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Two dimensional collisions combine vector motion with the conservation of momentum. Students need this cheat sheet because collision problems often look complicated when velocities point in different directions. Worked examples become easier when each vector is split into xx and yy components.

The goal is to organize information clearly before solving equations.

Key Facts

  • Momentum is a vector, so for each object p=mv\vec{p} = m\vec{v} and its components are px=mvxp_x = mv_x and py=mvyp_y = mv_y.
  • In every isolated collision, total momentum is conserved in both directions: px,i=px,f\sum p_{x,i} = \sum p_{x,f} and py,i=py,f\sum p_{y,i} = \sum p_{y,f}.
  • Velocity components from an angle θ\theta measured from the positive xx direction are vx=vcosθv_x = v\cos\theta and vy=vsinθv_y = v\sin\theta.
  • Speed and direction can be rebuilt from components using v=vx2+vy2v = \sqrt{v_x^2 + v_y^2} and θ=tan1(vyvx)\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right).
  • For a perfectly inelastic collision where objects stick together, use m1v1i+m2v2i=(m1+m2)vfm_1\vec{v}_{1i} + m_2\vec{v}_{2i} = (m_1 + m_2)\vec{v}_f.
  • For an elastic collision, kinetic energy is also conserved, so 12m1v1i2+12m2v2i2=12m1v1f2+12m2v2f2\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2.
  • If one object is initially at rest, its initial momentum components are px,i=0p_{x,i} = 0 and py,i=0p_{y,i} = 0 for that object.
  • A correct two dimensional solution must satisfy both component momentum equations, not just the equation in the direction of the initial motion.

Vocabulary

Momentum
Momentum is the vector quantity p=mv\vec{p} = m\vec{v} that measures an object's mass in motion.
Impulse
Impulse is the change in momentum, written J=Δp\vec{J} = \Delta \vec{p}.
Elastic collision
An elastic collision is a collision in which both momentum and kinetic energy are conserved.
Inelastic collision
An inelastic collision is a collision in which momentum is conserved but kinetic energy is not conserved.
Component
A component is the part of a vector along one axis, such as vxv_x or vyv_y.
Isolated system
An isolated system has no net external force, so its total momentum remains constant.

Common Mistakes to Avoid

  • Adding speeds instead of momentum vectors is wrong because momentum depends on both mass and direction, so use px=mvxp_x = mv_x and py=mvyp_y = mv_y.
  • Using only one conservation equation is wrong because two dimensional collisions require both px,i=px,f\sum p_{x,i} = \sum p_{x,f} and py,i=py,f\sum p_{y,i} = \sum p_{y,f}.
  • Forgetting signs on components is wrong because motion left or downward should usually be negative relative to the chosen axes.
  • Assuming kinetic energy is always conserved is wrong because only elastic collisions conserve kinetic energy, while inelastic collisions lose some mechanical energy.
  • Using tan1(vyvx)\tan^{-1}\left(\frac{v_y}{v_x}\right) without checking the quadrant is wrong because the signs of vxv_x and vyv_y determine the actual direction.

Practice Questions

  1. 1 A 2.0kg2.0\,\text{kg} puck moving east at 6.0m/s6.0\,\text{m/s} collides and sticks to a 3.0kg3.0\,\text{kg} puck moving north at 4.0m/s4.0\,\text{m/s}. Find the final speed and direction of the combined pucks.
  2. 2 A 0.50kg0.50\,\text{kg} ball moving at 10m/s10\,\text{m/s} along the +x+x direction breaks into two pieces. One 0.20kg0.20\,\text{kg} piece moves at 12m/s12\,\text{m/s} at 3030^\circ above +x+x. Find the velocity components of the 0.30kg0.30\,\text{kg} piece.
  3. 3 A 1.5kg1.5\,\text{kg} cart moving at 8.0m/s8.0\,\text{m/s} east collides with a 1.0kg1.0\,\text{kg} cart initially at rest. After the collision, the first cart moves at 5.0m/s5.0\,\text{m/s} at 2525^\circ north of east. Find the second cart's final velocity components.
  4. 4 In a two dimensional collision, why can the total kinetic energy decrease while total momentum in the xx and yy directions still remains conserved?