An oxygen atom (mass 16 u) moving at 733 m/s at 15.0^\circ15.0 ∘ with respect to the \hat{i} i ^ direction collides and sticks to an oxygen molecule (mass 32 u) moving at 528 m/s at 128^\circ128 ∘ with respect to the \hat{i} i ^ direction. The two stick together to form ozone. What is the direction of the final velocity of the ozone molecule?

Explanation:

Representing the velocity of oxygen atom in vector form

v₁ = 733 cos 15 i + 733 sin15 j

= 708 i +189.7 j

Similarly velocity of oxygen molecule

= 528 cos 128 i + 528 sin128 j

= -325 i + 416 j

Applying conservation of momentum

m₁ v₁ + m₂v₂ = ( m₁ +m₂ ) v

16 (  708 i +189.7 j ) + 32 ( -325 i + 416 j ) = 48 v

11328 i +3035.2 j - 10400 i + 13312 j = 48 v

928 i + 16347.2 j = 48 v

v = 19.33 i + 340.56 j

angle of this vector

tanθ = 340.56 / 19.33

= 17.61

87 °

the answer is 100. i just did this lesson and got a 100% on it. : )

-fishtank