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Kaplunovsky - practice 08

# Kaplunovsky - practice 08 - practice 08 Due noon Question 1...

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practice 08 – – Due: Mar 20 2007, noon 1 Question 1, chap -1, sect -1. part 1 of 3 8 points Two blocks of respective masses m 1 = 0 . 66 kg and m 2 = 3 . 9 kg stand without mo- tion on a frictionless horizontal table. A spring is compressed between the two blocks, but the blocks remain motionless because they are tied to each other by a cord. μ = 0 before m 1 m 2 The masses of the spring and the cord are neg- ligible compared to the masses of the blocks. Once the system is set up, the cord is burned by a match and the blocks are pushed apart by the spring. μ = 0 after m 1 m 2 v 1 v 2 Once the spring is completely relaxed, it de- taches from the m 1 block, and the blocks continue to move away from each other at respective speeds v 1 and v 2 . Given m 1 = 0 . 66 kg, m 2 = 3 . 9 kg, v 2 = 2 . 8 m / s, find the speed v 1 of the first block. Correct answer: 16 . 5455 m / s (tolerance ± 1 %). Explanation: The velocities of the two blocks are related by momentum conservation: There are no external horizontal forces on this system, so the net horizontal momentum is conserved. P net = 0 [before] = m 2 v 2 m 1 v 1 , [after] (the minus sing here is due to opposite direc- tions of the two blocks). Therefore, v 1 = m 2 m 1 × v 2 = 16 . 5455 m / s Question 2, chap -1, sect -1. part 2 of 3 6 points Find the net kinetic energy of the two blocks. Correct answer: 105 . 626 J (tolerance ± 1 %). Explanation: K = k 1 + k 2 = m 1 v 2 1 2 + m 2 v 2 2 2 = (0 . 66 kg)(16 . 5455 m / s) 2 2 + (3 . 9 kg)(2 . 8 m / s) 2 2 = 105 . 626 J . Question 3, chap -1, sect -1. part 3 of 3 6 points Given that the spring was initially com- pressed by 1 . 1 cm, find its spring constant k . Correct answer: 1 . 74589 × 10 6 N / m (tolerance ± 1 %). Explanation: The compressed spring had elastic energy U = k x 2 2 where x = 1 . 1 cm is the compression distance. After the spring its released, its energy con- verts into the kinetic energy of the two blocks U K = 105 . 626 J . Hence, before the spring was released, it had kx 2 2 = U = 105 . 626 J , which gives us k = 2 U x 2 = 1 . 74589 × 10 6 N / m .

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practice 08 – – Due: Mar 20 2007, noon 2 Question 4, chap 8, sect 1. part 1 of 1 10 points A spring with a spring-constant 3 . 9 N / cm is compressed 35 cm and released. The 9 kg mass skids down the frictional incline of height 44 cm and inclined at a 24 angle. The acceleration of gravity is 9 . 8 m / s 2 . The path is frictionless except for a dis- tance of 0 . 5 m along the incline which has a coefficient of friction of 0 . 3 . 9 kg 24 μ = 0 . 3 0 . 5 m 44 cm 35 cm k = 3 . 9 N / cm v f Figure: Not drawn to scale. What is the final velocity v f of the mass? Correct answer: 3 . 35358 m / s (tolerance ± 1 %). Explanation: Let : g = 9 . 8 m / s 2 = , k = 3 . 9 N / cm = 390 N / m , x = 35 cm = 0 . 35 m , μ = 0 . 3 , = 0 . 5 m , h = 0 . 44 m , m = 9 kg , and θ = 24 , Consider the kinetic energy of the mass. The mass receives its initial kinetic energy from the potential energy of the spring K i = U spring = 1 2 k x 2 (1) = 1 2 (390 N / m) (0 . 35 m) 2 = 23 . 8875 J .
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