Consider a rectangular block of wood moving with a velocity v₀ in a gas of mass density $ρ$ and at temperature T. Assuming the velocity is along x-axis and the area of cross-section of the block perpendicular to v₀ is A.The drag force acting on the block is $N ρ A v_{0} \sqrt{\frac{k T}{m}}$ , where m is mass of each molecule. Find N.

Very difficult

Solution

Using kinetic theory of ideal gases,
n = number density of molecules
m = mass of each gas molecule
$v_{R M S}$ = RMS velocity of each gas molecule
v = velocity of molecule that is moving along x-axis
As RMS velocity along each direction is equal $[{(v_{R M S})}_{x} = {(v_{R M S})}_{Y} = {(v_{R M S})}_{Z} = v]$
Now, $v_{R M S}^{2} = {(v_{R M S})}_{x}^{2} = {(v_{R M S})}_{y}^{2} = {(v_{R M S})}_{z}^{2} = 3 v^{2}$
$\Rightarrow v^{2} = \frac{v_{R M S}^{2}}{3}$
Let block move towards right. It faces “front” molecules that collide with the block while moving towards left. Also there are “back” molecules that collide with the block from behind while moving towards right.
Relative speed of “front” molecules before collision = $v + v_{0}$
Relative speed of “back” molecules before collision = $v - v_{0}$
Since, collisions are elastic, relative speed of molecules remains same before and after collision. Since, block is heavy, so it does not change its speed and direction of motion.
After collision, “front” molecules move towards right, and “back” molecules move towards left.
Let v₁ and v₂ be speeds of “front” and “back” molecules along x-axis respectively after collision.
So, $v + v_{0} = v_{1} - v_{0} \Rightarrow v_{1} = v + 2 v_{0}$
And, $v - v_{0} = v_{0} - (- v_{2}) \Rightarrow v_{2} = v - 2 v_{0}$
Magnitude of momentum transferred by each “front” molecule to the block = $(m a s s o f m o l e c u l e) (c h a n g e i n i t s v e l o c i t y) = m |\{(v + 2 v_{0}) - (- v)\}| = 2 m (v + v_{0})$
Its direction is towards left.
Magnitude of momentum transferred by each “back” molecule to the block = $(m a s s o f m o l e c u l e) (c h a n g e i n i t s v e l o c i t y) = m |\{- (v - 2 v_{0}) - (v)\}| = 2 m (v - v_{0})$
Its direction is towards right.
Consider an interval of time $Δ t$ . All “front” molecules within a distance (along x-axis) $(v + v_{0}) Δ t$ will collide with the block. And “back” molecules within a distance (along x-axis) $(v - v_{0}) Δ t$ will collide with the block.
Momentum transferred by “front” molecules in time $Δ t$ = $[h a l f t h e n u m b e r o f m o l e c u l e s i n v o l u m e A (v + v_{0}) Δ t] (m o m e n t u m t r a n s f e r r e d b y e a c h " f r o n t " m o l e c u l e)$ $[\frac{1}{2} n A (v + v_{0}) Δ t] (2 m (v + v_{0})) = n m A {(v + v_{0})}^{2} Δ t$

(The factor of half is considered because half of molecules move towards the block and other half towards the block, in each direction).
Similarly, momentum transferred by “back” molecules in time $Δ t$ = $[h a l f t h e n u m b e r o f m o l e c u l e s i n v o l u m e A (v - v_{0}) Δ t] (m o m e n t u m t r a n s f e r r e d b y e a c h " b a c k " m o l e c u l e)$
$[\frac{1}{2} n A (v - v_{0}) Δ t] (2 m (v - v_{0})) = n m A {(v - v_{0})}^{2} Δ t$
Total momentum transferred to the block in time $Δ t$ is = $n m A {(v + v_{0})}^{2} Δ t - n m A {(v - v_{0})}^{2} Δ t$
= $4 n m A v v_{0} Δ t$
Hence, drag force on the block = $\frac{m o m e n t u m t r a n s f e r r e d t o b l o c k}{t i m e} = 4 n m A v v_{0}$
But $m n = ρ$ (density of gas)
And $v_{R M S} = \sqrt{\frac{3 k T}{m}}$
So, $v^{2} = \frac{v_{R M S}^{2}}{3} = \frac{\frac{3 k T}{m}}{3} = \frac{k T}{m}$
Hence, drag force = $4 ρ A v_{0} \sqrt{\frac{k T}{m}}$ .

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