TRANSCRIBE THE FOLLOWING TEXT IN DIGITAL FORMAT A/C toCSconservationFinal momen8 X 22V₁ + V₂Kinetic energytum=> V₁? +V₂²V₁ == 2 mlsBefore explosion8 ку=Scanned withCamScanner311/2/2 mv ₁ ² + 1 m √₂² = 16x 4 x V₁ ² + + x4 X V₂² = 164ofInhal momentur=on solving eqn 0 %@4 X V₁momentum.= 2m/sRmls+ 4 V ₂& V/₂ = 2 mlswe get.mAfter explosion4kg|Akgamis

According to the conservation of angular momentum Final momentum = Initial momentum

A/C to conservation of Final momentum 8 X 2 V₁ + V₂ = 4 Kinetic energy =>> V₁² +Va 2 on solving eqn V₁ = 2 mls 1/2mv₁ ² + + m V₂² = 16 11/1/2 X 4 X V₁ ² + 1 X4 X √₂² = 16 늦 Before explosion 8ку 4 X V₁ + 4 V₂ O momentum Inhal momentu. Rmls & V₂ = 2m/s & = 2mls we = get. After explosion 4kg Avg amls

A/C to conservation of Final momentum 8 X 2 V₁ + V₂ = 4 Kinetic energy =>> V₁² +Va 2 on solving eqn V₁ = 2 mls 1/2mv₁ ² + + m V₂² = 16 11/1/2 X 4 X V₁ ² + 1 X4 X √₂² = 16 늦 Before explosion 8ку 4 X V₁ + 4 V₂ O momentum Inhal momentu. Rmls & V₂ = 2m/s & = 2mls we = get. After explosion 4kg Avg amls

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter33: Particle Physics

Section: Chapter Questions

Problem 19PE: (a) What is the uncertainty in the energy released in the decay of a due to its short lifetime? (b)...

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