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$\vec{F} = d\vec{p}/dt$, impulse $\int\vec{F}\,dt = \Delta\vec{p}$ as the area under a force–time graph, average and peak forces, and momentum as a vector.
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By the end of this lesson you will be able to compute impulses by integrating forces, apply the impulse–momentum theorem with vector momenta, and find average collision forces.
From Physics 1 you know momentum $p = mv$, impulse as force times time, and conservation of momentum in collisions. The earlier lessons of this course wrote Newton's second law as a differential equation and integrated forces over time. This lesson puts those together: the time integral of force is the change of momentum, whatever the force's shape, which is how brief, violent forces are measured and managed.
| Term | What it means |
|---|---|
| Momentum | $\vec{p} = m\vec{v}$, a vector with units of kg m/s. |
| Impulse | $\vec{J} = \int\vec{F}\,dt$, with units of N s, equal to kg m/s. |
| Impulse–momentum theorem | $\vec{J}_{\text{net}} = \Delta\vec{p}$. |
| Average force | $\bar{F} = J/\Delta t$, the steady force that would give the same impulse. |
| Force–time graph | A plot of force against time whose area is the impulse. |
| Collision time | The short interval during which two objects push on each other. |
| Rebound | A collision that reverses the velocity, doubling the change of momentum compared with stopping. |
Newton wrote his second law as $\vec{F}_{\text{net}} = d\vec{p}/dt$, with $\vec{p} = m\vec{v}$ the momentum. For constant mass it is $m\vec{a}$, but the momentum form is the more general. Integrating it over a time interval gives the impulse–momentum theorem:
$$\vec{J} = \int_{t_1}^{t_2}\vec{F}_{\text{net}}\,dt = \Delta\vec{p}.$$
The impulse $\vec{J}$ is the area under the force–time graph. The theorem is especially useful for collisions, where the force is large, brief and complicated: it may rise and fall in a thousandth of a second in a shape no one knows exactly, but its area must equal the change of momentum, which is easy to measure.
Dividing the impulse by the collision time gives the average force, $\bar{F} = \Delta p/\Delta t$. For a given change of momentum, a longer collision means a smaller average force. That single relation is the physics of airbags, padded helmets, bending knees on landing and crumple zones.
Another way: picture
Plot the force a bat exerts on a ball against time: a tall, narrow spike lasting a millisecond. Its area is the impulse. Now plot the force of a pillow stopping the same ball: a low, wide hump. If both stop the ball, their areas are equal. The pillow wins by spreading the same area over more time, which keeps the height, the force, small.
Another way: steps
Momentum points along the velocity, so direction matters in every calculation. Take a positive direction and give each momentum its sign. A $0.15$ kg ball arriving at $40$ m/s toward a wall has $p = -6.0$ kg m/s if away from the wall is positive; leaving at $30$ m/s it has $+4.5$ kg m/s. The change is $10.5$ kg m/s, not $1.5$.
In two dimensions, change each component separately. A puck deflected by a wall at an angle keeps its momentum along the wall, if the wall is smooth, and reverses the perpendicular part: the impulse is perpendicular to the wall. That is why the impulse from a smooth surface always points straight out of it, and why a pool ball bounces off a cushion at the angle it arrived.
When the force is known as a function of time, integrate it: $J = \int F\,dt$. A force rising linearly as $F = ct$ for a time $T$ gives $J = \tfrac{1}{2}cT^2$. When the force is given as a graph, find the area: a rectangle for a constant force, a triangle for a force rising steadily and falling back, a trapezoid for one that ramps and holds.
Real collision forces are measured with force plates or instrumented bats, which record thousands of readings per second. The software adds up the areas of thin slices to get the impulse. Sports scientists use this to measure a sprinter's push-off from the blocks or a jumper's takeoff: the impulse, divided by the body's mass, is the velocity gained.
The average force over a collision, $\bar{F} = J/\Delta t$, is the height of a rectangle with the same area as the real force curve. The real force always has a peak above the average, sometimes far above it: for a symmetric triangle the peak is twice the average, and for the smooth hump of a ball hitting a bat about one and a half times.
Injuries and breakage depend on the peak, so designers try to make force curves flat. A good padded surface does two things: it lengthens the collision, lowering the average, and it flattens the curve, bringing the peak close to the average. An egg dropped on foam survives for both reasons.
For a given change of momentum, $\bar{F}\Delta t$ is fixed, so doubling the time halves the average force. A person landing from a jump with locked knees stops in perhaps $10$ ms; bending the knees stretches that to $200$ ms, cutting the average force twentyfold. Catching a ball with the hands moving backward does the same.
Cars are built on this idea. A car's front end crumples over half a meter or more, stretching a crash that might last a few milliseconds against a rigid barrier into a tenth of a second. Seat belts stretch slightly, and airbags let the occupant sink in and vent their gas as they compress. Each adds time to the stop, and time is what turns a lethal force into a survivable one.
The impulse–momentum theorem uses the net force, so every force acting during the interval contributes. For a collision lasting a millisecond, gravity's impulse, $mg\Delta t$, is tiny beside the collision force and can be ignored. For longer intervals it cannot: a ball in flight for $2$ s receives an impulse of $mg \times 2$ s from gravity, which is exactly the change in its vertical momentum.
This is the impulse approximation: during a brief collision, forces of ordinary size are negligible compared with the collision force. It is why momentum is nearly conserved in collisions even when gravity or friction act, the subject of the collisions lesson, and it is a judgment that should be checked by comparing the sizes of the impulses.
Checking an answer. Units: N s and kg m/s are the same. A rebound must give a larger impulse than a stop at the same speed. The impulse must point the same way as the net force. And an average force must be less than the peak of any force curve that produced it.
The momentum form of the second law also covers systems whose mass changes, if applied carefully to the whole system. A conveyor belt onto which sand pours at a rate $dm/dt$, moving at a steady speed $v$, needs a force $F = v\,dm/dt$ just to bring each grain up to the belt's speed, even though nothing accelerates. A rope pulled up off the floor needs extra force for the same reason.
Rockets are the famous case, treated in full in the university course: exhaust carried away backward gives the rocket forward momentum. In all such problems the safe method is to write the total momentum of everything in the system at two instants and apply $\vec{F}_{\text{ext}} = d\vec{P}/dt$ to that whole system.
Force plates are standard in sports science labs and in many athletic programs. An athlete stands on the plate and jumps; the plate records the upward force at a thousand readings a second. Subtracting the body's weight and integrating over the push-off gives the net impulse, and dividing by the mass gives the takeoff speed, from which the jump height follows as $v^2/2g$.
Coaches use the shape of the force curve, not just its area. Two athletes with the same impulse, and so the same jump height, may produce it differently: one with a short, high push, the other with a longer, lower one. The impulse–momentum theorem says both jump equally high; the force curve says which is stronger and which uses technique, and that difference is what the training is built on.
At the Insurance Institute for Highway Safety's Vehicle Research Center in Ruckersville, Virginia, instrumented crash-test dummies ride cars into barriers. A $75$ kg dummy moving at $15$ m/s must lose $1125$ kg m/s of momentum however it is stopped. Striking a steering wheel, that might happen in $10$ ms, an average force of more than $100$ kN.
The seat belt and airbag stretch the stop to around $100$ ms, cutting the average force to about $11$ kN, which a body can survive. The airbag inflates in about $30$ ms and then vents as the occupant presses into it, deliberately lengthening the collision. Sensors in the dummy record the force against time, and engineers integrate to check that the impulse matches the momentum lost, confirming the sensors, and then judge the design by the peak force and how long it lasts.
A pitched fastball arrives at about $40$ m/s and, when squarely hit, leaves the bat at about $45$ m/s in the opposite direction. The $0.145$ kg ball's momentum changes by about $12$ kg m/s in roughly a millisecond of contact, an average force above $10$ kN, several tons, and a peak force half again as large.
High-speed cameras show the ball flattening to nearly half its diameter during that millisecond. Researchers, including physicists who study bats for Major League Baseball, measure the force against time with sensors in the bat. The impulse–momentum theorem connects those measurements to the exit speed, and comparing wood and aluminum bats comes down to how much each lengthens the contact and how much energy each returns to the ball.
When a ball bounces back at the speed it arrived, its speed has not changed, so it is tempting to think the wall gave it little or no impulse. But momentum is a vector. The wall must first remove the incoming momentum and then supply the outgoing momentum in the opposite direction: an impulse of $2mv$, twice what stopping the ball would take. That is why a bouncing object hits harder than one that sticks.
A second error is to confuse impulse with force. A small force acting for a long time can deliver the same impulse as a large force acting briefly.
A $0.15$ kg ball hits a wall at $40$ m/s and rebounds at $30$ m/s. Choose away from the wall as positive.
$v_1 = -40\ \text{m/s}, \quad v_2 = +30\ \text{m/s}$
Signs record direction.
Find the momentum before.
$p_1 = 0.15 \times (-40) = -6.0\ \text{kg m/s}$
Toward the wall.
Find the momentum after.
$p_2 = 0.15 \times 30 = 4.5\ \text{kg m/s}$
Away from the wall.
Find the impulse.
$J = p_2 - p_1 = 4.5 - (-6.0) = 10.5\ \text{N s}$
The speeds add because the direction reverses.
Find the average force over $2.0$ ms.
$\bar{F} = \dfrac{10.5}{2.0 \times 10^{-3}} = 5250\ \text{N}$
Away from the wall.
A force $F = 300t$ (N, s) acts on a $2.0$ kg cart at rest for $0.20$ s. Write the impulse.
$J = \displaystyle\int_0^{0.20}300t\,dt$
The force changes with time.
Evaluate the integral.
$J = 150t^2\Big|_0^{0.20} = 150 \times 0.040 = 6.0\ \text{N s}$
The area of a triangle.
Find the change of velocity.
$\Delta v = \dfrac{J}{m} = \dfrac{6.0}{2.0} = 3.0\ \text{m/s}$
From rest.
Find the average force.
$\bar{F} = \dfrac{6.0}{0.20} = 30\ \text{N}$
Impulse over time.
Compare with the final force.
$F(0.20) = 60\ \text{N} = 2\bar{F}$
For a linear rise, the average is half the final value.
A $0.17$ kg hockey puck hits the smooth boards at $12$ m/s, $30°$ from the line perpendicular to them, and bounces off at the same speed and angle. Resolve the incoming velocity.
$v_\perp = -12\cos 30° = -10.4, \quad v_\parallel = 12\sin 30° = 6.0\ \text{m/s}$
Perpendicular to and along the boards.
Write the outgoing velocity.
$v_\perp = +10.4, \quad v_\parallel = 6.0\ \text{m/s}$
Smooth boards reverse only the perpendicular part.
Find the change of momentum along the boards.
$\Delta p_\parallel = 0.17 \times (6.0 - 6.0) = 0$
No force along a smooth surface.
Find the change perpendicular to the boards.
$\Delta p_\perp = 0.17 \times (10.4 - (-10.4)) = 3.54\ \text{kg m/s}$
Twice the incoming perpendicular momentum.
Find the direction of the impulse.
$\vec{J} \perp \text{boards, pointing out of them}$
The boards can only push.
Find the average force over $5.0$ ms.
$\bar{F} = \dfrac{3.54}{5.0 \times 10^{-3}} = 708\ \text{N}$
Straight out of the boards.
Find the change of momentum.
$\Delta p = 0.050 \times 70 = 3.5\ \text{kg m/s}$
From rest.
Divide by the contact time.
$\bar{F} = \dfrac{3.5}{0.50 \times 10^{-3}}$
Impulse over time.
Evaluate the force.
A ball moving at speed $v$ hits a wall and stops, receiving an impulse of $3$ N s. An identical ball at the same speed hits a bouncy wall and rebounds at speed $v$. What impulse does the second ball receive?
Complete the worked solution: a $5$ kg cart at rest is struck by a force that rises steadily from zero to $58$ N and falls back to zero, over a total of $3$ s. Find the impulse in N s, the cart's final speed in m/s, and the average force in N.
Find the area of the triangle.
$J = \tfrac{1}{2}F_0T =$ j
The impulse is the area under the force–time graph.
Divide the impulse by the mass.
$\Delta v = \dfrac{J}{m} =$ v
Starting from rest.
Divide the impulse by the duration.
$\bar{F} = \dfrac{J}{T} =$ a
The steady force with the same effect.
Check the average against the peak.
$\text{average} = \text{half the peak for a triangle}$
Its area equals a rectangle of half the height.
Match each quantity to its expression.
| $m\vec{v}$ | $d\vec{p}/dt$ | $\int\vec{F}\,dt$ | $J/\Delta t$ | |
|---|---|---|---|---|
| momentum | ||||
| second law | ||||
| impulse | ||||
| average force |
A force pulse is $F(t) = 12t(4 - t)$ newtons for $0 \le t \le 4$ s and zero otherwise. Fill in its impulse in N s, its average over the pulse in N, and its peak in N.
| value | |
|---|---|
| impulse (N s) | |
| average force (N) | |
| peak force (N) |
A cart has momentum $3$ kg m/s at $t = 0$ and feels a net force $F = 2 + 6t$ (newtons, seconds). Write its momentum, in kg m/s, as a formula in $t$.
Answer:
A $0.15$ kg baseball arrives at the plate at $33$ m/s and leaves the bat at $48$ m/s straight back toward the pitcher. The bat and ball are in contact for $2$ ms. What average force does the bat exert, in newtons?
Answer: N average from the bat
In a crash test at the Insurance Institute for Highway Safety's center in Virginia, a $79$ kg dummy moving at $14$ m/s is brought to rest by the seat belt and airbag over $100$ ms. What average force acts on it, in newtons?
Answer: N on the dummy
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A cart has momentum $5$ kg m/s at $t = 0$ and feels a net force $F = 2 + 4t$ (newtons, seconds). Write its momentum, in kg m/s, as a formula in $t$.
Answer:
You can use impulse and momentum. Explain to someone why an airbag lowers the force on a driver even though it cannot change how much momentum the driver loses.
19. Your turn: a $0.050$ kg golf ball leaves the tee at $70$ m/s after $0.50$ ms of contact. Find the average force., step 3
$\bar{F} = 7000\ \text{N}$
About fourteen thousand times the ball's weight.