Answer:
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!!!
