AP Physics – Linear Momentum and Impulse‑Momentum Theorem

AP Physics – Linear Momentum and Impulse‑Momentum Theorem

AP Physics: Linear Momentum & Impulse-Momentum Theorem – Exam-Ready Study Guide

What This Is

Linear momentum (p) is the "oomph" an object has due to its mass and velocity—think of a bowling ball rolling down the lane vs. a ping-pong ball. The Impulse-Momentum Theorem connects force, time, and momentum change, explaining why airbags save lives (they increase the time of impact, reducing force) or why a baseball player "follows through" on a swing (to maximize impulse). This topic is heavily tested on the AP exam, especially in collision problems, free-response questions (FRQs), and conceptual multiple-choice.

Key Terms & Concepts

• Linear Momentum (p):
  p = m·v
• m = mass (kg), v = velocity (m/s).

• Momentum is a vector—direction matters! A 2 kg ball moving left at 3 m/s has p = –6 kg·m/s.

• Impulse (J):
  J = F·Δt = Δp

• F = average net force (N), Δt = time interval (s), Δp = change in momentum (kg·m/s).

• Impulse is the "push" that changes momentum. Example: A soccer player kicks a ball (large F over small Δt) vs. a gentle tap (small F over large Δt).

• Impulse-Momentum Theorem:
  The impulse on an object equals its change in momentum.
  F·Δt = m·Δv = m(v_f – v_i)

• Key idea: Force and time trade off. A small force over a long time can produce the same impulse as a large force over a short time.

• Conservation of Momentum:
  In a closed system (no external forces), total momentum before = total momentum after.
  m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f

• Applies to collisions and explosions (e.g., a cannon recoiling after firing a cannonball).

• Elastic Collision:
  Both momentum and kinetic energy are conserved.

• Example: Two billiard balls colliding and bouncing apart.

• Inelastic Collision:
  Momentum is conserved, but kinetic energy is not (some is lost as heat/sound).

• Example: A clay ball sticking to a wall.

• Perfectly Inelastic Collision:
  Objects stick together after collision (maximum kinetic energy loss).

• Example: Two train cars coupling together.

• Center of Mass (COM):
  The average position of all mass in a system. For a system with no external forces, the COM moves at constant velocity.

• x_COM = (m₁x₁ + m₂x₂) / (m₁ + m₂)

• Newton’s Second Law (Momentum Form):
  F_net = Δp / Δt

• Force is the rate of change of momentum. This is the foundation of the Impulse-Momentum Theorem.

• Area Under a Force-Time Graph:
  The area = impulse (J). Useful for variable forces (e.g., a bat hitting a ball).

Step-by-Step: Solving Momentum Problems

1. Collision/Explosion Problems (Conservation of Momentum)

Steps:
1. Draw a before/after diagram (label masses, velocities, and directions).
2. Define a coordinate system (e.g., right = +, left = –).
3. Write the conservation equation:
m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f
(For explosions, initial momentum is often zero.) 4. Solve for the unknown (e.g., final velocity).
5. Check units and signs (momentum is a vector!).

Example:
A 1000 kg car moving at +20 m/s collides with a 1500 kg truck at rest. If they stick together, what’s their final velocity? - Before: (1000)(20) + (1500)(0) = 20,000 kg·m/s - After: (1000 + 1500)v_f = 20,000 → v_f = +8 m/s

2. Impulse Problems (F·Δt = Δp)

Steps:
1. Identify initial and final momentum (p = mv).
2. Calculate Δp = p_f – p_i.
3. Use J = F·Δt = Δp to find the missing variable (force or time).
4. For variable forces, find the area under the F-t graph.

Example:
A 0.15 kg baseball is thrown at 40 m/s and hit back at 50 m/s. If the bat contacts the ball for 0.005 s, what’s the average force? - Δp = (0.15)(50) – (0.15)(–40) = 13.5 kg·m/s - F = Δp / Δt = 13.5 / 0.005 = 2700 N

3. Center of Mass Problems

Steps:
1. List masses and positions (x₁, x₂, etc.).
2. Use x_COM = (m₁x₁ + m₂x₂) / (m₁ + m₂).
3. For velocity, use v_COM = (m₁v₁ + m₂v₂) / (m₁ + m₂).

Example:
Two 2 kg masses are at x = 1 m and x = 3 m. Where’s the COM? - x_COM = (2·1 + 2·3) / (2 + 2) = 2 m

Common Mistakes

• Mistake: Forgetting momentum is a vector (ignoring direction).
  Correction: Always assign +/– signs based on your coordinate system. A ball bouncing off a wall reverses momentum sign.

• Mistake: Assuming kinetic energy is conserved in all collisions.
  Correction: Only in elastic collisions is KE conserved. In inelastic collisions, some KE is lost (e.g., sound, heat).

• Mistake: Using F = ma instead of F = Δp/Δt for impulse problems.
  Correction: For variable mass (e.g., a rocket ejecting fuel), F = Δp/Δt is the correct form of Newton’s 2nd Law.

• Mistake: Mixing up impulse (J = F·Δt) with work (W = F·d).
  Correction: Impulse changes momentum, work changes energy. They’re different!

• Mistake: Not checking units (e.g., using grams instead of kg).
  Correction: Always convert to kg, m/s, N·s for momentum/impulse.

AP Exam Insights

• FRQs often test:
• Conservation of momentum in 2D collisions (e.g., a car crash at an angle).
• Impulse from F-t graphs (calculate area under the curve).
• Center of mass for systems of particles (e.g., a person walking on a boat).

• Real-world applications (e.g., airbags, sports, rockets).

• Multiple-choice traps:

• Sign errors: A ball bouncing off a wall has Δp = –2mv (not zero!).
• Elastic vs. inelastic: If objects stick together, it’s perfectly inelastic (KE not conserved).

• External forces: Momentum is not conserved if external forces act (e.g., friction, gravity).

• Tricky distinction:

• Impulse (J) = change in momentum (Δp). They’re the same thing!
• Momentum (p) = mass × velocity. Don’t confuse it with kinetic energy (½mv²).

Quick Check Questions

1. Multiple Choice

A 0.2 kg hockey puck slides at 10 m/s and is hit by a stick, reversing its direction to 15 m/s. What is the impulse delivered to the puck? (A) –1 N·s (B) 1 N·s (C) –5 N·s (D) 5 N·s

Answer: (D) 5 N·s Explanation: Δp = m(v_f – v_i) = 0.2(–15 – 10) = –5 kg·m/s. Impulse = Δp = 5 N·s (magnitude is positive; direction is implied by sign).

2. Free-Response (Short)

A 1200 kg car moving at +15 m/s collides with a stationary 1800 kg truck. After the collision, the car moves at –5 m/s.
(a) What is the truck’s velocity after the collision? (b) Is this collision elastic, inelastic, or perfectly inelastic? Justify your answer.

Answer:
(a) v_truck = +13.3 m/s
- Conservation of momentum: (1200)(15) + (1800)(0) = (1200)(–5) + (1800)v_f - 18,000 = –6,000 + 1800v_f → v_f = 13.3 m/s

(b) Inelastic (but not perfectly inelastic) - KE before = ½(1200)(15)² = 135,000 J - KE after = ½(1200)(5)² + ½(1800)(13.3)² ≈ 171,000 J - KE is not conserved, so it’s inelastic. Objects don’t stick together, so it’s not perfectly inelastic.

Last-Minute Cram Sheet

1. Momentum (p) = mv (vector! direction matters).
2. Impulse (J) = F·Δt = Δp (area under F-t graph).
3. Conservation of momentum: m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f (no external forces).
4. Elastic collision: KE conserved. Inelastic: KE not conserved. Perfectly inelastic: objects stick.
5. Center of mass: x_COM = (m₁x₁ + m₂x₂) / (m₁ + m₂).
6. Newton’s 2nd Law (momentum form): F_net = Δp / Δt.
7. ⚠️ Momentum is a vector—always assign +/– signs!
8. ⚠️ Impulse ≠ work (J = F·Δt vs. W = F·d).
9. ⚠️ External forces break momentum conservation (e.g., friction, gravity).
10. ⚠️ For explosions, initial momentum is often zero (e.g., a cannon firing).