Kvant Physics Problem 175

Verified: yes
Verdicts: PASS + PASS
Solve time: 5m39s
Source on kvant.digital

Problem

This problem was mistakenly published as number 165 in the journal.

Estimate the duration of an elastic collision between two steel or two rubber spheres of equal radii $R=1~\text{см}$. For steel, 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 velocities $v = 1~\text{м}/\text{с}$.

What is the average force of interaction between the spheres?

G. L. Kotkin

Setup and Assumptions

Two identical solid elastic spheres of radius $R = 1~\text{cm} = 1.0 \cdot 10^{-2}\text{m}$ move head-on with equal speeds $v = 1\text{m/s}$. One case corresponds to steel with density $\rho_c = 7.8 \cdot 10^{3}\text{kg/m}^3$ and Young’s modulus $E_c = 2.1 \cdot 10^{11}\text{N/m}^2$, the other to rubber with density $\rho_r = 1.0 \cdot 10^{3}\text{kg/m}^3$ and Young’s modulus $E_r = 1.0 \cdot 10^{6}\text{N/m}^2$.

The contact interaction is described in the Hertz elastic collision approximation, where deformation is localised near the contact region and the collision time is determined by elastic wave propagation and nonlinear contact stiffness.

Each sphere is treated as homogeneous and isotropic, gravity and dissipative losses are neglected, and the motion is considered in the center-of-mass frame. The Poisson ratio dependence in the effective modulus is absorbed into an order-of-magnitude estimate, so the effective elastic scale is taken as $E$.

The unknowns are the collision duration $\tau$ and the characteristic average force $\overline{F}$ during contact.

Physical Principles

The mass of each sphere follows from volume and density,

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

In a head-on elastic collision of identical spheres, the impulse delivered to each sphere equals the change in momentum,

$J = \Delta p = 2 m v.$

The Hertz contact law gives a nonlinear force-displacement relation of the form

$F \sim E R^{1/2} \delta^{3/2},$

which leads, via dimensional analysis of the collision dynamics, to a contact time scaling

$\tau \sim \left(\frac{m^2}{R E^2 v}\right)^{1/5}.$

The average force is estimated from impulse balance,

$\overline{F} \sim \frac{J}{\tau} = \frac{2 m v}{\tau}.$

Derivation

The mass of one sphere is computed from geometry,

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

For $R = 1.0 \cdot 10^{-2}~\text{m}$, the volume is

$V = \frac{4}{3}\pi (10^{-2})^3 = 4.19 \cdot 10^{-6}~\text{m}^3.$

For steel,

$m_c = 4.19 \cdot 10^{-6} \cdot 7.8 \cdot 10^{3} = 3.27 \cdot 10^{-2}~\text{kg}.$

For rubber,

$m_r = 4.19 \cdot 10^{-6} \cdot 10^{3} = 4.19 \cdot 10^{-3}~\text{kg}.$

The collision time follows from Hertz scaling,

$\tau \sim \left(\frac{m^2}{R E^2 v}\right)^{1/5}.$

For steel,

$m_c^2 = (3.27 \cdot 10^{-2})^2 = 1.07 \cdot 10^{-3}~\text{kg}^2,$

$R E_c^2 v = (10^{-2})(2.1 \cdot 10^{11})^2 = (10^{-2})(4.41 \cdot 10^{22}) = 4.41 \cdot 10^{20},$

$\frac{m_c^2}{R E_c^2 v} = 2.43 \cdot 10^{-24}.$

Taking the fifth root,

$\tau_c = (2.43 \cdot 10^{-24})^{1/5} = 10^{-4.8} \cdot (2.43)^{1/5}\text{s} = 1.6 \cdot 10^{-5} \cdot 1.2 = 1.9 \cdot 10^{-5}\text{s}.$

For rubber,

$m_r^2 = (4.19 \cdot 10^{-3})^2 = 1.76 \cdot 10^{-5}~\text{kg}^2,$

$R E_r^2 v = (10^{-2})(10^{6})^2 = (10^{-2})(10^{12}) = 10^{10},$

$\frac{m_r^2}{R E_r^2 v} = 1.76 \cdot 10^{-15}.$

Taking the fifth root,

$\tau_r = (1.76 \cdot 10^{-15})^{1/5} = 10^{-3} \cdot (1.76)^{1/5} = 10^{-3} \cdot 1.12 = 1.1 \cdot 10^{-3}~\text{s}.$

The average force follows from impulse,

$\overline{F} = \frac{2 m v}{\tau}.$

For steel,

$\overline{F}_c = \frac{2 \cdot 3.27 \cdot 10^{-2} \cdot 1}{1.9 \cdot 10^{-5}} = \frac{6.54 \cdot 10^{-2}}{1.9 \cdot 10^{-5}} = 3.4 \cdot 10^{3}~\text{N}.$

For rubber,

$\overline{F}_r = \frac{2 \cdot 4.19 \cdot 10^{-3} \cdot 1}{1.1 \cdot 10^{-3}} = \frac{8.38 \cdot 10^{-3}}{1.1 \cdot 10^{-3}} = 7.6~\text{N}.$

Result

Collision duration:

$\tau_c \approx 1.9 \cdot 10^{-5}\text{s}, \quad \tau_r \approx 1.1 \cdot 10^{-3}\text{s}.$

Average force:

$\overline{F}_c \approx 3.4 \cdot 10^{3}\text{N}, \quad \overline{F}_r \approx 7.6\text{N}.$

$$\boxed{\text{Steel: } \overline{F} \approx 3 \cdot 10^{3}\text{N}, \quad \text{Rubber: } \overline{F} \approx 8\text{N}}$$

Sanity Checks

Dimensional consistency follows from $\tau \sim (m^2/(R E^2 v))^{1/5}$, where $m^2$ has units $\text{kg}^2$, $E^2$ has units $\text{N}^2/\text{m}^4 = \text{kg}^2/(\text{m}^2\text{s}^4)$, so the ratio inside the fifth root has units $\text{s}^5$, giving time after exponentiation.

The steel collision time is much shorter than the rubber collision time because $E$ enters quadratically inside the fifth root, producing a strong sensitivity to stiffness, consistent with the ratio of moduli $E_c/E_r \sim 10^5$.

The steel force is three orders of magnitude larger than the rubber force, consistent with both higher stiffness and shorter interaction time.

A dominant source of numerical sensitivity is the fifth-root dependence; a factor of $10$ error in modulus produces only a factor of about $10^{2/5} \approx 2.5$ error in time, but enters force both through mass and time scaling.