HW #8-solutions - wolz (cmw2833) HW #8 Antoniewicz (57420)...

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Unformatted text preview: wolz (cmw2833) HW #8 Antoniewicz (57420) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Three particles are placed in the xy plane. A m 1 = 36 g particle is located at ( x 1 , y 1 ), where x 1 = 3 . 1 m and y 1 = 4 . 1 m and a m 2 = 46 g particle is located at ( x 2 , y 2 ), where x 2 =- 7 . 7 m and y 2 =- 7 . 5 m. What must be the x coordinate of the m 3 = 25 g particle so that the center of mass of the three-particle system is at the origin? Correct answer: 9 . 704 m. Explanation: We know that the x coordinate of the center of mass is defined by x CM = m 1 x 1 + m 2 x 2 + m 3 x 3 m 1 + m 2 + m 3 Now, x CM is zero, and using the input values for m 1 , m 2 , m 3 and x 1 and x 2 , we find and solving for x 3 : x 3 =- m 1 x 1- m 2 x 2 m 3 = 9 . 704 m 002 (part 1 of 2) 5.0 points A square plate is produced by welding to- gether four smaller square plates, each of side a . The weight of each of the four plates is shown in the figure. x y 60 N 40 N 40 N 30 N (0 , 0) (2 a, 0) (0 , 2 a ) (2 a, 2 a ) Find the x-coordinate of the center of grav- ity (as a multiple of a ). Correct answer: 0 . 911765 a . Explanation: Let : x 1 = x 2 = 1 2 a , x 3 = x 4 = 3 2 a , W 1 = 60 N , W 2 = 40 N , W 3 = 40 N , and W 4 = 30 N . y x W 1 W 2 W 3 W 4 a 2 3 a 2 a 2 3 a 2 The total weight is W = W 1 + W 2 + W 3 + W 4 = 170 N . Applying the definition of center of gravity, x cg = i W i x i W 1 + W 2 + W 3 + W 4 = ( W 1 + W 2 ) a 2 + ( W 3 + W 4 ) 3 a 2 W = (60 N + 40 N) + 3 (40 N + 30 N) 2 (170 N) a = . 911765 a . 003 (part 2 of 2) 5.0 points Find the y-coordinate of the center of gravity (as a multiple of a ). Correct answer: 0 . 970588 a . Explanation: Let : y 1 = y 4 = 1 2 a and y 2 = y 3 = 3 2 a . wolz (cmw2833) HW #8 Antoniewicz (57420) 2 y cg = i W i y i W 1 + W 2 + W 3 + W 4 = ( W 1 + W 4 ) a 2 + ( W 2 + W 3 ) 3 a 2 W = (60 N + 30 N) + 3 (40 N + 40 N) 2 (170 N) a = . 970588 a . 004 (part 1 of 2) 5.0 points A(n) 2 . 3 kg object moving with a speed of 8 . 6 m / s collides with a(n) 4 kg object mov- ing with a velocity of 8 . 8 m / s in a direction 39 . 0135 from the initial direction of motion of the 2 . 3 kg object. 8 . 6 m / s 8 . 8 m / s 4 kg 2 . 3 kg 39 . 0135 What is the speed of the two objects after the collision if they remain stuck together? Correct answer: 8 . 26658 m / s. Explanation: Let : m 1 = 2 . 3 kg , m 2 = 4 kg , m f = m 1 + m 2 = 6 . 3 kg , v 1 = 8 . 6 m / s , v 2 = 8 . 8 m / s , p 1 = m 1 v 1 = 19 . 78 kg m / s , p 2 = m 2 v 2 = 35 . 2 kg m / s , 2 p 1 p 2 = 2 m 1 v 1 m 2 v 2 = 1392 . 51 kg 2 m 2 / s 2 , p x = p 1 + p 2 cos = 47 . 1303 kg m / s , p y = p 2 sin = 22 . 1585 kg m / s , = 39 . 0135 , and - = 140 . 987 ....
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HW #8-solutions - wolz (cmw2833) HW #8 Antoniewicz (57420)...

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