Proof that an inelastic collision with max. kinetic energy loss of 2 objects have equal final velocity, in 1D? [closed]

Question

Consider two objects with velocities $u_1$ and $u_2$ which collide together. Lets denote the final velocities by $v_1$ and $v_2$.

I know that if the loss of kinetic energy is maximal iff $v_1 = v_2$. I know how to prove it using simplified assumptions (see below), but how can I prove it in general using basic algebra and calculus?

Is there a way to use the coefficient of restitution in the proof?

The method I've attempted assumes that the masses are equal ($m_1 = m_2 = m$) and does not consider the possibility of $u_1/v_1 = 0$:

By Conservation of Momentum:

$$ m u_1 + m u_2 = m v_1 + m v_2 $$ i.e. $$ u_1 + u_2 = v_1 + v_2 $$

Let $v_2 = a v_1$, and $u_2 = b u_1$ where $a$ and $b$ are constants.

Therefore, $$ (b+1) u_1 = (a+1) v_1 \tag{i}\label{i} $$

$$ \frac{b+1}{a+1} = \frac{v_1}{u_1} $$

Dividing the initial kinetic Energy $\mathrm{KE}_i$ by the final kinetic Energy $\mathrm{KE}_f$, can help to find the kinetic efficiency $R$ of the collision. So,

$$ \begin{align} R &= \frac{\mathrm{KE}_f}{\mathrm{KE}_i}\\ &= \frac{\frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2}{\frac{1}{2} m u_1^2 +\frac{1}{2} m u_2^2} \\ &= \frac{v_1^2 + v_2^2}{u_1^2 + u_2^2}\\ &= \frac{v_1^2 + (a v_1)^2}{u_1^2 + (b u_1)^2} \\ &=\frac{(a^2+1)v_1^2}{(b^2+1)u_1^2} \\ \end{align} $$

Or with $\eqref{i}$:

$$ R = \frac{(a^2+1)(b+1)^2}{(b^2+1)(a+1)^2} = \frac{(b+1)^2}{(b^2+1)} \cdot \frac{a^2+1}{(a+1)^2} $$

Modelling this collision in a 2D graph, assume $\frac{(b+1)^2}{b^2+1}$ to be a real constant $k$. Also Let $R$ be a dependent variable $y$ and "$a$" be the independent variable $x$. The equation finally becomes,

$$ y = k \frac{x^2+1}{(x+1)^2} $$ To find the minimum KE ratio, differentiate It can be shown that for $y' = 0$ implies $x = 1$

This implies that only when $v_1 = v_2$ can lowest final KE can be achieved.

Answer 1

First, show that kinetic energy loss is independent of reference frame, using momentum conservation. Second, go to the center-of-mass frame, in which the situation you described corresponds to zero final velocity for both objects.

This is technically not an answer, but hope it helps.

Answer 2

We introduce inelastic collisions to students by using the case of two bodies colliding and sticking together in which case they have the same velocity by definition. But remember, the definition of an inelastic collision is that kinetic energy is not conserved $ \textit{that is all}$ . So if there is some other mechanism that transforms some kinetic energy out of the system then in a generic situation the two bodies will not have the same final velocity.

In the examples we do in class, one way of not conserving the kinetic energy is having the two bodies stick together, but that is just one way. Supposing one puts a shock absorber around the colliding masses. Then the two objects will not stick together and in a generic situation they will have different velocities and kinetic energy will not be conserved i.e it will be an inelastic collision. One could also think of scattering of atoms or molecules, in that case if a reaction occurs in the collision and the products after the collision are different, kinetic energy will not be conserved and we will have an inelastic collision.