8.4 Elastic Collisions in One Dimension

Elastic Collisions in One Dimension

An elastic collision is one where both momentum and kinetic energy are conserved. In one dimension, two objects collide along a straight line, bounce off each other, and no energy is lost to heat, sound, or deformation. This is an idealized model, but it closely describes collisions between hard objects like billiard balls or air track gliders, and it gives you the tools to predict exactly how fast each object moves after impact.

Elastic Collisions in One Dimension

What Makes a Collision Elastic

For a collision to be elastic, two conditions must hold:

1. Total momentum is conserved (true for all collisions in a closed system).
2. Total kinetic energy is conserved (this is what distinguishes elastic from inelastic collisions).

The kinetic energy and momentum of each individual object will generally change during the collision. What stays constant is the total for the system.

A perfectly elastic collision has a coefficient of restitution equal to 1, meaning the objects separate at the same relative speed they approached. Real collisions are never perfectly elastic, but some come very close (steel ball bearings, billiard balls, particles in ideal gas models).

Final Velocities After Elastic Collisions

To find the final velocities, you set up two equations and solve them simultaneously.

Conservation of momentum:

$m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$

Conservation of kinetic energy:

$\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2$

where:

• $m_1$, $m_2$ are the masses of the two objects
• $v_1$, $v_2$ are their initial velocities (before the collision)
• $v_1'$, $v_2'$ are their final velocities (after the collision)

Solving these two equations together (the algebra is shown in the problem-solving section below) gives you the general solutions:

$v_1' = \frac{m_1 - m_2}{m_1 + m_2}v_1 + \frac{2m_2}{m_1 + m_2}v_2$

$v_2' = \frac{2m_1}{m_1 + m_2}v_1 + \frac{m_2 - m_1}{m_1 + m_2}v_2$

These formulas are worth memorizing (or at least recognizing) because they save significant time on problems.

A useful shortcut: The kinetic energy equation is quadratic, which makes it harder to work with directly. Instead, you can rearrange both conservation equations and show that for elastic collisions, the relative velocity reverses:

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

This means the speed of approach equals the speed of separation. You can use this linear equation in place of the kinetic energy equation to make the algebra much easier.

Special Cases Worth Knowing

These come up frequently on exams:

• Equal masses ($m_1 = m_2$): The objects exchange velocities. If object 2 is initially at rest, object 1 stops and object 2 moves off with object 1's original velocity.
• Heavy object hits light stationary object ($m_1 \gg m_2$): The heavy object barely slows down, and the light object flies off at roughly twice the heavy object's speed.
• Light object hits heavy stationary object ($m_1 \ll m_2$): The light object bounces back at nearly its original speed, and the heavy object barely moves. Think of a tennis ball bouncing off a wall.

Conservation of Internal Kinetic Energy

The term "internal kinetic energy" refers to the kinetic energy of the objects within the system, as opposed to the kinetic energy associated with the center of mass motion. In an elastic collision, this internal kinetic energy is fully conserved.

In contrast, during an inelastic collision some of this internal kinetic energy converts into other forms (thermal energy, sound, permanent deformation). That's the key physical difference: elastic collisions preserve the total kinetic energy of the system, while inelastic collisions do not.

Center of Mass in Elastic Collisions

The velocity of the center of mass is:

$v_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$

Because total momentum is conserved and no external forces act on the system, $v_{cm}$ stays constant before, during, and after the collision. The individual objects change velocity around this center of mass, but the center of mass itself moves at a steady pace throughout.

Newton's cradle is a great demonstration of this principle: when one ball swings in and strikes the row, the ball on the opposite end swings out with the same speed, conserving both momentum and kinetic energy through a chain of elastic collisions.

Problem-Solving for Elastic Collisions

Here's a step-by-step approach:

1. List your knowns. Identify $m_1$, $m_2$, $v_1$, and $v_2$. Assign a positive direction (e.g., to the right) and use negative signs for objects moving the other way.

2. Write the momentum equation: $m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$

3. Write the relative velocity equation (easier than the KE equation): $v_1 - v_2 = -(v_1' - v_2')$

4. Solve the system. You now have two linear equations and two unknowns ($v_1'$ and $v_2'$). Solve by substitution:

   • Rearrange the relative velocity equation to express $v_1'$ in terms of $v_2'$ (or vice versa).
   • Substitute into the momentum equation and solve for the remaining unknown.
   • Back-substitute to find the other velocity.

5. Check your answer. Plug $v_1'$ and $v_2'$ back into both the momentum equation and the kinetic energy equation to verify that both quantities are conserved. If either total doesn't match, recheck your algebra and signs.

Example: A 2 kg cart moving at 3 m/s to the right collides elastically with a 1 kg cart at rest.

• Momentum: $(2)(3) + (1)(0) = 2v_1' + 1v_2'$ → $6 = 2v_1' + v_2'$
• Relative velocity: $3 - 0 = -(v_1' - v_2')$ → $v_2' - v_1' = 3$
• From the second equation: $v_2' = v_1' + 3$
• Substitute: $6 = 2v_1' + (v_1' + 3)$ → $3 = 3v_1'$ → $v_1' = 1$ m/s
• Then $v_2' = 4$ m/s

Check KE: initial = $\frac{1}{2}(2)(9) = 9$ J; final = $\frac{1}{2}(2)(1) + \frac{1}{2}(1)(16) = 1 + 8 = 9$ J. Kinetic energy is conserved.