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# s5 - Phys 263/Amath 261 Assignment 5 SOLUTIONS 1 First...

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Phys 263/Amath 261 Assignment 5 SOLUTIONS 1. First, consider the center of mass momentum, which reacts to two forces: the tension on the rope (by the hand), and gravity: ˙ P = T - Mg. (1) Now, the mass of the suspended rope is the mass density times the length x : M = μx (2) so that the momentum of the system is P = Mv 0 = μxv 0 (3) and ˙ P = μ ˙ xv 0 = μv 2 0 (4) since the speed v 0 is a constant. Equating these expressions gives μv 2 0 = T - Mg = T - μxg (5) which at x = a gives T = μ ( v 2 0 + ag ) (6) 2. (a) The total angle between the two particles in the LAB frame is given by Θ = ψ + ζ , where ζ is the angle of the emerging particle. From cos Θ = ( m 1 - m 2 ) v 2 2 m 1 v 1 (7) since m 1 = m 2 , we get cos Θ = 0 or Θ = π/ 2. Since ψ = π/ 4, this means ζ = π 4 = 45 (8) (b) To relate LAB and CM angles, use the formula derived in class tan ψ = sin θ cos θ + ( m 1 /m 2 ) = sin θ cos θ + 1 (9) and since tan π/ 4 = 1 sin θ = cos θ + 1 (10) which is true for θ = π/ 2, plus larger angles like π , etc. The smallest angle is the CM angle of the first ball, θ = π 2 = 90 (11) 1

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(c) We can write conservation of momentum in this case as: x -component mu 1 = mv 1 cos( π/ 4) + mv 2 cos( π/ 4) (12) u 1 = v
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s5 - Phys 263/Amath 261 Assignment 5 SOLUTIONS 1 First...

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