Module A.2 IBDP SL/HL Track

Forces and Momentum

Master dynamics, Newton's laws of motion, systems in equilibrium, and momentum interactions.

Forces and Momentum

Introduction

In Kinematics, we mastered describing how objects move. Now, we shift our focus to Dynamics: an inquiry into why objects move. Consider a common everyday interaction: sliding a heavy text book across a laboratory desk. When you stop pushing, it immediately slows down and comes to rest. For centuries, thinkers believed that force was required to sustain motion.

However, by changing our perspective, we realize that the textbook didn’t stop because of a lack of force—it stopped because an invisible, external force (friction) actively opposed its motion. This realization forms the absolute core of dynamics: forces do not cause motion; forces cause changes in motion.


Key Concepts

1. Newton’s Laws of Motion

The behavior of all macroscopic matter in the universe under the influence of forces is governed by three elegant axioms:

  • First Law (Law of Inertia): An object continues in its state of rest or uniform motion in a straight line at a constant velocity unless acted upon by a net external force. Inertia is a structural property of matter directly proportional to its mass.
  • Second Law (The Fundamental Law): The net external force acting on an object is directly proportional to, and in the same direction as, the rate of change of its linear momentum over time.

F = Δp / Δt

When mass remains constant, this simplifies to the famous relationship:

F = ma

  • Third Law (Action-Reaction Pairs): When body A exerts a force on body B, body B simultaneously exerts an equal and opposite force on body A. Crucially, these two forces always act on different objects, meaning they can never cancel each other out within a system analysis.

2. Solid Friction: Static vs. Kinetic

Friction acts parallel to contacting surfaces and directly opposes relative motion. We distinguish between two phases:

  • Static Friction (Ff ≤ μs * R): The force that resists the initial relative motion of two stationary surfaces. It is a self-adjusting force that matches your pulling force up to a precise maximum threshold.
  • Kinetic Friction (Ff = μk * R): Once the object breaks free and begins sliding, the resisting force drops slightly to a constant, stable value, independent of speed.

Where: μs = coefficient of static friction, μk = coefficient of kinetic friction (typically μk < μs), and R = Normal Reaction Force (N).

3. Systems in Translational Equilibrium

When an object is stationary or moving at a perfect, constant velocity, its acceleration is exactly zero. According to Newton’s Second Law, the vector sum of all forces acting on the body must be zero.

ΣF = 0

To resolve equilibrium problems on inclined planes, we break vectors into independent component tracks:

  • Component parallel to the incline: F = mg * sin(θ)
  • Component perpendicular to the incline: F = mg * cos(θ)

4. Linear Momentum and Impulse

  • Linear Momentum (p): Defined as the product of an object’s mass and its velocity. It is a vector quantity possessing the exact direction of the velocity vector.

p = mv (Units: kg m s⁻¹ or N s)

  • Impulse (J): The change in linear momentum experienced by an object. It is calculated as the product of the average net force applied and the duration of the time interval. On a Force–Time graph, the area enclosed underneath the curve represents the total Impulse.

Impulse = F * Δt = Δp

5. Conservation of Linear Momentum

In a closed, isolated system—meaning a system completely free from external forces like friction—the total linear momentum remains perfectly constant before, during, and after any interaction or collision.

Total Initial Momentum = Total Final Momentum m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2

  • Elastic Collisions: Both total linear momentum and total kinetic energy are perfectly conserved.
  • Inelastic Collisions: Total momentum is conserved, but kinetic energy is lost to the environment as thermal energy or structural deformation. If objects stick together after colliding, it is a perfectly inelastic collision.

Common Mistakes

  • Drawing the Normal Force Blindly Upwards: Students frequently assume the normal reaction force (R) is always exactly equal to weight (mg) and points straight up. Remember, “Normal” means perpendicular to the surface. On an inclined plane or when an extra force pushes down on an object, R changes completely and is usually equal to mg * cos(θ).
  • Treating Centripetal or Net Force as a Separate, Real Force: When drawing Free-Body Diagrams (FBDs), students often add an extra arrow labeled “Fnet” or “Fcentripetal”. A net force is never an independent force; it is simply the vector sum resulting from real physical forces (like gravity, tension, or friction). Never draw “Fnet” on a formal FBD.
  • Ignoring Vector Directions in Momentum: Since momentum is a vector, direction signs are paramount. If two balls move toward each other, one must be declared positive and the other negative. Forgetting to input a negative sign for an object moving left or rebounding backward is the number one cause of lost marks in Paper 2 dynamics questions.

Exam Tips

  • Isolate Bodies in Multi-Object Problems: When dealing with connected objects (like two masses linked over a pulley or blocks pushing against each other), write a separate, distinct Newton’s Second Law equation (F = ma) for each individual object before trying to solve them simultaneously.
  • Friction and Inequality Signs: Pay close attention to the inequality in the static friction formula (Ff ≤ μs * R). The maximum force of static friction occurs only at the exact instant the object is “on the verge of slipping”. If the object is resting comfortably without a pulling force, the friction force is simply zero.
  • Look for “Constant Speed”: The moment you spot the phrase “constant speed” or “terminal velocity” in a question stem, immediately write down a = 0 and ΣF = 0. This instantly unlocks the problem by turning a complex dynamics calculation into a simple balanced force equation.

Practice Questions

Question 1 (Multiple Choice)

A block of mass M is pushed along a rough horizontal surface by a constant horizontal force F. The block accelerates uniformly at a rate of a. If the force of friction acting between the block and the surface is f, which of the following expressions represents the true relationship according to Newton’s Second Law?

A. F = M * a B. F + f = M * a C. F - f = M * a D. f - F = M * a

Correct Answer: C

Solution Architecture

Step 1: Identify the directions of the horizontal forces. The applied force F acts in the direction of motion (positive), while the friction force *f* opposes it (negative).

Step 2: Calculate the net horizontal force: Fnet = F - f.

Step 3: Equate the net force to the product of mass and acceleration according to Newton's second law: Fnet = M * a. Therefore, F - f = M * a, making option C correct.


Question 2 (Structured Paper 2 Style)

A toy freight car of mass 0.50 kg travels at a constant speed of 2.0 m s⁻¹ along a horizontal, frictionless track. It collides with and couples to a second stationary freight car of mass 0.30 kg.

(a) Determine the final common velocity of the coupled freight cars immediately after the collision. [2 marks]

(b) Calculate the total loss of kinetic energy during this interaction, and state the nature of the collision. [3 marks]

Paper 2 Structured Problem

Official Markscheme Breakdown

Part (a) Solution:

State the principle of conservation of linear momentum since the track is frictionless: m1 * u1 + m2 * u2 = (m1 + m2) * v

Substitute the known variables: (0.50 * 2.0) + (0.30 * 0) = (0.50 + 0.30) * v

1.0 = 0.80 * v

v = 1.25 m s⁻¹

Award [1 mark] for correct substitution into the momentum balance, and [1 mark] for the final velocity value with correct units.

Part (b) Solution:

Calculate initial kinetic energy (before collision): Eki = 0.5 * m1 * u1² = 0.5 * 0.50 * (2.0)² = 1.0 J

Calculate final kinetic energy (after collision): Ekf = 0.5 * (m1 + m2) * v² = 0.5 * 0.80 * (1.25)² = 0.625 J

Determine the difference: Loss = Eki - Ekf = 1.0 - 0.625 = 0.375 J

Kinetic Energy Lost = 0.38 J (or 0.375 J)

Since kinetic energy is lost and the objects stick together, the collision is **inelastic**.

Award [1 mark] for initial and final energy calculations, [1 mark] for the correct energy loss difference value, and [1 mark] for identifying the collision explicitly as inelastic.

Dynamics & Newton's Laws Explained
Momentum and Collisions Lab Analysis