Set 54 Problem number 18

Problem

The particles in a cylindrical container have velocities randomly distributed in 3 dimensions.  The container contains 28 * 10^6 particles, each of mass 6.185741E-08 Kg.  Assume that the particles always have a uniform velocity of 100.6 m/s.  The particles collide with a wall which is one of the parallel walls of a rectangular 'box', separated by 79 meters from its counterpart.  What is the average force exerted on this wall?

Solution

When velocities are distributed randomly in 3 dimensions, the average force will be 1/3 of the force that would result if the velocities were all perpendicular to the wall.

First, then, assume that all velocities are 100.6 m/s, directed perpendicular to the wall. We find the change in momentum during the time required for all particles to make a collision, and the time required.

• The time required for all particles to collide with the wall is the time required for a complete round trip.
• The time is thus the time required to travel 2( 79 meters) at 100.6 m/s, which is time between collisions with specified wall:

• 2( 79 meters)/( 100.6 m/s) = 1.570577 s.

• When a collision occurs, the change in velocity is

• vel change at collision: 2( 100.6 m/s) = 201.2 m/s;

• the total mass of the particles is

• total mass of all particles: 28 * 10^6 ( 6.185741E-08 kg) = 1.732008 kg,

• so the change in momentum when all particles colliding with the wall is

• total momentum change: ( 201.2 m/s)( 1.732008 kg) = 348.4799 kg m/s.

• If this change takes place in 1.570577 seconds, the average force is

• ave force = ave rate of momentum change = ( 348.4799 kg m/s)/( 1.570577 sec) = 221.8802 `kgm/s^2 = 221.8802 Newtons.

• Now, we return to the assumption that the velocities are random in direction. The force will then be 1/3 of the 221.8802 Newtons, or

• actual force: 73.96008 Newtons.

Generalized Solution

It turns out that if velocities are randomly distributed among the three mutually perpendicular directions in space, the average force exerted on any surface is 1/3 of the force that would result if the velocity was always perpendicular to the surface. For the given situation, if the velocity was directed perpendicular to the wall, then all of the particles would collide with that wall in the time required to make a round trip to the opposite wall and back.

If N such particles, each of mass m, move at velocity v between walls separated by a distance L, then the round trip will cover distance 2 L and will require time `dt = 2 L / v.

The momentum change associated with each particle has magnitude m `dv = 2 m v (see preceding problem) and the total momentum change for all N particles is N ( 2 m v) = 2 N m v.

The average force is therefore

average force = momentum change/time interval = 2 N m v / `dt = 2 N m v / (2 L / v) = N m v^2 / L.