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Theme A — Space, Time and Motion

A.2.15 Collisions and Explosions in Two-Dimensions — IB Physics HL Revision Notes & Topical Questions

5 min read 8 parts Lesson 26 of 63 1 quiz

📚 IB Physics HL Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Lesson summary

A.2.15 Collisions and Explosions in Two-Dimensions — IB PHYSICS HL notes: Apply conservation of momentum to collisions and explosions in two dimensions using components.Momentum is a vector — it can be resolved into horizontal and vertical components.Each component of momentum is…

  • Collisions and explosions both obey conservation of momentum in each direction
  • A collision in 2D involves two objects interacting in a plane
  • x direction
  • y direction
  • Conservation of momentum and energy together gives θA + θB
  • Before the explosion

Part 1 of 6Momentum in two dimensions

Momentum in two dimensions
Momentum is a vector
  • Momentum is a vector it can be resolved into horizontal and vertical components.
Each component of momentum is conserved separately.
  • The sum of horizontal components is the same before and after.
  • The sum of vertical components is the same before and after.
Collisions and explosions both obey conservation of momentum in each direction
  • Collisions and explosions both obey conservation of momentum in each direction when no external resultant force acts.

Part 2 of 6Collisions in two dimensions

1A collision in 2D involves two objects interacting in a plane
  • A collision in 2D involves two objects interacting in a plane velocities have both horizontal and vertical components.
2
Ball A at speed uA hits stationary ball B.
  • Ball A leaves at vA and angle θA above the horizontal.
  • Ball B leaves at vB and angle θB below the horizontal.
3x direction
  • x direction mAuA + 0 = mAvAcos θA + mBvBcos θB.
4y direction
  • 0 + 0 = mAvAsin θA − mBvBsin θB.
  • The minus sign appears because B moves downward while positive y is upward.
  • The initial y momentum is zero because ball A has no vertical component and B is at rest.

Part 3 of 6Elastic collisions in 2D

1
If the collision is elastic with one body initially at rest, kinetic energy is also conserved.
2Conservation of momentum and energy together gives θA + θB
90°.
3Use tan θ
Use tan θ sin θ / cos θ to combine the two component equations.
4
Check whether the collision is elastic by comparing kinetic energies before and after.

Part 4 of 6Explosions in two dimensions

1
An explosion in 2D: one stationary object becomes two or more objects moving in different directions.
2Before the explosion
  • Before the explosion velocity and momentum are zero.
3After the explosion
  • After the explosion the vector sum of the momenta must still be zero.
  • The horizontal components sum to zero and the vertical components sum to zero.
4Kinetic energy is never conserved in an explosion
  • Kinetic energy is never conserved in an explosion it increases from zero.

Part 5 of 6Method for 2D problems

1
Draw before and after diagrams and label masses, speeds and angles with subscripts.
2
Choose x and y axes and a positive direction for each.
3
Resolve every velocity into components using sin and cos.
4
Write conservation of momentum for the x direction and for the y direction.
5
Solve — usually the y equation first when it contains fewer unknowns — then use x.
Method for 2D problems
Check the calculator is in degree mode and brackets close trigonometric functions.
Rearranging before substituting keeps track of many masses and velocities.

Part 6 of 6More from the specification

More from the specification
Collisions in two dimensions
  • Collisions in two dimensions linear momentum is conserved separately in each perpendicular direction, so components are resolved
Elastic and inelastic
  • Elastic and inelastic kinetic energy is conserved only in elastic collisions
Newton's laws
  • Newton's laws resultant force equals rate of change of momentum, and forces include weight, friction and fluid resistance
Circular motion
  • Circular motion an object moving in a circle needs a net centripetal force, with equal and opposite forces forming Newton third law pairs
Normal reaction
  • The contact force perpendicular to a surface.
Reaction force
  • Force from a support pushing up on an object.
Fluid resistance
  • Drag force from a fluid opposing motion.
Impulse
  • J = FΔt = Δp.
    What the symbols mean: F = force (N) · Δt = time the force acts (s) · Δp = change in momentum (kg m s−1); FΔt is the impulse (N s)
Radius
  • Distance from the centre to the circular path.
Normal
  • Line at 90° to a surface.
Tension
  • Force transmitted along a string or rope.
Centripetal force
  • Net force toward the centre, mv2/r.
Static friction
  • Friction between surfaces at rest relative to each other; fs ≤ μsFN.
Circular motion
  • Motion along a circular path with a force towards the centre.
Angular speed
  • Magnitude of the angular velocity.
Acceleration
  • Rate of change of velocity.
Friction
  • Force parallel to the surface opposing relative motion or its tendency.
Newton
  • Unit of force, kg m s-2.
Weight
  • Gravitational force on a mass.
Constant
  • Not changing.
Linear
  • A graph that is a straight line through the origin, so resistance is constant.

Key terms

TermMeaning
Component of momentumPart of momentum along a chosen axis.
ExplosionEvent in which an object at rest splits and the parts move apart.
Elastic collisionCollision in which kinetic energy is conserved.
Conservation of momentumTotal momentum is unchanged when no external force acts.
ResolveSplit a vector into perpendicular components.
RecoilBackward motion from momentum conservation when parts are expelled.

Avoid these mistakes

1
Conserving only the total speed instead of each momentum component.
2
Forgetting the minus sign for a component pointing the opposite way.
3
Setting final momentum to zero in a collision where the objects move off.
4
Using the angle with the wrong trigonometric function for each component.
5
Assuming kinetic energy is conserved in every collision.

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