Chapter 4: Problem 48
A particle of mass \(m_{1}\) and speed \(v_{1}\) collides with a second particle of mass \(m_{2}\) at rest. If the collision is perfectly inelastic (the two particles lock together and move off as one) what fraction of the kinetic energy is lost in the collision? Comment on your answer for the cases that \(m_{1} \ll m_{2}\) and that \(m_{2} \ll m_{1}\).
Short Answer
Step by step solution
Understand the concept of a perfectly inelastic collision
Write the conservation of momentum equation
Solve for final velocity after collision
Express initial kinetic energy
Express final kinetic energy after collision
Calculate the fraction of kinetic energy lost
Comment on special cases
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Kinetic Energy Loss
In the context of the exercise, the initial kinetic energy is held by the moving particle, and it's given by the expression \[ KE_\text{initial} = \frac{1}{2}m_{1}v_{1}^2 \].
Once the collision occurs and the objects combine, the final kinetic energy changes since the total mass and velocity have been altered. After solving for the final kinetic energy, you use the difference to determine the energy lost portion: \[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}}. \]
This difference represents the kinetic energy dissipated in other forms during the collision. Understanding this energy exchange helps in a variety of physics problems, illustrating the interplay between energy and motion.
Momentum Conservation
The conservation principle states that the total momentum before an interaction equals the total momentum afterward, assuming no external forces act on the system. For the given exercise, the initial momentum was entirely due to the particle in motion, described as: \[ m_{1}v_{1} + m_{2} \times 0. \]
After the collision, both particles move as a single entity, and their mutual final momentum becomes:\[ (m_{1} + m_{2})v. \] By setting these two expressions equal, we calculate the final velocity using: \[ v = \frac{m_{1}v_{1}}{m_{1} + m_{2}}. \]
This setup demonstrates how the velocity of the combined mass can be derived from the original momentum.
- **Key Point**: Momentum conservation provides insight into post-collision movement, despite the complex transformations of energy that occur in perfectly inelastic collisions.
Perfectly Inelastic Collision
The resulting single body, with mass \((m_{1} + m_{2})\), travels with a velocity derived from prior momentum calculations: \[ v = \frac{m_{1}v_{1}}{m_{1} + m_{2}}. \]
With their collective movement, the energy profile of the collision shifts markedly. If the first mass is much smaller than the second (\(m_{1} \ll m_{2}\)), almost all kinetic energy is drained, with most converting to internal energies or other forms. Conversely, if the second mass is minor compared to the first (\(m_{2} \ll m_{1}\)), minimal kinetic energy is lost, and post-collision movement closely reflects the initial conditions.
- **Perfectly Inelastic Collisions**: Often serve as models in physics problems illustrating how energy dissipation and momentum conservation work together uniquely.