Theory — Momentum in Two Dimensions
Momentum is a Vector
Momentum p = mv is a vector, so in two dimensions it has separate x- and y-components. In an isolated collision (no external forces), momentum is conserved independently in each direction: the total x-momentum before equals the total x-momentum after, and likewise for y.
Σp_y,before = Σp_y,after
pₓ = m·vₓ · p_y = m·v_y
Elastic vs. Inelastic
Momentum is always conserved in an isolated collision. Kinetic energy, however, is conserved only in an elastic collision. In an inelastic collision some kinetic energy is converted to heat or deformation; in a perfectly inelastic collision the objects stick together.
Elastic: KE conserved · Inelastic: KE decreases
Impulse
The impulse delivered to an object equals its change in momentum. For one ball, the impulse is a vector with x- and y-components.
Jₓ = m(v_fx − v_ix) · J_y = m(v_fy − v_iy)
x and y separate
Resolve every velocity into components; conservation holds in each direction independently.
Off-centre hit
An off-axis collision sends the balls off at angles, giving non-zero y-momentum for each — but the totals still match.
Newton's third law
The impulse ball 1 gives ball 2 is equal and opposite to the impulse ball 2 gives ball 1.
| Quantity | Relationship | Notes |
|---|---|---|
| Momentum components | pₓ = mvₓ, p_y = mv_y | vector, by direction |
| Conservation | Σpₓ, Σp_y each constant | independent in x and y |
| Kinetic energy | KE = ½m(vₓ²+v_y²) | conserved only if elastic |
| Impulse | J = Δp = m(v_f − v_i) | per ball, a vector |
Instructions — Running the Virtual Experiment
The 2D Collision tab lets you set the masses, velocities, and the vertical offset for an off-centre hit, then fire; the Components & Data tab records the x- and y-momentum (and KE) before and after so you can verify each is conserved. Record every reading in your lab report with screenshots.
Simulation — Two Balls on a Table
Setup
| Ball | m | vₓ before | v_y before | vₓ after | v_y after | pₓ before | p_y before | pₓ after | p_y after |
|---|---|---|---|---|---|---|---|---|---|
| Fire a collision on the 2D Collision tab to populate the data. | |||||||||
Team Questions
Example Lab Report
Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission, and include labelled screenshots of the before and after states.
2D Collisions
Physics | Section: [Your Section] | Date: [Date]
Lab Members: [Names of all members present]
Purpose
To verify that linear momentum is conserved independently in the x- and y-directions during a two-dimensional collision, to compare elastic and inelastic collisions through their kinetic energy, and to calculate the impulse delivered to one ball from its change in momentum.
Theory
Momentum is a vector, p = mv, so in 2D it resolves into pₓ = mvₓ and p_y = mv_y. In an isolated collision each component is conserved: Σpₓ and Σp_y are unchanged. Kinetic energy KE = ½m(vₓ²+v_y²) is conserved only for an elastic collision. The impulse on a ball equals its change in momentum, J = Δp = m(v_f − v_i).
KE = ½m(vₓ²+v_y²) · J = m(v_f − v_i)
Sample — Off-Centre Elastic Collision
Pink ball (1.0 kg) moving at vₓ = 2.0 m/s strikes a stationary 1.0 kg blue ball off-centre (offset 0.5). After impact: pink v = (0.45, −0.84) m/s, blue v = (1.55, +0.84) m/s.
| Component | Before | After |
|---|---|---|
| Σpₓ (kg·m/s) | 2.00 | (1.0)(0.45)+(1.0)(1.55) = 2.00 |
| Σp_y (kg·m/s) | 0.00 | (1.0)(−0.84)+(1.0)(+0.84) = 0.00 |
| Total KE (J) | 2.00 | 0.45 + 1.55 ≈ 2.00 |
Impulse on pink ball: Jₓ = m(v_fx − v_ix) = 1.0(0.45 − 2.0) = −1.55 kg·m/s; J_y = 1.0(−0.84 − 0) = −0.84 kg·m/s.
Discussion
The total x-momentum was 2.00 kg·m/s both before and after, and the total y-momentum was zero both before and after, confirming that momentum is conserved separately in each direction — the y-components of the two balls were equal and opposite (−0.84 and +0.84), summing to zero just as before the collision. For the elastic case the total kinetic energy was unchanged at 2.00 J, whereas repeating with an inelastic setting reduced the kinetic energy while momentum stayed conserved. The impulse on the pink ball, (−1.55, −0.84) kg·m/s, was equal and opposite to the impulse on the blue ball, as required by Newton's third law.
Small discrepancies in a real experiment would come from reading velocity components off the display and from the exact moment the motion is paused. Using component sums rather than a single speed makes the conservation check unambiguous.
Conclusion
Momentum was conserved independently in x and y for the 2D collision, kinetic energy was conserved only in the elastic case, and the impulse on each ball equalled its change in momentum, equal and opposite between the two balls.
Practice Questions
Show all work and include units. Resolve momentum into components: pₓ = mvₓ, p_y = mv_y.