Answer to Question #156645 in Classical Mechanics for Afroja Akhi

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

Answer

In this question

Mass of the 1st molecule=m kg

Speed of the molecule =300 m/sec(given)

Mass of the second molecule =m kg

Initial velocity of the 2nd molecule =0 m/sec (rest)

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

Now,Initial momentum of the molecule

$mv=300m$

After collision

$u_1=v_1\cos30+v_2\sin\theta$

And also

$0=v_1\sin30+v_2\cos\theta$ ---------eq1

Squaring and adding both above equation

$u_1^2+v_1^2u_1v_1\cos\theta=v_2^2--eq3$

$u_1^2=v_1^2+v_2^2---eq4$

Using eq 3 and 4

$v_1^2=2v_1u_1\cos30$

Finally solving above equation

$v_1=260m/s$

$v_2=150m/s$

Putting these value in equation 1

Angle

$\theta=60^°$