4.4 Elastic and Inelastic Collisions

Collisions in physics can be elastic or inelastic, affecting how energy is conserved or transformed. Elastic collisions maintain total kinetic energy, while inelastic collisions result in energy loss through heat, sound, or deformation.

Understanding these collision types is crucial for analyzing real-world scenarios. Elastic collisions are ideal, like billiard balls, while inelastic collisions, such as car crashes, involve energy transformations and help explain everyday phenomena.

Elastic vs Inelastic Interactions

Elastic Collision Kinetic Energy

In elastic collisions, the total kinetic energy of the system remains unchanged before and after the collision. This is a fundamental characteristic that distinguishes elastic collisions from other types. 🎯

• The kinetic energy is redistributed among the colliding objects, but the sum stays constant
• Real-world examples approach elastic behavior: billiard balls, marbles, and atomic particles
• Perfectly elastic collisions are theoretical idealizations with no energy loss

To calculate the total kinetic energy in an elastic collision:

$KE_{initial} = KE_{final}$ $\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$

Individual Object Kinetic Energy

While the system's total kinetic energy remains constant in elastic collisions, the kinetic energy of each individual object typically changes during the interaction.

• An object may gain kinetic energy at the expense of another object
• The energy exchange depends on the mass ratio of the colliding objects
• In a head-on collision between objects of equal mass, they can completely exchange velocities

To find the final velocities in an elastic collision, you must solve these two equations simultaneously:

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$ (conservation of momentum) $\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$ (conservation of kinetic energy)

Inelastic Collision Energy Decrease

Inelastic collisions are characterized by a decrease in the system's total kinetic energy after the collision occurs. 📉

• Some initial kinetic energy converts to thermal energy, sound, or deformation
• The amount of energy lost varies depending on the materials and collision speed
• Even though kinetic energy decreases, the total energy of the system (including all forms) remains conserved

The decrease in kinetic energy can be quantified as:

$KE_{lost} = KE_{initial} - KE_{final}$

Energy Transformation in Collisions

During collisions, energy can transform from one form to another depending on the collision type and object properties. 🔄

• Elastic collisions: kinetic energy is conserved with minimal transformation
• Inelastic collisions: kinetic energy transforms into various forms including:
  • Thermal energy (heat) from friction and deformation
  • Sound energy from vibrations and pressure waves
  • Potential energy stored in deformed materials

The principle of energy conservation still applies to all collisions, with the total energy (all forms combined) remaining constant:

$E_{total,initial} = E_{total,final}$

Perfectly Inelastic Collisions

A perfectly inelastic collision represents the extreme case where objects stick together after collision, moving as a single unit. 🤝

• Maximum possible kinetic energy is lost during the collision
• The objects share a common final velocity
• Examples include a dart sticking in a board or cars that crumple and lock together

The final velocity in a perfectly inelastic collision can be found using conservation of momentum:

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f$

The kinetic energy lost is calculated by:

$KE_{lost} = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 - \frac{1}{2}(m_1 + m_2)v_f^2$

Practice Problem 1: Elastic Collision

Solution

For an elastic collision, we need to apply both conservation of momentum and conservation of kinetic energy:

Conservation of momentum: $m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

$(2.0 \text{ kg})(5.0 \text{ m/s}) + (3.0 \text{ kg})(0 \text{ m/s}) = (2.0 \text{ kg})v_{1f} + (3.0 \text{ kg})v_{2f}$

$10.0 \text{ kg}\cdot\text{m/s} = 2.0v_{1f} + 3.0v_{2f}$

Conservation of kinetic energy: $\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$

$\frac{1}{2}(2.0)(5.0)^2 + \frac{1}{2}(3.0)(0)^2 = \frac{1}{2}(2.0)v_{1f}^2 + \frac{1}{2}(3.0)v_{2f}^2$

$25.0 \text{ J} = 1.0v_{1f}^2 + 1.5v_{2f}^2$

Solving these equations: From the first equation: $v_{2f} = \frac{10.0 - 2.0v_{1f}}{3.0}$

Substituting into the second equation: $25.0 = 1.0v_{1f}^2 + 1.5(\frac{10.0 - 2.0v_{1f}}{3.0})^2$

Solving this quadratic equation gives $v_{1f} = 0.0 \text{ m/s}$ or $v_{1f} = -3.0 \text{ m/s}$

Since the first ball was initially moving in the positive direction, $v_{1f} = 0.0 \text{ m/s}$ is not physically meaningful for this collision. Therefore, $v_{1f} = -3.0 \text{ m/s}$.

Substituting back: $v_{2f} = \frac{10.0 - 2.0(-3.0)}{3.0} = \frac{10.0 + 6.0}{3.0} = \frac{16.0}{3.0} = 5.33 \text{ m/s}$

Therefore, the first ball reverses direction with a speed of 3.0 m/s, and the second ball moves forward at 5.33 m/s.

Practice Problem 2: Perfectly Inelastic Collision

Solution

(a) For a perfectly inelastic collision, we use conservation of momentum to find the common final velocity:

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f$

$(1500 \text{ kg})(20 \text{ m/s}) + (2500 \text{ kg})(0 \text{ m/s}) = (1500 \text{ kg} + 2500 \text{ kg})v_f$

$30,000 \text{ kg}\cdot\text{m/s} = 4000 \text{ kg} \times v_f$

$v_f = \frac{30,000 \text{ kg}\cdot\text{m/s}}{4000 \text{ kg}} = 7.5 \text{ m/s}$

(b) To find the kinetic energy lost:

$KE_{lost} = KE_{initial} - KE_{final}$

$KE_{initial} = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}(1500)(20)^2 + \frac{1}{2}(2500)(0)^2 = 300,000 \text{ J}$

$KE_{final} = \frac{1}{2}(m_1 + m_2)v_f^2 = \frac{1}{2}(4000)(7.5)^2 = 112,500 \text{ J}$

$KE_{lost} = 300,000 \text{ J} - 112,500 \text{ J} = 187,500 \text{ J}$

Therefore, the common final velocity is 7.5 m/s, and 187,500 J of kinetic energy is lost during the collision.