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Collisions in two dimensions

AshPhys · Revision cheat sheet
AS Level9702 AS Ch.6Chapter 6: MomentumLesson 6.5Sections 6.5

Goal: Solve a glancing collision by conserving momentum in two perpendicular directions.

velocity

01The big picture

Momentum is a vector, so in a glancing collision it is conserved in every direction at once. Choose two perpendicular directions (xx along the original motion, yy across it) and conserve momentum in each separately.

Resolve each momentum into components with cos⁡\cos and sin⁡\sin. If nothing was moving in the yy direction before, the yy-components after must cancel.

Glancing collision [paraphrase]
A collision that is not head-on: the objects move off at angles.
Fig. 1 · A glancing collision
A glancing collisionPuck A moves right at 5.0 m/s to hit puck B at rest. After the collision A moves at 3.0 m/s at 53 degrees above the original direction and B moves at 4.0 m/s at 37 degrees below it.uA = 5.0BvA = 3.0vB = 4.053°37°
A and B move off at angles; their yy-momenta cancel and their xx-momenta add up to A’s momentum before.

02Key equations

Along x
∑px before=∑px after\textstyle\sum p_x \text{ before} = \sum p_x \text{ after}
Along y
∑py before=∑py after\textstyle\sum p_y \text{ before} = \sum p_y \text{ after}
Components
px=pcos⁡θ,py=psin⁡θp_x = p\cos\theta,\quad p_y = p\sin\theta
Symbols and units
  • pxp_xmomentum along xkg m s−1\mathrm{kg\,m\,s^{-1}}
  • θ\thetaangle to the x direction∘^\circ

03Sample problems with solutions

Sample 1 · A glancing collision
A 2.0 kg puck A at 5.0 m s⁻¹ hits an identical puck B at rest. A leaves at 3.0 m s⁻¹ at 53° to its first direction (Fig. 1). Find B’s velocity.

  1. Along xx (masses equal, so they cancel):

    5.0=3.0cos⁡53∘+vx  ⇒  vx=3.2 m s−15.0 = 3.0\cos 53^\circ + v_x \;\Rightarrow\; v_x = 3.2\ \mathrm{m\,s^{-1}}

  2. Along yy (zero before):

    0=3.0sin⁡53∘+vy  ⇒  vy=−2.4 m s−10 = 3.0\sin 53^\circ + v_y \;\Rightarrow\; v_y = -2.4\ \mathrm{m\,s^{-1}}

  3. Combine:

    v=3.22+2.42=4.0 m s−1,θ=tan⁡−12.43.2=37∘v = \sqrt{3.2^{2} + 2.4^{2}} = 4.0\ \mathrm{m\,s^{-1}}, \quad \theta = \tan^{-1}\frac{2.4}{3.2} = 37^\circ
Answer: 4.0 m s⁻¹ at 37° on the other side of the original direction

04Practice problems

  1. P1
    In a two-dimensional collision, what is conserved separately?
    Final answerMomentum in each of two perpendicular directions
    Answer: Momentum in each of two perpendicular directions
  2. P2
    A 1.0 kg ball moving east at 4.0 m s⁻¹ sticks to a 1.0 kg ball moving north at 4.0 m s⁻¹. Find their velocity.
    Final answer2.8 m s⁻¹ north-east (45°)
    Answer: 2.8 m s⁻¹ north-east (45°)
  3. P3
    A 3.0 kg object at rest splits into three 1.0 kg pieces; two go east and north at 6.0 m s⁻¹. Find the third.
    Final answer8.5 m s⁻¹ south-west
    Answer: 8.5 m s⁻¹ south-west
  4. P4
    Is the collision in the sample elastic?
    Final answerYes: 25 J of kinetic energy before and after
    Answer: Yes: 25 J of kinetic energy before and after

05Other useful information

Common mistakes

  • ✗ Wrong: Add the speeds after: 3.0 + 4.0 = 5.0?

    ✓ Right: Add momentum components, never speeds.

Exam tips

  • Draw the before and after momenta tip to tail: they close a triangle.

What the syllabus asks

  • 9702 topic 3: conservation of momentum in one and two dimensions. Coursebook 6.5.

Explore: Practice questions for this lesson