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HW4_Sol_F10

# HW4_Sol_F10 - Sept 20 2010 PHYS 2101 HW#4 WileyPlus Problem...

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Sept. 20, 2010 PHYS 2101 HW#4 WileyPlus Problem Solutions 1 HW #4 Solutions – or at least most of them… (USING BOOK VALUES 9 th Edition in problems; problem # indicated by []) 1) [9-1] . We use Eq. 9-5 to solve for (a) The x coordinate of the system’s center of mass is: Solving the equation yields x 3 = –1.50 m. (b) The y coordinate of the system’s center of mass is: Solving the equation yields y 3 = –1.43 m. 2) [9-5] . Since the plate is uniform, we can split it up into three rectangular pieces, with the mass of each piece being proportional to its area and its center of mass being at its geometric center. We’ll refer to the large 35 cm 10 cm piece (shown to the left of the y axis in Fig. 9-38) as section 1; it has 63.6% of the total area and its center of mass is at ( x 1 ,y 1 ) = ( 5.0 cm, 2.5 cm). The top 20 cm 5 cm piece (section 2, in the first quadrant) has 18.2% of the total area; its center of mass is at ( x 2 , y 2 ) = (10 cm, 12.5 cm). The bottom 10 cm x 10 cm piece (section 3) also has 18.2% of the total area; its center of mass is at ( x 3 , y 3 ) = (5 cm, 15 cm). (a) The x coordinate of the center of mass for the plate is x com = (0.636) x 1 + (0.182) x 2 + (0.182) x 3 = – 0.45 cm . (b) The y coordinate of the center of mass for the plate is y com = (0.636) y 1 + (0.182) y 2 + (0.182) y 3 = – 2.0 cm . 3) [9-14] (a) The phrase (in the problem statement) “such that it [ particle 2 ] always stays directly above particle 1 during the flight” means that the shadow (as if a light were directly above the particles shining down on them) of particle 2 coincides with the position of particle 1, at each moment. We say, in this case, that they are vertically aligned. Because of that alignment, v 2 x = v 1 = 10.0 m/s. Because the initial value of v 2 is given as 20.0 m/s, then (using the Pythagorean theorem) we must have = 300 m/s for the initial value of the y component of particle 2’s velocity. Equation 2-16 (or conservation of energy) readily yields y max = 300/19.6 = 15.3 m. Thus, we obtain H max = m 2 y max / m total = (3.00 g)(15.3 m)/(8.00 g) = 5.74 m. (b) Since both particles have the same horizontal velocity, and particle 2’s vertical component of velocity vanishes at that highest point, then the center of mass velocity then is simply (as one can verify using Eq. 9-17). (c) Only particle 2 experiences any acceleration (the free fall acceleration downward), so Eq. 9-18 (or Eq. 9-19) leads to a com = m 2 g / m total = (3.00 g)(9.8 m/s 2 )/(8.00 g) = 3.68 m/s 2

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Sept. 20, 2010 PHYS 2101 HW#4 WileyPlus Problem Solutions 2 for the magnitude of the downward acceleration of the center of mass of this system. Thus, .
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