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Question: A beam of gas molecules is incident normally on a plate AB. Each gas molecule has mass \(m\) and vel...

A beam of gas molecules is incident normally on a plate AB. Each gas molecule has mass mm and velocity vv . The incident beam falls on an area A on the plate and all the molecules strike the plate elastically. Number of molecules in the unit volume of the beam is nn. When the plate is moved to the right a force F1F_{1} is needed to keep it moving with constant velocity $u(

& A.8mAnvu \\\ & B.6mAnuv \\\ & C.4mAnuv \\\ & D.3mAnuv \\\ \end{aligned}$$
Explanation

Solution

A elastic collision occurs when the total kinetic energy of the two bodies interacting bodies is conserved. Here, we can consider the interaction of the molecules with the plate to be in one direction and solve the following.
Formula: F=N×ΔpF=N\times \Delta p

Complete answer:
Let us assume that the given plate is moving towards the right as shown in the figure with a velocity uu, then the particles, which are bombarding the plate with a velocity vv will get reflected back towards the left with some velocity v′v\prime.
Since it is given that the particle undergoes elastic collision, we can say that the coefficient of restitution e=1e=1.
Then, here, we have 1=u−(−v′)v−u1=\dfrac{u-(-v\prime)}{v-u}
  ⟹  v′+u=v−u\implies v\prime+u=v-u
  ⟹  v′=v−2u\implies v\prime=v-2u
Then the change in momentum is given as Δp=mv′−(−mv)\Delta p=mv\prime-(-mv)
  ⟹  Δp=m(v−2u)+mv\implies \Delta p=m(v-2u)+mv
  ⟹  Δp=2(mv−mu)\implies \Delta p=2(mv-mu)
Let us consider the number moles of the molecules which hit the plate per unit time, it is given as N=nAv=nA(v−u)N=nAv=nA(v-u), where nn is the number of moles, which are incident on area AA of the plate.
Then the force F1F_{1} acting on the plate is given as F1=N×ΔpF_{1}=N\times \Delta p
  ⟹  F1=nA(v−u)×2m(v−u)\implies F_{1}=nA(v-u)\times 2m(v-u)
  ⟹  F1=2mnA(v−u)2\implies F_{1}=2mnA(v-u)^{2}
Similarly, the force F2F_{2} due to which the plate moves to the left with a velocity uu, then the particles, which are bombarding the plate with a velocity vv will get reflected back towards the left with the some velocity v′′v\prime\prime.
Then, again, since the collision is elastic, we can say that, e=1=v′′−uv−(−u)e=1=\dfrac{v\prime\prime-u}{v-(-u)}
  ⟹  v+u=v′′−u\implies v+u=v\prime\prime- u
  ⟹  v′′=v+2u\implies v\prime\prime=v+2u
Then, the change in momentum is given as Δp=mv′′−(−mv)\Delta p=mv\prime\prime-(-mv)
  ⟹  Δp=m(v+2u)+mv\implies \Delta p=m(v+2u)+mv
  ⟹  Δp=2(mv+mu)\implies \Delta p=2(mv+mu)
Then the number moles of the molecules which hit the plate per unit time, it is given as N=nAv=nA(v+u)N=nAv=nA(v+u)
Then the force F2F_{2} acting on the plate is given as F2=N×ΔpF_{2}=N\times \Delta p
  ⟹  F2=nA(v+u)×2m(v+u)\implies F_{2}=nA(v+u)\times 2m(v+u)
  ⟹  F1=2mnA(v+u)2\implies F_{1}=2mnA(v+u)^{2}
Then F1−F2=2mnA[(v+u)2−(v−u)2]F_{1}-F_{2}=2mnA[(v+u)^{2}-(v-u)^{2}]
  ⟹  F1−F2=2mnA[(v+u−(v−u))(v+u+(v−u))]\implies F_{1}-F_{2}=2mnA[(v+u-(v-u))(v+u+(v-u))]
  ⟹  F1−F2=2mnA[(v+u−v+u)(v+u+v−u)]\implies F_{1}-F_{2}=2mnA[(v+u-v+u)(v+u+v-u)]
  ⟹  F1−F2=2mnA[(2u)(2v)]\implies F_{1}-F_{2}=2mnA[(2u)(2v)]
∴F1−F2=8mnAuv\therefore F_{1}-F_{2}=8mnAuv

Hence the correct option is A.8mAnvuA.8mAnvu

Note:
We know that coefficient of restitution is defined as the ratio between the velocity after collision to the velocity before collision. It is given ase=relative velocity after collisionrelative velocity before collision\text{e=}\dfrac{\text{relative velocity after collision}}{\text{relative velocity before collision}}. Here we are considering two situations, when the plate is moving towards left and right. Then using the two we can find the respective forces.