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Unformatted text preview: 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 = x (2) so that the momentum of the system is P = Mv = xv (3) and P = xv = v 2 (4) since the speed v is a constant. Equating these expressions gives v 2 = T- Mg = T- xg (5) which at x = a gives T = ( v 2 + ag ) (6) 2. (a) The total angle between the two particles in the LAB frame is given by = + , where is the angle of the emerging particle. From cos = ( m 1- m 2 ) v 2 2 m 1 v 1 (7) since m 1 = m 2 , we get cos = 0 or = / 2. Since = / 4, this means = 4 = 45 (8) (b) To relate LAB and CM angles, use the formula derived in class tan = sin cos + ( m 1 /m 2 ) = sin cos + 1 (9) and since tan / 4 = 1 sin = cos + 1 (10) which is true for = / 2, plus larger angles like , etc. The smallest angle is the CM angle of the first ball, = 2 = 90 (11) 1 (c) We can write conservation of momentum in this case as:...
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This note was uploaded on 09/20/2010 for the course AMATH 261 taught by Professor Rogermelko during the Spring '10 term at Waterloo.
- Spring '10