Question

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

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

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.

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:$$m_1{v}_1+m_{2}imes0=(m_1+m_2)v$$Where $v$ is the final velocity of the combined mass after the collision.

Solve for final velocity after collision

Using the momentum conservation equation from Step 2, solve for the final velocity $v$:$$v=\frac{m_1{v}_1}{m_1+m_2}$$

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:$$K{E}_{extinitial}=\frac{1}{2}{m}_1{v}_1^2$$

Express final kinetic energy after collision

Calculate the final kinetic energy (KE) of the combined mass moving with velocity $v$:$$K{E}_{extfinal}=\frac{1}{2}(m_1+m_2)v^2$$Substitute the expression for $v$ from Step 3:$$K{E}_{extfinal}=\frac{1}{2}(m_1+m_2){\left(\frac{m_1{v}_1}{m_1+m_2}\right)}^2=\frac{1}{2}\frac{m_1^2{v}_1^2}{m_1+m_2}$$

Calculate the fraction of kinetic energy lost

The kinetic energy lost $\mathrm{\Delta }KE$ is the difference between the initial and final kinetic energies:$$\mathrm{\Delta }KE=K{E}_{\text{initial}}-K{E}_{\text{final}}=\frac{1}{2}{m}_1{v}_1^2-\frac{1}{2}\frac{m_1^2{v}_1^2}{m_1+m_2}$$The fraction of kinetic energy lost is:$$\text{Fraction lost}=\frac{\mathrm{\Delta }KE}{K{E}_{\text{initial}}}=1-\frac{m_1}{m_1+m_2}$$

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.

Key concepts

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 $$K{E}_{\text{initial}}=\frac{1}{2}{m}_1{v}_1^2$$.

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: $$\mathrm{\Delta }KE=K{E}_{\text{initial}}-K{E}_{\text{final}}.$$

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.