Unit 4 · Topic 4.2

4.2 Change in Momentum and Impulse

A net external force changes a system's momentum, and the rate of change is the force: $\vec{F}_{net} = \frac{d\vec{p}}{dt}$ . Over a time interval, the impulse $\vec{J} = \int\vec{F}\,dt$ , the area under a force–time graph, equals the change in momentum. The same idea handles systems whose mass changes while they move at constant velocity.

Key terms

impulse
impulse–momentum theorem
force–time graph
change in momentum
rate of change of momentum

Force as the rate of change of momentum

Newton's second law in its most general form is:

$\vec{F}_{net} = \frac{d\vec{p}}{dt}$

For a constant mass, $\frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}$ , the familiar form. On a momentum–time graph, the slope is the net force. A flat section means zero net force; a steep section means a large force.

Impulse and the impulse–momentum theorem

Impulse measures a force's effect over time:

$\vec{J} = \int_{t_1}^{t_2}\vec{F}\,dt$

It's a vector, in N·s (the same as kg·m/s). For a constant force, $\vec{J} = \vec{F}\Delta t$ . For a changing force, it's the area under the force–time graph. Integrating $\vec{F}_{net} = \frac{d\vec{p}}{dt}$ over time gives the impulse–momentum theorem:

$\vec{J}_{net} = \Delta\vec{p}$

The average force during an impact is $\vec{F}_{avg} = \frac{\Delta\vec{p}}{\Delta t}$ . For the same change in momentum, a longer collision time means a smaller average force. That's the physics behind airbags, padded helmets, bending your knees when you land, and catching an egg by moving your hand back with it.

Bounces give bigger impulses

Because Δp is a vector change, direction matters. A ball that hits a wall and stops has $\Delta p = mv$ . A ball that bounces back at the same speed has $\Delta p = 2mv$ , twice as much, so the wall gives it twice the impulse. Always subtract with signs: $\Delta p = p_f - p_i$ .

Reading the graphs

Force–time graph: the area under the curve is the impulse. A triangle-shaped spike of peak force 400 N lasting 0.020 s gives an impulse of $\frac{1}{2}(400)(0.020) = 4.0$ N·s. Area below the time axis is impulse in the negative direction.
Momentum–time graph: the slope is the net force. Where the graph is steepest, the force is largest; where it's flat, the net force is zero.
Two different collisions can give the same impulse: a tall, narrow spike (hard surface, short time) and a low, wide hump (soft surface, long time) can have equal areas.

When the mass changes

Sometimes a system gains or loses mass while moving, like sand pouring onto a conveyor belt or rain falling into an open cart. Applying $\vec{F}_{net} = \frac{d\vec{p}}{dt}$ with $\vec{p} = m\vec{v}$ gives two terms: $m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt}$ .

In this course you only handle the case where the velocity stays constant, so the first term is zero and:

$\vec{F}_{net} = \frac{dm}{dt}\vec{v}$

Here dm/dt is the rate at which mass is added (kg/s). For a conveyor belt kept at constant speed, the motor must supply this extra force to bring each new bit of sand up to the belt's speed.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
A ball rebounding from a wall (classic trap)
A 0.15 kg ball moving at 30 m/s hits a wall head-on and rebounds at 25 m/s. The contact lasts 0.0050 s. Find the impulse on the ball and the average force from the wall.
Show the solutionHide the solution
1. Step 1: Take the ball's initial direction as positive: $v_i = +30$ m/s and $v_f = -25$ m/s.
2. Step 2: $\Delta p = m(v_f - v_i) = 0.15(-25 - 30) = -8.25$ kg·m/s. The trap is computing 0.15(25 − 30) = −0.75 kg·m/s, which ignores the reversal.
3. Step 3: Impulse = Δp = −8.25 N·s, pointing away from the wall.
4. Step 4: $F_{avg} = \frac{-8.25}{0.0050} = -1650$ N, about 1.7 kN away from the wall.
Answer: Impulse 8.25 N·s away from the wall; average force about 1650 N away from the wall.
Example 2
Integrating a force over time
A 3.0 kg object moving at +1.0 m/s along a frictionless track feels a force $F(t) = 6t^2$ N in the +x direction from t = 0 to t = 2.0 s. Find the impulse and the final velocity.
Show the solutionHide the solution
1. Step 1: $J = \int_0^{2.0} 6t^2\,dt = \left[2t^3\right]_0^{2.0} = 16$ N·s.
2. Step 2: $\Delta p = J$ , so $\Delta v = \frac{16}{3.0} \approx 5.33$ m/s.
3. Step 3: $v_f = 1.0 + 5.33 \approx 6.3$ m/s in the +x direction.
Answer: J = 16 N·s; $v_f \approx 6.3$ m/s.
Example 3
Sand on a conveyor belt
Sand drops vertically onto a horizontal conveyor belt at 5.0 kg/s. The belt keeps moving at a constant 2.0 m/s. What extra horizontal force must the motor supply to keep the belt's speed constant?
Show the solutionHide the solution
1. Step 1: Each second, 5.0 kg of sand goes from zero horizontal velocity to 2.0 m/s, so the horizontal momentum of the belt-plus-sand system grows at $\frac{dp}{dt} = \frac{dm}{dt}v$ .
2. Step 2: $F = (5.0)(2.0) = 10$ N, in the direction the belt moves.
Answer: 10 N in the belt's direction of motion.

Common mistakes

Forgetting the sign change when an object bounces. Δp is final minus initial, with directions.
Saying a longer collision means a smaller impulse. For the same Δp the impulse is the same; the average force is what gets smaller.
Reading the height of a force–time graph as the impulse. The impulse is the area under it.
Applying $F = \frac{dm}{dt}v$ when the velocity is also changing. It's only for constant velocity in this course.

On the exam

Expect force–time graphs (often triangles or curves) where you find the impulse from the area and then the change in velocity. On momentum–time graphs, find the force from the slope.
Conceptual questions ask why safety devices work. Name the impulse–momentum theorem: the same Δp over a longer time means a smaller average force.

Connected topics

Videos

Linear momentum and impulse | AP Physics | Khan Academy
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Time-dependent forces | AP Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 4.2 Change in Momentum and Impulse. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A bat hits a ball. The force on the ball rises steadily from 0 to 400 N and then falls steadily back to 0, over a total time of 0.010 s. What impulse does the bat deliver?

A 0.20 kg ball moving horizontally at 10 m/s hits a wall head-on and bounces straight back at 8.0 m/s. It is in contact with the wall for 0.020 s.

Described situation

Question 2 of 4Calculator allowed

What is the magnitude of the ball's change in momentum?

Question 3 of 4Calculator allowed

What is the average force the wall exerts on the ball?

Question 4 of 4Calculator allowed

A net force $F(t) = 3t^2$ (N, with t in s) acts on a 1.0 kg object that starts from rest. What is the object's speed at t = 2.0 s?

0 of 4 answered