Modelling an Inelastic Collision

I've attempted to model the inelastic collision between a moving mass and a large stationary object, e.g. dropping an ball on the floor, for analysing the approximate kinematics and heat dissipation throughout the collision. I modelled this using a mechanical vibrating system with one degree of freedom that consists of a mass $m$, attached to a single spring, of stiffness $k$, and a single dashpot damper, of damping rate $\lambda$, "connected in parallel", as shown below:

$x$ is the displacement of the mass from the point the collision just begins. The relationships between the tensile force, $T$, and the displacement, $x$, for the spring and dashpot is, respectively, assumed as the following:

$$T_{spring} = kx$$ $$T_{dashpot} = \lambda \dot x$$

BASIS FOR MODELLING AND ASSUMPTIONS

I selected this system as the spring represents the elastic nature of the collision (conversion between KE and elastic PE), and the damper represents the inelastic nature of the system (causes dissipation of mechanical energy). Therefore, this model will have parameters $k$ and $\lambda$ that will need to be determined by experimental means for a particular collision.

The model is assumed to be valid only for positive values off $x$, as negative $x$ implies the colliding masses are not in contact, so the effects of the spring and dashpot in the model will disappear. Also, it is assumed that the large mass is stationary and is massive enough to not move due to the collision. Just before the collision occurs, at time $t=0$, the mass $m$ will be travelling with initial velocity $u$ in the direction of $x$. It is assumed that the mass system has one degree of freedom, such that it cannot travel in any direction not parallel to $x$, and it cannot rotate, such that the extensions of the spring and dashpot must be the same. The springs and dashpots have negligible mass.

RESULTS

The differential equation of the system is obtained by Newton's 2nd Law:

$$\frac{m}{k}\ddot x + \frac{\lambda}{k}\dot x + x = 0$$

By using Laplace transforms:

$\ddot x(t) \to s^2 \bar x(s) - u$
$\dot x(t) \to s \bar x(s)$
$x(t) \to \bar x(s)$

I obtain the following:

$$x(t) = \begin{cases} \frac{u}{\omega} e^{-zt} \sin(\omega t), & \text{for z \ne \sqrt{ \frac{k}{m} }} \\[2ex] ut e^{-zt}, & \text{for z = \sqrt{\frac{k}{m}}} \end{cases}$$

where:
$z = \frac{\lambda}{2m}$ is a "dissipation factor"
$\omega = \sqrt{\frac{k}{m} - z^2}$ is the frequency of vibration

Note that if $\omega$ is not real, then the system is overdamped, and the following substitutions should be made:
$\omega = i \omega_{real}$
$\sin{\omega t} = \sin(i \omega_{real} t) = i \sinh(\omega_{real} t)$

If the system has $z=\sqrt{\frac{k}{m}}$, the system is critically damped, and mechanical energy will be dissipated at the maximum rate.

If the system is either critically damped, or overdamped, then all of the kinetic energy will be dissipated throughout the collision, so the mass will remain attached to the large mass after the collision.

For the case of a mass that is underdamped, the velocity is given (by differentiation of $x(t)$) as follows:

$$\dot x(t) = -\frac{uz}{\omega} e^{-zt} \sin(\omega t) + u e^{-zt} \cos(\omega t)$$

The time it takes for the mass to return to $x = 0$ is $t = \frac{\pi}{\omega}$. Therefore, the velocity of the mass after the collision is $-u e^{-\pi z / \omega}$. Therefore, the KE gain, i.e. ratio of KE after collision to KE before collision, is:

$$G_{KE} = e^{-2\pi z / \omega}$$

and the energy dissipated by the collision is:

$$E_{loss} = \frac{1}{2}mu^2\left( 1 - e^{-2\pi z / \omega} \right)$$

MY PROBLEM

1) Is this model sufficient in approximating inelastic collisions? Have I made any fatal assumptions, e.g. is the use of a viscous dashpot to model heat dissipation inaccurate? If you know the mechanics of heat loss in an inelastic collision, could you give me any insight to improving this model? Is this model valid for particular materials

2) It seems that for a particular collisions, the KE-gain is independent of the speed of impact of the smaller mass. If I repeated the experiment with the same two collision masses, is there reason to believe the collision parameters, $k$ and $\lambda$ would change? If not, as a bonus round, would anyone have the resources to experimentally determine that the KE-gain for collisions, (where the two colliding masses are kept the same, but the speed of the impacting mass is varied in the experiment) is independent of the impacting mass's speed?

• The validity of a model really depends on what you are asking of it. Your model may not predict instantaneous force well whilst correctly computing total energy absorbed, for example. There is a lot of detail here - what do you consider "sufficient"? – Floris Jan 28 '15 at 18:17
• I was attempting to model the changes of energy (KE to heat) as a result of the collision, and the displacement/velocity of the impacting mass as a function of time. For example, the model should be able to determine the maximum amount that the impacting mass embeds itself into the large mass due to deformation of the larger body in the collision. Also, what I mean by "sufficient" is that the displacements/energy changes can be determined by the model to an accuracy useful to a macroscopic observation (preferably accurate to less than 1% error) – Involute Jan 28 '15 at 18:29
• Very often these processes are nonlinear - in which case your model may not be sufficient. Can you explain what you mean in point (2) for "KE-gain"? The only KE that I see is being lost, and since one of the objects is stationary I'm not sure what you mean by the "smaller" mass (there is only one?). – Floris Jan 28 '15 at 19:48
• I made up the term "KE-gain" as something analogous to a voltage gain for an amplifier, i.e. the ratio of the output to the input. For a gain of less than one, the output will be lower than the input as it is in this case. As for what I mean by the smaller mass, I just saw the stationary object as the "larger mass", as a large enough stationary mass should have negligible motion resulting from the collision. Sorry for the confusion! – Involute Jan 28 '15 at 20:05
• The math looks sound, but any time you try to get a model to within 1% you are facing an uphill battle... – Floris Jan 28 '15 at 20:16