AP Physics 1 Unit 4: Mastering Linear Momentum

Momentum as a Vector Quantity

What momentum is (and why physicists care)

Linear momentum is a measure of “motion that’s hard to stop.” The more mass an object has and the faster it moves, the more momentum it has. Momentum matters because it is one of the most powerful bookkeeping tools in mechanics: when the net external influence on a system is small (or acts for a very short time), total momentum tends to stay the same even if the motion of individual objects changes dramatically—like in collisions and explosions.

Momentum is a vector, meaning it has both magnitude and direction. That direction is always the same as the object’s velocity direction.

For a single object with mass m and velocity v, the momentum p is

p = mv

• m is mass (kilograms, kg)
• v is velocity (meters per second, m/s)
• p is momentum (kilogram-meters per second, kg·m/s)

Because momentum is a vector, you often work with components:

p_x = mv_x

p_y = mv_y

A common mistake is to treat momentum like a scalar and forget the sign/direction. In 1D problems (motion along a line), you show direction with a plus/minus sign based on your chosen positive direction.

System momentum: adding momenta

AP Physics 1 momentum problems usually involve systems: collections of objects you choose to analyze together (two carts, a gun and bullet, two ice skaters, etc.). The key idea is that a system has a total momentum equal to the vector sum of each object’s momentum.

For multiple objects,

P_{\text{total}} = \sum p_i

In 1D,

P_{\text{total}} = m_1 v_1 + m_2 v_2 + \cdots

In 2D, you typically conserve momentum separately in the x and y directions because vectors add by components:

P_{x,\text{total}} = \sum m_i v_{x,i}

P_{y,\text{total}} = \sum m_i v_{y,i}

Interpreting momentum physically

Momentum connects directly to how hard it is to change motion.

• A fast tennis ball has noticeable momentum, but a fast bowling ball has much more because of its larger mass.
• A truck moving slowly can have the same momentum as a small car moving quickly.

This “tradeoff” between mass and speed shows up constantly in collision problems: a massive object may barely change speed in a collision, while a lighter object’s velocity may change drastically.

Worked example: 1D momentum with signs

A 0.50 kg ball moves to the right at 6.0 m/s. Another 0.50 kg ball moves to the left at 2.0 m/s. Take right as positive. Find each momentum and the total momentum.

Ball 1:

p_1 = mv = (0.50)(+6.0) = +3.0

Ball 2:

p_2 = (0.50)(-2.0) = -1.0

Total:

P_{\text{total}} = p_1 + p_2 = +3.0 + (-1.0) = +2.0

Interpretation: the system’s net momentum points right.

Worked example: 2D momentum components

A 0.20 kg puck moves at 5.0 m/s at 30° above the +x axis. Find p_x and p_y.

First find velocity components:

v_x = v\cos(30^\circ) = 5.0\cos(30^\circ)

v_y = v\sin(30^\circ) = 5.0\sin(30^\circ)

Momentum components:

p_x = mv_x = (0.20)\cdot 5.0\cos(30^\circ)

p_y = mv_y = (0.20)\cdot 5.0\sin(30^\circ)

You usually do not need to compute the final decimals unless asked; AP questions often emphasize setup and direction.

Exam Focus

• Typical question patterns:
  • Compute momentum (or momentum components) from mass and velocity, including sign conventions.
  • Compare momenta of different objects and reason conceptually about which has more and why.
  • Find total system momentum before/after an event as a vector sum.
• Common mistakes:
  • Dropping direction: treating momentum as positive when the velocity is negative.
  • Mixing up speed and velocity: using magnitudes where components (or signs) are required.
  • Adding momenta as scalars in 2D instead of conserving separately in x and y.

Impulse and the Momentum Theorem

Why impulse is introduced

Momentum tells you how much “motion” an object has, but physics problems often ask how motion changes due to forces—especially when forces act for short times, like during a collision. The tool that bridges force and momentum is impulse.

Impulse captures the combined effect of a force and the time interval over which it acts. A large force for a short time can have the same effect as a smaller force for a longer time.

Defining impulse

For a constant net force F_{\text{net}} acting over a time interval \Delta t, the **impulse** J is

J = F_{\text{net}}\Delta t

Impulse has the same units as momentum (N·s is equivalent to kg·m/s), which hints at the core theorem of this unit.

The impulse-momentum theorem (momentum theorem)

The central relationship is that net impulse equals change in momentum:

J = \Delta p

and

\Delta p = p_f - p_i

So for a single object,

F_{\text{net}}\Delta t = m v_f - m v_i

This is essentially Newton’s Second Law written in a form that is extremely useful when forces vary in time or act briefly.

Variable forces and force-time graphs

In real collisions, the force is rarely constant: it spikes up and down as objects deform and rebound. AP Physics 1 commonly represents this with a force vs. time graph. The key idea:

• Impulse is the area under the net force vs. time curve.

If you can compute the area (rectangle, triangle, trapezoid, or combination), you can find impulse and therefore the change in momentum.

Even without calculus, you can express the “area idea” conceptually as

J = \text{area under } F(t) \text{ vs. } t

And you can define an average force F_{\text{avg}} so that

J = F_{\text{avg}}\Delta t

Why “softening” collisions reduces injury

The impulse-momentum theorem explains real safety designs.

• If your momentum must drop from some value to (approximately) zero, then \Delta p is fixed.
• Since J = F_{\text{avg}}\Delta t = \Delta p, increasing the stopping time \Delta t decreases the average force.

That is why airbags, crumple zones, bending your knees when landing, and “giving” when catching a ball all reduce force on the body.

A common misconception is to think “longer collision time means bigger force because they’re interacting longer.” The theorem shows the opposite when the required \Delta p is fixed.

Worked example: average force from momentum change

A 0.15 kg baseball moving at 40 m/s is caught and brought to rest in 0.020 s. Estimate the magnitude of the average force on the ball.

Change in momentum:

\Delta p = m v_f - m v_i = (0.15)(0) - (0.15)(40) = -6.0

The impulse magnitude is 6.0 N·s. Average force magnitude:

F_{\text{avg}} = \frac{J}{\Delta t} = \frac{6.0}{0.020} = 300

Direction: The force is opposite the initial motion (it must slow the ball).

Worked example: impulse from a triangular force-time graph

A force increases linearly from 0 N to 400 N over 0.010 s, then decreases linearly back to 0 N by 0.020 s. Find the impulse.

The graph is a triangle with base 0.020 s and height 400 N. Area:

J = \frac{1}{2}(0.020)(400) = 4.0

So the object’s momentum changes by 4.0 kg·m/s in the direction of the net force.

Exam Focus

• Typical question patterns:
  • Use J = \Delta p to find average force, collision time, or change in velocity.
  • Compute impulse from the area under a force-time graph (often piecewise shapes).
  • Explain qualitatively how increasing collision time reduces force for the same momentum change.
• Common mistakes:
  • Using the applied force instead of the net force (impulse uses net force on the object).
  • Confusing “force-time area” with “force-time slope” (slope is not impulse).
  • Forgetting direction: impulse and \Delta p are vectors, so signs matter in 1D.

Systems, Internal Forces, and When Momentum Is Conserved

The idea of a “system” is a physics superpower

Momentum conservation is not magic—it depends on what you include in your system and whether external influences can change the system’s total momentum.

A system is whatever objects you decide to group together for analysis. For momentum problems, a good system choice often turns messy contact forces into “internal forces” that cancel out in pairs.

Internal vs. external forces

• Internal forces are forces that objects in the system exert on each other.
• External forces are forces exerted on the system by objects outside the system.

Newton’s Third Law is what makes internal forces so helpful: when two objects in the system interact, they exert equal-magnitude, opposite-direction forces on each other. Those internal forces can change each object’s momentum, but they do not change the system’s total momentum.

What can change total system momentum is net external impulse.

Momentum conservation condition (impulse version)

Momentum is conserved for a system when the net external impulse is zero (or negligible):

J_{\text{ext}} = 0

Using the impulse-momentum theorem for a system,

J_{\text{ext}} = \Delta P_{\text{system}}

So if J_{\text{ext}} = 0, then

\Delta P_{\text{system}} = 0

which implies

P_{\text{system},f} = P_{\text{system},i}

This is the deep reason collisions on a nearly frictionless track conserve momentum: external forces like friction are tiny, and the normal force and gravity often cancel in pairs or act perpendicular to motion.

“Isolated” vs “closed” in AP language

AP Physics 1 typically uses the idea of an isolated system to mean “net external force is zero” or “net external impulse is zero over the interaction time.” In collision problems, you’ll often see: “neglect friction,” which is code for “treat horizontal external impulse as zero.”

A subtle but important point: momentum can be conserved in one direction even if external forces exist in another direction.

• Example: A puck sliding on ice has external forces (gravity and normal), but these are vertical and (approximately) cancel. So horizontal momentum is conserved.

Worked example: deciding whether to conserve momentum

Two carts collide on a horizontal track. There is noticeable friction with the track, but the collision lasts 0.05 s. Should you conserve momentum during the collision?

Reasoning: Friction is external, so in principle it can change system momentum. But you must compare the external impulse from friction to the internal impulses during the collision.

• During the brief collision, the contact force between carts is typically large.
• The friction force is relatively small and acts for only 0.05 s.

So the external impulse due to friction during the collision is often negligible, and momentum is approximately conserved during the collision, even if momentum is not conserved over longer times.

This is a common AP modeling assumption: short, high-force interactions can be treated as “isolated” for momentum.

Exam Focus

• Typical question patterns:
  • Identify the best system so that momentum conservation applies (e.g., “cart + cart” instead of one cart).
  • Decide whether momentum is conserved in a given direction based on external forces/impulses.
  • Explain using Newton’s Third Law why internal forces do not change total system momentum.
• Common mistakes:
  • Conserving momentum for a single object that experiences an external force (momentum conservation is for systems).
  • Assuming momentum is never conserved if any external force exists, even when external impulse is negligible.
  • Forgetting that momentum conservation can apply in one axis but not the other.

Conservation of Linear Momentum in Interactions (Collisions and Explosions)

Conservation law statement

When the net external impulse on a system is zero, total momentum before an interaction equals total momentum after:

P_i = P_f

In 1D for two objects,

m_1 v_{1,i} + m_2 v_{2,i} = m_1 v_{1,f} + m_2 v_{2,f}

This applies to collisions (objects push on each other and may bounce or stick) and to explosions (objects push apart due to internal stored energy like a spring or chemical energy).

Why collisions are the “natural home” of momentum

During a collision:

• Forces are large.
• Interaction times are short.
• External impulses are often negligible compared with internal impulses.

That combination makes momentum conservation extremely reliable in many setups, even when mechanical energy is not conserved.

Explosions and recoil: momentum starts at zero (often)

A very common AP situation: an object at rest breaks into two pieces or two objects push apart from rest. If the system starts at rest, then initial total momentum is zero:

P_i = 0

So after the push apart,

P_f = 0

For two pieces in 1D,

m_1 v_{1,f} + m_2 v_{2,f} = 0

This means the momenta are equal in magnitude and opposite in direction:

m_1 v_{1,f} = -m_2 v_{2,f}

A key interpretation: the lighter piece generally moves faster (to have the same momentum magnitude).

Worked example: recoil (gun and bullet style)

A 2.0 kg launcher fires a 0.020 kg projectile. The projectile leaves at 200 m/s to the right. Neglect external horizontal impulse. Find the launcher’s recoil velocity.

Initial total momentum is zero. Final momentum:

0 = (0.020)(200) + (2.0)v_{\text{launcher}}

Compute projectile momentum:

(0.020)(200) = 4.0

Solve:

2.0 v_{\text{launcher}} = -4.0

v_{\text{launcher}} = -2.0

So the launcher recoils left at 2.0 m/s.

Worked example: “push off” on frictionless ice

Two skaters at rest push off each other. Skater A has mass 50 kg, skater B has mass 80 kg. After pushing off, A moves at 3.2 m/s to the left. Find B’s speed.

Let left be negative. Initial total momentum is zero:

0 = (50)(-3.2) + (80)v_B

80v_B = 160

v_B = 2.0

B moves right at 2.0 m/s.

Momentum conservation does not mean “velocity stays the same”

A frequent conceptual error is to assume conservation means each object keeps its momentum or velocity. Conservation applies to the total momentum of the system. Individual momenta can change drastically due to internal forces.

Exam Focus

• Typical question patterns:
  • Set up and solve P_i = P_f for collisions, recoil, explosions, and “push-off” scenarios.
  • Use “initially at rest” to set P_i = 0 and deduce opposite final momenta.
  • Explain qualitatively why a smaller mass moves faster after an explosion-like separation.
• Common mistakes:
  • Forgetting to include all objects in the system momentum equation.
  • Using speeds instead of velocities (dropping signs), leading to wrong directions.
  • Treating momentum conservation as an energy statement (momentum can be conserved even when kinetic energy changes).

Collisions in One Dimension: Inelastic vs Elastic

What makes collisions tricky

A collision problem usually gives you masses and initial velocities and asks for final velocities (or a shared velocity if objects stick). Momentum conservation gives you one equation in 1D. But in many collisions you have two unknown final velocities, so you need additional information.

That additional information comes from the collision type.

Inelastic collisions (including “perfectly inelastic”)

In an inelastic collision, objects may deform, heat up, or produce sound. Kinetic energy is not conserved, but momentum can still be conserved (if external impulse is negligible).

A perfectly inelastic collision is a special inelastic collision where the objects stick together and move with a common final velocity. This is extremely common on AP exams because it turns the two-unknown problem into a one-unknown problem.

If objects stick:

v_{1,f} = v_{2,f} = v_f

Momentum conservation becomes

m_1 v_{1,i} + m_2 v_{2,i} = (m_1 + m_2) v_f

Solve for the shared velocity:

v_f = \frac{m_1 v_{1,i} + m_2 v_{2,i}}{m_1 + m_2}

The numerator is the initial total momentum; the denominator is the total mass.

Worked example: perfectly inelastic (stick together)

A 1.0 kg cart moving right at 4.0 m/s hits and sticks to a 3.0 kg cart at rest. Neglect friction during the collision. Find the final velocity.

Use momentum conservation:

(1.0)(4.0) + (3.0)(0) = (1.0 + 3.0)v_f

4.0 = 4.0 v_f

v_f = 1.0

So the combined carts move right at 1.0 m/s.

Kinetic energy check (why it decreases)

It’s often helpful (and sometimes required) to determine whether kinetic energy is conserved.

Initial kinetic energy:

K_i = \frac{1}{2}m_1 v_{1,i}^2 + \frac{1}{2}m_2 v_{2,i}^2

Final kinetic energy for stuck carts:

K_f = \frac{1}{2}(m_1 + m_2) v_f^2

For the example:

K_i = \frac{1}{2}(1.0)(4.0^2) = 8.0

K_f = \frac{1}{2}(4.0)(1.0^2) = 2.0

Kinetic energy decreased; the “missing” energy went into deformation, heat, and sound. Momentum conservation does not prevent this because momentum and kinetic energy are different physical quantities.

Elastic collisions in 1D

In an elastic collision, both momentum and kinetic energy are conserved (again assuming negligible external impulse). Many real-world macroscopic collisions are not perfectly elastic, but AP Physics 1 uses elastic collisions as an idealized model.

So you have two conservation equations:

m_1 v_{1,i} + m_2 v_{2,i} = m_1 v_{1,f} + m_2 v_{2,f}

\frac{1}{2}m_1 v_{1,i}^2 + \frac{1}{2}m_2 v_{2,i}^2 = \frac{1}{2}m_1 v_{1,f}^2 + \frac{1}{2}m_2 v_{2,f}^2

This gives two equations for the two unknown final velocities.

AP Physics 1 may also use (or allow) the standard 1D elastic collision results (derived from the two conservation laws). If you use them, make sure the collision is truly stated to be elastic.

The standard formulas (1D elastic collision) are:

v_{1,f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1,i} + \frac{2m_2}{m_1 + m_2}v_{2,i}

v_{2,f} = \frac{2m_1}{m_1 + m_2}v_{1,i} + \frac{m_2 - m_1}{m_1 + m_2}v_{2,i}

These look intimidating, but they encode intuitive limiting cases.

Important special case: equal masses in an elastic collision

If m_1 = m_2, the formulas simplify dramatically:

• The objects exchange velocities in 1D elastic collisions.

So if one is initially at rest, the moving one stops and the initially-stationary one takes off with the original speed (like billiard balls).

Worked example: elastic collision (one initially at rest)

A 2.0 kg cart moving right at 3.0 m/s collides elastically with a 1.0 kg cart at rest. Find both final velocities.

Given:

• m_1 = 2.0, v_{1,i} = 3.0
• m_2 = 1.0, v_{2,i} = 0

Use the standard elastic formulas:

v_{1,f} = \frac{2.0 - 1.0}{2.0 + 1.0}(3.0) + \frac{2(1.0)}{2.0 + 1.0}(0)

v_{1,f} = \frac{1}{3}(3.0) = 1.0

And

v_{2,f} = \frac{2(2.0)}{2.0 + 1.0}(3.0) + \frac{1.0 - 2.0}{2.0 + 1.0}(0)

v_{2,f} = \frac{4}{3}(3.0) = 4.0

So cart 1 continues right at 1.0 m/s, and cart 2 moves right at 4.0 m/s.

A quick check: initial momentum is 2.0\cdot 3.0 = 6.0, final momentum is 2.0\cdot 1.0 + 1.0\cdot 4.0 = 6.0. Initial kinetic energy is \frac{1}{2}(2.0)(9) = 9, final kinetic energy is \frac{1}{2}(2.0)(1) + \frac{1}{2}(1.0)(16) = 1 + 8 = 9.

Exam Focus

• Typical question patterns:
  • Perfectly inelastic: objects stick and you solve for a shared final velocity using momentum conservation.
  • Elastic: use both momentum and kinetic energy conservation (sometimes with provided/recognized standard results).
  • Conceptual: determine whether kinetic energy increases/decreases and explain where energy goes in inelastic collisions.
• Common mistakes:
  • Assuming kinetic energy is conserved whenever momentum is conserved.
  • Using the “elastic formulas” when the collision is not stated to be elastic.
  • Forgetting that in perfectly inelastic collisions, both objects share the same final velocity.

Two-Dimensional Momentum: Components, Angles, and “Glancing” Events

Why 2D momentum is conceptually the same (but looks harder)

In two dimensions, momentum conservation is not a new law—it’s the same vector conservation. The main difference is that you must treat momentum as a vector and usually break it into components.

The guiding strategy is:

1. Choose coordinate axes.
2. Write conservation of momentum in the x direction.
3. Write conservation of momentum in the y direction.
4. Solve the resulting system.

In 2D, total momentum conservation can be written as a vector equation:

\vec{P}_i = \vec{P}_f

which is equivalent to conserving components:

P_{x,i} = P_{x,f}

P_{y,i} = P_{y,f}

A common AP scenario: one object initially at rest

Many 2D collision/explosion questions involve one object moving initially and one at rest, or an explosion from rest. These setups reduce algebra and emphasize vector reasoning.

Worked example: explosion from rest in 2D

A 3.0 kg object at rest explodes into two pieces: piece A is 1.0 kg and moves at 6.0 m/s along +x. Piece B is 2.0 kg. Find the velocity components of piece B.

Initial momentum is zero in both directions:

P_{x,i} = 0

P_{y,i} = 0

After explosion:

P_{x,f} = m_A v_{A,x} + m_B v_{B,x}

P_{y,f} = m_A v_{A,y} + m_B v_{B,y}

Given piece A moves purely along +x, so v_{A,y} = 0.

Conserve x:

0 = (1.0)(6.0) + (2.0)v_{B,x}

2.0 v_{B,x} = -6.0

v_{B,x} = -3.0

Conserve y:

0 = (1.0)(0) + (2.0)v_{B,y}

v_{B,y} = 0

So piece B moves left at 3.0 m/s.

Worked example: 2D collision with a right-angle deflection

A 0.50 kg puck moves east at 4.0 m/s and collides with a 0.50 kg puck initially at rest. After the collision, the first puck moves north at 2.0 m/s. Find the velocity of the second puck (components and speed).

Let east be +x, north be +y.

Initial momentum:

P_{x,i} = (0.50)(4.0) = 2.0

P_{y,i} = 0

Final momentum:

P_{x,f} = (0.50)(0) + (0.50)v_{2,x}

because puck 1 final has no x-component if it moves purely north.

Conserve x:

2.0 = (0.50)v_{2,x}

v_{2,x} = 4.0

Now y:

0 = (0.50)(2.0) + (0.50)v_{2,y}

0 = 1.0 + 0.50 v_{2,y}

v_{2,y} = -2.0

So puck 2 moves with components (4.0 m/s east, 2.0 m/s south). Its speed:

v_2 = \sqrt{v_{2,x}^2 + v_{2,y}^2} = \sqrt{4.0^2 + (-2.0)^2} = \sqrt{20}

Momentum conservation in only one direction

A frequent AP twist: external forces may exist in one direction but not the other.

Example: A collision on a horizontal surface typically has negligible external impulse horizontally (friction neglected), so P_x is conserved. Vertically, the normal force and gravity may provide external impulses, so P_y may not be conserved unless the interaction time is short and vertical motion is constrained.

That’s why many 2D momentum questions involve motion on a frictionless table: it makes both horizontal directions “safe” for conservation.

Exam Focus

• Typical question patterns:
  • Use separate conservation equations for x and y to solve for unknown velocity components.
  • Explosion/collision from rest: use P_i = 0 to reason about opposite momentum vectors.
  • Combine momentum conservation with geometry (angles, components) to find direction or magnitude.
• Common mistakes:
  • Writing one momentum equation with magnitudes instead of two component equations.
  • Mixing up sine and cosine when breaking velocities into components.
  • Conserving momentum in a direction where there is significant external impulse (especially in vertical direction).

Center of Mass and Its Connection to Momentum

What “center of mass” means

The center of mass is the mass-weighted “balance point” of a system. For momentum, the center of mass matters because the system’s total momentum is directly tied to how the center of mass moves.

Even if parts of a system move in complicated ways (like an exploding firework), the center of mass can move smoothly according to external forces only.

Center of mass for discrete objects

For objects with masses m_1, m_2, \dots at positions x_1, x_2, \dots along a line, the center of mass position is

x_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i}

In 2D, you treat each coordinate separately:

x_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i}

y_{\text{cm}} = \frac{\sum m_i y_i}{\sum m_i}

The key momentum relationship: total momentum and center-of-mass velocity

For a system with total mass M and center-of-mass velocity v_{\text{cm}}, the total momentum P satisfies

P = M v_{\text{cm}}

This is incredibly useful conceptually:

• If P = 0, then v_{\text{cm}} = 0 and the center of mass stays at rest.
• If external impulse is zero, then P is constant, so v_{\text{cm}} is constant.

This is why, in an explosion with no external impulse, pieces fly apart but the center of mass continues on the same path it had before the explosion.

External forces control center-of-mass motion

A deep unifying idea across mechanics is:

• Internal forces can change relative motion within the system.
• External forces determine the motion of the system as a whole (center of mass).

If the net external force is zero, the center of mass moves at constant velocity. That’s the “system-level” version of Newton’s First Law.

Worked example: center of mass position (1D)

Two carts on a line: cart A has mass 2.0 kg at x = 1.0 m, cart B has mass 6.0 kg at x = 5.0 m. Find x_{\text{cm}}.

Total mass:

M = 2.0 + 6.0 = 8.0

Center of mass:

x_{\text{cm}} = \frac{(2.0)(1.0) + (6.0)(5.0)}{8.0} = \frac{2.0 + 30.0}{8.0} = 4.0

So the center of mass is at 4.0 m, closer to the heavier cart.

Worked example: center-of-mass motion in an explosion

A 4.0 kg cart moves right at 3.0 m/s on a frictionless track. It explodes into two pieces: 1.0 kg and 3.0 kg. Immediately after, the 1.0 kg piece moves right at 9.0 m/s. Find the 3.0 kg piece’s velocity and verify the center-of-mass idea.

Conserve momentum (no external impulse horizontally):

P_i = (4.0)(3.0) = 12

Final momentum:

12 = (1.0)(9.0) + (3.0)v

12 = 9.0 + 3.0v

v = 1.0

So the 3.0 kg piece moves right at 1.0 m/s.

Now center-of-mass velocity should remain 3.0 m/s because total momentum and mass are unchanged:

v_{\text{cm}} = \frac{P}{M} = \frac{12}{4.0} = 3.0

Check using final pieces:

P_f = (1.0)(9.0) + (3.0)(1.0) = 12

So v_{\text{cm}} remains 3.0 m/s as expected.

Exam Focus

• Typical question patterns:
  • Compute center of mass location for discrete masses (often 1D or simple 2D).
  • Use P = M v_{\text{cm}} to connect momentum conservation to center-of-mass motion.
  • Explain qualitatively why explosions do not change center-of-mass motion without external forces.
• Common mistakes:
  • Averaging positions without mass-weighting (using a simple mean instead of a weighted mean).
  • Assuming the center of mass must lie on a physical object (it can be in empty space).
  • Thinking internal forces can change the motion of the center of mass when external forces are negligible.

Putting It All Together: Multi-Concept Momentum Problems

How real AP problems combine ideas

Many AP Physics 1 questions are not “just momentum.” They layer momentum with other units:

• Momentum during a collision, then energy or kinematics afterward.
• Impulse from a force-time graph, then relate impulse to stopping distance using kinematics.
• Two-dimensional momentum to find a post-collision speed, then projectile motion (in later units) or frictional stopping (work-energy).

The skill is choosing the right tool for each phase.

A powerful way to structure your thinking is to separate an event into phases:

1. A short interaction phase (collision/explosion) where momentum is conserved.
2. A longer motion phase where external forces (friction, gravity components) matter and you use energy or kinematics.

Worked example: collision then friction stopping (two phases)

A 0.20 kg puck slides right at 6.0 m/s on a horizontal surface and collides with a 0.30 kg puck at rest. They stick together. The surface has kinetic friction coefficient 0.10 (assume it acts after the collision). Find the distance they slide after the collision before stopping.

Phase 1: collision (perfectly inelastic)
Momentum conservation during the collision:

(0.20)(6.0) + (0.30)(0) = (0.20 + 0.30)v_f

1.2 = 0.50 v_f

v_f = 2.4

So immediately after collision, the combined mass 0.50 kg moves at 2.4 m/s.

Phase 2: sliding to stop under friction
Now friction provides an external force, so momentum is not conserved. Use work-energy or kinematics.

Friction force magnitude:

F_k = \mu_k N = \mu_k mg = (0.10)(0.50)(9.8) = 0.49

Acceleration magnitude:

a = \frac{F_k}{m} = \frac{0.49}{0.50} = 0.98

Direction is opposite motion, so a = -0.98 m/s² if right is positive.

Use kinematics to stop: v^2 = v_0^2 + 2a\Delta x with v = 0 and v_0 = 2.4:

0 = (2.4)^2 + 2(-0.98)\Delta x

0 = 5.76 - 1.96\Delta x

\Delta x = 2.94

So they slide about 2.9 m after the collision.

This kind of two-phase reasoning is extremely common: conserve momentum only in the brief interaction, then switch tools.

Worked example: impulse from a force-time graph then change in speed

A 1.5 kg cart initially moves right at 1.0 m/s. A force acts on it in the +x direction for 0.40 s. The force-time graph is a rectangle at 6.0 N for the entire 0.40 s. Find the final speed.

Impulse:

J = F\Delta t = (6.0)(0.40) = 2.4

Impulse equals change in momentum:

\Delta p = 2.4

Initial momentum:

p_i = (1.5)(1.0) = 1.5

Final momentum:

p_f = p_i + \Delta p = 1.5 + 2.4 = 3.9

Final speed:

v_f = \frac{p_f}{m} = \frac{3.9}{1.5} = 2.6

So the cart’s speed increases to 2.6 m/s.

Exam Focus

• Typical question patterns:
  • Multi-stage scenarios: conserve momentum during collision, then use energy/kinematics afterward.
  • Use force-time graph area to get impulse, then connect to velocity change.
  • Explain modeling choices: why momentum is conserved during a short collision but not over longer intervals with friction.
• Common mistakes:
  • Conserving momentum across a time interval where a significant external impulse acts (like friction over seconds).
  • Using kinetic energy conservation for inelastic collisions.
  • Mixing phases: using post-collision speed as if it were the pre-collision speed (or vice versa).