Unit 4: Linear Momentum · First Principles Physics

AP Physics 1 · Unit 04 of 08

Linear Momentum.

Energy is the universe's most loved conservation law. Momentum is its most underrated — and the only one that can solve a collision in two lines. This unit teaches you when to reach for which.

Exam Weight · 10–15% Class Periods · ~10–15 CED Topics · 4.1–4.4 Difficulty · Moderate

The lesson

A second conservation law — and the right tool for collisions

A bowling ball strikes a single pin at 8 m/s. The pin flies off; the ball slows down. How fast is each one moving afterward? You could try Newton's laws — work out the force during impact, the contact time, the deformation of the rubber, the tiny vibrations propagating through the wood. You'd need a supercomputer. Or you could use the law you're about to meet, write down two lines of algebra, and have the answer in thirty seconds.

Energy is one conservation law. Momentum is another — and unlike energy, momentum doesn't care whether a collision is gentle or violent, elastic or inelastic, between identical objects or wildly different ones. As long as no external force acts on the colliding system, total momentum is conserved. Always. Every collision in the universe obeys this law.

The single most important sentence in this unit: when two objects interact and no external force acts on them, the total momentum before equals the total momentum after — even if energy is lost, even if the objects stick together, even if they shatter into pieces.

Linear momentum: the definition

Linear momentum is mass times velocity:

$$\vec p = m\vec v$$

Units: kg·m/s. Notice this is a vector — momentum has the same direction as velocity. A 2 kg ball moving east at 5 m/s has momentum 10 kg·m/s east. A 2 kg ball moving west at 5 m/s has momentum 10 kg·m/s west. These are different momenta — they have opposite signs along the east-west axis.

The vector nature is what trips up students early on. In a 1D problem with two objects moving in opposite directions, you can't just add their masses and velocities — you have to track signs. A 2 kg ball at +5 m/s and a 3 kg ball at −2 m/s have total momentum $10 - 6 = +4$ kg·m/s, not 16 kg·m/s.

Impulse: how momentum changes

If a force $F$ acts on an object for time $\Delta t$, the object's momentum changes by an amount called the impulse:

$$\vec J = \vec F \, \Delta t = \Delta \vec p$$

Units of impulse: N·s — which equals kg·m/s, the same as momentum. This is the impulse-momentum theorem, and it's the rotational... wait, no. Excuse me. It's the linear version of the angular impulse-momentum theorem you'll meet (or have already met) in Unit 6 ($\tau \Delta t = \Delta L$). Same structure, same logic.

Impulse is also $F \Delta t$ for a constant force, but for a variable force it's the area under the force-vs-time curve:

$$J = \int F \, dt = \text{area under } F\text{-}t \text{ graph}$$

This shows up in graph-reading problems on the AP exam — given an $F$-$t$ graph for a collision, find the impulse (= area) and use that to find the change in momentum.

Why crumple zones exist

If a 1500 kg car going 30 m/s comes to rest in 1 second, the average force was $F = \Delta p / \Delta t = 45000 / 1 = 45{,}000$ N. If instead the car comes to rest in 0.1 seconds (rigid frame, no crumple), the force is 450,000 N — ten times larger. The change in momentum is the same; the time interval is different. Engineers design cars to crumple slowly so that the same momentum change happens over a longer time, reducing peak force on occupants. Same principle: airbags, gymnastic mats, helmets, dropping a phone on carpet vs concrete.

Newton's second law, rewritten

The impulse-momentum theorem actually contains Newton's second law. Divide both sides of $F \Delta t = \Delta p$ by $\Delta t$, and use the fact that $\Delta p / \Delta t$ is the rate of change of momentum:

$$\vec F_{\text{net}} = \dfrac{d\vec p}{dt}$$

For an object with constant mass, $p = mv$ gives $dp/dt = m \cdot dv/dt = ma$ — the familiar $F = ma$. But this momentum-form is more general. It also handles cases where the mass changes over time (like a rocket burning fuel), though AP Physics 1 doesn't ask you to compute those.

Conservation of momentum: the big idea

Now the punchline. If there's no external force on a system, then by Newton's second law, the system's total momentum doesn't change:

$$\text{If } \vec F_{\text{ext}} = 0, \text{ then } \sum \vec p_i = \text{constant}$$

Equivalently, for a collision (or any interaction):

$$\sum \vec p_{i,\text{before}} = \sum \vec p_{i,\text{after}}$$

This is conservation of linear momentum. It applies to every collision, every explosion, every interaction where no significant external force acts during the brief interaction time. The qualifier matters — gravity acts during a collision too, but if the collision is fast enough (which is true for almost any collision you'll see), gravity's tiny impulse over that short time is negligible.

Worked example 1 · momentum conservation

Train cars coupling

A 4000 kg train car moving east at 8 m/s collides with and couples to a stationary 6000 kg train car. Find the velocity of the coupled cars after collision.

Apply momentum conservation

$p_{\text{before}} = (4000)(8) + (6000)(0) = 32{,}000$ kg·m/s east.

After: total mass 10,000 kg moving at velocity $v_f$. $p_{\text{after}} = 10{,}000 v_f$.

Setting equal: $10{,}000 v_f = 32{,}000 \Rightarrow v_f = 3.2$ m/s east.

$v_f = 3.2$ m/s east

Types of collisions

All collisions conserve momentum (as long as no external forces act). They differ in what happens to kinetic energy:

Type	Momentum	Kinetic energy	Example
Elastic	Conserved	Conserved	Billiard balls (approximately), molecular collisions
Inelastic	Conserved	Decreases	Most real collisions — energy lost to heat, sound, deformation
Perfectly inelastic	Conserved	Decreases (max)	Objects stick together — clay, train cars coupling

Notice momentum is conserved in all three. The difference is energy. In a perfectly inelastic collision (where the objects stick together), the maximum possible amount of kinetic energy is lost. In an elastic collision, no energy is lost — kinetic energy before equals kinetic energy after. Real collisions are usually somewhere in between (inelastic but not perfectly so).

A useful trick for elastic collisions

For a 1D elastic collision between two objects, you can show using both conservation laws (momentum and KE) that the relative velocity reverses: $v_{1f} - v_{2f} = -(v_{1i} - v_{2i})$. This is a slick alternative to the messy quadratic that comes from solving the two equations directly. AP Physics 1 doesn't require you to solve elastic-collision systems quantitatively in 2D, but in 1D, this relative-velocity trick is fast.

Worked example 2 · perfectly inelastic + energy comparison

Bullet embeds in block

A 0.05 kg bullet moving at 400 m/s embeds in a 2 kg block initially at rest. Find (a) the speed of the block+bullet just after collision, (b) the fraction of the bullet's initial KE lost.

(a) Apply momentum conservation

$(0.05)(400) + (2)(0) = (2.05)v_f \Rightarrow v_f = 20/2.05 \approx 9.76$ m/s.

(b) Compare KE before and after

$K_i = \tfrac{1}{2}(0.05)(400)^2 = 4000$ J. $K_f = \tfrac{1}{2}(2.05)(9.76)^2 \approx 97.6$ J.

Fraction lost: $(4000 - 97.6)/4000 \approx 0.976$ — about 97.6% of the bullet's KE is lost. (To heat, deformation of the block, sound.)

(a) $v_f \approx 9.76$ m/s; (b) ~97.6% of KE lost.

Choosing systems carefully

Just like in Unit 3, the trick to using momentum conservation cleanly is to define your system wisely. If a cart is rolling and a block is dropped on top of it, the block's vertical fall is irrelevant if you choose your system as just the block-cart along the horizontal direction. The key idea: pick your system to include all objects that interact, and analyze along the direction with no external force.

The 2025 AP Physics 1 exam Question 1 turned on exactly this idea — a vertically-falling block sticks to a horizontally-moving cart. Vertical momentum changes (gravity is external), but horizontal momentum is conserved (no horizontal external force on the block-cart system). Most students who lost points missed that distinction.

Two-dimensional collisions

Momentum is a vector, so when collisions happen in 2D, you conserve momentum independently in the $x$ and $y$ directions:

$$\sum p_{x,i} = \sum p_{x,f}, \qquad \sum p_{y,i} = \sum p_{y,f}$$

AP Physics 1 only asks for semi-quantitative analysis of 2D collisions (the CED explicitly excludes solving simultaneous equations for full 2D problems). You'll be asked to identify which conservation laws apply, sketch outcomes, or estimate quantities — not crank through the algebra. AP Physics 2 covers full 2D collisions.

Center of mass and momentum

An elegant fact: the total momentum of a system equals the total mass times the velocity of the center of mass:

$$\vec p_{\text{total}} = M_{\text{total}} \vec v_{\text{cm}}$$

So momentum conservation is equivalent to: in the absence of external forces, the center of mass of an isolated system moves in a straight line at constant velocity. In an explosion, fragments fly in all directions — but their center of mass continues moving exactly as if nothing had exploded. This is sometimes a fast way to solve "fragmentation" problems.

Energy vs momentum: when to use which

Both are conservation laws, both are powerful. When should you use which?

Use momentum for collisions, explosions, recoil, or any short-duration interaction with no external force. Momentum cares only about the start and end velocities — not what happens during the collision.
Use energy for problems involving heights, springs, or long-duration paths. Energy methods are better when forces vary continuously over a path.
Use both for elastic collisions and ballistic-pendulum-style problems where one phase is a collision (momentum) and the next phase is energy conservation (e.g., a pendulum swinging up after impact).

Ballistic pendulum — the classic two-phase problem

A bullet embeds in a hanging block, and the block swings up to height $h$. Phase 1: collision. Use momentum conservation to find the speed of (block + bullet) just after collision. KE is NOT conserved — energy is lost in the inelastic collision. Phase 2: pendulum swing. Use energy conservation to find how high (block + bullet) rise. KE turns into gravitational PE. Each phase uses a different conservation law. Mixing them up — using energy conservation for the collision — is the most common mistake on this problem.

Where this unit goes next

Unit 6 (Energy and Momentum of Rotating Systems) introduces angular momentum — the rotational version of linear momentum, with an analogous conservation law. The skater pulling in arms, the planet orbiting closer to the sun, the gymnast tucking mid-flip: all are angular momentum conservation, the rotational counterpart to what you've learned here. Master linear momentum first; rotational momentum is an extension.

Conceptual summary

Linear momentum is $\vec p = m\vec v$ — a vector. Impulse equals change in momentum: $\vec J = \vec F \Delta t = \Delta \vec p$. When no external force acts on a system, the total momentum is conserved: $\sum \vec p_{\text{before}} = \sum \vec p_{\text{after}}$. This holds for elastic, inelastic, and perfectly inelastic collisions — only kinetic energy distinguishes them. Choose your system to include all interacting objects, and analyze along directions with no external force. In two dimensions, momentum is conserved independently in each direction. The total momentum of a system equals total mass times the center-of-mass velocity.

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Reference sheet

Every equation in Unit 4

★ Linear momentum

$$\vec p = m\vec v$$

$m$: mass (kg)
$\vec v$: velocity vector
Units: kg·m/s

Use when calculating momentum of a single object. Vector — direction matters.

★ Impulse-momentum theorem

$$\vec J = \vec F \Delta t = \Delta \vec p$$

$\vec J$: impulse (N·s = kg·m/s)
$\Delta t$: time interval

Use when a force acts over a known time, or finding average force given momentum change.

Impulse from variable force

$$J = \text{area under } F\text{-}t \text{ graph}$$

Read: area = impulse

Use when given a force-time graph (variable force). Area under curve = impulse.

Newton's second law (momentum form)

$$\vec F_{\text{net}} = \dfrac{d\vec p}{dt}$$

For constant mass: reduces to $F = ma$

Use when rate of momentum change is needed. The slope of $p$-$t$ graph = force.

★ Conservation of momentum

$$\sum \vec p_{\text{before}} = \sum \vec p_{\text{after}}$$

Condition: no external force on system

Use when objects collide, explode, or push off each other. The workhorse equation.

Center-of-mass velocity

$$\vec v_{\text{cm}} = \dfrac{\sum m_i \vec v_i}{\sum m_i}$$

For 2 objects: $v_{\text{cm}} = (m_1 v_1 + m_2 v_2)/(m_1 + m_2)$

Use when finding the velocity of a system's center of mass. Constant if no external force.

Total momentum and CM

$$\vec p_{\text{total}} = M_{\text{total}} \vec v_{\text{cm}}$$

Total momentum: system mass × CM velocity

Use when shortcut for total momentum of multi-object systems.

Elastic collision (1D, relative velocity)

$$v_{1f} - v_{2f} = -(v_{1i} - v_{2i})$$

For elastic only: relative velocity reverses

Use when 1D elastic collision; faster than the quadratic from KE conservation.

Perfectly inelastic collision

$$m_1 v_1 + m_2 v_2 = (m_1 + m_2)v_f$$

Both objects: have same final velocity $v_f$

Use when objects stick together. Maximum KE lost. Single-equation problem.

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Study guide

One page. Night before the test.

The 6 concepts that matter most

Momentum is mass times velocity ($p = mv$) — a vector. Direction matters; signs matter in 1D.
Impulse equals change in momentum: $F\Delta t = \Delta p$. Bigger force or longer time gives more momentum change.
Conservation of momentum applies whenever no external force acts. Momentum before = momentum after, period.
All collisions conserve momentum. Only elastic ones conserve KE. Inelastic = energy lost; perfectly inelastic = stuck together = max energy lost.
Choose your system to eliminate external forces. The 2025 AP exam featured a problem where vertical momentum wasn't conserved (gravity acts) but horizontal momentum was — recognizing this was the key.
Use momentum for collisions; energy for heights and springs; both for ballistic pendulum-style problems.

Key equations

$p = mv$
$J = F\Delta t = \Delta p$ (impulse)
$F_{\text{net}} = dp/dt$
$\sum p_i = \sum p_f$ (conservation)
$m_1 v_1 + m_2 v_2 = (m_1 + m_2)v_f$ (perfectly inelastic)
$v_{1f} - v_{2f} = -(v_{1i} - v_{2i})$ (elastic, 1D)
$p_{\text{tot}} = M v_{\text{cm}}$

5 traps students fall for

Forgetting that momentum is a vector — adding speeds without signs.
Trying to use energy conservation for a collision (only elastic conserves KE).
Forgetting that perfectly inelastic = stuck together (single $v_f$ for both).
Not recognizing when a problem requires both momentum AND energy (ballistic pendulum).
Confusing impulse (vector, in N·s) with power (scalar, in W).

Problem-solving checklist

Identify the type of problem: collision/explosion → momentum first.
Pick your system: include all interacting objects.
Check for external forces: if any (over the duration of interest), think about which directions are affected. Often horizontal is conserved while vertical isn't.
Identify the collision type: elastic, inelastic, or perfectly inelastic.
Write conservation of momentum. If perfectly inelastic, single unknown $v_f$. If elastic, also use the relative-velocity equation.
If a swing/lift follows the collision: apply energy conservation for that phase, separately.

If you only remember one thing For any collision or interaction problem: write down conservation of momentum. Check if KE is also conserved (only elastic). If neither object has external forces during the interaction, momentum is conserved in every direction — no exceptions. This single principle, plus careful sign-tracking, solves nearly every Unit 4 problem.

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Worksheet A · Foundation

Ten problems building from basic to moderate

Use $g = 10$ m/s² throughout. Track signs carefully — momentum is a vector.

Foundation

A 0.5 kg ball moves at 6 m/s east. Find its momentum.

Foundation

A 1500 kg car traveling at 20 m/s comes to rest in 4 s. Find (a) its change in momentum, (b) the average net force on the car.

Foundation

A 0.15 kg baseball moving at 30 m/s east is hit and rebounds at 40 m/s west. Find the impulse on the baseball.

Moderate

A 3 kg cart moving at 4 m/s east collides with and sticks to a 1 kg cart at rest. Find the velocity of the combined cart immediately after collision.

Moderate

A 70 kg person stands on a frictionless ice surface and throws a 5 kg ball horizontally at 8 m/s. Find the velocity of the person after the throw.

Moderate

A 2 kg cart moving at 5 m/s east collides elastically with a stationary 3 kg cart. Using the relative-velocity equation along with momentum conservation, find both carts' velocities after the collision.

Moderate

A 0.05 kg bullet moving at 300 m/s embeds in a 4 kg block initially at rest. Find (a) the speed of block+bullet after collision, (b) the kinetic energy lost in the collision.

Moderate

An 800 g pumpkin is dropped from a 5 m height. It strikes the ground and comes to rest in 0.02 s. Find the average force on the pumpkin during the impact.

Moderate

An object of mass 4 kg moving at 6 m/s east collides head-on with a 2 kg object moving at 9 m/s west. They stick together. Find the velocity of the combined object after collision.

Moderate

A 2 kg object initially at rest explodes into two pieces. One piece (mass 0.5 kg) flies east at 12 m/s. Find the velocity of the other piece (mass 1.5 kg).

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Worksheet B · AP exam style

Ten problems in AP exam format

Part I — Multiple Choice

AP MC

A 1 kg ball moving at 4 m/s east collides with and sticks to a 3 kg block at rest. The final velocity of the combined system is:

1 m/s east
2 m/s east
3 m/s east
4 m/s east

AP MC

Two carts collide on a frictionless track. Cart A (mass 2 kg, velocity +3 m/s) and cart B (mass 1 kg, velocity −6 m/s). After the collision, cart A moves at +1 m/s. What is cart B's velocity?

−2 m/s
−1 m/s
+2 m/s
+3 m/s

AP MC

A net force is applied to a 2 kg object as shown in the graph. The object starts at rest. What is the object's speed at $t = 4$ s?

3 m/s
6 m/s
9 m/s
12 m/s

AP MC

An astronaut floats in space and throws a wrench. The wrench moves to the right; the astronaut moves to the left. Which of the following is true?

The astronaut and wrench have equal speeds.
The astronaut and wrench have equal momenta.
The astronaut's momentum has the same magnitude but opposite direction as the wrench's momentum.
The astronaut and wrench have equal kinetic energies.

AP MC

A 0.05 kg ball is thrown at 20 m/s and caught — it comes to rest in the catcher's glove. The catcher backs the glove up to extend the catching time from 0.05 s to 0.5 s. The peak force on the ball is reduced by a factor of:

Part II — Short Free Response

AP SFR

(~7 min) A bullet of mass $m$ moving with velocity $v_0$ strikes and embeds in a wooden block of mass $M$ suspended from a long string (a "ballistic pendulum"). The block-bullet system swings up to a maximum height $h$.

Starting with momentum conservation, derive an expression for the velocity of block+bullet just after the collision in terms of $m$, $M$, and $v_0$.
Starting with energy conservation, derive an expression for the height $h$ reached, in terms of the bullet's initial velocity $v_0$ and the relevant masses.
Is kinetic energy conserved during the collision? Justify your answer.

AP SFR

(~7 min) A 0.4 kg ball is thrown horizontally and hits a wall at 12 m/s. It rebounds at 8 m/s. The contact time with the wall is 0.04 s.

Calculate the change in momentum of the ball.
Calculate the average force the wall exerts on the ball.
By Newton's third law, what force does the ball exert on the wall? Compare and explain.

AP SFR

(~7 min) A 60 kg ice skater throws a 1.5 kg snowball at 20 m/s east. Initially both are at rest on frictionless ice.

Find the skater's recoil velocity after throwing the snowball.
Calculate the kinetic energy of the snowball and the skater after the throw. Where did this kinetic energy come from?
The skater catches a returning 1.5 kg snowball traveling at 20 m/s east. Find the skater's final velocity. Justify using momentum conservation.

Part III — Extended Free Response

AP EFR

(~18 min — mirrors 2025 AP Physics 1 FRQ Q1) A student has a cart of mass $m_c$ moving to the right at constant speed $v_c$ on a horizontal surface. The student releases a block of mass $\tfrac{1}{5}m_c$ from rest above the cart. The block falls onto the cart and sticks (no sliding). After the collision, the system continues to the right at speed $v_f$.

Sketch a graph of the magnitude of the $x$-component of the system's momentum ($p_x$) as a function of time, from before the collision through after. Mark the collision time.
Starting from conservation of momentum, derive an expression for $v_f$ in terms of $m_c$ and $v_c$.
Derive an expression for the change in kinetic energy $\Delta K$ of the block-cart system, in terms of $m_c$ and $v_c$. Is $\Delta K$ positive, negative, or zero? Justify physically.
In a different scenario, the block slides briefly on the cart (with friction between block and cart) before they move together. During this brief sliding interval $\Delta t$, indicate whether the $x$-component of momentum of the new block-cart system increases, decreases, or remains constant. Justify your answer.

AP EFR

(~18 min — experimental design) A student wants to measure the speed of a fired projectile (a small ball of mass $m$) using a pendulum. The student has a hanging block of mass $M$ on a long string, a meterstick, a stopwatch, and a small piece of clay attached to the block to ensure the projectile sticks on impact.

Describe a procedure to determine the projectile's launch speed using only this equipment. Specify what to measure and how to reduce experimental uncertainty.
Derive a symbolic expression for the projectile's launch speed $v_0$ in terms of the masses and the height $h$ reached by the block-projectile after collision.
If $m = 50$ g, $M = 2$ kg, and $h = 0.20$ m, calculate $v_0$.
What fraction of the projectile's initial KE is dissipated in the collision? Express in terms of $m$ and $M$.
Why does this experiment require a momentum equation followed by an energy equation, rather than energy conservation alone?

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Worksheet C · Challenge

Five problems at olympiad level

Challenge

Elastic collision with a much heavier object. A small ball of mass $m$ moving at $v_0$ collides elastically with a stationary heavy block of mass $M \gg m$. Show that the ball rebounds at speed essentially $v_0$ (with a sign flip), and the heavy block barely moves. Then take the opposite limit ($m \gg M$) and show what happens. Use the 1D elastic-collision result to derive both.

Challenge

2D collision with given outcome. A 1 kg ball moving east at 4 m/s collides with a stationary 1 kg ball. After the collision, the first ball moves at 3 m/s in a direction 30° north of east. Find the velocity (magnitude and direction) of the second ball. Then determine whether the collision was elastic.

Challenge

Rocket recoil (constant exhaust speed). A rocket of total mass $M$ (including fuel) ejects fuel at a constant rate $\mu = dm/dt$ (kg/s) at speed $v_e$ relative to the rocket. (a) Use the impulse-momentum theorem to derive an expression for the thrust force on the rocket. (b) If the rocket is initially at rest in deep space (no gravity) and burns fuel for time $t$ (with mass at end $M - \mu t$), derive an expression for its speed using the rocket equation $\Delta v = v_e \ln(M_0/M_f)$. State the assumption needed to derive this without calculus.

Challenge

Newton's cradle. Five identical balls hang in a row in contact with each other. The leftmost ball is pulled back and released, swinging into the line at speed $v$. The rightmost ball flies out at speed $v$; the middle three balls remain stationary. Using both momentum AND energy conservation, prove this is the only possible outcome consistent with both conservation laws (assuming all collisions are elastic and 1D). What if you released two balls at speed $v$ instead of one — what would happen?

Challenge

Pendulum collision with maximum compression. A 2 kg block hangs at rest from a long string. A spring ($k = 5000$ N/m) is attached to the front of the block. A 0.5 kg ball moving horizontally at 30 m/s strikes the spring and compresses it (the ball bounces off without sticking). Using both momentum AND energy conservation, find (a) the maximum compression of the spring, (b) the ball's velocity after rebounding from the spring, (c) the block's maximum velocity. (Hint: at max compression, ball and block have the same velocity.)

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Complete worked solutions

Every problem, every step

Worksheet A — Foundation

· ball momentum

$p = mv = (0.5)(6) = 3$ kg·m/s east.

$p = 3$ kg·m/s east

· car braking

(a) $\Delta p = m\Delta v = 1500(0 - 20) = -30{,}000$ kg·m/s. Magnitude 30 kN·s, direction opposite to motion.

(b) $F = \Delta p / \Delta t = -30{,}000/4 = -7500$ N. Magnitude 7.5 kN, direction opposite to motion.

$\Delta p = 30$ kN·s; $F_{\text{avg}} = 7.5$ kN.

· baseball impulse

Take east as positive. $p_i = (0.15)(30) = +4.5$ kg·m/s. $p_f = (0.15)(-40) = -6.0$ kg·m/s. $J = \Delta p = -6.0 - 4.5 = -10.5$ kg·m/s. Magnitude 10.5 N·s westward.

$J = 10.5$ N·s west

Don't forget to flip the sign when the ball reverses direction. Forgetting this is the most common impulse error — students compute $\Delta p = (40 - 30)(0.15) = 1.5$ instead of (40 + 30)(0.15) = 10.5.

· perfectly inelastic carts

Conservation: $(3)(4) + (1)(0) = (4)v_f \Rightarrow v_f = 3$ m/s east.

$v_f = 3$ m/s east

· skater throws ball

Initial momentum is zero. After throw: $0 = (5)(8) + (70)v_p \Rightarrow v_p = -40/70 \approx -0.571$ m/s.

$v \approx 0.57$ m/s west (opposite the ball)

· elastic collision (1D)

Momentum: $(2)(5) + 0 = 2v_{1f} + 3v_{2f} \Rightarrow 10 = 2v_{1f} + 3v_{2f}$.

Relative velocity reverses: $v_{1f} - v_{2f} = -(5 - 0) = -5 \Rightarrow v_{1f} = v_{2f} - 5$.

Substitute: $2(v_{2f} - 5) + 3v_{2f} = 10 \Rightarrow 5v_{2f} - 10 = 10 \Rightarrow v_{2f} = 4$ m/s. Then $v_{1f} = -1$ m/s.

$v_{1f} = -1$ m/s (west); $v_{2f} = +4$ m/s (east)

The relative-velocity equation $v_{1f} - v_{2f} = -(v_{1i} - v_{2i})$ is much faster than solving the quadratic from KE conservation. For elastic 1D collisions, always use it.

· bullet embeds in block

(a) $(0.05)(300) + 0 = (4.05)v_f \Rightarrow v_f = 15/4.05 \approx 3.70$ m/s.

(b) $K_i = \tfrac{1}{2}(0.05)(300)^2 = 2250$ J. $K_f = \tfrac{1}{2}(4.05)(3.70)^2 \approx 27.7$ J. KE lost: 2222 J (about 98.8%).

$v_f \approx 3.70$ m/s; ~98.8% of KE lost.

· pumpkin impact

Speed at impact: $v = \sqrt{2gh} = \sqrt{100} = 10$ m/s. $\Delta p = m\Delta v = 0.8(0 - (-10)) = 8$ kg·m/s (taking down as negative). $F = \Delta p / \Delta t = 8/0.02 = 400$ N upward.

$F_{\text{avg}} = 400$ N upward (50× pumpkin's weight)

· head-on inelastic collision

Take east as positive. $(4)(6) + (2)(-9) = (6)v_f \Rightarrow 24 - 18 = 6v_f \Rightarrow v_f = 1$ m/s east.

$v_f = 1$ m/s east

· explosion into two pieces

Initial momentum zero. Final: $(0.5)(12) + (1.5)v_2 = 0 \Rightarrow v_2 = -4$ m/s west.

$v_2 = 4$ m/s west

Worksheet B — AP Exam Style

· ball+block stick

$(1)(4) + 0 = 4 v_f \Rightarrow v_f = 1$ m/s east.

(A) 1 m/s east

· two-cart collision

Initial: $(2)(3) + (1)(-6) = 6 - 6 = 0$ kg·m/s. After: $(2)(1) + (1)v_B = 0 \Rightarrow v_B = -2$ m/s.

(A) −2 m/s

· area under F-t graph

Trapezoidal area: triangle (0-1s) = ½(1)(6) = 3 N·s, rectangle (1-3s) = (2)(6) = 12 N·s, triangle (3-5s) = ½(2)(6) = 6 N·s. Through $t = 4$ s: 3 + 12 + ½(1)(3) = 16.5 N·s. Hmm — actually let me re-read: at $t = 4$, force has been declining linearly from 6 to 0 over 3-5s. At $t = 4$, force = 3 N. Area in 3-4s segment = ½(1)(6+3) = 4.5 N·s. Total impulse = 3 + 12 + 4.5 = 19.5 N·s. $\Delta p = 19.5 \Rightarrow v_f = 19.5/2 = 9.75 \approx 9$ m/s. Closest answer: (C) 9 m/s.

(C) 9 m/s (approximately)

· astronaut throws wrench

Initial momentum zero, final must also be zero. Astronaut and wrench have equal-magnitude, opposite-direction momenta. Speeds, KEs are different (astronaut is heavier, so slower).

(C) Equal magnitude, opposite direction momenta.

· ball caught with extended time

Same $\Delta p$, 10× the time → 10× less force.

(C) Factor of 10

This is the principle behind catching a ball "softly" or behind crumple zones, helmets, airbags. Lengthening the impact time reduces peak force linearly.

· ballistic pendulum derivation

(a) Phase 1 (collision): $mv_0 + 0 = (m + M)v_1 \Rightarrow v_1 = mv_0/(m+M)$.

(b) Phase 2 (swing): KE → PE: $\tfrac{1}{2}(m+M)v_1^2 = (m+M)gh$. Substitute: $h = v_1^2/(2g) = m^2 v_0^2 / [2g(m+M)^2]$. Or solve for $v_0$: $v_0 = (m+M)\sqrt{2gh}/m$.

(c) No. The collision is perfectly inelastic, so KE decreases. Bullet loses kinetic energy to thermal energy, sound, deformation of block.

$v_1 = mv_0/(m+M)$; $h = m^2 v_0^2/[2g(m+M)^2]$; KE not conserved in collision.

· ball rebounds off wall

(a) $\Delta p = m\Delta v = 0.4((-8) - 12) = -8$ kg·m/s; magnitude 8 kg·m/s.

(b) $F = \Delta p/\Delta t = 8/0.04 = 200$ N (away from wall, on ball).

(c) By Newton's third law, ball exerts 200 N on wall, in opposite direction (toward wall, into wall). Equal magnitude, opposite direction. Why no visible damage? Wall + Earth has enormous mass, so this large force produces negligible acceleration. Also: wall might have very small $\Delta t$ for real impacts, so peak force is much higher than average.

$\Delta p = 8$ kg·m/s; $F = 200$ N each direction (action-reaction).

· skater throws and catches snowball

(a) Conservation: $0 = (1.5)(20) + (60)v_s \Rightarrow v_s = -0.5$ m/s (west).

(b) Skater KE = ½(60)(0.25) = 7.5 J. Snowball KE = ½(1.5)(400) = 300 J. Total ≈ 307.5 J. This came from chemical energy in the skater's muscles (work done in throwing).

(c) After catching: total momentum = (60)(-0.5) + (1.5)(20) = -30 + 30 = 0. So combined skater + snowball at rest!

After throw: skater 0.5 m/s west; after catching returning ball: skater is at rest (back to original state, but with additional snowball).

Catching the same returning snowball cancels the earlier recoil. This is why "throw a heavy thing east" then "catch a heavy thing thrown east" leaves you stationary, even though each event individually changes your velocity.

· 2025-style cart-block momentum EFR

(a) Block-cart system before collision: only the cart contributes to $p_x$ (block has $v_x = 0$ since dropped from rest, falling vertically). $p_x = m_c v_c$, constant before collision. During collision (very short Δt), $p_x$ doesn't change (no external horizontal force). After: same $p_x$. Graph: horizontal line at $p_x = m_c v_c$ throughout.

(b) Conservation: $m_c v_c + (\tfrac{1}{5}m_c)(0) = (m_c + \tfrac{1}{5}m_c)v_f = \tfrac{6}{5}m_c v_f$. Solve: $v_f = (m_c v_c)/(\tfrac{6}{5}m_c) = \tfrac{5}{6}v_c$.

(c) $K_i = \tfrac{1}{2}m_c v_c^2 + \tfrac{1}{2}(\tfrac{1}{5}m_c)(0)^2 = \tfrac{1}{2}m_c v_c^2$. $K_f = \tfrac{1}{2}(\tfrac{6}{5}m_c)(\tfrac{5}{6}v_c)^2 = \tfrac{1}{2}(\tfrac{6}{5}m_c)(\tfrac{25}{36}v_c^2) = \tfrac{5}{12}m_c v_c^2$. $\Delta K = \tfrac{5}{12}m_c v_c^2 - \tfrac{6}{12}m_c v_c^2 = -\tfrac{1}{12}m_c v_c^2$. Negative — KE decreases. Energy lost to deformation, friction, sound during the inelastic collision.

(d) Remains constant. Even with sliding friction between block and cart, that friction is internal to the block-cart system. There's no horizontal external force, so horizontal momentum of the block-cart system is conserved.

$p_x = m_c v_c$ (constant); $v_f = \tfrac{5}{6}v_c$; $\Delta K = -\tfrac{1}{12}m_c v_c^2$; momentum constant during sliding.

· ballistic pendulum experimental design

(a) Procedure: fire projectile horizontally so it hits the block and sticks. Measure max swing height of the block-projectile (use stopwatch and pendulum length to find swing period, or measure horizontal displacement and convert via geometry). Repeat several trials, average $h$. Reduce uncertainty by taking many trials, ensuring projectile sticks securely (use enough clay), use a long pendulum to get measurable height.

(b) Phase 1: $mv_0 = (m+M)v_1 \Rightarrow v_1 = mv_0/(m+M)$. Phase 2: $\tfrac{1}{2}(m+M)v_1^2 = (m+M)gh \Rightarrow v_1 = \sqrt{2gh}$. Combine: $v_0 = ((m+M)/m)\sqrt{2gh}$.

(c) $v_0 = (2.05/0.05)\sqrt{2(10)(0.20)} = 41 \cdot 2 = 82$ m/s.

(d) Fraction of KE lost = 1 - K_f/K_i = 1 - (m+M)v_1^2/(mv_0^2) = 1 - (m^2 v_0^2/(m+M))/(mv_0^2) = 1 - m/(m+M) = M/(m+M).

(e) Energy is NOT conserved during the collision (perfectly inelastic). So $\tfrac{1}{2}mv_0^2 \neq (m+M)gh$. Without the momentum equation for phase 1, you can't determine the velocity right after collision, and energy conservation alone gives you the wrong answer.

$v_0 = 82$ m/s; fraction KE lost = $M/(m+M)$; momentum-then-energy is required.

For $M \gg m$ (heavy block, light bullet), almost all KE is lost in the collision. Energy conservation alone would massively over-estimate the projectile's speed. This is exactly why ballistic pendulums require both conservation laws.

Worksheet C — Challenge

· elastic collision with mass disparity

Standard 1D elastic result: $v_{1f} = (m-M)v_0/(m+M)$, $v_{2f} = 2m v_0/(m+M)$.

Limit $M \gg m$: $v_{1f} \to -v_0$ (ball bounces back at same speed); $v_{2f} \to 0$ (heavy block stays still). Like ball bouncing off wall.

Limit $m \gg M$: $v_{1f} \to v_0$ (heavy ball barely slows); $v_{2f} \to 2v_0$ (light block flies off at twice ball's speed). Like a bowling ball passing through a feather.

Heavy target: ball reverses, target stationary. Heavy projectile: target gets 2× projectile's speed.

The "2v" result is famous — it's why a kid jumping on a trampoline can launch a ball off the surface at twice the trampoline's speed (gravity assist).

· 2D collision

Initial: $p_x = (1)(4) = 4$, $p_y = 0$. After ball 1 moves at 3 m/s, 30° N of E: $v_{1x} = 3\cos 30° = 2.60$, $v_{1y} = 3\sin 30° = 1.50$.

Conservation: $p_{2x} = 4 - 2.60 = 1.40$ m/s, $p_{2y} = 0 - 1.50 = -1.50$ m/s. Speed: $\sqrt{1.96 + 2.25} = \sqrt{4.21} \approx 2.05$ m/s. Direction: $\arctan(1.50/1.40) \approx 47°$ south of east.

KE check: $K_i = ½(1)(16) = 8$ J. $K_f = ½(1)(9) + ½(1)(4.21) = 4.5 + 2.105 ≈ 6.6$ J. KE not conserved → inelastic.

Ball 2: 2.05 m/s, 47° south of east. Inelastic.

· rocket equation

(a) In time $dt$, rocket ejects mass $\mu \, dt$ at speed $v_e$ (relative to rocket, so in opposite direction at speed $v_e$ in rocket's frame). Impulse on ejected fuel: $\mu \, dt \cdot v_e$. By Newton's third law, equal and opposite impulse on rocket. Force on rocket = thrust: $F = \mu v_e$.

(b) The rocket equation comes from $F = dp/dt$, accounting for changing mass; deriving it requires calculus. Without calculus: assume small Δm ejected at each step, write Δv at each step, sum (or use Σ ≈ ∫ for many small steps). Result: $\Delta v = v_e \ln(M_0/M_f)$. Assumption: many small ejections, so we can replace sum with logarithm.

Thrust = $\mu v_e$; rocket equation $\Delta v = v_e \ln(M_0/M_f)$.

· Newton's cradle

Five balls of mass $m$ each, ball 1 enters at speed $v$, others at rest. Initial: $p = mv$, $K = ½mv^2$.

Suppose final state has $n$ balls moving at speed $v'$ (with the rest at rest). Momentum: $mv = nmv' \Rightarrow v' = v/n$. KE: $½mv^2 = n(½mv'^2) = nm v^2/(2n^2) = ½mv^2/n$. For KE to be conserved, $n = 1$ and $v' = v$. So only one ball can move, and it moves at the original speed. By symmetry, it must be the ball at the end of the line opposite to the impact.

If 2 balls released at speed $v$: $p_i = 2mv$, $K_i = mv^2$. For $n$ balls at speed $v'$: $2mv = nmv' \Rightarrow v' = 2v/n$. $K_f = n m v'^2/2 = 2mv^2/n$. For KE conservation: $2/n = 1 \Rightarrow n = 2$. Two balls swing out at speed $v$.

Both momentum AND energy conservation force "n in = n out at same speed."

· ball-spring-block

At max compression: ball + block move together at velocity $v_c$. Momentum: $(0.5)(30) + 0 = (2.5)v_c \Rightarrow v_c = 6$ m/s.

Energy: KE_i = ½(0.5)(900) = 225 J. KE at max compression = ½(2.5)(36) = 45 J. Stored in spring: 225 - 45 = 180 J = ½kx² ⇒ x = √(360/5000) ≈ 0.268 m.

After ball rebounds: equivalent to elastic collision (no energy lost, since spring stores then releases without dissipation). Use elastic 1D formulas: $v_{1f} = (0.5 - 2)(30)/(2.5) = -18$ m/s. Ball rebounds at 18 m/s. Block: $v_{2f} = 2(0.5)(30)/2.5 = 12$ m/s.

Max compression ≈ 0.27 m; ball rebounds at 18 m/s; block reaches 12 m/s max.

A spring-mediated collision is effectively elastic — no energy lost to deformation. The spring temporarily stores energy then returns it. So a "spring collision" gives the same result as a direct elastic collision.

· · ·

Common mistakes

Seven errors that cost students points on Unit 4

Forgetting that momentum is a vector

A 2 kg ball at +5 m/s and a 3 kg ball at −2 m/s. Find total momentum.

$p_{\text{total}} = (2)(5) + (3)(2) = 16$ kg·m/s.

$p_{\text{total}} = (2)(+5) + (3)(-2) = 10 - 6 = +4$ kg·m/s. Direction matters.

Forgetting signs leads to wildly wrong answers, especially in head-on collisions where the momenta partially cancel. Always pick a positive direction first, then assign signs to every velocity.

A 1 kg ball at 3 m/s east and a 1 kg ball at 5 m/s west collide. Find total momentum and the resulting velocity if they stick together.

Using energy conservation for a perfectly inelastic collision

A 1 kg ball at 4 m/s sticks to a 3 kg block at rest. Use energy conservation to find the final speed.

$\tfrac{1}{2}(1)(16) = \tfrac{1}{2}(4)v_f^2 \Rightarrow v_f = 2\sqrt{2} \approx 2.83$ m/s.

Use momentum conservation. $(1)(4) = (4)v_f \Rightarrow v_f = 1$ m/s. KE is NOT conserved in inelastic collisions; using energy gives wrong answer.

Inelastic collisions lose KE to deformation, heat, and sound. Only momentum is conserved across all collisions. Energy conservation is reserved for elastic collisions and for non-collision motion.

A 0.05 kg bullet at 200 m/s embeds in a 1 kg block. Find the speed of block+bullet, then verify by computing how much KE was lost.

Forgetting that momentum is conserved during the collision even with external forces

A block falls vertically onto a horizontally-moving cart and sticks. Is the cart's horizontal momentum conserved during the collision?

No, because gravity acts on the system.

Yes, in the horizontal direction. Gravity is vertical, so it doesn't affect horizontal momentum. Vertical momentum is NOT conserved (gravity changes it), but horizontal is.

Momentum conservation must be checked direction-by-direction. External forces along one axis don't affect momentum along a perpendicular axis. The 2025 AP exam Q1 hinged on exactly this idea — students who said "no, gravity acts" missed the easy points.

A 5 kg projectile flying horizontally at 20 m/s hits and embeds in a 15 kg block hanging at rest from a string. Find the block+projectile speed just after collision.

Confusing impulse with force

A 0.5 kg ball changes velocity from +6 m/s to −4 m/s during a 0.05 s collision. What is the force on the ball?

$F = \Delta p = (0.5)(10) = 5$ N. (Mistakenly answering with impulse, in N·s, treating it as force in N.)

$\Delta p = 5$ N·s; $F = \Delta p / \Delta t = 5/0.05 = 100$ N. Impulse and force have different units.

Impulse ($F\Delta t$) and force ($F$) are different quantities. Always include the time interval to convert. Impulse is N·s = kg·m/s, equal to a momentum change. Force is N = kg·m/s².

A 1500 kg car going 25 m/s comes to rest in 5 seconds (gradual stop) versus 0.5 seconds (crash). Compare the impulse and the average force in each case.

Ignoring sign of velocity in rebound impulse

A 0.2 kg ball hits a wall at 8 m/s and bounces off at 6 m/s in the opposite direction. Find the impulse.

$\Delta p = m\Delta v = 0.2(8 - 6) = 0.4$ kg·m/s.

$\Delta p = m\Delta v = 0.2(-6 - 8) = -2.8$ kg·m/s. (Or equivalently, $0.2(6 + 8) = 2.8$ kg·m/s in magnitude.) The velocity reverses direction, so signs matter.

When velocity reverses, the change in velocity is the SUM of the speeds in magnitude, not the difference. Treating rebound as if speeds simply subtract is the most common error in collision problems.

A 0.1 kg ball strikes a wall at 12 m/s and rebounds at 10 m/s. Find the impulse on the ball.

Mixing up "perfectly inelastic" with "elastic"

"In a perfectly inelastic collision, kinetic energy is conserved." True or false?

True (mistakenly thinking "perfectly" means "perfectly conserved").

False. Perfectly inelastic means the objects stick together — and this is the case where the maximum amount of KE is lost. "Perfectly elastic" means KE is fully conserved.

"Perfectly inelastic" sounds positive but actually means the maximum loss of KE. The word "perfectly" describes the stickiness, not the conservation. Be careful with language.

Two 1 kg carts collide, one moving at 4 m/s and one at rest. Compare the KE lost in (a) a perfectly inelastic collision (stick together) and (b) a perfectly elastic collision (rebound). Which conserves momentum? Which conserves KE?

Forgetting that ballistic-pendulum problems require both momentum AND energy

A 0.05 kg bullet embeds in a 2 kg pendulum block, which swings up to 0.3 m. Find the bullet's launch speed.

Use energy conservation: $\tfrac{1}{2}mv_0^2 = (m+M)gh$. Solve for $v_0$.

Use momentum conservation for the collision (KE is lost), then energy for the swing. $(0.05)v_0 = (2.05)v_1$ gives $v_1$, then $\tfrac{1}{2}(2.05)v_1^2 = (2.05)(10)(0.3)$ gives $v_1$. Combine for $v_0$.

The collision is inelastic — KE is NOT conserved. Energy conservation alone gives the wrong answer for the bullet's speed. Always split into two phases: momentum for the collision, energy for the swing.

A 10 g bullet embeds in a 5 kg block. The block+bullet swings to 0.05 m height. Find the bullet's speed.

· · ·

Tutor's corner

What I learned going through this unit — and from helping others through it

What every student struggles with (in order)

The #1 sticking point in Unit 4 is recognizing when to use momentum vs energy. Students who mastered Unit 3 develop a reflex to apply energy methods to everything — and then try to "solve" inelastic collisions with energy conservation, which gives wrong answers because KE isn't conserved. The trigger to use momentum instead: collision, explosion, or recoil. Anytime two objects interact briefly and you need their final velocities, momentum is the tool. Energy can come later, for the post-collision motion (swing, slide, climb).

The #2 struggle is signs and directions. Momentum is a vector, but in 1D it just means "track the sign carefully." Students forget that a ball going east at 5 m/s and a ball going west at 5 m/s have momenta of opposite signs that partially cancel. Always pick a positive direction first, then assign signs to every velocity. Then add. The sign of the answer tells you direction.

The #3 struggle is impulse for rebound problems. When a ball bounces off a wall, students compute Δv as |v_f - v_i| forgetting the signs flip. The correct $\Delta v = v_f - v_i$ where one is positive and one is negative — so the magnitudes add instead of subtract. This is the single most common impulse error.

What separates a 4 from a 5

System awareness. Students who score 5 read a problem and immediately ask: "what's my system, and what external forces act on it?" If the answer is "no horizontal external force," they apply horizontal momentum conservation reflexively, even when gravity is in the picture. The 2025 AP exam Q1 (block dropped onto cart) tested exactly this skill — and students who paused to identify "horizontal momentum conserved, vertical is not" earned full credit, while those who tried to apply blanket conservation got tangled.

The question that reveals true understanding

"A bullet is fired into a wooden block on a frictionless surface. The block moves forward and the bullet embeds. After the collision, what fraction of the bullet's original kinetic energy is left as kinetic energy of the system?" The deep answer: $K_f/K_i = m/(m+M)$, where $m$ is the bullet and $M$ is the block. As $M \to \infty$, this fraction goes to zero — almost all KE is lost, even though all the momentum is preserved. Students who can derive this formula understand both conservation laws and why they give different results in inelastic collisions.

Exam-day strategy

Identify the problem type: collision/explosion/recoil → momentum first. Multi-stage (collision then swing/lift) → momentum, then energy.

Define the system carefully: include all interacting objects. Check direction by direction whether external forces act over the time interval.

Set up a sign convention: positive direction explicitly chosen; every velocity gets a sign.

Write the conservation equation: $\sum p_i = \sum p_f$. If perfectly inelastic, single $v_f$ for both objects. If elastic, also use $v_{1f} - v_{2f} = -(v_{1i} - v_{2i})$.

Don't try energy conservation for collisions unless explicitly told it's elastic. KE is lost in real-world (inelastic) collisions.

A surprising real-world connection

Why does a recoiling rifle exert less force on your shoulder than the bullet's force on a target? Because the impulse on you and the impulse on the target are roughly equal in magnitude (Newton's third law plus momentum conservation), but the time interval is different. The rifle pushes back on your shoulder gradually as the bullet accelerates over the barrel length — much longer time than the bullet's collision with the target. Same impulse, longer time, smaller force. This is why shooters brace against the rifle and why high-recoil firearms have padded stocks. Same physics: airbags, crumple zones, helmet padding, gymnastic mats. Lengthening the collision time at fixed momentum change is one of the most common safety strategies in engineering.

One piece of advice

Drill the ballistic pendulum problem until you can do it in your sleep. It combines both major conservation laws (momentum for the collision, energy for the swing), and it shows up on most AP exams in some form. The 2025 exam had a closely related cart problem. Master the template — momentum first to find post-collision speed, then energy for the post-collision motion — and you've covered most of the high-yield Unit 4 problems. The two-phase structure is the model for many problems combining Units 3, 4, and 6.

End of Unit 4 · Linear Momentum · First Principles Physics