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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

Expert verified
The fraction of kinetic energy lost is \(1 - \frac{m_{1}}{m_{1} + m_{2}}\). For \(m_{1} \ll m_{2}\), most kinetic energy is lost; for \(m_{2} \ll m_{1}\), little is lost.

Step by step solution

01

Understand the concept of a perfectly inelastic collision

A perfectly inelastic collision is one in which two objects collide and move together as a single object after the collision. In such a collision, kinetic energy is not conserved, but momentum is conserved.
02

Write the conservation of momentum equation

The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. For our particles, this can be written as:\[ m_{1}v_{1} + m_{2} imes 0 = (m_{1} + m_{2})v \]Where \(v\) is the final velocity of the combined mass after the collision.
03

Solve for final velocity after collision

Using the momentum conservation equation from Step 2, solve for the final velocity \(v\):\[ v = \frac{m_{1}v_{1}}{m_{1} + m_{2}} \]
04

Express initial kinetic energy

Calculate the initial kinetic energy (KE) of the system, which is only due to the first particle since the second one is initially at rest:\[ KE_{ ext{initial}} = \frac{1}{2}m_{1}v_{1}^2 \]
05

Express final kinetic energy after collision

Calculate the final kinetic energy (KE) of the combined mass moving with velocity \(v\):\[ KE_{ ext{final}} = \frac{1}{2}(m_{1} + m_{2})v^2 \]Substitute the expression for \(v\) from Step 3:\[ KE_{ ext{final}} = \frac{1}{2}(m_{1} + m_{2})\left(\frac{m_{1}v_{1}}{m_{1} + m_{2}}\right)^2 = \frac{1}{2}\frac{m_{1}^2v_{1}^2}{m_{1} + m_{2}} \]
06

Calculate the fraction of kinetic energy lost

The kinetic energy lost \( \Delta KE \) is the difference between the initial and final kinetic energies:\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = \frac{1}{2}m_{1}v_{1}^2 - \frac{1}{2}\frac{m_{1}^2v_{1}^2}{m_{1} + m_{2}} \]The fraction of kinetic energy lost is:\[ \text{Fraction lost} = \frac{\Delta KE}{KE_{\text{initial}}} = 1 - \frac{m_{1}}{m_{1} + m_{2}} \]
07

Comment on special cases

For \(m_{1} \ll m_{2}\):- \(\text{Fraction lost} \approx 1\). Almost all kinetic energy is lost as the smaller mass is absorbed.For \(m_{2} \ll m_{1}\):- \(\text{Fraction lost} \approx 0\). The kinetic energy remains mostly intact as the larger mass continues moving with nearly the same velocity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy Loss
When two objects collide, especially in a perfectly inelastic collision where they stick together, a portion of the initial kinetic energy does not remain in the system. This energy could be transformed into other forms, such as sound, heat, or internal energy. The energy transformation arises because kinetic energy is not conserved during such a collision, though momentum is.

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
In any type of collision, including perfectly inelastic ones, the total momentum of the system is conserved.

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
In a perfectly inelastic collision, the key characteristic is the complete merging and sticking together of the colliding masses after impact. This concept underscores why kinetic energy does not remain completely conserved, despite momentum being intact.

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.

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Most popular questions from this chapter

(a) Consider an electron (charge \(-e\) and mass \(m\) ) in a circular orbit of radius \(r\) around a fixed proton (charge \(+e\) ). Remembering that the inward Coulomb force \(k e^{2} / r^{2}\) is what gives the electron its centripetal acceleration, prove that the electron's KE is equal to \(-\frac{1}{2}\) times its \(\mathrm{PE}\); that is, \(T=-\frac{1}{2} U\) and hence \(E=\frac{1}{2} U\). (This result is a consequence of the so-called virial theorem. See Problem 4.41.) Now consider the following inelastic collision of an electron with a hydrogen atom: Electron number 1 is in a circular orbit of radius \(r\) around a fixed proton. (This is the hydrogen atom.) Electron 2 approaches from afar with kinetic energy \(T_{2} .\) When the second electron hits the atom, the first electron is knocked free, and the second is captured in a circular orbit of radius \(r^{\prime} .\) (b) Write down an expression for the total energy of the three-particle system in general. (Your answer should contain five terms, three PEs but only two KEs, since the proton is considered fixed.) (c) Identify the values of all five terms and the total energy \(E\) long before the collision occurs, and again long after it is all over. What is the KE of the outgoing electron 1 once it is far away? Give your answers in terms of the variables \(T_{2}, r,\) and \(r^{\prime}\).

Consider the Atwood machine of Figure \(4.15,\) but suppose that the pulley has radius \(R\) and moment of inertia \(I\). (a) Write down the total energy of the two masses and the pulley in terms of the coordinate \(x\) and \(\dot{x}\). (Remember that the kinetic energy of a spinning wheel is \(\frac{1}{2} I \omega^{2}\).) (b) Show (what is true for any conservative one-dimensional system) that you can obtain the equation of motion for the coordinate \(x\) by differentiating the equation \(E=\) const. Check that the equation of motion is the same as you would obtain by applying Newton's second law separately to the two masses and the pulley, and then eliminating the two unknown tensions from the three resulting equations.

Prove that if \(f(\mathbf{r})\) and \(g(\mathbf{r})\) are any two scalar functions of \(\mathbf{r},\) then \(\nabla(f g)=f \nabla g+g \nabla f\)

Both the Coulomb and gravitational forces lead to potential energies of the form \(U=\gamma / | \mathbf{r}_{1}-\) \(\mathbf{r}_{2} |,\) where \(\gamma\) denotes \(k q_{1} q_{2}\) in the case of the Coulomb force and \(-G m_{1} m_{2}\) for gravity, and \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) are the positions of the two particles. Show in detail that \(-\nabla_{1} U\) is the force on particle 1 and \(-\nabla_{2} U\) that on particle 2.

Consider a head-on elastic collision between two particles. (since the collision is head-on, the motion is confined to a single straight line and is therefore one-dimensional.) Prove that the relative velocity after the collision is equal and opposite to that before. That is, \(v_{1}-v_{2}=-\left(v_{1}^{\prime}-v_{2}^{\prime}\right),\) where \(v_{1}\) and \(v_{2}\) are the initial velocities and \(v_{1}^{\prime}\) and \(v_{2}^{\prime}\) the corresponding final velocities.

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