A container holds .4 x 10^6 particles, each with velocity always directed perpendicular to the axis of the container and perpendicular to its ends. The particles all have constant velocity 100.1 m/s, and require .93 seconds to make a round trip from a given wall to the other wall and back. Each particle has mass 7.183442E-07 Kg.
When a collision occurs, the particle reverses direction. So it experiences a change in velocity of
The total mass of the particles is .4 * 10^6 ( 7.183442E-07 kg) = .2873377 kg, so the change in momentum change when all particles collide with the wall is total
If this change takes place in .93 seconds, the Impulse-Momentum Theorem guarantees that average force is
Recall that the Impulse-Momentum Theorem is a restatement of Newton's Second Law, and states that the change in momentum `d(mv) of an object is equal to the impulse Fave*`dt of the force on the object.
In an elastic collision of a particle of mass m and a fixed wall, with the particle's velocity directed perpendicular to the wall at velocity v, the particle's velocity will change from v to -v. This will result in a momentum change from p1 = m v to p2 = m(-v) = - mv, a change of
The magnitude of this momentum change is 2 m v. If there are N such particles, the total momentum change will be n * (2 m v) = 2 N m v.
If N such collisions happen every `dt seconds, then the average force is equal to the average rate of momentum change: