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exam2_2007s_sol - Physics 221 2007S Exam 2 SOLUTIONS[28...

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Physics 221 2007S Exam 2 SOLUTIONS Page 1 of 24 [28] Assume that the moon can be modeled as a solid sphere of uniform density. The mass of the moon is 22 7.4 10 kg × and the radius of the moon is 1800 km. Which is the best estimate of the moment of inertia of the moon about an axis through the moon’s center of mass? (A) 35 2 2.4 10 kg m × (B) 35 2 1.6 10 kg m × (C) 35 2 1.2 10 kg m × (D) 34 2 9.6 10 kg m × (E) 34 2 4.8 10 kg m × Answer [D]: For a solid sphere of uniform density, 2 2 5 I MR = , so plugging in the numbers from the problem: 2 22 2 34 2 2 5 0.4(7.4 10 kg)(1800000m) 9.6 10 kg m I MR = = × = × [29] Three objects of mass m are dropped from a height h . One falls straight down, one slides down a frictionless incline and one swings at the end of a pendulum. What is the relationship between their speeds when they reach the ground? Neglect air resistance. h v F v I v P Fall Incline Pendulum
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Physics 221 2007S Exam 2 SOLUTIONS Page 2 of 24 [30] A truck with mass 2000kg is driving along a road. It collides with a stationary car with mass 1000kg. The two vehicles fuse together and the moment after the collision, the fused wreckage of the two vehicles is moving at a speed of 10m/s. What was the initial speed of the truck? ( ) 2000 kg+1000 kg 10 2000 kg m 15 s i truck i f truck car f truck car i f truck p m v p m m v m m v v m m s = = = + + = = = [31] ] Particle B has twice the mass of particle A and particle B is moving with linear momentum twice as big as particle A. What is the ratio: Kinetic energy of B : Kinetic energy of A? 2 2 p K m = Thus 2 2 2 2 2 2 : : : 2 2 (2 ) : (2 ) 4: 2 2:1 B A B A B A A B B A A A A p p K K p m p m m m p p m = = = = =
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Physics 221 2007S Exam 2 SOLUTIONS Page 3 of 24 [32] Consider the assembly depicted below of nine 1m×1m tiles. Six of the tiles have mass 1kg while the other three have mass 2kg as shown. Using the axes in the figure, what is the location of the center of mass for the system? (A) (2.2m , 1.5m) (B) (1,5m , 2.2m) (C) (1.5m , 1.5m) (D) (1.5m , 1.7m) (E) (1.7m , 1.5m) Answer [E]: The total mass of the system is 12 kg. The simplest way to approach the problem is do divide it up into three pieces where the CM is obvious and then combine those results into a grand center of mass. The division is shown in red. The grand center of mass for the whole system is thus ˆ ˆ ˆ ˆ ˆ ˆ (4kg)(1.5m 2.5m ) (4kg)(1.0m 1.0m ) (4kg)(2.5m 1.0m ) (4kg) (4kg) (4kg) ˆ ˆ (1.7m 1.5m ) A A B B C C cm A B C m r m r m r r m m m i j i j i j i j + + = + + + + + + + = + + = + G G G G Another way of simplifying the problem is by noting that the y component of the CM has to be in the middle ( y CM =1.5 m) because the system is made of three horizontal slices of
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