Answer to Question #100726 in Mechanics | Relativity for PETER

Question

A gas molecule having a speed of 300m/s horizontally collides elastically with another molecule of the same mass is initially at rest. After collision the first molecule moves at an angle of 30° to it's initial direction. Find the speed of each molecule after collision and the direction of the second molecule.

Answer 1

As per the given question,

Mass of the first molecule=m

Speed of the molecule =300 m/sec

Mass of the second molecule =m

Initial velocity of the second molecule =0 m/sec

After the collision, first molecule is moving at an angle = 30

We have to determine the velocity of each molecule and direction of the second molecule?

Now,

Initial momentum of the molecule $= mv= 300m kg-m/sec$

$u_1 = v_1\cos30 + v_2\cos\theta_2-------(1)$

$v_1\sin30 = v_2\sin\theta_2----------(2)$

and $u^2_1=v^2_1+v^2_2-----------(3)$

Now, squaring and adding equation (i) and (ii),

$\Rightarrow u^2_1+v^2_1−2u_1v_1\cos30=v^2_2-----------(4)$

Now, from the equation (3) and (4)

$2v^2_1=2u_1v_1\cos30$

from the above,

$v_1=u_1\cos30=260m/s$

$v_2 = 150 m/s$

from the equation(ii)

$\theta=60^\circ$