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Homework 12 Solutions

# Homework 12 Solutions - homework 12 ALIBHAI ZAHID Due 4:00...

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homework 12 – ALIBHAI, ZAHID – Due: Apr 18 2007, 4:00 am 1 Question 1 part 1 of 1 10 points A block of mass 4 kg is hung from a spring, causing it to stretch 8 cm at equilibrium, as shown below. The 4 kg block is then replaced by a 6 kg block, and the new block is released from the position shown below, at which point the spring is unstretched. The acceleration of gravity is 9 . 8 m / s 2 . 8 cm y max 4 kg equilbrium 6 kg How far will the 6 kg block fall before its direction is reversed? Hint: For the new block case, the total distance of fall is twice the amplitude of the oscillation. Correct answer: 24 cm (tolerance ± 1 %). Explanation: Let : m 4 = 4 kg , m 6 = 6 kg , y 4 = 8 cm , and g = 9 . 8 m / s 2 . When the 4 kg block is replaced by 6 kg block, the equilibrium point is y 6 = z 4 m 4 m 6 = (8 cm) (4 kg) (6 kg) = 12 cm . below the unstretched point. However, the point at which the block re- verses its direction is at its maximum am- plitude in the downward direction, which is two times the amplitude from its equilibrium point y amplitude = 12 cm . Thus, the block will go y max = 2 (12 cm) = 24 cm , before its direction is reversed. Alternative solution: The equilibrium position for the 4 kg is 8 cm, as given. The maxium stretched position for 6 kg after re- leased from rest is y max . The spring constant is k = m 4 g x 4 . Conservation of energy implies that m 6 g y max = 1 2 k y 2 max so y max = 2 m 6 g k = 2 m 6 m 4 x 4 = 2 (6 kg) (4 kg) (8 cm) = 24 cm . Question 2 part 1 of 1 10 points Quantum concepts are critical in explaining all of the following EXCEPT 1. Compton scattering. 2. Bohr’s theory of the hydrogen atom. 3. Rutherfold’s scattering experiments. correct 4. the blackbody spectrum. 5. the photoelectric effect. Explanation: Rutherfold’s scattering experiments re- vealed that the atoms consist of a positively

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homework 12 – ALIBHAI, ZAHID – Due: Apr 18 2007, 4:00 am 2 charged heavy nucleus and surrounding light electrons. Classical electromagnetic theory can explain the results without referring to quantum concepts. In all of the other four choices, quantum concepts are necessary to explain the phe- nomena. Question 3 part 1 of 2 10 points A 1 . 24 kg block at rest on a tabletop is at- tached to a horizontal spring having constant 16 . 1 N / m. The spring is initially unstretched. A constant 27 . 7 N horizontal force is applied to the object causing the spring to stretch. The acceleration of gravity is 9 . 8 m / s 2 . 27 . 7 N 1 . 24 kg 16 . 1 N / m Find the speed of the block after it has moved 0 . 26 m from equilibrium if the surface between block and tabletop is frictionless. Correct answer: 3 . 27695 m / s (tolerance ± 1 %). Explanation: Given : m = 1 . 24 kg , x f = 0 . 26 m , k = 16 . 1 N / m , and F = 27 . 7 N . Applying the work kinetic energy theorem, F x f = 1 2 m v 2 f + 1 2 k x 2 f m v 2 f = 2 F x f k x 2 f v 2 f = 2 F x f k x 2 f m = 2 (27 . 7 N) (0 . 26 m) 1 . 24 kg (16 . 1 N / m) (0 . 26 m) 2 1 . 24 kg = 10 . 7384 m 2 / s 2 so that v f = radicalBig 10 . 7384 m 2 / s 2 = 3 . 27695 m / s .
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Homework 12 Solutions - homework 12 ALIBHAI ZAHID Due 4:00...

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