Press Next (or Play) to walk through each story one small step at a time. The numbers come last. Tap a round play button to hear a step read aloud. After every three stories there is one quick check.
1. A long push and a short push
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Your science class is doing an egg drop. Two teams drop the same egg from the same balcony. One egg lands on the concrete and splats. The other lands on a thick pillow and survives.
Both eggs hit the ground at the same speed, and both end up stopped. So both lose exactly the same momentum. Why does one break and the other not? And why does a car have airbags, and why do you bend your knees when you jump off a wall?
2. Catching an egg softly
3. Bounce or stick?
Quick check. One question before the next stories.
When you catch a fast ball, you pull your hands back with it. Why does that hurt less?
Another example: Landing on a mat
4. A quick, hard kick
5. Pushing a stalled car
6. Braking a bike
Quick check. One question before the next stories.
A steady 10 N push acts for 3 s on a cart that starts at rest (no friction). How much momentum does the cart gain?
Another example: Same impulse, two ways
7. Check yourself
Think of your answer first, then tap to see it.
a) Why does a gymnast land on a thick, soft mat instead of the hard floor?
Show answer
She loses the same momentum either way. The soft mat makes the stop take longer, so the force on her is smaller (impulse = force × time stays the same).
b) A 0.5 kg ball moving at 4 m/s hits a wall and stops in 0.1 s. What impulse acts on it? What average force?
Show answer
Change in momentum = 0.5 kg × 4 m/s = 2 kg·m/s, so the impulse is 2 N·s (pointing away from the wall). Force = 2 N·s ÷ 0.1 s = 20 N.
c) The same ball bounces straight back at 4 m/s instead. What is the impulse now?
Show answer
Toward the wall is positive. Δp = 0.5 × (−4) − 0.5 × (+4) = −2 − 2 = −4 kg·m/s. The impulse is 4 N·s away from the wall: twice as much as just stopping.
Already know this?
Three quick questions. Get all three right on the first try and you can skip ahead to the simulation.
Why do gym mats make landing from a jump safer?
A 2 kg ball moving right at 3 m/s bounces straight back at 3 m/s. Right is positive. What is its change in momentum?
A steady 20 N force acts on a box for 0.5 s. What is the impulse?
The idea
Picture an egg drop. Two eggs fall from the same balcony. They hit the ground at the same speed.
One lands on concrete. It stops in a split second, and it splats. The other lands on a thick pillow. It sinks in slowly and survives.
Both eggs lose the same motion. The difference is the stopping time. A short stop needs a big push. A long stop needs only a small push.
To change an object's momentum, you must push or pull on it for some time. The push times the time is called the impulse.
J = Favg Δt
Units: newton seconds, N·s, which is the same as kg·m/s. Impulse is a vector: it points the same way as the force.
Real forces in a crash or a kick are not constant. They rise and fall fast. Then the impulse is the area under the force-time graph. (Area above the time axis is positive, area below is negative.)
The big rule is the impulse-momentum theorem: the net impulse on an object equals its change in momentum.
Jnet = Δp = m vf − m vi
Divide both sides by Δt and you get Newton's second law in its momentum form: Fnet = Δp / Δt. The net force is how fast momentum changes. (On a p-t graph, the slope is the net force.)
The egg answer. Both eggs need the same Δp to stop. If the stopping time Δt is 50 times longer on the pillow, the average force is 50 times smaller. Same impulse, spread out over more time, means a gentler force. Airbags, crumple zones, helmets, and bent knees all work this way: they make Δt longer.
Trap: saying the pillow (or an airbag) makes the change in momentum smaller. Instead: Δp is the same, because the egg stops either way; the cushion makes Δt longer, so the average force is smaller.
Bouncing takes more impulse than stopping. To bounce a ball back, the wall must first stop it and then throw it back. If a ball arrives at +v and leaves at −v, then Δp = m(−v) − m(v) = −2mv, twice as much as just stopping.
The same idea with numbers
Read the steps as text
A 0.060 kg egg hits the ground at 5.0 m/s. On concrete it stops in 0.0020 s. On a pillow it stops in 0.10 s. Find the impulse and the average net force in each case. Take up as positive.
Velocities with signs: vi = −5.0 m/s (moving down), vf = 0. Why: up is +, and the egg is falling.
Δp = m vf − m vi = 0 − (0.060 kg)(−5.0 m/s) = +0.30 kg·m/s. Why: change = final minus initial. The change points up, because the ground pushes up.
So the net impulse is J = +0.30 N·s in both cases. Why: impulse-momentum theorem, J = Δp. Same start, same end, same impulse.
Pillow: Favg = 0.30 N·s ÷ 0.10 s = 3.0 N. Why: 50 times the time gives 1/50 of the force.
These are net forces. The ground's push is the net force plus the egg's weight (mg = 0.060 × 9.8 = 0.59 N): about 151 N on concrete and 3.6 N on the pillow. Why: Fnet = Fground − mg, so Fground = Fnet + mg. Here the weight is tiny next to 150 N, but not next to 3.0 N.
Trap: plugging in milliseconds as seconds (2.0 ms used as 2.0 s gives 0.15 N). Instead: change to seconds first: 2.0 ms = 0.0020 s, so F = 0.30 / 0.0020 = 150 N.
Lab 1: a cart hits a wall
The cart has a soft orange bumper. Set the mass, the speed, the contact time Δt and how much it bounces. Predict the impulse, then play. The purple shaded area under the force graph grows during the hit. It always ends equal to the change in momentum you see on the p-t graph.
Try this: keep m, v₀ and the bounce fixed, and change only the contact time from 0.20 s to 0.02 s. The shaded area (the impulse) does not change at all. The peak force gets 10 times bigger. That is why a stiff bumper hurts. Then set the bounce to 0 and to 1 and compare the areas: 1 × p₀ and 2 × p₀.
Lab 2: two carts push on each other
When two carts bump, each pushes the other. Make one cart heavy and one light, predict, then play. Look at the force graph: the two curves are mirror images, so the two impulses are equal and opposite. The momentum one cart loses, the other gains.
Examples
Example 1: a car brakes basic
A 1200 kg car moving at 20 m/s brakes to a stop in 4.0 s. Find the impulse on the car and the average net force. Take the direction of motion as positive.
Show solutionRead the steps as text
Δp = m vf − m vi = 0 − (1200 kg)(20 m/s) = −24 000 kg·m/s. Why: the car loses all its momentum.
Favg = J / Δt = −24 000 ÷ 4.0 = −6000 N, so 6000 N backward. Why: J = FavgΔt. The minus sign means the force points opposite the motion, as braking should.
Example 2: stick or bounce? medium
A 0.50 kg ball hits a wall at 6.0 m/s. In case A it is a clay ball and sticks. In case B it is a rubber ball and bounces straight back at 4.0 m/s. Each hit lasts 0.050 s. Find the impulse and the average force on the ball in each case. Take toward the wall as positive.
Show solutionRead the steps as text
Case A: Δp = 0 − (0.50)(6.0) = −3.0 kg·m/s. So J = −3.0 N·s. Why: the ball goes from +6.0 m/s to 0.
Favg = −3.0 ÷ 0.050 = −60 N (away from the wall). Why: F = J / Δt.
Case B: Δp = (0.50)(−4.0) − (0.50)(6.0) = −2.0 − 3.0 = −5.0 kg·m/s. So J = −5.0 N·s. Why: the final velocity is negative because the ball comes back. Forgetting this sign gives the wrong answer −1.0.
Favg = −5.0 ÷ 0.050 = −100 N. Why: same time, bigger impulse, so bigger force. The bouncing ball pushes the wall harder (Newton's third law: the wall gets +100 N).
Trap: using +4.0 m/s for the bounce: Δp = 0.50(4.0 − 6.0) = −1.0 kg·m/s. Instead: the ball comes back, so vf = −4.0 m/s and Δp = 0.50(−4.0 − 6.0) = −5.0 kg·m/s.
Example 3: reading impulse from a force-time graph AP
A 0.40 kg ball at rest is kicked. The force on it is shown below: it rises steadily from 0 to 400 N in 0.010 s, stays at 400 N for 0.020 s, then falls steadily to 0 in 0.010 s. Find the impulse, the ball's speed afterwards, and the average force.
Show solutionRead the steps as text
Split the area into two triangles and a rectangle. Each triangle: ½ × 0.010 s × 400 N = 2.0 N·s. Rectangle: 0.020 s × 400 N = 8.0 N·s. Why: impulse is the area under the F-t graph.
J = 2.0 + 8.0 + 2.0 = 12 N·s. Why: add the pieces. (Trapezoid formula check: ½ × (0.040 + 0.020) × 400 = 12 N·s.)
Δp = J, so m vf − 0 = 12 → vf = 12 ÷ 0.40 = 30 m/s. Why: the ball starts at rest.
Favg = J / Δt = 12 ÷ 0.040 = 300 N. Why: the average force is the height of a rectangle with the same area over the same time. It is less than the 400 N peak.
Trap: multiplying the peak force by the whole time: 400 N × 0.040 s = 16 N·s. Instead: impulse is the area under the graph, 12 N·s; the average force (300 N) is less than the peak.
Example 4: landing from a jump AP
A 70 kg person jumps off a wall and lands at 4.0 m/s. With stiff legs she stops in 0.020 s. With bent knees she stops in 0.20 s. Find the average force from the ground in each case. Take up as positive.
Show solutionRead the steps as text
Δp = 0 − (70)(−4.0) = +280 kg·m/s. Why: she moves down (negative) and ends at rest.
Ground force = Fnet + mg, with mg = 70 × 9.8 = 686 N. Stiff: about 14 700 N. Bent: about 2090 N. Why: the net force is the ground's push up minus her weight down.
Bending the knees makes the ground force about 7 times smaller (14 700 ÷ 2090). Why: the weight part does not shrink, so the ratio is a bit less than the time ratio of 10.
Practice (AP style)
1. A 0.20 kg ball moves toward a wall at 8.0 m/s and bounces straight back at 6.0 m/s. Taking toward the wall as positive, what impulse does the wall give the ball?
Show answer
J = Δp = (0.20)(−6.0) − (0.20)(8.0) = −1.2 − 1.6 = −2.8 N·s, pointing away from the wall. The answer −0.40 comes from forgetting that the final velocity is negative.
2. In a crash, an airbag brings a driver to rest. Compared with hitting the steering wheel, how does the airbag change the impulse on the driver and the average force on the driver?
Show answer
The driver goes from the same speed to rest either way, so Δp, and therefore the impulse, is the same. The airbag makes Δt longer, so Favg = Δp/Δt is smaller.
3. A 0.30 kg hockey puck is at rest. A stick hits it with a force that rises steadily from 0 to 600 N and falls steadily back to 0, a triangle lasting 0.020 s in total. What is the puck's speed right after the hit? (Ignore friction.)
Show answer
Area of the triangle: J = ½ × 0.020 s × 600 N = 6.0 N·s. Then v = J/m = 6.0 ÷ 0.30 = 20 m/s. Using the peak force for the whole time (600 × 0.020 = 12 N·s) gives the wrong answer 40 m/s.
4. The momentum of a cart increases steadily from 2.0 kg·m/s to 8.0 kg·m/s in 3.0 s. What is the net force on the cart?
Show answer
Fnet = Δp/Δt = (8.0 − 2.0) ÷ 3.0 = 2.0 N. It is the slope of the p-t graph. (2.7 N comes from using 8.0 instead of the change.)
5. Two balls with the same mass hit a wall at the same speed. Ball X sticks to the wall. Ball Y bounces back with half its speed. Which statement is correct?
Show answer
X: |Δp| = mv. Y: |Δp| = m(½v) + mv = 1.5 mv. Reversing direction always needs more impulse than just stopping. Twice as large would need a full-speed bounce.
6. A 2.0 kg cart moves at +3.0 m/s. A constant force of −5.0 N acts on it for 4.0 s. What is the cart's velocity at the end?
Show answer
J = FΔt = (−5.0)(4.0) = −20 N·s. pf = pi + J = (2.0)(3.0) − 20 = −14 kg·m/s. vf = −14 ÷ 2.0 = −7.0 m/s. The cart slows, stops, and speeds up the other way.
7. (Free response, experimental design) You have a cart, a track, a force sensor with a spring bumper fixed to the end of the track, and a motion detector. Describe an experiment to test the claim "the area under the force-time graph equals the cart's change in momentum." Say what you measure, how you use the data, and what result supports the claim.
Show answer
Measure the cart's mass m with a balance.
Give the cart a push toward the force sensor. The motion detector records its velocity just before the hit (vi) and just after it bounces off (vf). The force sensor records F against t during the hit.
Find the area under the F-t graph (the software can integrate, or count squares) to get J.
Compute Δp = m(vf − vi), using signs (vf is negative if it bounces).
Repeat for several push speeds (and maybe several masses). Plot J against Δp. The claim is supported if the points lie on a straight line through the origin with slope close to 1.
Sources of error: friction on the track before and after the hit, and reading v just before/after; keep the track level and measure v close to the bumper.
8. (Free response, qualitative to quantitative) A student says: "A soft mat makes landing safer because it reduces your change in momentum." (a) Explain what is wrong with the claim. (b) A 50 kg gymnast lands at 6.0 m/s. A hard floor stops her in 0.050 s and a mat in 0.40 s. Find the average net force in each case and show that your numbers agree with your explanation.
Show answer
(a) She goes from the same speed to rest either way, so Δp is the same. The mat makes the stopping time longer, and since Fnet = Δp/Δt, the force is smaller.
(b) Δp = 50 × 6.0 = 300 kg·m/s (up) in both cases.
Floor: F = 300 ÷ 0.050 = 6000 N. Mat: F = 300 ÷ 0.40 = 750 N.
Same Δp, 8 times the time, 1/8 of the net force. That matches (a).
9. (Short free response: mathematical routines) A 0.15 kg baseball arrives at a bat at 40 m/s and leaves in the opposite direction at 50 m/s. The contact lasts 1.5 ms. Take the direction the ball leaves as positive. (a) Calculate the change in momentum of the ball. (b) Calculate the average force on the ball. (c) If the force-time graph is a triangle, what is the peak force? (d) State the size and direction of the average force the ball exerts on the bat.
(b) 1.5 ms = 0.0015 s. Favg = 13.5 / 0.0015 = 9000 N, in the + direction.
(c) Triangle area = ½ Fpeak Δt = Favg Δt, so Fpeak = 2 × 9000 = 18 000 N.
(d) Third law: 9000 N on the bat, in the − direction (back toward where the ball came from).
Point guide (4 points):
1 point: signs on the velocities so Δp uses 90 m/s, not 10 m/s.
1 point: converts ms to s and gets 9000 N.
1 point: peak 18 000 N from the triangle area.
1 point: 9000 N opposite direction on the bat.
10. (Short free response: translation between representations) A 2.0 kg cart starts at rest. The net force on it (right positive) is: +6.0 N from 0 to 2.0 s; 0 from 2.0 s to 3.0 s; −4.0 N from 3.0 s to 5.0 s. (a) Calculate the impulse in each interval and the total impulse. (b) Calculate the velocity at t = 2.0 s and t = 5.0 s. (c) Describe the momentum-time graph from 0 to 5.0 s.
Show answer
(a) Areas under the F-t graph: 0 to 2 s: 6.0 × 2.0 = +12 N·s; 2 to 3 s: 0; 3 to 5 s: −4.0 × 2.0 = −8.0 N·s. Total: +4.0 N·s.
(b) t = 2.0 s: p = 12 kg·m/s, v = 12 / 2.0 = 6.0 m/s. t = 5.0 s: p = 4.0 kg·m/s, v = 2.0 m/s (still moving right).
(c) Straight line rising from 0 to 12 kg·m/s (slope +6.0 N) from 0 to 2.0 s; flat at 12 from 2.0 to 3.0 s; straight line falling to 4.0 kg·m/s (slope −4.0 N) from 3.0 to 5.0 s. The slope of p-t is the net force.
Point guide (4 points):
1 point: impulses +12, 0, −8.0 N·s.
1 point: 6.0 m/s and 2.0 m/s.
1 point: correct p-t shape with values.
1 point: links the slope of p-t to the force.
Common mistakes
The mistake: forgetting the sign when something bounces (Δp = m × 6 − m × 8 instead of m × (−6) − m × 8).
Why it is wrong: velocity is a vector. A ball coming back has the opposite sign.
How to spot it: a bounce must give a bigger |Δp| than a stop. If yours is smaller, a sign is missing.
The mistake: "the airbag (or pillow, or bent knees) reduces the impulse."
Why it is wrong: the impulse is fixed by the start and end momentum. Only the time and the force change.
How to spot it: ask "same start speed, same end speed?" If yes, the impulse is the same.
The mistake: using the peak force times the whole time to find the impulse.
Why it is wrong: the impulse is the area under the graph. A triangle has half the area of the rectangle around it.
How to spot it: sketch the shape. Use ½ × base × height for triangles, and add pieces for trapezoids.
The mistake: calling F = Δp/Δt "the force from the ground".
Why it is wrong: Δp/Δt is the net force. If gravity also acts, the ground force is the net force plus mg (when the ground pushes up).
How to spot it: draw a quick force diagram. If there are two forces, Δp/Δt is their sum, not either one alone.
The mistake: thinking the lighter object in a collision gets the bigger force or the bigger impulse.
Why it is wrong: Newton's third law: equal and opposite forces for the same time, so equal and opposite impulses. The lighter one only gets the bigger change in velocity.
How to spot it: in Lab 2 the two force curves are always mirror images, whatever the masses.