hw5_sol_s11

# hw5_sol_s11 - Phys 2101 Homework 5 Solution Spring `11 The...

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Phys 2101 Homework 5 Solution Spring `11 The numerical answers to the problems are calculated with the values used in the problems found in the printed textbook; the values for your particular WebAssign assignment will be different. 1.P9-1.We use Eq. 9-5 to solve for 33 ( , ). xy (a) The x coordinate of the system s center of mass is:      3 1 1 2 2 3 3 com 1 2 3 (2.00 kg)( 1.20 m) 4.00 kg 0.600 m 3.00 kg 2.00 kg 4.00 kg 3.00 kg 0.500 m. x m x m x m x x m m m    Solving the equation yields x 3 = 1.50 m. (b) The y coordinate of the system s center of mass is:      3 1 1 2 2 3 3 com 1 2 3 (2.00 kg)(0.500 m) 4.00 kg 0.750 m 3.00 kg 2.00 kg 4.00 kg 3.00 kg 0.700 m. y m y m y m y y m m m Solving the equation yields y 3 = 1.43 m. 2. P9-4. We will refer to the arrangement as a “table.” We locate the coordinate origin at the left end of the tabletop (as shown in Fig. 9-37). With + x rightward and + y upward, then the center of mass of the right leg is at ( x,y ) = (+ L , L /2), the center of mass of the left leg is at ( x,y ) = (0, L /2), and the center of mass of the tabletop is at ( x,y ) = ( L /2, 0). (a) The x coordinate of the (whole table) center of mass is       com 0 3 / 2 32 M L M M L L x M M M . With L = 22 cm, we have x com = (22 cm)/2 = 11 cm. (b) The y coordinate of the (whole table) center of mass is       com / 2 / 2 3 0 35 M L M L M L y M M M   , or y com = (22 cm)/5 = 4.4 cm.

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From the coordinates, we see that the whole table center of mass is a small distance 4.4 cm directly below the middle of the tabletop. 3. We use the constant-acceleration equations of Table 2-1 (with + y downward and the origin at the release point), Eq. 9-5 for y com and Eq. 9-17 for v com . (a) The location of the first stone (of mass m 1 ) at t = 300 10 3 s is y 1 = (1/2) gt 2 = (1/2)(9.8 m/s 2 ) (300 10 3 s) 2 = 0.44 m, and the location of the second stone (of mass m 2 = 2 m 1 ) at t = 300 10 3 s is y 2 = (1/2) gt 2 = (1/2)(9.8 m/s 2 )(300 10 3 s 100 10 3 s) 2 = 0.20 m. Thus, the center of mass is at m m m m m m m m m y m y m com y 28 . 0 1 2 1 ) 20 . 0 ( 1 2 ) 44 . 0 ( 1 2 1 2 2 1 1 (b) The speed of the first stone at time t is v 1 = gt , while that of the second stone is v 2 = g ( t 100 10 3 s). Thus, the center-of-mass speed at t = 300 10 3 s is       2 3 2 3 3 11 1 1 2 2 com 1 2 1 1 9.8 m/s 300 10 s 2 9.8 m/s 300 10 s 100 10 s 2 2.3 m/s. mm m v m v v m m m m   4. (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.
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## hw5_sol_s11 - Phys 2101 Homework 5 Solution Spring `11 The...

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