φPhysicsLeaving Certificate

Topic 03 · Forces and Motion

Forces & momentum

Student revision notes covering the full two-year LC Physics course. This unit covers Newton's three laws of motion, the distinction between mass and weight, and the conservation of momentum in collisions.

Learning goals

What students should be able to explain, calculate and investigate after completing Forces and Momentum.

  • State and apply Newton's first, second and third laws of motion.
  • Distinguish mass from weight.
  • Draw correct free-body diagrams showing forces acting ON an object.
  • Define momentum and state the principle of conservation of momentum.
  • Solve one-dimensional collision problems using conservation of momentum.
  • Higher Level: Model two-dimensional collisions using perpendicular and parallel momentum components.
  • Describe an experiment verifying Newton's Second Law and one verifying conservation of momentum.

Key definitions

Precise terms, symbols and SI units used in Forces and Momentum.

The centre of mass is the point at which the mass of an object or system can be treated as concentrated when analysing its overall motion. For a uniform, symmetrical object it is at the geometric centre; for an irregular object it may be elsewhere.

Mass vs Weight

Mass and weight are often confused.

Mass is a scalar measured in kilograms and is the same everywhere in the universe. Weight is a force (a vector), measured in newtons, and depends on the gravitational field strength g at that location: W = mg. An astronaut has the same mass on the Moon as on Earth, but weighs about one-sixth as much, because the Moon's gravitational field strength is about one-sixth of Earth's.

W = mg

Newton's Laws of Motion

Newton's First Law describes inertia: objects resist changes to their state of motion.

Newton's Second Law quantifies how a resultant (net) force changes an object's motion: F = ma, where F is the resultant force, not any single applied force. Newton's Third Law describes force pairs: these always act on two different objects, never on the same object, so they can never cancel each other out for a single object's motion.

F = ma

For a trolley accelerating to the right on a horizontal surface, draw only forces acting on the trolley: weight downward, normal reaction upward, driving or tension force to the right, and any friction or resistance to the left. The vertical forces balance, but the horizontal forces do not; the resultant horizontal force points right, so the trolley accelerates right.

  • Do not draw velocity or acceleration arrows as forces on a free-body diagram.
  • If the object accelerates, the forces in the direction of acceleration cannot be balanced.
  • Use Fresultant = ma only after combining all forces on that object with signs or components.

Free-Body Diagrams

A free-body diagram shows only the forces acting ON a chosen object — never the forces that object exerts on other things.

For an object resting on a table, the diagram shows its weight (down) and the normal force from the table (up), plus any applied force and friction, but not the force the object exerts on the table.

Momentum and Collisions

Momentum, p = mv, is conserved in any collision or explosion within a closed system (no external resultant force acts).

During a collision, object 1 exerts a force on object 2 and object 2 exerts an equal, opposite force on object 1: F₁₂ = −F₂₁. The forces act for the same interaction time, so the impulses are equal and opposite. The two momentum changes therefore cancel, leaving total momentum unchanged when no resultant external force acts.

p = mv
total momentum before = total momentum after
  • Equal and opposite interaction forces: F₁₂ = −F₂₁.
  • Same interaction time gives equal and opposite impulses.
  • The momentum gained by one object equals the momentum lost by the other.
  • Therefore total momentum before = total momentum after.

Worked example

A 60 kg student at rest throws a 3.0 kg bag horizontally at 8.0 m s⁻¹. Find the student’s recoil velocity.

  1. Initial total momentum = 0.
  2. After the throw: (3.0 × 8.0) + (60 × v) = 0.
  3. 24 + 60v = 0, so v = −0.40 m s⁻¹.

Answer: The student moves at 0.40 m s⁻¹ opposite to the bag.

Diagrams and formulae

Key diagrams and equations for reviewing Forces and Momentum.

W = mg
F = ma
p = mv
m1u1 + m2u2 = m1v1 + m2v2

Free-body diagram: only forces acting ON the box are shown (weight, normal force, applied force, friction) — never the forces the box exerts on the table or on anything else.

Newton's Third Law: the book pushes down on the table, and the table pushes up on the book with an equal and opposite force. These two forces act on two different objects, so they never cancel for the same object.

Momentum diagram: masses and velocities are shown before and after a collision, with directions indicated by the arrows.

  • Weight: W = mg, where W is weight in newtons (N), m is mass in kilograms (kg), and g is gravitational field strength in newtons per kilogram (N kg⁻¹).
  • Newton’s second law: F = ma, where F is resultant force in newtons (N), m is mass in kilograms (kg), and a is acceleration in metres per second squared (m s⁻²).
  • Momentum: p = mv, where p is momentum in kilogram metres per second (kg m s⁻¹), m is mass in kilograms (kg), and v is velocity in metres per second (m s⁻¹).
  • Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, where u is initial velocity and v is final velocity; subscripts identify the two objects.

Worked Example

A step-by-step calculation applying the central relationships in Forces and Momentum.

A trolley of mass 2 kg moving at 3 m s⁻¹ collides with a stationary trolley of mass 1 kg. After the collision they move off together. Find their common velocity.

  • Momentum before = momentum after (they stick together, so this is one combined mass afterward)
  • m₁u₁ + m₂u₂ = (m₁ + m₂)v
  • (2 × 3) + (1 × 0) = (2 + 1)v
  • 6 = 3v → v = 2 m s⁻¹

This is an inelastic collision because the trolleys stick together after impact.

Practical / Investigation

Experimental planning, measurement and evaluation for Forces and Momentum.

Experiment: Verifying Newton's Second Law (F ∝ a at constant mass)

  • Apparatus: trolley, runway (may be compensated for friction by slightly tilting), string over a pulley, slotted masses, light gates or ticker timer.
  • Method: keep the total mass of the trolley + hanging masses constant throughout by transferring small masses from the trolley to the hanger for each run — this changes the accelerating force (weight of hanging mass) without changing the total mass being accelerated.
  • Independent variable: force applied (weight of hanging masses). Dependent variable: acceleration. Controlled variable: total mass of the system.
  • Plot acceleration (y-axis) against force (x-axis); a straight line through the origin supports F ∝ a. The slope of this line equals 1/m.
  • Sources of error: friction and air resistance are rarely fully compensated; the mass of the string/pulley is usually assumed negligible.
  • Conclusion wording: 'Within experimental error, the graph of acceleration against force is a straight line through the origin, supporting F ∝ a at constant mass, as predicted by Newton's Second Law.'

Experiment: Verifying conservation of momentum

  • Apparatus: two dynamics trolleys or air-track gliders, light gates or motion sensors, balance.
  • Method: measure the masses of both trolleys. Measure their velocities before and after collision. Calculate total momentum before and total momentum after. Compare the two values.
  • Result: within experimental error, total momentum before equals total momentum after, provided external forces such as friction are small.
  • Sources of error: friction, inaccurate velocity measurement, trolleys not moving in a straight line, collision not clean.
  • Improvement: use an air track or low-friction runway and light gates.
  • Keep the accelerating force constant and increase the total moving mass by adding known masses to the trolley for each run.
  • Measure the acceleration for each total mass. Plot acceleration a against 1/m; a straight line through the origin supports a ∝ 1/m when the resultant force is constant.
  • Use the same runway and hanging force, compensate for friction, and include the trolley, hanger and added masses when calculating the total moving mass.
  • Newton II, constant mass: transfer masses from trolley to hanger so accelerating force changes while total moving mass remains constant; plot a against F.
  • Newton II, constant force: add masses to the trolley while keeping the accelerating force fixed; plot a against 1/m.
  • Momentum: measure both trolley masses and signed velocities immediately before and after collision; compare total vector momentum, not speed alone.
  • Level/compensate the runway, use light gates or motion sensors, repeat runs and identify friction as an external-force limitation.

Quick-Check Questions

Short retrieval questions covering the essential ideas in Forces and Momentum.

1. Define momentum and state its unit.

2. State Newton's Third Law.

3. Why does an astronaut's weight change on the Moon but not their mass?

4. In an investigation of a ∝ F, why must the total mass of trolley + hanger stay constant?

5. What quantity is conserved in all collisions, regardless of whether they are elastic or inelastic?

Common mistakes and exam tips

Common misconceptions and exam-technique checks for Forces and Momentum.

  • Avoid: Putting forces exerted by the object onto its free-body diagram. Correct idea: A free-body diagram contains only forces acting on the chosen object.
  • Avoid: Saying Newton’s third-law pair cancels on one object. Correct idea: The two forces act on different objects, so they do not cancel on a single free-body diagram.
  • Avoid: Ignoring direction in momentum conservation. Correct idea: Momentum is a vector; choose a positive direction and use signed velocities.
  • Interpret F in F = ma as the resultant force.
  • Choose and state a positive direction before a collision calculation.
  • For a Newton’s second-law graph, use the plotted axes to decide whether the slope is m or 1/m.

Exam-style questions

Five exam-style questions with worked answers on Forces and Momentum.

Higher Level — Required Content

Required Higher Level content for this topic.

Momentum is conserved separately in perpendicular directions. Choose x- and y-axes, resolve every momentum, apply conservation in each direction, then recombine components if a resultant is required.

  • Write a sign convention before calculating.
  • Momentum is a vector: direction matters.
  • Check both component equations independently.

End-of-unit revision

Formulae, definitions, experiments and mixed revision questions for Forces and Momentum.

  • W = mg
  • F = ma (resultant force)
  • p = mv
  • m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
  • Newton's First, Second and Third Laws
  • Mass vs weight
  • Momentum and its conservation
  • A correctly labelled free-body diagram (forces ON the object only)
  • Newton's Third Law force pairs shown on two separate objects
  • Momentum before/after diagram with directions shown
  • Verifying Newton's Second Law (a ∝ F at constant total mass, using the trolley-to-hanger mass transfer method)
  • Verifying conservation of momentum in a collision (e.g. using trolleys or air track and light gates)
  • Mass/weight confusion
  • Free-body diagrams showing the wrong object's forces
  • Believing Newton's Third Law pairs cancel
  • Changing system mass while testing a ∝ F

1. A ball of mass 0.5 kg travelling at 4 m s⁻¹ strikes a wall and rebounds at 3 m s⁻¹. Calculate the change in momentum of the ball.

2. Explain, using Newton's Third Law, why a swimmer is able to move forward by pushing water backward.

3. A trolley of mass 1.5 kg is accelerated from rest to 2 m s⁻¹ in 0.5 s by a constant force. Calculate the force applied.

4. Describe how you would keep the 'total accelerating mass' constant in a Newton's Second Law experiment while increasing the applied force.

1. Taking the initial direction as positive: Δp = m(v − u) = 0.5 × (−3 − 4) = −3.5 kg m s⁻¹ (magnitude 3.5 kg m s⁻¹, direction reversed).

2. The swimmer pushes water backward (action); by Newton's Third Law, the water pushes the swimmer forward with an equal and opposite force (reaction) — these forces act on different objects (water and swimmer), so the swimmer accelerates forward.

3. F = ma = m(v−u)/t = 1.5 × (2−0)/0.5 = 6 N.

4. Move a small mass from the trolley onto the hanger for each run — this increases the hanging mass (and so the accelerating force) while keeping the combined mass of trolley + hanger + masses constant throughout the experiment.

The table below shows which equations from this unit appear in the official State Examinations Commission Formulae and Tables book, and where.

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