OCR Specification focus:
‘Apply the principle of conservation of momentum to closed systems during collisions or separations.’
Collisions involve the interaction of objects exchanging momentum. Understanding one- and two-dimensional collision problems requires applying vector principles and conservation laws to analyse post-collision motion precisely.
Conservation of Momentum in Collisions
The principle of conservation of momentum is a fundamental law governing all collision and separation events within a closed system, meaning no external forces act on it. When two or more objects collide or separate, the total momentum before interaction equals the total momentum after interaction. This principle applies regardless of whether the collision occurs in one or two dimensions.
Momentum: The product of an object’s mass and velocity; a vector quantity that represents the quantity of motion possessed by an object.
In all analyses, momentum must be treated as a vector, having both magnitude and direction. This means that the components of momentum along each perpendicular axis (usually x and y) must be considered separately in two-dimensional problems.
EQUATION
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Conservation of Momentum: Σp₍before₎ = Σp₍after₎
p = mv
m = mass (kg)
v = velocity (m s⁻¹)
—-----------------------------------------------------------------
In a closed system, if two bodies A and B collide, then:
mₐuₐ + mᵦuᵦ = mₐvₐ + mᵦvᵦ
where u and v represent the initial and final velocities, respectively.
Conditions for Conservation
For momentum to be conserved:
The system must be closed — no external forces like friction or air resistance act significantly during the collision.
The interaction forces between colliding objects must be internal and obey Newton’s Third Law (equal and opposite).
The vector sum of momenta remains constant in both magnitude and direction across all dimensions.
One-Dimensional Collision Problems
A one-dimensional collision occurs when objects move along a straight line, such as head-on or rear-end impacts. In these cases, motion is confined to a single axis, simplifying analysis.
During a one-dimensional collision:
The total momentum before collision equals total momentum after collision.
Depending on whether the collision is elastic or inelastic, kinetic energy may or may not be conserved.
Direction is crucial: velocities in opposite directions are assigned opposite signs.
EQUATION
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Linear Momentum in One Dimension: p = mv
p = linear momentum (kg m s⁻¹)
m = mass (kg)
v = velocity, including direction (m s⁻¹)
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If two objects collide and one comes to rest, or they move together after impact, the equations simplify according to the specific interaction type. The analysis remains rooted in the conservation of momentum along a single line of motion.
Key Analytical Steps
To analyse a one-dimensional collision:
Assign a positive direction (usually to the right or forward).
Write momentum equations for all objects before and after the collision.
Substitute known quantities for mass and velocity.
Solve for the unknown final velocities using simultaneous equations when necessary.
For one-dimensional collisions, choose a positive direction, assign signs, and apply conservation of momentum along that line only.
A head-on 1D collision with masses m₁ and m₂ showing initial (u₁, u₂) and final (v₁, v₂) velocities along a single axis. The diagram reinforces sign conventions and the use of a common positive direction. It is intentionally minimal to focus on the quantities needed for OCR-style momentum equations. Source
In inelastic collisions, objects may stick together, forming a combined mass with a single velocity. In elastic collisions, they rebound, and kinetic energy is also conserved.
Two-Dimensional Collision Problems
In reality, many collisions occur in two dimensions, such as when billiard balls collide or particles scatter. These require a vector approach using component analysis.
Component of Momentum: The part of an object’s momentum resolved along a specific direction, typically horizontal (x) or vertical (y).
In two-dimensional collisions, the total momentum is conserved independently in each perpendicular direction:
EQUATION
—-----------------------------------------------------------------
Conservation in x-direction: Σpₓ(before) = Σpₓ(after)
Conservation in y-direction: Σpᵧ(before) = Σpᵧ(after)
pₓ, pᵧ = momentum components in x and y (kg m s⁻¹)
—-----------------------------------------------------------------
The overall momentum vector can be reconstructed by combining these perpendicular components using vector addition.
Step-by-Step Process for Two-Dimensional Problems
To solve two-dimensional collision problems:
Draw a clear vector diagram showing the velocities before and after collision, with angles to a reference axis.
Resolve each velocity vector into horizontal and vertical components using trigonometric functions (cosine and sine).
Apply conservation of momentum separately to the x- and y-components.
Combine results to determine the magnitude and direction of the final velocities.
In two-dimensional elastic collisions of equal masses with one initially at rest, the outgoing velocities are perpendicular.
Diagram of an incident sphere striking an identical stationary sphere; after an elastic collision, the two spheres separate with orthogonal velocity vectors. This visual supports resolving components and recognising the right-angle result for equal masses. Labels distinguish laboratory and centre-of-momentum viewpoints without unnecessary clutter. Source
When collisions occur at oblique angles, the vector sum of momenta in both directions must remain constant. Kinetic energy analysis, if relevant, is applied only after momentum conservation has been satisfied.
Vector Diagrams and Representations
A vector diagram visually represents momentum before and after a collision. Each vector’s length corresponds to the momentum’s magnitude, and the orientation shows its direction. By aligning pre- and post-collision momentum vectors head-to-tail, one can verify whether total momentum has been conserved geometrically.
Important notes for accurate diagrams:
The scale should be consistent across both axes.
Vectors should be drawn for each object involved in the collision.
The vector sum of the post-collision momenta should equal the initial total momentum vector.
Resolve all momenta into x and y components and apply conservation of momentum independently to each direction.
Momentum-component layout for a 2D collision where one object is initially at rest. The axes (î, ĵ) and angles (θ₁,f, θ₂,f) are indicated to guide writing separate x- and y-component equations. The original page includes surrounding worked text; the figure alone is what you need for the OCR method. Source
Relationship with Newton’s Laws
Two-dimensional collision analysis is closely linked to Newton’s Laws of Motion. Newton’s Third Law ensures that forces during collision are equal and opposite, leading directly to conservation of momentum. Newton’s Second Law, when expressed as the rate of change of momentum, underlies how internal forces cause momenta of individual bodies to change while the total remains constant.
Closed System: A system in which no external forces act, ensuring conservation of total momentum.
In open systems, external influences such as friction or external impulses cause deviations from ideal conservation behaviour, leading to loss of momentum in a specific direction.
Practical Applications
Understanding one- and two-dimensional collisions is crucial for analysing real-world phenomena, including:
Vehicle crash reconstruction, where direction and post-impact velocities determine cause and severity.
Particle interactions in physics experiments, where conservation of momentum helps identify unknown particles.
Sports dynamics, such as the rebound angles in snooker or football deflections.
In every scenario, the same principles apply — the vector sum of momentum before interaction equals that after, provided the system is closed and free of external interference.
FAQ
Momentum is always conserved in a closed system, regardless of whether the collision is elastic or inelastic. This is because it depends only on Newton’s Third Law and the absence of external forces.
Kinetic energy, however, is conserved only in perfectly elastic collisions. In inelastic or partially elastic collisions, some kinetic energy is transformed into other forms, such as heat, sound, or deformation energy.
Thus, momentum conservation is a universal rule, while energy conservation applies selectively depending on the nature of the collision.
For equal masses where one object is initially stationary, the two objects move away at right angles (90°) after an elastic collision.
When the masses are unequal, this right-angle result no longer holds. The angles depend on the ratio of the masses and the initial velocity of the moving object. In general:
The lighter object deflects more sharply.
The heavier object continues closer to the original direction of motion.
This outcome arises directly from conserving both momentum components and kinetic energy simultaneously.
Two-dimensional collisions involve motion in both x and y directions, meaning momentum must be treated as a vector quantity.
Vector diagrams help:
Visually represent the direction and magnitude of each object’s momentum before and after collision.
Verify momentum conservation by ensuring the head-to-tail vector sum remains unchanged.
Identify the angles and relationships between final velocities, especially when objects scatter or rebound obliquely.
Without vector diagrams, it is difficult to confirm the accuracy of calculated directions and maintain consistent vector relationships.
External forces can be ignored when their effects are negligible during the short duration of a collision. Examples include:
Collisions between gliders on a nearly frictionless air track.
Molecular collisions in gases where external fields are weak.
Vehicle impacts analysed over very brief timescales, before external resistances (e.g. friction) significantly alter total momentum.
In these cases, the system behaves approximately as closed, allowing accurate use of momentum conservation principles even though minor external forces exist.
Uncertainties in measuring mass, velocity, or angle can significantly impact momentum calculations.
Key sources of uncertainty include:
Timing errors from motion sensors or video frame rates.
Frictional effects altering measured velocities.
Angle measurement inaccuracies, particularly when collisions occur at shallow or near-perpendicular angles.
To minimise uncertainty:
Repeat trials and take mean velocity values.
Use high-speed cameras or light gates for precision.
Carefully align measuring apparatus to maintain consistent angle references.
Reliable data ensure that momentum conservation appears valid within experimental limits.
Practice Questions
Question 1 (2 marks)
Two gliders of masses 0.4 kg and 0.6 kg move towards each other on a frictionless air track. The 0.4 kg glider moves at 3.0 m s⁻¹ to the right and the 0.6 kg glider moves at 2.0 m s⁻¹ to the left. They collide and stick together.
Calculate the velocity of the combined gliders immediately after the collision, stating its direction.
Mark Scheme
Correct use of conservation of momentum:
(0.4 × 3.0) + (0.6 × –2.0) = (0.4 + 0.6)v (1 mark)
Correct calculation of final velocity: v = 0 m s⁻¹ (no movement) (1 mark)
Accept sign convention used clearly; accept “stationary” as a valid final answer.
Question 2 (5 marks)
A smooth horizontal surface allows two identical 0.25 kg pucks to collide elastically. Puck A moves at 4.0 m s⁻¹ due east and strikes stationary puck B. After collision, puck A moves off at 30° north of east.
(a) State the principle that applies to the total momentum of the two pucks during the collision. (1 mark)
(b) Write the equations expressing conservation of momentum in the east (x) and north (y) directions. (2 marks)
(c) Determine the speed of puck B after the collision. (2 marks)
Mark Scheme
(a)
Correct statement: Total momentum of a closed system is conserved if no external forces act. (1 mark)
(b)
East (x): (0.25 × 4.0) = (0.25 × vA × cos30°) + (0.25 × vB × cosθ) (1 mark)
North (y): 0 = (0.25 × vA × sin30°) – (0.25 × vB × sinθ) (1 mark)
(c)
Recognition that collision is elastic and that the outgoing velocities are perpendicular for equal masses, therefore vA = vB = 4.0 m s⁻¹ (1 mark)
Correct reasoning that puck B moves at 60° south of east (perpendicular to puck A’s path). (1 mark)
Award full marks for clear use of vector resolution and accurate recognition of the right-angle relationship between outgoing velocities.
