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AP Physics C: Mechanics • Linear Momentum

Change in Momentum and Impulse

How forces acting over time produce changes in motion—the bridge between Newton's laws and momentum conservation.

Explore the lesson 18 min read
A motion gauge with symbols for mass and movement.
Inside this lesson 10 sections 5 questions 8 figures

SECTION 1

Historical Context & Motivation

Long before physicists formulated the modern concept of momentum, natural philosophers wrestled with a fundamental question: what quantity best characterizes a body in motion? The ancient notion of impetus—a kind of internal motive force imparted to a projectile—dominated medieval thinking. It was not until the Scientific Revolution that rigorous, quantitative treatments replaced these intuitive but imprecise ideas. The concept of impulse, the product of force and the time interval over which it acts, arose naturally once Newton cast his second law in its most general form—a form that deals not with acceleration alone, but with the rate of change of momentum.

1644 • Descartes' Quantity of Motion

René Descartes introduced quantitas motus (quantity of motion) as mass × speed, laying groundwork for the modern momentum concept despite lacking a vector formulation.

1687 • Newton's Principia

Isaac Newton published the Principia Mathematica, defining force as the rate of change of momentum (Lex II), implicitly introducing the impulse–momentum theorem.

1743 • d'Alembert's Principle

Jean le Rond d'Alembert reformulated Newton's laws in terms of virtual work, reinforcing that force × time yields a change in the momentum vector—a core idea in analytical mechanics.

1900s • Modern Applications

The impulse–momentum framework became essential in automotive crash testing, rocket propulsion analysis, and sports biomechanics, wherever forces vary rapidly over short time intervals.

The central question this lesson addresses is deceptively simple: if we know the net force on an object as a function of time, how do we calculate the resulting change in its momentum? And conversely, if we observe a momentum change, what can we infer about the force that produced it? These two perspectives—force-to-motion and motion-to-force—form the heart of the impulse–momentum theorem, and mastering them is essential for every AP Physics C student.

SECTION 2

Core Principles & Definitions

Before diving into the mathematics, it is crucial to establish a precise vocabulary. In everyday language "momentum" and "impulse" are used loosely, but in physics each term carries a sharply defined meaning rooted in Newton's second law. The following foundational ideas form the scaffolding for everything that follows.

1

Linear Momentum (p⃗)

Defined as p⃗ = mv⃗, momentum is a vector quantity with the same direction as velocity. Its SI unit is kg·m/s. Momentum quantifies 'how hard it is to stop' a moving object.
2

Net Force as dp⃗/dt

Newton's second law in its most general form states F⃗_net = dp⃗/dt. This reduces to F⃗ = ma⃗ only when mass is constant. The general form is indispensable for variable-mass systems like rockets.
3

Impulse (J⃗)

Impulse is defined as J⃗ = ∫F⃗ dt over the interaction interval. It is the time-integrated effect of a force and has the same SI unit as momentum (kg·m/s or equivalently N·s).
4

Impulse–Momentum Theorem

Directly integrating F⃗_net = dp⃗/dt yields J⃗ = Δp⃗ = p⃗_f − p⃗_i. The net impulse on a system equals the change in its momentum—no exceptions, no approximations.
5

Average Force

When the exact F(t) is unknown, the average force is defined by F⃗_avg = Δp⃗ / Δt. This yields the constant force that would produce the same impulse over the same interval.
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KEY TAKEAWAY

Think of impulse like filling a swimming pool: the total water delivered depends on both the flow rate (force) and how long you leave the hose running (time). A garden hose running for an hour can deliver the same total water as a fire hose running for a few seconds. Likewise, a small force applied for a long time can produce the same impulse—and therefore the same momentum change—as a large force applied briefly.

SECTION 3

Force–Time Diagrams & Impulse as Area

One of the most powerful representations in momentum analysis is the force versus time graph. Because impulse is defined as the integral J = ∫F dt, the impulse delivered to an object is geometrically equal to the area under the F(t) curve. This graphical interpretation is indispensable on the AP exam, where force–time data are frequently given in graph form and you must extract the impulse by computing that area.

Force vs. Time — Impulse as Area Under the CurveTime (ms)Force (N)05001000150020000246810Rising rampPlateau: F = 2000 NFalling rampJ = Area = 8.0 N·s
A typical collision force profile. The force rises from zero (at t = 2 ms), peaks at 2000 N, remains constant for 2 ms, then drops back to zero. The shaded area under the curve equals the impulse J = 8.0 N·s, which is identically equal to the change in momentum Δp of the struck object.

In the diagram above, the force profile is trapezoidal: a linear ramp from 0 to 2000 N over 2 ms, a 2 ms plateau at 2000 N, and a linear ramp back to 0 over another 2 ms. The total area consists of two triangles (each with area ½ × 2 × 10⁻³ s × 2000 N = 2.0 N·s) and one rectangle (2 × 10⁻³ s × 2000 N = 4.0 N·s), giving J = 2.0 + 4.0 + 2.0 = 8.0 N·s. This graphical approach—decomposing the area into simple geometric shapes—is the method you will use most often on the AP exam when the force is given as a piecewise-linear function of time. When the force is given as an analytic function, you will instead evaluate the integral directly.

SECTION 4

Mathematical Framework

The impulse–momentum theorem follows directly from Newton's second law through a straightforward integration. Starting from the vector form of the second law and integrating both sides over the time interval of interest, we arrive at an exact relationship between the net impulse and the resulting momentum change. Let us develop this derivation carefully, because variations of it appear routinely on the AP Physics C exam.

NEWTON'S SECOND LAW (GENERAL FORM)

F⃗net=dp⃗/dt=d(mv⃗)/dtF⃗_net = dp⃗/dt = d(mv⃗)/dt

F⃗_net is the net external force, p⃗ is the linear momentum, m is the mass, and v⃗ is the velocity. When m is constant, this reduces to F⃗_net = ma⃗.

IMPULSE DEFINITION (INTEGRAL FORM)

J⃗=∫(ti→tf)F⃗netdtJ⃗ = ∫(t_i → t_f) F⃗_net dt

J⃗ is the net impulse vector. The integral is taken over the full interaction time from tit_i to tft_f. Units: N·s = kg·m/s.

IMPULSE–MOMENTUM THEOREM

J⃗=Δp⃗=p⃗f−p⃗i=mv⃗f−mv⃗iJ⃗ = Δp⃗ = p⃗_f − p⃗_i = mv⃗_f − mv⃗_i

Derived by integrating both sides of F⃗_net = dp⃗/dt: ∫F⃗ dt = ∫dp⃗ = p⃗_f − p⃗_i. This is an exact vector equation valid for any time-varying force, assuming constant mass.

AVERAGE FORCE

F⃗avg=Δp⃗/Δt=(mv⃗f−mv⃗i)/(tf−ti)F⃗_avg = Δp⃗ / Δt = (mv⃗_f − mv⃗_i) / (t_f − t_i)

When you cannot resolve F(t) in detail, the average force provides the constant-force equivalent. This is especially useful in collision analysis where the force profile is unknown but the duration and velocity change are measurable.

It is worth emphasizing that the impulse–momentum theorem is not a separate postulate; it is a mathematical consequence of Newton's second law. Any problem solvable via F = ma can, in principle, be recast as an impulse problem, but the impulse formulation is particularly advantageous in two scenarios: (1) when the force varies with time in a way that makes direct acceleration analysis cumbersome, and (2) when the interaction time is very short (collisions), so that the net impulse captures the entire effect of a complicated force profile in a single integral.

💡

AP Exam Tip

On the free-response section, always begin impulse problems by writing J⃗ = Δp⃗ and defining your sign convention. Graders award points for clear setup, and a consistent sign convention prevents errors when velocities reverse direction during a collision.

SECTION 5

Impulse in Collisions — Extending Contact Time

One of the most practically significant consequences of the impulse–momentum theorem is its implication for collision safety. When an object undergoes a given change in momentum Δp, the impulse J = Δp is fixed regardless of how the collision unfolds. Since J = F_avg × Δt, increasing the contact time Δt necessarily decreases the average force experienced by the object. This principle underlies automotive crumple zones, airbags, helmet padding, and even the technique of "giving" with a catch in baseball—all of which work by extending the collision duration to reduce peak forces.

Same Impulse, Different Forces: Hard vs. Soft CollisionHard collision (short Δt)Soft collision (long Δt)Time (ms)Force (N)05101504k8k12kF_peak ≈ 12 000 NJ = ½ × 4 ms × 12 kN= 24 N·sTime (ms)Force (N)05101504k8k12kF_peak ≈ 4 000 NJ = ½ × 12 ms × 4 kN= 24 N·s
Both collisions deliver identical impulse (24 N·s) and therefore identical Δp. Left: a hard, short collision (Δt ≈ 4 ms) produces a peak force of ~12 000 N. Right: a soft, long collision (Δt ≈ 12 ms) produces only ~4 000 N peak force. Crumple zones, airbags, and padding exploit this relationship.

The side-by-side comparison above illustrates this vividly. Both triangular force profiles enclose the same area—24 N·s—so the momentum change is identical. However, the hard collision concentrates that impulse into 4 ms, generating a peak force three times larger than the soft collision, which spreads the same impulse over 12 ms. This is precisely why the AP exam might ask you to explain how a design modification (e.g., adding padding) reduces injury risk: the change in momentum is dictated by the initial and final velocities alone, so the only way to reduce F_avg is to increase Δt.

🛡️

DESIGN PRINCIPLE

For a fixed Δp, peak force scales inversely with contact time. Every safety device—from a catcher's mitt to a car's crumple zone—works by trading force magnitude for collision duration. This is an immediate, physical consequence of J = F_avg × Δt = Δp.

SECTION 6

Worked Example — Variable Force Impulse

Let us work through a problem that requires integration of a time-dependent force, a standard AP Physics C skill. This example illustrates the full mathematical procedure from force function to final velocity.

A 2.0 kg block, initially at rest on a frictionless surface, is struck by a force F(t) = (600 N/s)t − (100 N/s²)t² for 0 ≤ t ≤ 4.0 s. Find (a) the impulse delivered, (b) the final velocity, and (c) the average force during the interaction.

  1. 1

    Step 1 — Identify Given Values

    Mass m = 2.0 kg, initial velocity viv_i = 0 m/s, force function F(t) = 600t − 100t² (in SI units), and the time interval is t = 0 to t = 4.0 s. The motion is one-dimensional, so we work with scalar equations along the direction of F.
  2. 2

    Step 2 — Compute the Impulse via Integration

    J = ∫₀⁴ (600t − 100t²) dt = [300t² − (100/3)t³] evaluated from 0 to 4. Substituting: J = 300(16) − (100/3)(64) = 4800 − 6400/3 = 4800 − 2133.3 = 2666.7 N·s.
    J ≈ 2667 N·s
  3. 3

    Step 3 — Apply the Impulse–Momentum Theorem

    By the impulse–momentum theorem, J = Δp = mv_f − mv_i. Since viv_i = 0: vfv_f = J / m = 2666.7 / 2.0 = 1333.3 m/s.
    vfv_f ≈ 1333 m/s
  4. 4

    Step 4 — Calculate the Average Force

    F_avg = J / Δt = 2666.7 N·s / 4.0 s = 666.7 N. This is the constant force that, if applied for 4.0 s, would produce the same impulse and therefore the same momentum change.
    F_avg ≈ 667 N
  5. 5

    Step 5 — Reasonableness Check

    The force function F(t) = 600t − 100t² equals zero at t = 0 and t = 6 s, reaching a maximum of F(3) = 900 N at t = 3 s. Our average of 667 N is below the peak but above zero, which is consistent. The impulse is positive, yielding a positive final velocity in the direction of the applied force—also consistent.
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Common Mistake

Do not confuse F_avg with F_max. Students frequently compute the peak force and multiply by Δt, which overestimates the impulse whenever the force is not constant. Always integrate F(t) to find J, then use F_avg = J/Δt afterward if needed.

Separate idealized Physics C example: impulse is the force–time integral. With mass 2 kg starting from rest, the model implies a final speed of 4000/3 m/s.Separate idealized Physics C example: impulse is the force–time integral. With mass 2 kg starting from rest, the model implies a final speed of 4000/3 m/s.

SECTION 7

Impulse vs. Work — Two Sides of Force

Students often conflate impulse and work because both involve force. However, they capture fundamentally different aspects of a force's effect. Impulse (J = ∫F dt) measures how much a force changes an object's momentum, whereas work (W = ∫F·dx) measures how much a force changes an object's kinetic energy. The following table clarifies the distinction.

FeatureImpulse (J⃗)Work (W)
Definition∫F⃗ dt∫F⃗ · dr⃗
Scalar or Vector?VectorScalar
Quantity changedMomentum (p⃗)Kinetic energy (K)
Relevant theoremImpulse–MomentumWork–Energy
Integration variableTime (dt)Displacement (dr⃗)
UnitsN·s = kg·m/sN·m = J (joules)
Best used when...Time interval or velocity change is knownDisplacement or speed change is known

Impulse vs. Work: contrasting two fundamental effects of force

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STRATEGIC CHOICE ON THE EXAM

When a problem gives you time information (duration of a collision, F(t) graph), think impulse–momentum. When it gives you displacement information (distance traveled, F(x) graph), think work–energy. Choosing the right theorem is itself a skill the AP exam tests explicitly.

SECTION 8

Connections to Conservation Laws & Beyond

The impulse–momentum theorem is the single-object version of a far more powerful statement: the conservation of linear momentum for a system. When the net external force on a system is zero, the total impulse on the system is zero, and therefore total momentum is conserved. This link between impulse and conservation is the conceptual backbone of collision analysis, which occupies a significant portion of the AP Physics C curriculum.

ConceptImpulse–Momentum (This Lesson)Conservation of Momentum (Advanced)
Applies toA single object under a known net forceA system of objects with zero net external force
Key equationJ⃗ = Δp⃗ = mv⃗_f − mv⃗_ip⃗_system,i = p⃗_system,f
When to useExternal force is known (or its time integral)Internal forces dominate; external forces negligible
Derivation rootNewton's 2nd Law: F⃗ = dp⃗/dtNewton's 3rd Law + 2nd Law for each body
Variable-mass extensionRocket equation (thrust = vev_e × dm/dt)Still holds for the rocket + exhaust system

From impulse–momentum to conservation of momentum

Looking beyond the AP C curriculum, the impulse concept generalizes elegantly into Lagrangian and Hamiltonian mechanics, where the generalized impulse (∫QiQ_i dt) drives changes in generalized momenta. In relativistic mechanics, the four-vector formulation of impulse accounts for the velocity-dependent mass (Lorentz factor γ), but the core theorem J = Δp retains its structure. Even in quantum field theory, the transfer of four-momentum between particles in scattering events is essentially an impulse concept—testifying to the extraordinary durability of the ideas you are learning in this lesson.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A ball of mass m is thrown against a wall and bounces back with the same speed. A second identical ball is thrown at the same speed but sticks to the wall (does not bounce). Which ball experiences the greater impulse from the wall?
  • A. The ball that sticks, because it transfers all of its kinetic energy to the wall
  • B. The ball that bounces, because its momentum reverses direction, yielding |Δp| = 2mv
  • C. Both experience the same impulse, because they have the same mass and initial speed
  • D. It depends on the elasticity of the wall, which is not specified
Show answer (B)
The ball that bounces experiences the greater impulse. Define the positive direction as toward the wall. For the bouncing ball, viv_i = +v and vfv_f = −v, so Δp = m(−v) − m(+v) = −2mv, giving |J| = 2mv. For the ball that sticks, viv_i = +v and vfv_f = 0, so Δp = 0 − mv = −mv, giving |J| = mv. The bouncing ball's momentum reverses direction, requiring twice the impulse magnitude of the ball that merely stops. This is a common AP conceptual question that tests whether students recognize that direction changes amplify the magnitude of Δp.

Question 2: BASIC CALCULATION

A 0.145 kg baseball traveling at 40.0 m/s is hit by a bat and leaves at 55.0 m/s in the opposite direction. The bat and ball are in contact for 1.20 ms. What is the magnitude of the average force exerted on the ball by the bat?
  • A. 6 650 N
  • B. 7 960 N
  • C. 11 500 N
  • D. 13 800 N
Show answer (C)
Define the initial direction of the ball as positive. Then viv_i = +40.0 m/s and vfv_f = −55.0 m/s (opposite direction). Δp = m(vfv_f − viv_i) = 0.145(−55.0 − 40.0) = 0.145(−95.0) = −13.775 kg·m/s. The magnitude of the impulse is |J| = 13.775 N·s. Using F_avg = |J|/Δt = 13.775 / 1.20 × 10⁻³ ≈ 11 480 N, or approximately 11.5 kN. This large force illustrates why bat–ball collisions involve forces far exceeding the ball's weight.

Question 3: INTERMEDIATE

A 3.0 kg object initially moving at 4.0 m/s in the +x-direction is acted upon by a force F(t) = (12 − 3t) N for 0 ≤ t ≤ 4.0 s. What is the velocity of the object at t = 4.0 s?
  • A. 8.0 m/s
  • B. 10.0 m/s
  • C. 12.0 m/s
  • D. 16.0 m/s
Show answer (C)
Compute the impulse: J = ∫₀⁴ (12 − 3t) dt = [12t − 1.5t²] from 0 to 4 = (48 − 24) − 0 = 24 N·s. Apply the impulse–momentum theorem: J = m(vfv_f − viv_i), so 24 = 3.0(vfv_f − 4.0). Solving: vfv_f − 4.0 = 8.0, therefore vfv_f = 12.0 m/s in the +x direction. Note that the force is positive for the entire interval (F(4) = 0), so the object accelerates in the +x direction throughout.

Question 4: APPLIED

4 pts
An automotive engineer is testing a crash barrier. A 1400 kg test vehicle traveling at 15.0 m/s strikes the barrier and comes to rest. In Design A, the rigid barrier stops the car in 0.080 s. In Design B, a deformable barrier stops the car in 0.45 s. (a) Calculate the impulse delivered to the car in each case. (b) Calculate the average force on the car for each design. (c) Explain, using the impulse–momentum theorem, why Design B is safer for passengers. (d) If the force in Design B varies as F(t) = F₀ sin(πt / 0.45) during 0 ≤ t ≤ 0.45 s, determine F₀.
Show answer
(a) [1 pt] The impulse is the same in both cases: J = Δp = m(vfv_f − viv_i) = 1400(0 − 15.0) = −21 000 N·s. The magnitude is 21 000 N·s. The impulse depends only on the momentum change, not on the collision duration. (b) [1 pt] Design A: F_avg = |J|/Δt = 21 000 / 0.080 = 262 500 N ≈ 263 kN. Design B: F_avg = 21 000 / 0.45 = 46 667 N ≈ 46.7 kN. (c) [1 pt] Since J = Δp is identical (fixed by the velocity change), and J = F_avg × Δt, increasing Δt by a factor of 0.45/0.080 ≈ 5.6 reduces the average force by the same factor. Lower forces mean lower decelerations (a = F/m), reducing injury risk to passengers. Design B is safer because it extends the stopping time. (d) [1 pt] The impulse magnitude equals |∫₀^{0.45} F₀ sin(πt/0.45) dt| = F₀ [−(0.45/π) cos(πt/0.45)] from 0 to 0.45 = F₀ (0.45/π)[−cos(π) + cos(0)] = F₀ (0.45/π)(2) = 0.9F₀/π. Setting this equal to 21 000: F₀ = 21 000π / 0.9 ≈ 73 300 N ≈ 73.3 kN.

Question 5: CRITICAL THINKING

4 pts
A 0.50 kg cart on a frictionless track starts from rest. A net force is applied to the cart as shown in the graph below: • From t = 0 to t = 2.0 s: F increases linearly from 0 to 6.0 N. • From t = 2.0 s to t = 5.0 s: F = 6.0 N (constant). • From t = 5.0 s to t = 7.0 s: F decreases linearly from 6.0 N to −3.0 N. (a) Calculate the total impulse delivered from t = 0 to t = 7.0 s. (b) Determine the velocity of the cart at t = 7.0 s. (c) At what time between t = 5.0 s and t = 7.0 s does the cart reach its maximum velocity? Justify your answer. (d) Determine the maximum velocity of the cart.
Show answer
(a) [1 pt] Break the F(t) graph into three regions. Region 1 (0 to 2 s): triangle with area ½(2.0)(6.0) = 6.0 N·s. Region 2 (2 to 5 s): rectangle with area (3.0)(6.0) = 18.0 N·s. Region 3 (5 to 7 s): the force is linear from 6.0 N to −3.0 N, forming a trapezoid with area ½(2.0)(6.0 + (−3.0)) = ½(2.0)(3.0) = 3.0 N·s. Total impulse: J = 6.0 + 18.0 + 3.0 = 27.0 N·s. (b) [1 pt] Using J = Δp = mv_f − mv_i and viv_i = 0: vfv_f = J/m = 27.0 / 0.50 = 54.0 m/s at t = 7.0 s. (c) [1 pt] The cart reaches maximum velocity when its acceleration is zero, which occurs when F(t) = 0. In Region 3, the force is linear: F(t) = 6.0 − (9.0/2.0)(t − 5.0) = 6.0 − 4.5(t − 5.0). Setting F = 0: 6.0 = 4.5(t − 5.0), so t − 5.0 = 4/3 s, giving t = 5.0 + 4/3 ≈ 6.33 s. Before this time the force is positive (accelerating the cart); after this time the force is negative (decelerating the cart). Therefore the velocity is maximum at t ≈ 6.33 s. (d) [1 pt] The impulse from t = 0 to t = 6.33 s is the sum of the first two regions (24.0 N·s) plus the impulse from 5.0 to 6.33 s. In this sub-interval, the force goes from 6.0 N to 0 (a triangle): area = ½(4/3)(6.0) = 4.0 N·s. Total impulse to t = 6.33 s: 6.0 + 18.0 + 4.0 = 28.0 N·s. Maximum velocity: v_max = 28.0 / 0.50 = 56.0 m/s.

SUMMARY

Summary — Change in Momentum and Impulse

Summary

The impulse–momentum theorem states that the net impulse on an object equals its change in momentum: J⃗ = Δp⃗ = mv⃗_f − mv⃗_i. Impulse is computed as the time integral of force, J⃗ = ∫F⃗ dt, which corresponds geometrically to the area under a force–time curve. When the exact F(t) is unknown, the average force is defined as F_avg = Δp/Δt.

For a fixed momentum change, extending the contact time reduces the peak and average forces—the principle behind airbags, crumple zones, and protective padding. Unlike work (which changes kinetic energy via ∫F·dx), impulse changes momentum via ∫F dt. On the AP exam, choose the impulse approach whenever time-domain information is provided, and always establish a clear sign convention before computing Δp.

Varsity Tutors • AP Physics C: Mechanics • Change in Momentum and Impulse