Exam 1 Spring 2006 Solutions

Exam 1 Spring 2006 Solutions - Physics 221 2006 S Exam 1...

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Physics 221 2006 S Exam 1 Solutions Page 1 of 27 Answer Key 1 Solutions 2 Questions 11 Answer Key 1 B 11 E 21 C 2 E 12 E 22 A 3 E 13 A 23 C 4 C 14 D 24 B 5 D 15 A 25 B 6 D 16 C 26 A 7 B 17 A 27 C 8 A 18 E 28 C 9 C 19 C 29 D 10 D 20 B 30 B
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Physics 221 2006 S Exam 1 Solutions Page 2 of 27 [1](B) The weight is W=mg. First calculate the mass from the data given: 22 2 8.5 in 11 in 2.54 cm 1 m g 1 (500 sheets) 75 sheet 1 in 100 cm m 1000 =2.26kg  × =   kg m g The weight is therefore 2 (2.26 kg)(9.81 ) 22.2 m s Wm g N == = [2](E) The total mass of the system is 6kg. The acceleration of all the blocks is thus equal to (6N)/(6kg)=1m/s². This is therefore the acceleration of block P. [3](E) The net force accelerating P at the above acceleration is T 1 . The net force accelerating the PQ system is T 2 . All blocks are undergoing the same acceleration so T 1 :T 2 =m P a:(m P +m Q )a=m P :(m P +m Q )=1:3. [4](C) By the work energy theorem , the net work done on the block is the change in kinetic energy. Since the speed does not change , the kinetic energy does not change and so the net work is 0. [5](D) Both the circle and the rectangle are moving to the right relative to the background. Around t = 2s, the spacing between the circles is smaller than the spacing between the rectangles, which means that the velocities of the circle and the rectangle relative to the background are something like this: Therefore, the velocity of the circle relative to the rectangle, C,R C,bg R,bg = G GG vvv , points to the left. The same is true at three seconds (even more so). Thus at both times the relative velocity is to the left. Also, we can look at the position of the circle relative to the rectangle at different times. At t = 1 s, the circle is quite ahead of the rectangle. At t = 2 s, still ahead but not so much. At t = 3, they are leveled. At t = 4 s, the circle is behind the rectangle. So from the point of view of the rectangle, the circle is moving left. v C,background v R,background
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Physics 221 2006 S Exam 1 Solutions Page 3 of 27 [6](D) The angle between the two vectors is (15º+90º+15º)=120º. The dot product is 1 2 cos 4 3 ( ) 6 AB AB θ == × × = G G i [7](B) The magnitude of D is: 222 15 16 12 25 D =+ + = Thus () 1 25 ˆˆ ˆ ˆ ˆ 15 16 12 0.60 0.64 0.48 D Di j k i j k D ==− + = + G [8](A) Since Sally is moving at a constant speed and not accelerating, the net force on her is 0. [9](C) We will apply the work-kinetic energy theorem twice, in the two situations where a force is doing work. The kinetic energy of the block after it leaves the spring is just the work that the spring does on the block: 2 1 2 spring KW kx The (negative) work done on the block by gravity must be equal to that kinetic energy if the block is brought momentarily to rest. If h is the height of the ramp then, on the one hand: ˆ ( ) gravity ramp W W r mgj x i hj mgh =∆ = += G G ii while on the other hand 2 1 2 2 2 gravity WK mgh k x mgh k x = − = Plugging in the numbers: 2 22 2(0.1 )(9.81 )(1 ) 2 196 (0.1 ) m s kg m mgh N k x mm =
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Physics 221 2006 S Exam 1 Solutions Page 4 of 27 [10](D) The change in velocity is the integral of acceleration as a function of time.
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Exam 1 Spring 2006 Solutions - Physics 221 2006 S Exam 1...

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