction, what happened, the thrust and mass, and pass or miss.
- My work offers Copy my work and Turn in a PDF. Nothing is submitted automatically. Review written reasoning separately from game success; blank responses remain visible.
- On shared devices, use Clear my work and tap again to confirm. The game's Start over does not clear worksheet answers. If browser storage is blocked, export before leaving.
Numbers like "#4" match the printed worksheet. Questions marked "After a miss" appear in the game only after a missed launch; on paper they are combined into one question near the end.
Check 1: Engine test
- Target: match the drone's 2.0 m/s². Correct thrust = mass × 2.0 m/s².
- Thrust by ship mass: 2,000 kg → 4,000 N; 1,600 → 3,200; 2,400 → 4,800; 3,000 → 6,000; 1,800 → 3,600; 2,200 → 4,400. (The mass changes after each miss.)
- Prediction: the acceleration their own thrust gives, (thrust − opposing force) ÷ mass. With the correct thrust it is 2.0 m/s².
#1. Show your calculation for the thrust: F = m × a, with your numbers. HS-PS2-1
F = m × a = 2,000 kg × 2.0 m/s² = 4,000 N (use the mass on their screen).
| Common wrong answer | What it shows |
|---|---|
| 1,000 N for a 2,000 kg ship (m ÷ a) | Rearranged a = F ÷ m incorrectly. |
| 2 N or 200 N | Treated the acceleration as if it were the force. |
After a miss: A run missed. What went wrong? Use a = Fnet ÷ m in your answer. HS-PS2-1
Example: "Our thrust was too small, so the net force was too small and the acceleration came out under 2.0 m/s². The drone pulled ahead. We needed F = m × a = 4,000 N."
Check 2: Cargo loaded
- Same 2.0 m/s² drone, heavier ship.
- Thrust by ship mass: 5,000 kg → 10,000 N; 4,500 → 9,000; 5,500 → 11,000; 6,000 → 12,000; 4,000 → 8,000.
#2. Show your calculation for the thrust: F = m × a. HS-PS2-1
F = 5,000 kg × 2.0 m/s² = 10,000 N.
| Common wrong answer | What it shows |
|---|---|
| 4,000 N (the check 1 thrust) | Expects the same force to give the same acceleration no matter the mass. |
#3. The drone's acceleration didn't change, but the ship got heavier. How did the thrust you needed change, and why? HS-PS2-1
It needed more thrust (10,000 N instead of 4,000 N). More mass means the same force gives less acceleration (a = F ÷ m), so keeping 2.0 m/s² takes a bigger force. With the acceleration held constant, force is proportional to mass: 2.5 times the mass took 2.5 times the force.
| Common wrong answer | What it shows |
|---|---|
| "It needed the same thrust, since the drone is the same." | Thinks the force sets the motion directly, ignoring mass. |
After a miss: A run missed. What went wrong? Use a = Fnet ÷ m in your answer. HS-PS2-1
Example: "We used the check 1 thrust. With 5,000 kg, 4,000 N only gives 0.80 m/s², so the drone pulled ahead."
Check 3: Speed gate
- The engine burns 5.0 s, then the ship coasts. Needed acceleration: a = v ÷ t = 10.0 ÷ 5.0 = 2.0 m/s².
- Thrust by ship mass: 4,000 kg → 8,000 N; 3,000 → 6,000; 3,500 → 7,000; 4,500 → 9,000; 2,500 → 5,000.
- Prediction with the correct thrust: 10.0 m/s.
#4. What acceleration does the ship need during the 5.0 s burn to reach 10.0 m/s? HS-PS2-1
a = v ÷ t = 10.0 m/s ÷ 5.0 s = 2.0 m/s².
| Common wrong answer | What it shows |
|---|---|
| 50 m/s² | Multiplied speed by time. |
| 0.5 m/s² | Divided time by speed. |
#5. What thrust gives that acceleration? Show F = m × a. HS-PS2-1
F = 4,000 kg × 2.0 m/s² = 8,000 N.
| Common wrong answer | What it shows |
|---|---|
| 40,000 N (m × v) | Used speed instead of acceleration: thinks force sets speed. |
| 200,000 N (m × v × t) | Multiplied by the time instead of dividing. |
#6. After the engine shut off, the ship kept moving at a steady 10.0 m/s. Which idea explains that? HS-PS2-1
With no net force, the velocity stays the same (Newton's first law). Zero net force means zero acceleration (a = Fnet ÷ m).
| Common wrong answer | What it shows |
|---|---|
| "Something must still be pushing" | Motion needs a force (the most common preconception). |
| "The force gets used up" | Impetus idea: force is stored in a moving object. |
| "The exhaust keeps pushing" | Mixes up third-law pairs with an ongoing force. |
After a miss: A run missed. What went wrong? Look at the velocity graph and use a = Fnet ÷ m. HS-PS2-1
Example: "We used m × v. Force sets the acceleration, not the speed, so the ship sped up too fast and crossed the gate over 10 m/s."
Check 4: Stuck thruster
- Net force needed: Fnet = m × 1.5 m/s². Engine thrust = Fnet + stuck thruster force.
- Thrust by mass and stuck thruster: 3,000 kg, 1,500 N → 4,500 + 1,500 = 6,000 N; 2,000 kg, 1,000 N → 4,000 N; 2,400 kg, 2,000 N → 5,600 N; 3,600 kg, 500 N → 5,900 N; 4,000 kg, 1,000 N → 7,000 N.
- Prediction with the correct thrust: 1.5 m/s².
#7. Show your math: the net force the ship needs, then the engine thrust. HS-PS2-1
Fnet = 3,000 kg × 1.5 m/s² = 4,500 N. Engine = 4,500 N + 1,500 N = 6,000 N.
| Common wrong answer | What it shows |
|---|---|
| 4,500 N | Found the net force but forgot the stuck thruster cancels part of the engine's push. |
| 3,000 N (4,500 − 1,500) | Subtracted the opposing force instead of adding it. |
#8. Why does the engine have to push harder than the net force the ship needs? HS-PS2-1
The stuck thruster pushes backward, so the net force is the engine's push minus that force. Only the net force accelerates the ship, so the engine has to supply the net force plus whatever the stuck thruster cancels.
After a miss: A run missed. What went wrong? Use Fnet = F engine − F opposing. HS-PS2-1
Example: "We set 4,500 N, but the stuck thruster took away 1,500 N, so the net force was only 3,000 N and a = 1.0 m/s²."
Check 5: Docking
- Deceleration needed: a = v² ÷ (2d) = 4.0² ÷ (2 × 20.0) = 0.40 m/s².
- Braking force by ship mass: 6,000 kg → 2,400 N; 4,000 → 1,600; 5,000 → 2,000; 7,000 → 2,800; 8,000 → 3,200.
- Prediction with the correct force: 20.0 m.
#9. What deceleration stops the ship from 4.0 m/s in exactly 20.0 m? HS-PS2-1
a = v² ÷ (2d) = 16 ÷ 40 = 0.40 m/s².
| Common wrong answer | What it shows |
|---|---|
| 0.20 m/s² (v ÷ d) | Didn't square the speed. |
| 0.80 m/s² (v² ÷ d) | Left out the 2. |
#10. What braking force gives that deceleration? Show F = m × a. HS-PS2-1
F = 6,000 kg × 0.40 m/s² = 2,400 N.
| Common wrong answer | What it shows |
|---|---|
| 24,000 N (m × v) | Thinks force sets speed. |
| 1,200 N or 4,800 N | Carried the v ÷ d or missing-2 error into the force. |
After a miss: A run missed. What went wrong? Use a = v² ÷ (2d) and F = m × a. HS-PS2-1
Example: "We forgot the 2, so we braked twice as hard as needed and stopped 10 m short."
Bonus: Mystery cargo
- Mass from the test burn: m = Fnet ÷ a, rounded to the nearest 500 kg (masses run 2,500 to 8,000 kg). Thrust = m × 2.0 m/s².
#11. How did your test burn tell you the hidden mass? Show your math. HS-PS2-1
m = Fnet ÷ a. Example: a 10,000 N test burn gave 2.5 m/s², so m = 10,000 ÷ 2.5 = 4,000 kg. Then F = 4,000 kg × 2.0 m/s² = 8,000 N.
When a launch missed (#12)
Any of the After a miss answers above. Look for the rule named correctly and the specific number that was off.
Wrap-up: Newton's second law
#13. How are net force, mass, and acceleration related? Use numbers from at least two flight checks as your evidence. HS-PS2-1
Claim: Acceleration equals net force divided by mass (a = Fnet ÷ m). More net force gives more acceleration, and more mass gives less acceleration for the same force. Evidence: In check 1, 4,000 N on a 2,000 kg ship gave 2.0 m/s². In check 2, a 5,000 kg ship needed 10,000 N for the same 2.0 m/s², so 2.5 times the mass took 2.5 times the force. In check 4, the engine pushed 6,000 N but only the 4,500 N net force accelerated the 3,000 kg ship, at 1.5 m/s². Reasoning: In every check, net force ÷ mass matched the acceleration we measured, so Newton's second law describes the relationship. Opposing forces subtract, so only the net force counts.