Question

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 :

• A. $2mv{v}^2\cos \theta$
• B. $2mv{v}^2{\cos }^2\theta$
• C. $2mvvcos\theta$
• D. $2mvv{\cos }^2\theta$

Answer 1

The correct option is A $2mv{v}^2\cos \theta$

Let's take the case of only one particle first.

So, one particle impinging normal on wall with velocity= $vcos{\theta }_1$

Also, let ${\theta }_2$ be the angle with which it rebounds & $v_2$ be the velocity considering momentum conservation, we get $2$ components of the motion.

(1) $m{v}_{1}cos{\theta }_1=-m{v}_{2}cos{\theta }_2$ {Normal to wall}

(2) $m{v}_{1}sin{\theta }_1=-m{v}_{2}sin{\theta }_2$ {Parallel to wall component}

From squaring and adding equation (1) & (3) we get; $v_1=v_2$ & ${\theta }_1={\theta }_2$

For an elastic collision;

So, change in $\stackrel{⟶}{p}$ normal to wall= $m{v}_{1}ccos{\theta }_1-(-m{v}_{2}cos{\theta }_2)=2m{v}_{1}cos{\theta }_1$

For $\nu$ number of particle= $2m\nu v_{1}cos{\theta }_1$

(1) Rate of change of momentum/volume= Farce/volume

$\Rightarrow$ Farce F= $\nu \left[2m{v}_{1}cos{\theta }_1\right][area\times dx]/dt$

So, Pressure= $\frac{Farce}{area}=\frac{\nu 2m{v}_{1}cos{\theta }_{1}areadx}{area\times dt}$

$=\nu 2m{v}_1^{2}cos\theta$

$=2m\nu v^{2}cos\theta$ .