Question

A beam of hydrogen molecules (h2) is directed toward a wall, at an angle of 55 with the normal to the wall. each molecule in the beam has a speed of 1.0 km/s and a mass of 3.3 1024 g.the beam strikes the wall over an area of 2.0 cm2, at the rate of 1023 molecules per second.what is the beam’s pressure on the wall?

Answer 1

The change in momentum of the particle upon hitting the wall is expressed as:

Change in momentum = Δp = 2 m v cosθ

where m = 3.3E-24 g = 3.3E-27 kg, v = 1.0 km/s = 1000 m/s, θ = 55°

Dividing both sides by Δt:

Δp / Δt = 2 (Δm / Δt) v cosθ

By definition, the force applied to a particle is equal to the change in momentum per second of the particle (by Newton's Second Law). Therefore:

Force on wall = Δp / Δt = 2 (Δm / Δt) v cosθ

We can get or calculate the value of (Δm / Δt) from the given data. That is:

Δm / Δt = m * particles per second = (3.3E-27 kg/particle) (1023 particle/s)

Δm / Δt = 3.3759 E-24 kg/s

Therefore the force is:

Total force on wall = 2 (3.3759 E-24 kg/s) (1000 m/s) cos(55)

Total Force on wall = 1.494E-22 N

Pressure = Total Force / Area = 1.494E-22 N / 2.0E-4 m^2

Pressure = 7.47E-19 Pascals

Therefore the pressure is 7.47*10^-19 Pa.