Mass of an Oxygen molecule is 5.35×10−26kg. and that of a Nitrogen molecule is 4.65×10−26kg.During their Brownian motion (random motion) in air, an Oxygen molecule travelling with a velocity of 400 m/s collides elastically with a nitrogen molecules travelling with a velocity of 500 m/s in the exactly opposite direction. Calculate the impulse received by each of them during collision. Assuming that the collision lasts for 1 ms, how much is the average force experienced by each molecule ?
Answers
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Answer:
Given = mass of oxygen i.e. m1 = 5.35×10⁻²⁶
mass of nitogen i.e. m2 = 4.65 x 10⁻²⁶
velocity (u1) = 400m/s
velocityy (u2) = 500m/s
to find = impulse
solution =
V₁ = (m1 - m2 / m1 + m2)u1 + (2m2 /m1+m2)u2
V₂= ( m2 - m1/ m1 + m2)u2 + (2m1 /m1+m2)u1
put the values in above formula,
V₁ = (5.35 - 4.65 )x 10⁻²⁶/ (5.35 + 4.65)x10⁻²⁶x 400
V2 = (4.65-5.35)x10⁻²⁶x(5.35+4.65)x10⁻²⁶x400
V₁=(0.07 x 400 ) - (0.93x500)= -437m/s
V₂= 35 +428= 463m/s
now,
for impulse J₀= m₀(v₁-u₁)
= 5.35 x 10⁻²⁶ x (-437 -400)
= -4.478 x 10⁻²³Ns
impulse JN= mN(V2- u2 )
= 4.65X10⁻²⁶ x (463 +500)
= +4.478x10⁻23Ns
net impulse = J₀/Δt = -4.478x10⁻²³ / 10⁻³
= -4.478x10⁻²⁰N