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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.
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Solution
Consider the diagram
Let n = number of molecules per unit volume
Vrms = 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`.
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