Part 3 of 5 - Analyze (a) With three particles, the total final momentum of the system is m₁₁f + m₂√₂f + m3√3f and it mu

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Part 3 Of 5 Analyze A With Three Particles The Total Final Momentum Of The System Is M F M F M3 3f And It Mu 1 (305.67 KiB) Viewed 29 times

Part 3 of 5 - Analyze (a) With three particles, the total final momentum of the system is m₁₁f + m₂√₂f + m3√3f and it must be zero to equal the original momentum. The mass of the third particle is m3 (15.5-5.03 -27 - 8.41) x 10- kg kg. We have Solving for 3f gives Submit = 2.06 m₁₁f + m₂√₂f + m3V3f = 0 1f V3f V3f x 10-27 = and in unit-vector notation, we have the following. 3.018 Ĵ+ 3.36 m₁₁f + m₂√₂f m3 2.06 x 10-20 kg. m/s Skip (you cannot come back) x 10-27 kg - 14.6 X = Your response differs from the correct answer by more than 10%. Double check your calculations. Î Your response differs from the correct answer by more than 10%. Double check your calculations. ĵ) x 106 m/s