Question

A molecule of gas in a container hits one wall (1) normally and rebounds back. It suffers no collision and hits the opposite wall (2) which is at an angle of ${30}^{\circ }$ with wall 1.
Assuming the collisions to be elastic and the small collision time to be the same for both the walls, the magnitude of average force by wall 2. ( $F_2$ ) provided to the molecule during collision satisfy :

• A. $F_1>F_2$
• B. $F_1<F_2$
• C. $F_1=F_2$, bothnon-zero
• D. $F_1=F_2=0$

Answer 1

The correct option is A $F_1>F_2$

Work done by the wall during a collision is equal to the change in momentum of the collided particle per unit time of collision.

Let the collision time in both cases be $t$, the mass of the particle be $m$ and the speed be $u$.

Change in momentum for the collision at Wall1 is $\Delta p_1=mu-(-mu)=2mu$

The final momentum of the gas molecule after its collision with wall 2 makes ${120}^o$ angle with the momentum just before its collision with wall 2.

Change in momentum for the collision at Wall2 is $\Delta p_2=\sqrt{(mu{)}^2+(mu{)}^2-2\left(mu\right)\left(mu\right)\cos \left({120}^{\circ }\right)}=\sqrt{3}mu$

Force is given by $F=\frac{p}{t}$

Thus, $F_1=2mu/t$ and $F_2=\sqrt{3}mu/t$.

Hence, $F_1>F_2$