A helium atom (m = 6.646 × 10−27) collides elastically with an oxygen atom (m = 2.656 × 10−26) at rest. After the collision, the helium atom is found to be moving with a velocity of 6.636×106 m/s in a direction at an angle of 84.7◦ relative to its original direction. The oxygen atom is observed to move at an angle of -40.4◦.

(a) Find the velocity of the oxygen atom after the collision.
(b) Find the speed of the helium atom before the collision.

Explanation:

Let velocity of helium atom before the collision be v and velocity of oxygen atom be V

Applying the law o conservation of momentum along the line of collision

6.646 x 10⁻²⁷v + 0 = 6.646 x 10⁻²⁷ x 6.636×10⁶ cos 84.7 + 2.656 x 10⁻²⁶ V cos40.4

6.646 x 10⁻²⁷v + 0 = 4.07  x 10⁻²¹  + 2.02 x 10⁻²⁶ V

6.646 x 10⁻⁻²⁷ v = 4.07  x 10⁻²¹ + 2.02 x 10⁻⁻²⁶ V

Applying law of conservation of momentum in a direction perpendicular to original direction of motion

6.646 x 10⁻²⁷ x 6.636×10⁶ sin 84.7 - 2.656 x 10⁻²⁶ V sin40.4 = 0

43.91 x 10⁻²¹ = 1.7214   x 10⁻²⁶V

V = 43.91 / 1.7214 x 10⁵ m /s

= 25.5 x 10⁵ m /s

Putting this value in the earlier equation

6.646 x 10⁻⁶ v = 4.07  x 10⁻²¹ + 2.02 x 10⁻⁵ V

6.646 x 10⁻⁶ v = 4.07  x 10⁻²¹ + 2.02 x10⁻²⁶  x 25.5 x 10⁵

6.646 x 10⁻²⁷ v = 4.07  x 10⁻²¹ + 51.51 x10⁻²¹ = 55.58 x 10⁻²¹

v =  55.58 x 10⁻²¹ / 6.646 x 10⁻²⁷

= 8.36 x 10⁶ m /s

the results is a decrease in air temperature with an increase of altitude.

explanation:

prolly a beacuse the momentuem the car has increases while going down hill