An oxygen atom (mass 16 u) moving at 777 m/s at 14.0° with respect to the î-direction collides and sticks to an oxygen molecule (mass 32 u) moving at 517 m/s at 120° with respect to the î-direction. The two stick together to form ozone. What is the final velocity (magnitude in m/s and direction in degrees counterclockwise from the +î-axis) of the ozone molecule? (Assume given angles are measured counterclockwise from the +î-axis.)

Velocity of ozone after collision

= (78.97î + 361.15ĵ) m/s

Magnitude of the velocity of ozone after collision = 369.68 m/s = 370 m/s

Direction of the velocity of ozone after collision = 77.7° counter-clockwise from the î-axis.

Explanation:

According to the law of conservation of momentum, momentum is conserved in this inelastic collision between the two oxygen atom and the oxygen molecule to give ozone.

Working in vector notations.

Mass of the oxygen atom = 16 amu

Note that amu = atomic mass unit

Velocity of the oxygen atom

= (777 cos 14°î + 777 sin 14°ĵ) m/s

= (753.92î + 187.97ĵ) m/s

Momentum of oxygen atom before collision

= mv = 16(753.92î + 187.97ĵ)

= (12,062.72î + 3,007.52ĵ) amu.m/s

Mass of the oxygen molecule = 32 amu

Velocity of the oxygen molecule

= (517 cos 120°î + 517 sin 120°ĵ) m/s

= (-258.50î + 447.74ĵ) m/s

Momentum of oxygen molecule before collision = mv = 32(-258.50î + 447.74ĵ)

= (-8,272.00î + 14,327.52ĵ) amu.m/s

(Momentum of the oxygen atom before collision) + (Momentum of the oxygen molecule before collision) = (Momentum of ozone after collision)

Momentum of ozone after collision

= (12,062.72î + 3,007.52ĵ) + (-8,272.00î + 14,327.52ĵ) = (3,790.72î + 17,335.04ĵ) amu.m/s

Mass of ozone = 48 amu

Momentum of ozone = (Mass of ozone)(velocity of ozone)

(3,790.72î + 17,335.04ĵ) = (48)(velocity of ozone)

v = (1/48)(3,790.72î + 17,335.04ĵ) = (78.97î + 361.15ĵ) m/s

Magnitude of velocity of ozone

= √[78.97² + 361.15²] = 369.68 m/s = 370 m/s

Direction = tan⁻¹ (361.15/78.97) = 77.7° counter-clockwise from the î-axis.

Hope this Helps!!!

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explanation:

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