Find the magnitude of the average force ⟨Fx⟩⟨Fx⟩ in the x direction that the particle exerts on the right-hand wall of the container as it bounces back and forth. Assume that collisions between the wall and particle are elastic and that the position of the container is fixed. Be careful of the sign of your answer. Express the magnitude of the average force in terms of mmm, vxvxv_x, and LxLxL_x.

Newton's second law is the change of the momentum with respect to time and the momentum is defined as the product of the mass and the velocity of the body.

[tex]F = \frac{dp}{dt} \\p = m v[/tex]

where F is the force, p the moment, m the mass, t the time and v is the velocity.

Ask for the average force, therefore we change the differentials for variations.

[tex]<F> = m \frac{\Delta v}{\Delta t}[/tex]

They indicate that the velocity in the direction of the wall is vₓ and the mass of the container is much greater than the mass of the particle, they also indicate that the collision is elastic, therefore the speed of the particle before and after the collision is equal , but its address changes.

Δv = vₓ - (-vₓ)

Δv = 2 vₓ

The change in velocity occurs during the collision, in the rest of the motion the particle has a constant velocity, using the uniform motion relation.

[tex]v = \frac{d}{t} \\t = \frac{d}{v}[/tex]

The particle travels a distance Lₓ from inside the container to the wall and bounces, we can find the total time for the particle where the distance of the entire route is:

d = 2 Lₓ

t = [tex]\frac{2L_x}{v_x}[/tex]

Let's substitute in Newton's second law.

[tex]<F>= \frac{m \ 2v_x}{ \frac{2L_x}{v_x} } \\<F>= \frac{m v_x^2}{L_x}[/tex]

In conclusion, using Newton's second law we can find the average force for the collision of a particle against the wall of a container is:

[tex]<F> = \frac{m \ v_x^2}{L_x}[/tex]

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