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Unformatted text preview: Ph1a Solutions 5 Chefung Chan ([email protected]), Fall 2007 Each homework problem is worth 5 points. Please disregard the point values listed on the problem itself. Use these instead. 5.1 QP14 (5 points) 5.1.a (1 point) Use the definition of the center of mass: y CM = ∑ i m i y i ∑ i m i = m 1 y 1 + m 2 y 2 m 1 + m 2 = md + (2 m )(0) m + 2 m = 1 3 d. 5.1.b (1 point) Use conservation of energy to find v 1 i by equating the energies at the top and bottom of the arc: E i = mgd = 1 2 mv 2 1 i = E f ⇒ v 1 i = √ 2 gd = v. The lower mass m 2 is at rest, so v 2 i = 0. Momentum is conserved in the collision. Because the collision is elastic the kinetic energy is also conserved. The initial momenta and energies are p 1 i = mv p 2 i = p ≡ p 1 i + p 2 i = mv K 1 i = 1 2 mv 2 K 2 i = K ≡ K 1 i + K 2 i = 1 2 mv 2 The final momentum of the two pendula are given by p 1 f = mv 1 f and p 2 f = 2 mv 2 f and the final kinetic energies by K 1 f = 1 2 mv 2 1 f and K 2 f = mv 2 2 f . Applying the conservations laws: p 1 f + p 2 f = p mv 1 f + 2 mv 2 f = mv K 1 f + K 2 f = K 1 2 mv 2 1 f + mv 2 2 f = 1 2 mv 2 Solving these equations yields v 1 f = 1 3 v and v 2 f = + 2 3 v . 5.1.c (2 points) The energies of the masses after the collision are E 1 f = 1 2 mv 2 1 f = 1 18 mv 2 E 2 f = mv 2 2 f = 4 9 mv 2 1 Apply conservation of energy to each mass, noting that v 2 = 2 gd : E 1 f = 1 18 mv 2 = mgh 1 E 2 f = 4 9 mv 2 = 2 mgh 2 h 1 = ( 1 3 v ) 2 2 g = 1 9 d h 2 = ( 2 3 v ) 2 2 g = 4 9 d....
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This note was uploaded on 06/05/2008 for the course PH 1a taught by Professor Goodstein during the Fall '07 term at Caltech.
 Fall '07
 Goodstein
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