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

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.

• Are you not overthinking this? "perfectly inelastic" has to mean they stick together after the collision, so they have the same velocity. Because in the rest frame (zero net momentum) if they move relative to each other they have higher energy than when they are both at rest. – Floris Aug 28 '17 at 0:24
• depending on the way you define "perfectly inelastic" in the first place this is either tautological (because the definition is equivalent to the desired outcome) or relatively difficult. – dmckee Aug 28 '17 at 0:24
• Perfectly inelastic collisions occur even when the masses differ, though that shouldn't change the a=1 minimum. Not allowing u1 to be 0 also limits you in the frames you can do solve the problem in. – Johnathan Gross Aug 28 '17 at 0:32
• $v_1 = v_2$ is the definition of what a "perfectly inelastic collision" means. You can't prove a definition is true. That is as nonsensical, as trying to prove that the word "cat" actually means the same as "the animal called a cat in English". – alephzero Aug 28 '17 at 0:56
• @alephzero It is possible to adopt a different definition with the same result. Say 'minimizing the bulk kinetic energy after the collision'. Of course this has the same result as can be shown immediately in the CoM frame or with a moderate amount of calculus in the general frame. – dmckee Aug 28 '17 at 1:50

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.
• @rubyduby your question does not make much sense. There are no conditions on kinetic energy loss and if artificially imposed them, in a general they would not lead to equal velocities. My statements apply to the situation you are interested in. Also your proof should really be a 3D graph since $a$ and $b$ can very independently with some restriction arising from momentum conservation. So you really don't even have the proof you think you have. – Amara Aug 29 '17 at 4:20