Question

A gas molecule having a speed of 300 m/sec collides elastically with another molecule of same mass which is initially at rest. After the collision the first molecule moves at an angle of 30 degree to its initial direction. Find the speed of each molecule after collision and the angle made with the incident direction by the recoiling target molecule.

Answer 1

conservation of momentum in horizontal direction-

$mv_1+mv_2=mv_1'cos30+mv_2'cos\theta$

$300=v_1'cos30+v_2'cos\theta~~~~~~-(1)$

Momentum in Y-direction

$mv_1'sin30=mv_2sin\theta \\ \dfrac{v_1'}{2}=v_2'sin\theta~~~~~~~-(2)$

From conservation of ene2gy-

$\dfrac{1}{2}mu_1^2=\dfrac{1}{2}mv_1'^2+\dfrac{1}{1}mv_2'^2 \\ v_1^2=v_1'^2+v_2'^2 \\ (300)^2=v_1'^2+v_2'^2~~~~~-(3)$

Equation 1 can be written as-

$v_2'cos\theta=300-v_1'cos30 \\ v_2'^2cos^2\theta=(300-v_1'cos30)^2\\ v_2'^2cos^2\theta=90000+v_1^rcos^2\theta-600v_1'cos\theta30~~~~~-(4)$

Equation 2 can be writtten as-

$v_2'sin\theta=\dfrac{v_1'^2}{2} \\v_2'^2sin^2\theta=\dfrac{v_1^2}{4}~~~~~~-(5)$

Adding equation 4 and 5-

$v_2'^2=90000+v_1'^2cos^2300-600v_1'cos30+\dfrac{v_1'^r}{4}~~~~~~~-(6)$

from equation 3 and 6-

$90000=v_1'^r+90000+v_1'^2cos^2\theta-600v_1'cos\theta+\dfrac{v_1^2}{4}$

$2v_1=600cos30\Rightarrow v_1'=259.81m/s$

From eqn(3)-

$(300)^2=(259.81)^2+v_2'^2\\ \Rightarrow v_2'=150m/s$

from equation 2-

$sin\theta=\dfrac{259.81}{2\times 150}$

$\theta=60^{\circ}$