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Consider a head-on collision between two particles of masses m1 and m2 The initial speed of the particles are u1 and u2 in the same direction. The collision starts at t = 0 and the particles interact for a time interval Δt. During the collision, the speed of the first particle varies asv(t) = u1 + (t/Δt)(v1- u1) Find the speed of the second particle as a function of time during the collision. |
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Answer» THANKS for asking the question! ANSWER:: USING law of CONSERVATION of MOMENTUM , m₁u₁ + m₂u₂ = m₁v(t) + m₂v' (v' = speed of second particle during collision) v' = (m₁/m₂)u₁ + u₂ - (m₁/m₂).v(t) v' = (m₁/m₂)u₁ + u₂ - (m₁/m₂).{u₁ + (t/Δt)(v₁- u₁)} v' = u₂ - (m₁/m₂).(t/Δt)(v₁- u₁) Hope it HELPS! |
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