C. Energy and Momentum · Momentum and CollisionsImpulse and the Impulse–Momentum Theorem
C2.1 C2.5About 65 minutes
Learning goals- I can derive the impulse–momentum theorem, F_net Δt = Δp, from Newton's second law.
- I can calculate impulse, average force, contact time or final velocity using J = F_net Δt = Δp.
- I can explain why bouncing gives a larger impulse than stopping.
- I can use impulse to explain safety devices and follow-through in sports.
Words to know
impulse (J)
The product of the net force on an object and the time it acts, J = F_net Δt. A vector, measured in N·s.
impulse–momentum theorem
The impulse of the net force on an object equals its change in momentum: F_net Δt = Δp. The curriculum calls it the change in momentum–impulse theorem.
average force
The constant force that would give the same impulse over the same time interval: F_avg = Δp ÷ Δt.
contact time
The time interval during which two objects touch in a collision.
follow-through
Keeping a stick, club, bat or racket in contact with the object for longer, to increase the impulse.
Half a millisecond
When a golf club hits a ball, the club and ball touch for only about half a millisecond. In that tiny time the ball goes from rest to about 250 km/h. How big must the force be? And why do golfers, hockey players and tennis players all talk about follow-through?
Remember from Grade 11: in the SPH3U Forces unit you saw that a longer stopping time means a smaller force, using F = mΔv ÷ Δt. In this lesson you will turn that idea into one of the most useful tools in physics: the impulse–momentum theorem.
From Newton's second law to impulse
Start with Newton's second law, with a constant (or average) net force:
F_net = ma = m(v_f − v_i) ÷ Δt
Multiply both sides by Δt:
F_net Δt = mv_f − mv_i = Δp
The left side, net force × time interval, is called the impulse, symbol J:
J = F_net Δt and J = Δp
This is the impulse–momentum theorem (your curriculum calls it the change in momentum–impulse theorem): the impulse of the net force on an object equals the object's change in momentum.
- Impulse is a vector in the direction of the net force.
- Its unit is the newton-second (N·s), which equals kg·m/s.
- If the force changes during the interaction, use the average net force, or find the impulse as an area under a force–time graph (next lesson).
Rearranged, F_net = Δp ÷ Δt. This is actually how Newton stated his second law: net force is the rate of change of momentum. This form is more general than F = ma, because it still works when the mass changes, as for a rocket burning fuel (you will use this in the rockets lesson).
Two ways to get the same impulse
The same change in momentum can come from a large force for a short time or a small force for a long time. To make a stop gentle, stretch out the time (airbags, gym mats, bending your knees). To make a hit powerful, apply as much force as you can and keep contact for as long as you can (follow-through).
Worked exampleExample 1: Driving a golf ball
A 0.046 kg golf ball is hit from rest and leaves the club at 70 m/s. The contact lasts 5.0 × 10⁻⁴ s. Find (a) the impulse on the ball and (b) the average force of the club on the ball.
- Take the direction of the ball's motion as positive.
- (a) J = Δp = m(v_f − v_i) = (0.046)(70 − 0) = 3.22 N·s.
- (b) F_avg = J ÷ Δt = 3.22 ÷ (5.0 × 10⁻⁴) = 6440 N.
- For comparison, the ball's weight is mg = (0.046)(9.8) ≈ 0.45 N. The club's push is about 14 000 times larger, so we can ignore gravity during the hit.
(a) J ≈ 3.2 N·s in the direction of the ball's motion. (b) F_avg ≈ 6.4 × 10³ N.
Worked exampleExample 2: Returning a tennis ball
A 0.058 kg tennis ball arrives at a racket at 25 m/s and leaves in the opposite direction at 30 m/s. The ball is in contact with the strings for 4.0 ms. Find the average force of the racket on the ball.
- Let the direction the ball leaves be positive. Then v_i = −25 m/s and v_f = +30 m/s.
- Δp = m(v_f − v_i) = (0.058)(30 − (−25)) = (0.058)(55) = 3.19 N·s.
- Δt = 4.0 ms = 4.0 × 10⁻³ s.
- F_avg = Δp ÷ Δt = 3.19 ÷ (4.0 × 10⁻³) ≈ 798 N.
About 8.0 × 10² N, in the direction the ball leaves. Reversing the ball's direction needs much more impulse than just stopping it (0.058 × 25 = 1.45 N·s).
Bouncing needs a bigger impulse than stopping. For the same mass and incoming speed, a ball that bounces straight back at the same speed has twice the change in momentum of a ball that sticks.Worked exampleExample 3: Final velocity from an impulse
A 0.50 kg cart moves at 2.0 m/s [E]. A constant net force of 3.0 N [W] acts on it for 0.80 s. Find its final velocity.
- East positive: v_i = +2.0 m/s and F_net = −3.0 N.
- J = F_net Δt = (−3.0)(0.80) = −2.4 N·s.
- J = m(v_f − v_i), so v_f = v_i + J ÷ m = 2.0 + (−2.4 ÷ 0.50) = 2.0 − 4.8 = −2.8 m/s.
v_f = 2.8 m/s [W]. The westward impulse first stopped the cart, then sped it up toward the west.
Worked exampleExample 4: How long to stop?
A 1200 kg car moving at 25 m/s brakes with an average net force of 6.0 × 10³ N. How long does it take to stop?
- The required impulse is Δp = 0 − (1200)(25) = −3.0 × 10⁴ N·s (opposite the motion).
- Δt = |Δp| ÷ |F_net| = (3.0 × 10⁴) ÷ (6.0 × 10³) = 5.0 s.
5.0 s. Notice we did not need the acceleration or any kinematics equation: impulse links force and time directly to momentum.
Why follow-through works
In a slapshot, a golf drive or a baseball swing, the athlete tries to keep the stick, club or bat in contact with the object for as long as possible while pushing hard. A longer contact time Δt with the same large force means a larger impulse, so a larger change in momentum and a faster ball or puck. Hockey sticks also flex: the blade stays on the puck slightly longer while the stick unbends.
Why bending your knees works
When you land from a jump, the floor must remove all your downward momentum. Landing stiff-legged stops you in a few hundredths of a second; bending your knees stretches the stop to a few tenths of a second. Same Δp, ten times the Δt, so roughly a tenth of the force on your joints.
Just for funA golf ball spends about half a millisecond touching the club. In one tenth of a second, about as long as a quick blink, you could fit about 200 of those hits.
Hands-on activitySoft hands: feel the impulse
You need: a tennis ball (or a soft ball), a partner, and a phone that can record slow-motion video. Optional: one raw egg in a sealed zip-top bag, used outdoors on grass. Grown-up note: wash hands after handling raw egg.
- Stand about 3 m apart and toss the ball back and forth. First catch it with stiff hands that do not move. Then catch it with soft hands that move back with the ball as you catch.
- Film both kinds of catch in slow motion. Count frames to estimate the stopping time Δt for each (time = frames ÷ frames per second, often 240 frames/s).
- The ball's change in momentum is about the same each time. Use F = Δp ÷ Δt to explain which catch should sting less, and estimate the ratio of the two forces from your two values of Δt.
- Optional egg toss: toss the bagged egg gently to your partner, who catches it with soft hands. Step back one pace after each successful catch. When the egg breaks, discuss what changed.
What to look for: soft-hands catches usually last several times longer, so the average force is several times smaller. In the egg toss, a longer toss means a larger speed, so a larger Δp; the catcher must stretch Δt even more to keep the force below what the shell can take.
Your turn
1A 0.16 kg puck goes from rest to 35 m/s in 0.010 s. Find the impulse and the average force on it.
J = (0.16)(35) = 5.6 N·s. F = 5.6 ÷ 0.010 = 560 N, in the direction of the puck's motion.
2A net force of 250 N acts on a 5.0 kg object for 0.20 s. Find the impulse and the change in velocity.
J = (250)(0.20) = 50 N·s. Δv = J ÷ m = 50 ÷ 5.0 = 10 m/s, in the direction of the force.
3A 0.40 kg soccer ball moving at 12 m/s [W] is kicked so it moves at 18 m/s [E]. The foot touches the ball for 0.025 s. Find the impulse and average force on the ball.
East positive: Δp = (0.40)(18 − (−12)) = (0.40)(30) = 12 N·s [E]. F = 12 ÷ 0.025 = 480 N [E].
4What impulse is needed to stop a 2000 kg SUV moving at 20 m/s?
J = 0 − (2000)(20) = −4.0 × 10⁴. 4.0 × 10⁴ N·s opposite to the motion.
5A catcher stops a 0.145 kg baseball moving at 40 m/s in 0.020 s. Find the average force on the ball.
Δp = (0.145)(40) = 5.8 N·s. F = 5.8 ÷ 0.020 = 290 N, opposite to the ball's motion.
6A 3.0 kg object moves at 4.0 m/s [S]. A net force of 12 N [N] acts on it for 2.0 s. Find its final velocity.
North positive: v_i = −4.0 m/s, J = (12)(2.0) = 24 N·s. v_f = −4.0 + 24 ÷ 3.0 = −4.0 + 8.0 = 4.0 m/s [N].
7Use the impulse–momentum theorem to explain why gymnasts land on thick mats.
The gymnast's change in momentum (from moving to stopped) is the same on any surface. The mat increases the stopping time Δt, so the average force F = Δp ÷ Δt is smaller.
More practice: warm-up
1State the impulse–momentum theorem in words.
The impulse of the net force on an object (net force × time) equals the change in the object's momentum.
2A force of 20 N acts for 0.50 s. What is the impulse?
3What are the units of impulse?
N·s, which are the same as kg·m/s.
4An object's momentum changes by 6.0 kg·m/s [N]. What impulse acted on it?
6.0 N·s [N], equal to the change in momentum.
5The change in momentum stays the same but the collision time doubles. What happens to the average force?
It is halved, because F = Δp ÷ Δt.
More practice: on your own
1A 50 kg person lands on a trampoline at 6.0 m/s [down] and leaves it at 6.0 m/s [up]. The contact time is 0.40 s. Find (a) the average net force and (b) the average force of the trampoline on the person. Use g = 9.8 N/kg.
Up positive: Δp = (50)(6.0 − (−6.0)) = 600 N·s. (a) F_net = 600 ÷ 0.40 = 1500 N [up]. (b) F_net = F_tramp − mg, so F_tramp = 1500 + (50)(9.8) = 1500 + 490 = 1990 N [up] (about 2.0 × 10³ N). Gravity matters here because the contact time is long.
2A 1500 kg car slows from 100 km/h to rest in 4.0 s. Find the average net force.
v = 100 ÷ 3.6 ≈ 27.8 m/s. Δp = (1500)(27.8) ≈ 4.17 × 10⁴ N·s. F = 4.17 × 10⁴ ÷ 4.0 ≈ 1.0 × 10⁴ N, opposite the motion.
3A 0.046 kg golf ball leaves the tee at 60 m/s. The average force of the club was 5500 N. How long were the club and ball in contact?
Δt = Δp ÷ F = (0.046)(60) ÷ 5500 = 2.76 ÷ 5500 ≈ 5.0 × 10⁻⁴ s (half a millisecond).
4A ball of clay and a rubber ball have the same mass and hit a wall at the same speed. The clay sticks; the rubber ball bounces back. Which receives the larger impulse from the wall? Explain.
The rubber ball. The clay's momentum changes from mv to 0 (size mv). The rubber ball's changes from mv to about −mv (size up to 2mv). A larger Δp means a larger impulse.
5In a serve, a racket pushes on a 0.058 kg tennis ball (momentarily at rest at the top of the toss) with an average force of 600 N for 5.0 ms. Find the ball's speed as it leaves the racket.
J = (600)(5.0 × 10⁻³) = 3.0 N·s. v = J ÷ m = 3.0 ÷ 0.058 ≈ 52 m/s.
Challenge
1During a hailstorm, 120 hailstones hit a car roof each second. Each has a mass of 0.0050 kg, falls at 15 m/s, and bounces up at 5.0 m/s. (a) Find the average force of the hail on the roof. (b) Rain with the same mass per second and speed does not bounce. Find its average force. Ignore the weight of the hail or rain sitting on the roof.
(a) For each stone, Δp = (0.0050)(5.0 − (−15)) = 0.10 N·s upward (roof on stone). Per second: 120 × 0.10 = 12 N·s each second, so the average force is 12 N (stones push down on the roof by Newton's third law). (b) Rain: each 0.0050 kg of water has Δp = (0.0050)(15 − 0) = 0.075 N·s. Per second: 120 × 0.075 = 9.0 N·s, so the force is 9.0 N. Bouncing increases the force.
2Derive the impulse–momentum theorem from Newton's second law and the definition of acceleration, for a constant net force.
F_net = ma and a = (v_f − v_i) ÷ Δt. So F_net = m(v_f − v_i) ÷ Δt. Multiplying by Δt: F_net Δt = mv_f − mv_i = Δp. The left side is the impulse J.
3A garden hose sprays 0.50 kg of water per second at 8.0 m/s straight at a wall. The water splashes sideways, so it loses all its momentum toward the wall. Find the average force of the water on the wall.
Each second, the momentum toward the wall that is removed is (0.50 kg)(8.0 m/s) = 4.0 kg·m/s. F = Δp ÷ Δt = 4.0 ÷ 1.0 = 4.0 N toward the wall. (The same idea gives the thrust of a rocket.)
Quiz
1. The impulse of the net force on an object is equal to its:
2. An airbag reduces the average force on a person in a collision mainly because it:
3. A 50 N net force acts for 0.20 s. What is the impulse?
4. Ball A and ball B have the same mass and hit a wall at the same speed. A sticks; B bounces straight back at the same speed. The impulse on B is:
5. A 2.0 kg object at rest receives an impulse of 8.0 N·s. What is its speed afterward?
6. Which equation is Newton's second law written in its more general momentum form?
Big idea (C2.1, C2.5): impulse J = F_net Δt equals the change in momentum Δp. The same change in momentum can come from a large force for a short time or a small force for a long time. This explains airbags, gym mats, soft catches and follow-through.
How to help: Toss a ball with your teen and catch it with stiff hands, then with soft hands. Ask them to explain the difference using Δp and Δt. Ask: "Why does a golfer follow through?" (longer contact time → bigger impulse → faster ball).
Common difficulties:
- Signs when an object reverses. A ball that bounces back has Δp = m(v_f − v_i) with v_i and v_f of opposite signs, so the speeds add.
- Net force vs one force. J = F_net Δt uses the net force. When the contact time is long (trampoline, landing), gravity matters: the surface's force = F_net + mg.
- Milliseconds. 4.0 ms = 4.0 × 10⁻³ s; forgetting to convert makes forces 1000 times too small.
- Thinking the airbag changes Δp. It does not; it changes Δt.