Physics • April 2026

Solving 1D Elastic
Collisions Between Two Gliders

A
Written By Archive Editorial
Reading Time 5 Min Read

Understanding Elastic Collisions

In physics, an elastic collision is a collision in which both total momentum and total kinetic energy are conserved. When two objects collide on a frictionless surface, we can determine their final velocities by setting up a system of equations based on these two physical laws.

Problem Setup

We have two gliders on an air track:

  • Glider 1: Mass $m_1 = 0.15, \text{kg}$, Initial velocity $u_1 = +0.80, \text{m/s}$ (moving right).
  • Glider 2: Mass $m_2 = 0.30, \text{kg}$, Initial velocity $u_2 = -2.2, \text{m/s}$ (moving left).

We need to find final velocities $v_1$ and $v_2$.

Step-by-Step Solution

1. Conservation of Momentum

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ $(0.15)(0.8) + (0.30)(-2.2) = 0.15v_1 + 0.30v_2$ $0.12 - 0.66 = 0.15v_1 + 0.30v_2$ $-0.54 = 0.15v_1 + 0.30v_2$ --- (Equation 1)

2. Elastic Collision Property

For an elastic collision, the relative velocity of approach equals the relative velocity of separation: $u_1 - u_2 = v_2 - v_1$ $0.80 - (-2.2) = v_2 - v_1$ $3.0 = v_2 - v_1 \Rightarrow v_2 = v_1 + 3.0$ --- (Equation 2)

3. Solving the Equations

Substitute Equation 2 into Equation 1: $-0.54 = 0.15v_1 + 0.30(v_1 + 3.0)$ $-0.54 = 0.15v_1 + 0.30v_1 + 0.90$ $-1.44 = 0.45v_1$ $v_1 = -3.2, \text{m/s}$

Now, solve for $v_2$: $v_2 = -3.2 + 3.0 = -0.2, \text{m/s}$

Final Result

  • Glider 1 (0.15 kg): Final velocity is $3.2, \text{m/s}$ to the left.
  • Glider 2 (0.30 kg): Final velocity is $0.2, \text{m/s}$ to the left.
Editorial Note

Automated explainer generated from student question.

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