A parallel beam of particles of mass m moving | vedclass

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(B) Consider a small area $A$ on the wall. The number of particles hitting this area in time $\Delta t$ is equal to the number of particles contained

Vedclass · Accepted Answer

$2\, mn\, v^2 \,\cos^2 	heta$

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(B) Consider a small area $A$ on the wall. The number of particles hitting this area in time $\Delta t$ is equal to the number of particles contained in a cylinder of base area $A$ and length $v \Delta t \cos 	heta$.Volume of this cylinder $= A \cdot v \Delta t \cos 	heta$.Number of particles $N = n \cdot A \cdot v \Delta t \cos 	heta$.The number of particles hitting per unit area per unit time is $N' = \frac{N}{A \Delta t} = nv \cos 	heta$.When a particle of mass $m$ hits the wall elastically at an angle $	heta$ to the normal,its velocity component parallel to the wall remains unchanged,while its velocity component perpendicular to the wall reverses direction.Change in momentum of one particle $= m(v \cos 	heta) - m(-v \cos 	heta) = 2mv \cos 	heta$.Pressure $P$ is the force per unit area,which is equal to the rate of change of momentum per unit area.$P = N' 	imes (	ext{change in momentum of one particle}) = (nv \cos 	heta) 	imes (2mv \cos 	heta) = 2mnv^2 \cos^2 	heta$.

$A$ parallel beam of particles of mass $m$ moving with velocity $v$ impinges on a wall at an angle $\theta$ to its normal. The number of particles per unit volume in the beam is $n$. If the collision of particles with the wall is elastic,then the pressure exerted by this beam on the wall is

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