Conserve momentum in two dimensions
Resolve each object's momentum into perpendicular components and conserve the total separately along each axis.
Section 3 · Dynamics
AS Level · Cambridge International Physics 9702, 2025–2027 syllabus, Topic 3. These original explanations and practice support theory study; they do not establish supervised practical competence or replace timed assessment and teacher review. Drag is treated qualitatively, and coefficient of restitution is outside this syllabus section.
Before you begin
- Use vectors and signed values consistently, and distinguish mass from weight.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How is momentum conservation applied to a two-dimensional interaction?
What you will learn
- Apply momentum conservation to two-dimensional interactions
- Resolve velocity and momentum into perpendicular components
- Reconstruct an unknown velocity from its components
Momentum conservation is a vector relationship
For an isolated system, the vector sum of momentum before an interaction equals the vector sum after it. In two dimensions, resolve each momentum vector into horizontal and vertical components.
Apply conservation separately along the x and y axes. Use one consistent sign convention on each axis. The two scalar equations share the same collision and can be combined to determine an unknown object's velocity components.
Combine components only after conservation is applied
Momentum is p = mv, so an object's momentum components are m vₓ and m vᵧ. If objects stick together, use the combined mass for the final velocity in both component equations.
Once both components of velocity are known, speed is the magnitude √(vₓ² + vᵧ²) and direction follows from the component ratio. Momentum is conserved whether an interaction is elastic or inelastic; kinetic energy requires a separate check.
Pause and explain
Two objects stick together after a two-dimensional collision. Which mass is used with their shared final velocity?
Worked example
A 0.20 kg puck moves east at 6.0 m s⁻¹ and sticks to a stationary 0.10 kg puck. Find the combined velocity after impact.
Show the worked solution
- East is the x direction. Initial x momentum is 0.20 × 6.0 = 1.2 kg m s⁻¹; initial y momentum is zero.
- The combined mass is 0.30 kg, so vₓ = 1.2/0.30 = 4.0 m s⁻¹ and vᵧ = 0.
- The result is entirely eastward.
Answer 4.0 m s⁻¹ east. Both momentum components are conserved.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 0.30 kg object moves east at 4.0 m s⁻¹ and sticks to a 0.20 kg object moving north at 3.0 m s⁻¹. Find the x and y components of their combined velocity.
Give me a hint
Conserve east-west and north-south momentum separately, then divide by the combined mass.
Compare my reasoning
- x momentum before = 0.30(4.0) = 1.2 kg m s⁻¹; y momentum before = 0.20(3.0) = 0.60 kg m s⁻¹.
- Combined mass = 0.50 kg.
- vₓ = 1.2/0.50 = 2.4 m s⁻¹; vᵧ = 0.60/0.50 = 1.2 m s⁻¹ north.
Velocity components: 2.4 m s⁻¹ east and 1.2 m s⁻¹ north.
Look for these in your work
- Components conserved independently
- Combined mass used
- Both directions and values stated
After a collision, a 0.50 kg object has velocity components 3.0 m s⁻¹ east and 4.0 m s⁻¹ north. Find its speed and direction.
Give me a hint
Treat the perpendicular components as a right triangle.
Compare my reasoning
- Speed = √(3.0² + 4.0²) = 5.0 m s⁻¹.
- Angle north of east = tan⁻¹(4.0/3.0).
- Report the direction from the east axis.
5.0 m s⁻¹ at about 53° north of east.
Look for these in your work
- Perpendicular components combined
- Magnitude calculated
- Angle reference stated
A 0.40 kg puck moves east at 5.0 m s⁻¹ and hits a stationary 0.60 kg puck. They stick and move at 2.0 m s⁻¹ east afterward. Determine the second puck's velocity components immediately before impact.
Give me a hint
Write x and y conservation separately; the first object's velocity has no y component.
Compare my reasoning
- Initial x momentum must equal final x momentum: 0.40(5.0) + 0.60vₓ = (1.00)(2.0).
- Thus 2.0 + 0.60vₓ = 2.0, so vₓ = 0.
- Initial y momentum equals final y momentum, which is zero; 0.60vᵧ = 0, so vᵧ = 0.
The second puck was stationary: vₓ = 0 and vᵧ = 0 m s⁻¹.
Look for these in your work
- Both component equations written
- x component solved
- y component solved
Common mistakes
- Applying only one scalar momentum equation to a two-dimensional collision
- Forgetting component signs
- Using separate final masses when objects stick together
What can you explain now?
How is momentum conservation applied to a two-dimensional interaction?
Compare with the explanation
Resolve and conserve momentum separately along perpendicular axes
A vector equation gives one component equation for each axis.
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Make it stick
Why do two-dimensional momentum problems need separate equations for two perpendicular components?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.