Question

Consider a rectangular block of wood moving with a velocity v₀ in a gas at temperature T and mass density ρ. Assume the velocity is along x-axis and the area of cross-section of the block perpendicular to v₀ is A. Show that the drag force on the block is `4ρAv_0 sqrt((KT)/m)`, where m is the mass of the gas molecule.

Answer 1

Consider the diagram

Let n = number of molecules per unit volume

V_rms = rms speed of the gas molecules

When a block is moving with speed v₀, relative speed of molecules w.r.t. front face = v + v₀

Coming head-on, momentum transferred to block per collision = 2m (v + v₀), where, m = mass of the molecule.

The number of collisions in time Δt = m(v + v₀)²nAΔt, where, A = area of cross-section of block and factor of 1/2 appears due to particles moving towards the block.

∴ The momentum transferred in time Δt = m(v + v₀)²nAΔt from the front surface.

Similarly, momentum transferred in time Δt = m(v – v₀)²nAΔt   ......(From the back surface)

∴ Net force  (drag force) = mnA[(v + v₀)² – (v – v₀)²]  .....[From front]

= mnA (4vv₀)

= (4 mnAv)v₀

= (4ρAv)v₀  ......(i)

Where we have assumed ρ = `(mn)/V = M/V`

If v = velocity along the x-axis

Then, we can write KE = `1/2 mv^2 = 1/2 K_BT`

⇒ `v = sqrt((K_BT)/m)`  .....`[(K_B = "Boltzmann constant"),(KE = "Kinetic energy"),(T = "Temperature")]`

∴ From equation (i), Drag force = (4ρAv)v₀ = `4ρA sqrt((K_BT)/m) v_0`.