Kvant Physics Problem 175

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Problem

It was mistakenly published in the journal under number 165.

Estimate the duration of an elastic collision of two steel or two rubber spheres of equal radius $R=1~\text{см}$. For steel, the Young’s modulus is $E_с = 2{,}1 \cdot 10^{11}\text{Н}/\text{м}^2$, and the density of steel is $\rho_с = 7{,}8 \cdot 10^3\text{кг}/\text{м}^3$. The Young’s modulus of rubber is $E_р \approx 10^{6}\text{Н}/\text{м}^2$, and its density is $\rho_р \approx 10^3\text{кг}/\text{м}^3$. The spheres move toward each other with speeds $v = 1~\text{м}/\text{с}$.

What is the average force of interaction between the spheres?

G. L. Kotkin

Setup and Assumptions

Two identical spheres of radius $R=1.0\times 10^{-2},\text{m}$ move directly toward each other with equal speeds $v=1.0,\text{m/s}$. Two material cases are considered: steel with Young’s modulus $E_{\mathrm{c}}=2.1\times 10^{11},\text{N/m}^2$ and density $\rho_{\mathrm{c}}=7.8\times 10^{3},\text{kg/m}^3$, and rubber with Young’s modulus $E_{\mathrm{r}}\approx 10^{6},\text{N/m}^2$ and density $\rho_{\mathrm{r}}\approx 10^{3},\text{kg/m}^3$.

The collision is treated within Hertz contact theory for elastic deformation of spheres. The contact is assumed purely elastic, with no energy dissipation, no plastic deformation, and no adhesion. The spheres are assumed homogeneous and isotropic. The motion is taken along the line of centers, so the problem reduces to one-dimensional relative motion with an effective mass equal to the reduced mass of the two-sphere system.

The quantity of interest is the characteristic collision duration $\tau$ and the average interaction force during contact $F_{\mathrm{avg}}$.

Physical Principles

The relative motion of two colliding spheres can be reduced to a single particle of reduced mass

$\mu=\frac{m}{2}$

for identical spheres of mass $m$, moving in the effective Hertz contact potential.

The Hertz contact force for elastic deformation of two identical spheres is

$F = k,\delta^{3/2},$

where $\delta$ is the compression and the effective stiffness is

$k=\frac{4}{3}E^\sqrt{R^}.$

For identical spheres,

$R^*=\frac{R}{2},$

and the effective elastic modulus is approximated for order-of-magnitude estimates by $E^*\sim E$.

The collision time scale for a Hertzian contact with initial relative velocity $u$ follows from dimensional analysis of the equation

$\mu \ddot{\delta} = -k \delta^{3/2},$

giving

$\tau \sim C \left(\frac{\mu^2}{k^2 u}\right)^{1/5},$

where $C\approx 2.9$ is a dimensionless constant of order unity.

The average force is estimated from impulse balance,

$F_{\mathrm{avg}} \sim \frac{\Delta p}{\tau},$

where the momentum change of each sphere is

$\Delta p = 2mv.$

Derivation

The mass of one sphere is

$m = \rho \frac{4}{3}\pi R^3.$

For steel,

$m_{\mathrm{c}} = 7.8\times 10^{3}\cdot \frac{4}{3}\pi (10^{-2})^3 ,\text{kg}.$

The volume is

$\frac{4}{3}\pi (10^{-2})^3 = 4.19\times 10^{-6},\text{m}^3,$

so

$m_{\mathrm{c}} \approx 3.27\times 10^{-2},\text{kg}.$

For rubber,

$m_{\mathrm{r}} \approx 10^{3}\cdot 4.19\times 10^{-6} = 4.19\times 10^{-3},\text{kg}.$

The effective radius for two identical spheres is

$R^*=\frac{R}{2}=5.0\times 10^{-3},\text{m},$

so

$\sqrt{R^*}=7.07\times 10^{-2},\text{m}^{1/2}.$

The stiffness parameter is

$k=\frac{4}{3}E\sqrt{R^*}.$

For steel,

$k_{\mathrm{c}} \approx \frac{4}{3}(2.1\times 10^{11})(7.07\times 10^{-2}),\text{N/m}^{3/2}.$

This gives

$k_{\mathrm{c}} \approx 2.0\times 10^{10},\text{N/m}^{3/2}.$

For rubber,

$k_{\mathrm{r}} \approx \frac{4}{3}(10^{6})(7.07\times 10^{-2}) \approx 9.4\times 10^{4},\text{N/m}^{3/2}.$

The reduced mass for identical spheres is

$\mu=\frac{m}{2}.$

For steel,

$\mu_{\mathrm{c}} \approx 1.64\times 10^{-2},\text{kg}.$

For rubber,

$\mu_{\mathrm{r}} \approx 2.10\times 10^{-3},\text{kg}.$

The relative approach speed is

$u=2v=2.0,\text{m/s}.$

The collision time is

$\tau = 2.9\left(\frac{\mu^2}{k^2 u}\right)^{1/5}.$

For steel,

$\frac{\mu_{\mathrm{c}}^2}{k_{\mathrm{c}}^2 u} \approx \frac{(1.64\times 10^{-2})^2}{(2.0\times 10^{10})^2\cdot 2}.$

This yields

$\frac{\mu_{\mathrm{c}}^2}{k_{\mathrm{c}}^2 u} \approx 3.4\times 10^{-25}.$

Taking the fifth root,

$\left(3.4\times 10^{-25}\right)^{1/5} \approx 1.3\times 10^{-5},$

so

$\tau_{\mathrm{c}} \approx 3.7\times 10^{-5},\text{s}.$

For rubber,

$\frac{\mu_{\mathrm{r}}^2}{k_{\mathrm{r}}^2 u} \approx \frac{(2.10\times 10^{-3})^2}{(9.4\times 10^{4})^2\cdot 2} \approx 2.5\times 10^{-16}.$

Taking the fifth root,

$\left(2.5\times 10^{-16}\right)^{1/5} \approx 7.6\times 10^{-4},$

so

$\tau_{\mathrm{r}} \approx 2.2\times 10^{-3},\text{s}.$

The average force follows from

$F_{\mathrm{avg}}=\frac{2mv}{\tau}.$

For steel,

$F_{\mathrm{avg,c}}=\frac{2(3.27\times 10^{-2})\cdot 1}{3.7\times 10^{-5}},\text{N},$

which gives

$F_{\mathrm{avg,c}} \approx 1.8\times 10^{3},\text{N}.$

For rubber,

$F_{\mathrm{avg,r}}=\frac{2(4.19\times 10^{-3})\cdot 1}{2.2\times 10^{-3}},\text{N},$

which gives

$F_{\mathrm{avg,r}} \approx 3.9,\text{N}.$

Result

For steel spheres,

$\tau_{\mathrm{c}} \approx 3.7\times 10^{-5},\text{s}, \quad F_{\mathrm{avg,c}} \approx 1.8\times 10^{3},\text{N}.$

For rubber spheres,

$\tau_{\mathrm{r}} \approx 2.2\times 10^{-3},\text{s}, \quad F_{\mathrm{avg,r}} \approx 3.9,\text{N}.$

Sanity Checks

The contact time increases sharply as the stiffness decreases, consistent with the dependence $\tau \propto E^{-2/5}$. The rubber case produces a time roughly two orders of magnitude larger than steel, reflecting the six orders of magnitude smaller Young’s modulus.

Dimensional consistency follows from $\mu^2/(k^2 u)$ having units of $s^5$, since $k$ carries units $\text{N/m}^{3/2}$ and the Hertz equation yields a nonlinear oscillator with a $3/2$ power law.

The steel collision time is on the order of tens of microseconds, consistent with typical ultrasonic contact times for rigid solids at meter-per-second impact speeds. The rubber case lies in the millisecond regime, consistent with visibly soft impacts.

The largest sensitivity arises in the estimate of $k$, since any omission of the effective radius factor or elastic modulus prefactor changes $\tau$ by a factor that propagates as a fifth root, making the final time relatively stable but still systematically shifted.