AP Physics 1 Impulse-Momentum Theorem

AP Physics 1· AP Physics 1 CED — Momentum· 14 min read

1. Derivation and Core Statement of the Theorem★★☆☆☆⏱ 3 min

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Impulse is defined as the integral of net force over the time interval the force acts:

J = \int_{t_1}^{t_2} F_{\text{net}}(t) dt

For constant net force, this simplifies to $J = F_{net} Δ t$ . Impulse is a vector quantity, with direction matching the direction of the net force that produces it. To derive the theorem, we start with Newton's second law:

F_{\text{net}} = ma = m\frac{\Delta v}{\Delta t} = \frac{\Delta (mv)}{\Delta t} = \frac{\Delta p}{\Delta t}

Rearranging terms gives the core impulse-momentum theorem:

J = \Delta p = p_{\text{final}} - p_{\text{initial}}

This means the net impulse on an object is exactly equal to the change in the object's linear momentum, matching everyday intuition: pushing for twice as long or with twice the force doubles the momentum change for the same interaction.

📐 Worked Example

A 0.15 kg baseball is thrown toward a batter at 35 m/s to the left. The batter hits it, and the ball leaves the bat at 45 m/s to the right. Contact time between bat and ball is 0.002 s. What is the average net force exerted on the ball by the bat?

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1. Define a coordinate system with right as the positive direction, giving signed velocities:
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$v_i = -35 \text{ m/s}, \quad v_f = +45 \text{ m/s}$
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1. Calculate the change in momentum:
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$\Delta p = m v_f - m v_i = m(v_f - v_i) = 0.15 \left(45 - (-35)\right) = 12 \text{ kg·m/s}$
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1. From the impulse-momentum theorem, for constant average force $J = F_{avg} Δ t = Δ p$
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1. Solve for average force:
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$F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{12}{0.002} = 6000 \text{ N}$
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The positive sign confirms the force points to the right, matching the direction of the momentum change.

Exam tip:

Always explicitly define your coordinate system before calculating momentum change to avoid sign errors for objects that reverse direction.

2. Impulse from Force-Time Graphs★★☆☆☆⏱ 3 min

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One of the most frequently tested AP Physics 1 applications of the impulse-momentum theorem is calculating impulse and momentum change from a plot of net force versus time. By definition, impulse is the integral of force over time, which equals the net area between the force curve and the time axis on the graph.

📐 Worked Example

The net force acting on a 2.0 kg block initially at rest increases linearly from 0 N to 10 N at $t = 2$ s, then decreases linearly back to 0 N at $t = 4$ s. What is the speed of the block at $t = 4$ s?

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1. By the impulse-momentum theorem, total impulse equals the area under the F-t graph, which equals the change in momentum.
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1. The graph forms a triangle with base = 4 s and height = 10 N. Calculate the area (impulse):
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$J = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 10 = 20 \text{ N·s}$
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1. Initial momentum is zero because the block starts at rest, so $ Delta p = p_f = m v_f = 20 ext{ kg·m/s}$
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1. Solve for final speed:
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$v_f = \frac{20}{2.0} = 10 \text{ m/s}$

Exam tip:

Always assign the correct sign to areas where force is negative (below the time axis). AP exam questions regularly include negative force regions to test your understanding of this convention.

3. Impulse-Momentum Theorem for Systems of Objects★★★☆☆⏱ 4 min

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The impulse-momentum theorem applies not just to single objects, but to entire systems of multiple interacting objects. For a system, we separate forces into two categories: internal forces (exerted by objects inside the system on other objects inside the system) and external forces (exerted by objects outside the system on objects inside the system).

By Newton's third law, every internal force has an equal and opposite internal reaction force. These paired forces act for the same amount of time, so their impulses are equal and opposite, adding up to zero total impulse for the system. Only external forces contribute to the net impulse of the system, leading to the system form of the theorem:

J_{\text{external, net}} = \Delta P_{\text{total}}

This is the direct foundation for conservation of momentum: if net external impulse is zero, total momentum of the system does not change.

📐 Worked Example

A 70 kg astronaut floating in space throws a 5 kg tool away from her at 12 m/s relative to her spaceship. Both are initially at rest. What is the astronaut's speed after throwing the tool? Assume no external forces act on the astronaut-tool system.

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1. Define the system as astronaut + tool. No external forces act, so net external impulse $J_{ext, net} = 0$ , which means $Δ P_{total} = 0$ , so total final momentum equals total initial momentum.
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1. Initial total momentum $P_{i} = 0$ , since both objects are at rest.
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1. Let the direction the tool is thrown be positive, so:
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$P_f = m_a v_a + m_t v_t = 0$
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1. Rearrange to solve for the astronaut's velocity:
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$v_a = - \frac{m_t v_t}{m_a} = - \frac{(5 \text{ kg})(12 \text{ m/s})}{70 \text{ kg}} \approx -0.86 \text{ m/s}$
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The negative sign means the astronaut moves in the opposite direction of the tool, with speed 0.86 m/s.

Exam tip:

Always identify internal vs external forces before applying the theorem to a system; forgetting to include an external impulse like friction will lead to an incorrect assumption that momentum is conserved when it is not.

4. AP-Style Practice Worked Examples★★★☆☆⏱ 4 min

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📐 Worked Example

A 0.5 kg rubber ball is dropped from rest from a height of 1.25 m onto a hard flat floor. The ball bounces back up to a maximum height of 0.80 m. What is the approximate magnitude of the net impulse exerted on the ball by the floor during the bounce? Use $g \approx 10 m/s^{2}$ . Options: A) 1.0 N·s, B) 4.5 N·s, C) 2.5 N·s, D) 9.0 N·s

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1. Use kinematics $v^{2} = v_{0}^{2} + 2 g h$ to find velocities just before and after the bounce, with up as the positive direction:
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Before impact, $v_{i}^{2} = 2 g h_{1} = 2 (10) (1.25) = 25$ , so $v_{i} = - 5$ m/s. After impact, $v_{f}^{2} = 2 g h_{2} = 2 (10) (0.80) = 16$ , so $v_{f} = + 4$ m/s.
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1. By the impulse-momentum theorem, impulse equals change in momentum:
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$\Delta p = m(v_f - v_i) = 0.5(4 - (-5)) = 4.5 \text{ N·s}$
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The correct answer is B.

📐 Worked Example

A force sensor measures the force applied to a 3.0 kg cart initially at rest on a frictionless horizontal track. Force is 0 N for $t < 0$ s, 6 N for $0 \leq t \leq 2$ s, 3 N for $2 < t \leq 5$ s, and 0 N for $t > 5$ s. (a) Calculate the total impulse applied to the cart over the 5 second interval. (b) Calculate the speed of the cart at $t = 5$ s. (c) Suppose an additional constant force of friction of 2 N acts on the cart opposite the direction of motion. Will the speed at $t = 5$ s be half of your answer to part (b)? Justify your answer.

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Part (a): Total impulse equals the area under the force-time graph, split into two rectangular regions:
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$J_1 = 6 \times 2 = 12 \text{ N·s}, \quad J_2 = 3 \times 3 = 9 \text{ N·s}, \quad J_{\text{total}} = 21 \text{ N·s}$
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Part (b): Initial velocity is zero, so $J = Δ p = m v_{f}$ :
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$v_f = \frac{21}{3.0} = 7 \text{ m/s}$
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Part (c): No, the speed will not be half. The impulse from friction is $J_{f} = - f Δ t = - 2 N \times 5 s = - 10$ N·s. Net impulse becomes $21 - 10 = 11$ N·s, so new speed is $\frac{11}{3} \approx 3.7$ m/s, which is greater than half of 7 m/s (3.5 m/s). Friction acts over the entire 5 second interval, so only 10 N·s is subtracted from the original impulse, not half of the original impulse.

📐 Worked Example

A model water rocket generates a thrust force that can be approximated as $F (t) = 12 t - 3 t^{2}$ for $0 \leq t \leq 4$ s, where $F$ is in newtons and $t$ is in seconds. The empty rocket has a mass of 0.2 kg. Use the impulse-momentum theorem to find the speed of the rocket at burnout (t=4 s), assuming the rocket starts from rest, ignore the mass of the water, and neglect air resistance.

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Impulse is the integral of thrust force (assumed net force, neglecting gravity and air resistance) from 0 to 4 s:
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$J = \int_{0}^{4} (12t - 3t^2) dt = 6t^2 - t^3 \bigg|_{0}^{4}$
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Evaluate at the bounds:
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$J = \left(6(4^2) - 4^3\right) - 0 = 96 - 64 = 32 \text{ N·s}$
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By the impulse-momentum theorem, $J = Δ p = m v_{f} - 0$ , so:
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$v_f = \frac{J}{m} = \frac{32}{0.2} = 160 \text{ m/s}$

5. Common Pitfalls

Wrong move:

Calculating change in momentum as $m (v_{f} + v_{i})$ when velocity reverses direction, forgetting to flip the sign of initial velocity

Why:

Students treat speed (a scalar) instead of velocity (a vector) when calculating momentum change, leading to half the correct value of impulse

Correct move:

Always define your coordinate system before calculating $ Delta p = p_f - p_i$, and substitute signed velocities for both initial and final momentum

Wrong move:

Adding magnitudes of areas regardless of sign when calculating total impulse from a force-time graph with positive and negative force regions

Why:

Students assume all area is positive as it is in standard geometry, so they ignore negative impulse from negative force regions

Correct move:

Mark all areas above the time axis as positive, all areas below as negative, then add the signed areas to get total impulse

Wrong move:

Treating the applied contact force as equal to net force for any collision, ignoring the object's weight

Why:

Students incorrectly assume contact force is always much larger than weight, so net force equals applied force, which fails when contact time is not extremely small

Correct move:

Always use $F_{net}$ for impulse calculations; if contact time is long enough that gravity contributes non-negligible impulse, include weight in your net force

Wrong move:

Adding impulses from internal forces when calculating total impulse for a system

Why:

Students confuse internal and external forces, so they add impulses from paired internal forces to the system's total impulse

Correct move:

For total impulse on a system, only add impulses from external forces; internal forces cancel out due to Newton's third law and contribute nothing

Wrong move:

Using the rectangle area formula ( $ba se \times h e i g h t$ ) for triangular regions on a force-time graph

Why:

Students rush through area calculation and mix up shape area formulas

Correct move:

Label each region of the F-t graph with its shape before calculating area, and write the correct area formula next to each shape

6. Quick Reference Cheatsheet

Category	Formula	Notes
Impulse (variable force)	$J = \int_{t_{1}}^{t_{2}} F_{net} (t) d t$	Equals net area under force vs. time graph; positive for F above axis, negative for F below
Impulse (constant net force)	$J = F_{net} Δ t$	Simplification for constant net force over the interaction interval
Impulse-Momentum Theorem (single object)	$J = Δ p = p_{f} - p_{i}$	Net impulse equals change in object's momentum; valid for all inertial reference frames
Linear Momentum	$p = m v$	Vector quantity; direction matches velocity direction
Impulse-Momentum Theorem (systems)	$J_{ext, net} = Δ P_{total}$	Internal forces cancel, so only external impulses contribute to total momentum change
Conservation of Momentum (special case)	$P_{initial} = P_{final}$	Applies when net external impulse on the system equals zero
Average Force from Impulse	$F_{avg} = \frac{Δ p}{Δ t}$	Used to find average contact force for short collision interactions

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Impulse calculation from force-time graph
2022 · FRQ
Average force on a colliding baseball

Going deeper

official syllabusAP Physics 1 Course and Exam DescriptionOfficial topic framework and learning objectives

What's Next

Mastering the impulse-momentum theorem is an essential prerequisite for the next core topics in AP Physics 1 Unit 5, starting with conservation of momentum, which we derived here as a special case of the theorem when net external impulse is zero. Without a solid understanding of how internal and external forces contribute to impulse, you cannot correctly apply conservation of momentum to collisions and explosions, a major free-response topic on the AP exam. Beyond Unit 5, this theorem forms the foundation for understanding rocket propulsion, and it connects to energy conservation when distinguishing between elastic and inelastic collisions. The graphical reasoning skills you practiced here with force-time graphs also transfer directly to work-energy theorem problems with force-displacement graphs, a key topic in Unit 3.