A3-4. One-Dimensional Collisions

A collision occurs when two or more bodies interact strongly over a very short time interval, changing their momentum. In a closed system with no external forces, Total Momentum is ALWAYS conserved in any collision.

1. Types of Collisions

1. Inelastic Collisions (Most Common)

Momentum is conserved, but Kinetic Energy is NOT conserved. Some kinetic energy is transformed into thermal energy, sound, or permanent deformation of the objects.
Special Case: In a Perfectly Inelastic Collision, the two objects stick together and move with a common final velocity.

2. Elastic Collisions (Ideal / Microscopic)

Both Momentum AND Kinetic Energy are perfectly conserved.
A unique mathematical consequence of this dual-conservation in 1D is that the relative velocity of approach equals the relative velocity of separation:

$$v_1 - v_2 = -(v_1' - v_2')$$

2. Deriving the 1D Elastic Collision Formulas

Instead of memorizing the final formulas, IB HL students must understand how to derive them by solving the momentum and kinetic energy conservation equations simultaneously.

Step 1: Write down the two conservation laws.
Conservation of Momentum:
$$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \quad \text{--- (Eq. 1)}$$
Conservation of Kinetic Energy:
$$\dfrac{1}{2}m_1 v_1^2 + \dfrac{1}{2}m_2 v_2^2 = \dfrac{1}{2}m_1 v_1'^2 + \dfrac{1}{2}m_2 v_2'^2 \quad \text{--- (Eq. 2)}$$

Step 2: Group the masses.
Rearrange Eq. 1 to group $m_1$ on the left and $m_2$ on the right:
$$m_1(v_1 - v_1') = m_2(v_2' - v_2) \quad \text{--- (Eq. 3)}$$
Cancel the $\frac{1}{2}$ in Eq. 2 and do the same grouping:
$$m_1(v_1^2 - v_1'^2) = m_2(v_2'^2 - v_2^2)$$
Factor the difference of squares:
$$m_1(v_1 - v_1')(v_1 + v_1') = m_2(v_2' - v_2)(v_2' + v_2) \quad \text{--- (Eq. 4)}$$

Step 3: Divide the equations.
Divide Eq. 4 by Eq. 3. The masses and subtraction terms magically cancel out!
$$v_1 + v_1' = v_2' + v_2$$
Rearrange to show the relative velocity rule: $\mathbf{v_1 - v_2 = -(v_1' - v_2')}$

Step 4: Substitute back to find final velocities.
Isolate $v_2'$ from the relative velocity equation: $v_2' = v_1 + v_1' - v_2$.
Substitute this $v_2'$ back into Eq. 1 and solve for $v_1'$. Doing the algebra yields the master formulas:

$$v_1' = \left(\dfrac{m_1 - m_2}{m_1 + m_2}\right)v_1 + \left(\dfrac{2m_2}{m_1 + m_2}\right)v_2$$
$$v_2' = \left(\dfrac{2m_1}{m_1 + m_2}\right)v_1 + \left(\dfrac{m_2 - m_1}{m_1 + m_2}\right)v_2$$

Example 1: 1D Elastic Collision

Problem: A block of mass $m_1 = 2.0\text{ kg}$ moving right at $v_1 = 5.0\text{ m/s}$ makes a perfectly elastic collision with a stationary block of mass $m_2 = 3.0\text{ kg}$ ($v_2 = 0\text{ m/s}$). Calculate the final velocities of both blocks and verify that kinetic energy is conserved.

BEFORE COLLISION 2kg 5 m/s 3kg v = 0 AFTER COLLISION 2kg v₁' = ? 3kg v₂' = ?

Solution:

  • 1. Calculate Final Velocity of block 1 ($v_1'$):
    Since target is stationary ($v_2 = 0$), the second term of the formula drops out.
    $$v_1' = \left(\dfrac{m_1 - m_2}{m_1 + m_2}\right)v_1 = \left(\dfrac{2 - 3}{2 + 3}\right)(5.0) = \left(\dfrac{-1}{5}\right)(5.0) = \mathbf{-1.0\text{ m/s}}$$ (The negative sign indicates the lighter block bounces backwards!)
  • 2. Calculate Final Velocity of block 2 ($v_2'$):
    $$v_2' = \left(\dfrac{2m_1}{m_1 + m_2}\right)v_1 = \left(\dfrac{2(2)}{2 + 3}\right)(5.0) = \left(\dfrac{4}{5}\right)(5.0) = \mathbf{4.0\text{ m/s}}$$
  • 3. Verification (Is Kinetic Energy Conserved?):
    Initial KE: $$ E_{k,i} = \dfrac{1}{2}(2.0)(5.0)^2 + 0 = 25\text{ J} $$ Final KE: $$ E_{k,f} = \dfrac{1}{2}(2.0)(-1.0)^2 + \dfrac{1}{2}(3.0)(4.0)^2 = 1.0 + 24 = 25\text{ J} $$ $25\text{ J} = 25\text{ J}$. Perfect elastic collision confirmed!