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Collision Problems in Centre of Mass Frame

Key idea: Solve 1D collision questions faster using the centre of mass frame: transform, apply elastic/inelastic rules in COM, then transform back.

  • GCE A-Level H3 Physics 2027
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Learning objectives

  • state that a frame of reference is a set of coordinates that can be used to determine positions and times of events in that frame
  • show an understanding that Newton’s laws of motion are obeyed in all inertial frames of reference
  • recall and apply the Galilean transformation equations to solve problems relating observations in different inertial frames of reference
  • show an understanding that the centre of mass frame (or zero momentum frame) is the inertial frame in which the total linear momentum of the system is zero
  • solve one-dimensional collision problems by considering velocities relative to the centre of mass of the system (i.e. in the zero-momentum frame)

Solving 1D collisions in the centre of mass (COM) frame often turns a messy algebra problem into a symmetry problem. In the COM frame, the total momentum is zero, which gives you a strong constraint before and after the collision.

1. Definitions (Must Know)

  • Total momentum (1D): Pₜₒₜ = ∑ mᵢ vᵢ (kg m s⁻¹).
  • COM velocity (1D): V_cm = (∑ mᵢ vᵢ)/(∑ mᵢ)
  • COM-frame velocity: vᵢ' = vᵢ - V_cm.
  • COM frame / zero-momentum frame: the inertial frame where Pₜₒₜ' = 0.
  • Perfectly elastic (1D): total momentum and total kinetic energy are conserved.
  • Completely inelastic: objects stick together and move with the same final velocity.
  • Kinetic energy: K = (1/2)mv² (J).

Prerequisite: 1D momentum conservation from H2/H3 foundations:

2. Key Ideas (What Earns Marks)

  • Transforming to the COM frame does not change relative velocities: (v₂-v₁) = (v₂'-v₁')
  • In the COM frame, total momentum is zero both before and after: P'_before = P'_after = 0.
  • For a perfectly elastic 1D collision, the COM-frame velocities simply reverse: v₁' = -u₁', v₂' = -u₂'
  • After solving in COM, convert back to the lab frame: vᵢ = vᵢ' + V_cm

Quick comparison (1D, COM frame viewpoint):

Collision typeWhat happens in COM frameWhat you do
Perfectly elasticSpeeds keep same magnitudes, directions reversev₁' = -u₁', v₂' = -u₂'
Completely inelastic (stick)Both end with the same velocity (often 0 in COM)Set final COM velocities equal, then transform back

3. Detailed Explanations

A. Workflow (the COM-frame method)

  1. Choose a sign convention (e.g. right is +).
  2. Compute the COM velocity: V_cm = (m₁u₁ + m₂u₂)/(m₁ + m₂)
  3. Transform each initial velocity into the COM frame using uᵢ' = uᵢ-V_cm.
  4. Apply the collision rule in the COM frame (elastic: reverse; stick: common velocity).
  5. Transform each result back using vᵢ = vᵢ' + V_cm.

B. Why elastic collisions are so simple in the COM frame (1D)

In the COM frame, the total momentum is zero both before and after the collision. For a perfectly elastic collision, kinetic energy is also conserved. In 1D, that forces the magnitudes of the velocities to stay the same, but the objects must separate (swap directions), giving: v₁' = -u₁', v₂' = -u₂'

Two objects approach in the COM frame with opposite momenta, then leave with each velocity reversed after a one-dimensional elastic collision.
For a one-dimensional elastic collision, both COM-frame velocities reverse; total momentum remains zero and kinetic energy is unchanged.

4. Common Mistakes

  • Mixing lab and COM symbols (keep primes for COM-frame quantities).
  • Sign errors when finding V_cm (use signed velocities).
  • Forgetting the final step: transforming back with vᵢ = vᵢ' + V_cm.
  • Assuming “elastic” without it being stated or justified.

5. Exam Tips

  • Do a quick check in the COM frame: m₁u₁' + m₂u₂' should equal 0 (within rounding).
  • For perfectly elastic collisions, reversal in COM is usually the fastest route.
  • Sanity check after transforming back:
    • if m₁ ≫ m₂, the heavier object’s speed should change less.

6. Worked Examples

Modelled example 1

Perfectly elastic collision (COM frame shortcut)

Core

Problem

A 2.0 kg object at u₁ = +3.0 m s⁻¹ collides elastically in one dimension with a 1.0 kg object at u₂ = -1.0 m s⁻¹. Use the COM-frame method to find v₁ and v₂.
Study the worked solution

  1. Find the COM speed

    Method

    V_cm = 5/3 = 1.67 m s⁻¹.

    Reason

    It is the total signed momentum divided by total mass.

    Working

    V_cm = ((2)(3) + (1)(-1))/3 = 5/3

  2. Transform incoming velocities

    Method

    u'₁ = 4/3 and u'₂ = -8/3 m s⁻¹.

    Reason

    Subtract V_cm from each lab velocity.

    Working

    u'₁ = 3-5/3 = 4/3, u'₂ = -1-5/3 = -8/3

  3. Use elastic reversal

    Method

    v'₁ = -4/3 and v'₂ = +8/3 m s⁻¹.

    Reason

    In a one-dimensional elastic collision, each COM-frame velocity reverses.

    Working

    v'ᵢ = -u'ᵢ

  4. Transform back

    Method

    v₁ = 0.33 and v₂ = 4.33 m s⁻¹.

    Reason

    Add the same COM velocity to both outgoing primed velocities.

    Working

    v₁ = -4/3 + 5/3 = 1/3, v₂ = 8/3 + 5/3 = 13/3

Guided practice 2

Completely inelastic collision (objects stick)

About 5 min

Problem

A 2.0 kg object at + 4.0 m s⁻¹ collides and sticks to a 3.0 kg object at -2.0 m s⁻¹. Find their final common velocity.

Try this before viewing the solution

Hints

Hint 1: use the COM interpretation
Objects that stick are both at rest after collision in their COM frame.
Hint 2: calculate that frame speed
Divide the total signed momentum by total mass.
View solution step by step

  1. Find total momentum

    Method

    Pₜₒₜ = +2.0 kg m s⁻¹.

    Reason

    The incoming momenta oppose.

    Working

    Pₜₒₜ = (2)(4) + (3)(-2) = 2

  2. Find COM velocity

    Method

    V_cm = +0.40 m s⁻¹.

    Reason

    Total mass is 5.0 kg.

    Working

    V_cm = 2.0/5.0 = 0.40 m s⁻¹

  3. Apply sticking condition

    Method

    The common lab velocity is + 0.40 m s⁻¹.

    Reason

    After sticking, both objects are stationary in the COM frame.

    Working

    v = V_cm

Common misconception 3

Elastic collision with a target initially at rest

Find and correct the mistake

Learner claim

A 1.0 kg object at + 6.0 m s⁻¹ collides elastically with a stationary 2.0 kg target. A learner reverses the lab velocities directly. Explain why that step is wrong and use the COM method to find the outgoing velocities.

Try this before viewing the solution

Velocities that reverse

View solution step by step

  1. Find COM velocity

    Method

    V_cm = 2.0 m s⁻¹.

    Reason

    The initial total momentum is 6.0 kg m s⁻¹ and mass is 3.0 kg.

    Working

    V_cm = 6.0/3.0 = 2.0

  2. Transform then reverse

    Method

    u'₁ = +4.0, u'₂ = -2.0, so v'₁ = -4.0, v'₂ = +2.0 m s⁻¹.

    Reason

    Elastic reversal is a COM-frame symmetry.

    Working

    v'ᵢ = -u'ᵢ

  3. Return to the lab

    Method

    v₁ = -2.0 and v₂ = +4.0 m s⁻¹.

    Reason

    Add V_cm = 2.0 m s⁻¹ to both primed velocities.

    Working

    v₁ = -4.0 + 2.0 = -2.0, v₂ = 2.0 + 2.0 = 4.0

Examiner practice 4

Using COM to explain the “relative speed” rule for elastic collisions

4 marks

Examination question

For Example C, where (u₁,u₂) = (6.0,0) and (v₁,v₂) = (-2.0,4.0) m s⁻¹, verify the relative-speed rule and explain it using COM-frame reversal. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Approach speed

    1 mark

    Method

    The relative speed of approach is 6.0 m s⁻¹.

    Reason

    Before collision, use u₁-u₂ for this ordering.

    Working

    u₁-u₂ = 6.0-0 = 6.0

  2. Separation speed

    1 mark

    Method

    The relative speed of separation is 6.0 m s⁻¹.

    Reason

    After collision, use v₂-v₁.

    Working

    v₂-v₁ = 4.0-(-2.0) = 6.0

  3. COM explanation

    2 marks

    Method

    The relative velocity reverses but keeps its magnitude.

    Reason

    Each object’s COM-frame velocity reverses in a one-dimensional elastic collision, and subtracting the common frame speed does not change relative velocity.

    Working

    v'₂-v'₁ = -(u'₂-u'₁)

Challenge 5

Kinetic energy loss in a completely inelastic collision

Minimal support

Independent transfer

For Example B, the 2.0 kg and 3.0 kg objects initially move at + 4.0 and -2.0 m s⁻¹ and stick at + 0.40 m s⁻¹. Calculate the kinetic energy lost.

Try this before viewing the solution

Hints

Hint 1: energy uses speed squared
Calculate each initial kinetic energy with the squared velocity, then use the combined mass for the final state.
View solution step by step

  1. Initial kinetic energy

    Method

    Kᵢ = 22 J.

    Reason

    Kinetic energies add and are non-negative despite opposite velocity directions.

    Working

    Kᵢ = (1/2)(2)(4²) + (1/2)(3)(2²) = 16 + 6 = 22 J

  2. Final kinetic energy

    Method

    K_f = 0.40 J.

    Reason

    The stuck objects move as a combined 5.0 kg mass.

    Working

    K_f = (1/2)(5.0)(0.40²) = 0.40 J

  3. Energy lost

    Method

    21.6 J of kinetic energy is lost.

    Reason

    Loss is initial minus final kinetic energy.

    Working

    Kₗₒₛₜ = 22-0.40 = 21.6 J

7. Mind Stretchers

Mind stretcher 1: When is the COM method worth it?Extension

Compare two methods for a perfectly elastic 1D collision:

  1. lab-frame: momentum + relative-speed rule, and
  2. COM-frame: transform → reverse → transform back.

When (in terms of algebra complexity) does the COM method feel simpler?

One reasonable answer

When the lab-frame equations would force you into simultaneous equations with multiple unknowns, the COM method can reduce the “collision physics” step to a one-line reversal and push the algebra into the (usually simpler) frame transforms.

Mind stretcher 2: Extreme mass ratio intuitionExtension

For a perfectly elastic 1D collision where m₁ ≫ m₂ and m₂ is initially at rest, what qualitative outcome should you expect for the speeds after the collision?

A good qualitative answer

You expect m₁ to continue forwards with almost the same speed, while the initially stationary light mass m₂ moves forwards at almost twice m₁‘s initial speed. In the limit m₁/m₂ → ∞, the exact elastic-collision results give v₁ → u₁ and v₂ → 2u₁. The light object does not rebound in this stated setup.

8. Optional/Enrichment: Partial Restitution (If Given)

If a question gives a coefficient of restitution e, you can still use the COM method, but the COM-frame “reversal” becomes “rebound with reduced relative speed”. This is beyond what you need unless e is explicitly provided.

See: Elastic & Inelastic Collisions

Next step

Return to the Frames of Reference hub. Then continue to Rotational Motion, or revisit the Centre of Mass Frame if the zero-momentum check is not yet automatic.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H3 Physics
Edition
GCE A-Level H3 Physics 2027