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HW_05_S

# HW_05_S - Ph1a Solutions 5 Chefung Chan([email protected]

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Ph1a Solutions 5 Chefung Chan ([email protected]), Fall 2004 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 0 p 2 i = 0 p p 1 i + p 2 i = mv 0 K 1 i = 1 2 mv 2 0 K 2 i = 0 K K 1 i + K 2 i = 1 2 mv 2 0 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 0 K 1 f + K 2 f = K 1 2 mv 2 1 f + mv 2 2 f = 1 2 mv 2 0 Solving these equations yields v 1 f = - 1 3 v 0 and v 2 f = + 2 3 v 0 . 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 0 E 2 f = mv 2 2 f = 4 9 mv 2 0 1

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Apply conservation of energy to each mass, noting that v 2 0 = 2 gd : E 1 f = 1 18 mv 2 0 = mgh 1 E 2 f = 4 9 mv 2 0 = 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. To find the height of the center of mass, apply the definition again: y CM = i m i y i i m i = m 1 h 1 + m 2 h 2 m 1 + m 2 = 1 9 md + 4 9 (2 m ) d 3 m = 1 3 d.
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HW_05_S - Ph1a Solutions 5 Chefung Chan([email protected]

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