Stopping a moving object changes its momentum, and that takes a force acting for some time. Day 1 gives you the rule and a way to estimate how long a stop takes. Day 2 shows why a bounce is harder than a stop, then asks how much cushioning an egg really needs.
🎯 Goals
Use the impulse-momentum theorem, F·Δt = Δp, to find a force, a time, or a change in velocity.
Estimate a stopping time and an average force from a stopping distance.
Compare a ball that bounces with one that sticks, and explain the difference in impulse.
Explain why a longer stop means a smaller force, and apply it to an egg.
🧰 Materials
🔩Day 1: a small steel ball or heavy marble (about 15–70 g)
🥣Day 1: three pans: a modeling-clay slab at least 3 cm thick, packed sand, and loose flour
🏀Day 2: a rubber or tennis ball, a thick foam pad or carpet piece, and the clay slab
📏Meter stick, a skewer or ruler for measuring depth, and masking tape
⚖️A scale that reads in grams
📱Optional: a phone with slow-motion video, for reading rebound heights
🧮Calculator
📓Physics notebook
DAY 1Impulse and the crater lab
Section 1 of 4
Impulse: Force Over Time
1
Read the rule.
A net force changes an object’s momentum, and what matters is the force times how long it acts. That product is called impulse.
J = F × Δt (impulse, in N·s) F × Δt = Δp = m(v_f − v_i) (the impulse-momentum theorem) Since 1 N·s = 1 kg·m/s, impulse and momentum are measured in the same unit.
2
Warm up with the theorem.
Take the direction of the initial motion as positive, and give signs to forces and changes in momentum. A correct answer is within 2%.
Impulse Warm-Up
Use Δp = m(v_f − v_i) and F = Δp ÷ Δt.
A 0.50 kg ball moves at +6.0 m/s and is brought to rest. What is its change in momentum?
A 1,000 kg car brakes from +20 m/s to rest in 4.0 s. What is the average force on it?F = Δp ÷ Δt
The same car hits a wall and stops in 0.10 s instead. What is the average force now?
A 2.0 N force acts on a cart for 3.0 s. What is the impulse?J = F × Δt
The 2.0 kg cart started from rest. What is its speed after that impulse?J = Δp = m(v_f − 0)
Section 2 of 4
Crater Lab: Estimating Stop Time and Force
3
Set up three landing pans.
Mass the ball in kilograms. Smooth each surface flat: a modeling-clay slab, packed sand, and loose flour. Mark a drop height of 1.00 m, measured from the bottom of the ball to the surface.
4
Drop the ball into each surface three times and measure the crater depth.
Release the ball from rest and don’t throw it. Use a fresh spot each time. After each drop, measure how far the ball sank below the original surface, then average your three depths for each material.
⚠️ Drop over the pan and keep faces clear. Flour makes the floor slippery, so wipe up spills.
5
Estimate the stopping time and the average force.
The impact speed is v = √(2gh). While the ball slows from v to 0 over the crater depth d, its average speed is v ÷ 2, so the stop takes Δt = 2d ÷ v. Then the average force is F = mv ÷ Δt.
Sample data (use only if you can’t run the lab)
Steel ball, mass 0.066 kg, dropped from 1.00 m. Crater depth: clay 0.008 m, sand 0.015 m, flour 0.025 m.
Surface
Ball mass m (kg)
Impact speed v = √(2gh) (m/s)
Crater depth d (m)
Stop time Δt = 2d ÷ v (s)
Average force F = mv ÷ Δt (N)
Crater Checker
Work a row by hand first. Then enter the ball’s mass, the drop height, and the crater depth to check it.
📓 Physics Notebook
Which surface gave the longest stop and which gave the largest force? Describe the pattern between crater depth and force. The ball’s momentum at impact was the same every time, so what changed?
✅ End of Day 1 check: the warm-up is all green, the crater table is complete, and one row is verified with the Crater Checker.
DAY 2Bounce, force-time, and the egg
Section 3 of 4
Bounce or Stick?
6
Drop a bouncy ball onto three surfaces.
Mass the ball in kilograms. Drop it from h₁ = 1.00 m onto the hard floor, then onto a thick foam pad or carpet, then onto the clay slab, where it should stick. Read how high it rebounds (h₂) against a meter stick, or from slow-motion video. Do three drops per surface and average h₂. On the clay, h₂ = 0.
7
Calculate the change in momentum for each surface.
Take up as positive. The ball arrives moving down, so its velocity is −v_before, and it leaves moving up at +v_after. That makes Δp = m(v_after + v_before), with each speed found from v = √(2gh). A ball that sticks has v_after = 0.
Sample data (use only if you can’t run the lab)
Ball mass 0.050 kg, dropped from 1.00 m. Rebound: hard floor 0.62 m, foam pad 0.30 m, clay 0 m.
Surface
Drop height h₁ (m)
Rebound height h₂ (m)
v before = √(2gh₁) (m/s)
v after = √(2gh₂) (m/s)
Δp = m(v_after + v_before) (kg·m/s)
Bounce Checker
Work a row by hand first. Then enter it to check your speeds and your change in momentum.
📓 Physics Notebook
Which surface needed the biggest impulse, and which needed the smallest? Explain why a ball that bounces back needs more impulse than one that simply stops. If the contact time were the same, which would feel the larger force?
Section 4 of 4
Same Impulse, Different Force
8
Explore a force-time graph.
Drag the sliders. The area of each box is the impulse, which is the egg’s momentum at impact. Find a crush distance that gets the force under the 40 N limit at 3.0 m.
Same Impulse, Different Force
A 0.058 kg egg is dropped and stops in the crush distance you choose. The crimson box is a hard stop of just 0.5 cm. The cyan box is yours. Both have the same area, because both are the same impulse. The dashed line is a rough 40 N limit for an egg.
v = 7.67 m/s · p = 0.445 kg·m/s Δt = 2d ÷ v = 5.2 ms · F = p ÷ Δt = 85 N
Over the 40 N limit. At this height the egg needs at least 4.3 cm of crush distance.
9
Work out what an egg needs, by hand.
A large egg has a mass of about 0.058 kg. Treat 40 N as a rough limit for the average force it can take. This is the calculation you’ll adapt for Project 5-1-4.
Egg Drop Preview
Drop height 3.0 m. Use g = 9.8 N/kg.
A 0.058 kg egg is dropped from 3.0 m. What is its speed just before impact?v = √(2gh), g = 9.8 N/kg
What is the egg’s momentum at impact?
To keep the average force at or below 40 N, how long must the stop take at least?Δt = Δp ÷ F
How far must the egg’s cushioning let it travel while stopping, at least?d = v × Δt ÷ 2
📓 Physics Notebook
A padded car bumper and a rigid steel bumper bring a car to rest from the same speed in the same collision. Use F·Δt = Δp to explain why the padded one protects you more. Which quantity is the same for both, and which one changes?
Question 1
Using your crater data, which surface produced the largest force and why? Describe the relationship between crater depth, stop time, and force.
Question 2
Explain in words why F = Δp ÷ Δt means that stretching out the stop reduces the force, even though Δp is fixed.
Question 3
Using your bounce data, why is the impulse bigger when the ball bounces than when it sticks? Why does that matter for a car hitting a wall that doesn’t give?
Question 4
Describe a feature that would give an egg container more stopping distance, and use your egg numbers to say about how much it needs to supply.
📤 Turn In
Your Impulse Warm-Up results
Completed crater table, with your notebook note on depth, time, and force
Completed bounce table, with your note on why a bounce needs more impulse
Your Egg Drop Preview results and your padded-vs.-rigid bumper explanation