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·?t = ?p

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

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

• Impulse-Momentum Theorem: The impulse on an object equals its change in momentum. F·?t = m·?v = 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. m?v + m?v = m?v?f + m?v?f

• 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 = (m?x? + m?x?) / (m? + m?)

• Newton’s Second Law (Momentum Form): F_net = ?p / ?t

• 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: m?v + m?v = m?v?f + m?v?f (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·?t = ?p)

Steps:
1. Identify initial and final momentum (p = mv).
2. Calculate ?p = p_f – p_i.
3. Use J = F·?t = ?p 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? - ?p = (0.15)(50) – (0.15)(–40) = 13.5 kg·m/s - F = ?p / ?t = 13.5 / 0.005 = 2700 N

3. Center of Mass Problems

Steps:
1. List masses and positions (x?, x?, etc.).
2. Use x_COM = (m?x? + m?x?) / (m? + m?).
3. For velocity, use v_COM = (m?v? + m?v?) / (m? + m?).

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 = ?p/?t for impulse problems. Correction: For variable mass (e.g., a rocket ejecting fuel), F = ?p/?t is the correct form of Newton’s 2nd Law.

• Mistake: Mixing up impulse (J = F·?t) 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 ?p = –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 (?p). 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: ?p = m(v_f – v_i) = 0.2(–15 – 10) = –5 kg·m/s. Impulse = ?p = 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·?t = ?p (area under F-t graph).
3. Conservation of momentum: m?v + m?v = m?v?f + m?v?f (no external forces).
4. Elastic collision: KE conserved. Inelastic: KE not conserved. Perfectly inelastic: objects stick.
5. Center of mass: x_COM = (m?x? + m?x?) / (m? + m?).
6. Newton’s 2nd Law (momentum form): F_net = ?p / ?t.
7. Momentum is a vector—always assign +/– signs!
8. Impulse-work (J = F·?t 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).