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Physics 303k homework 8

# Physics 303k homework 8 - ang(nza67 – HW#8 –...

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Unformatted text preview: ang (nza67) – 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 = 78 g particle is located at ( x 1 , y 1 ), where x 1 = 4 . 6 m and y 1 = 9 . 1 m and a m 2 = 72 g particle is located at ( x 2 , y 2 ), where x 2 =- 9 . 4 m and y 2 =- 5 . 5 m. What must be the x coordinate of the m 3 = 38 g particle so that the center of mass of the three-particle system is at the origin? Correct answer: 8 . 36842 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 = 8 . 36842 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 90 N 20 N 80 N 80 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: 1 . 09259 a . Explanation: Let : x 1 = x 2 = 1 2 a , x 3 = x 4 = 3 2 a , W 1 = 90 N , W 2 = 20 N , W 3 = 80 N , and W 4 = 80 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 = 270 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 = (90 N + 20 N) + 3 (80 N + 80 N) 2 (270 N) a = 1 . 09259 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 . 87037 a . Explanation: Let : y 1 = y 4 = 1 2 a and y 2 = y 3 = 3 2 a . ang (nza67) – 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 = (90 N + 80 N) + 3 (20 N + 80 N) 2 (270 N) a = . 87037 a . 004 (part 1 of 2) 5.0 points A(n) 3 . 8 kg object moving with a speed of 6 . 8 m / s collides with a(n) 3 . 6 kg object mov- ing with a velocity of 10 . 4 m / s in a direction 61 . 7788 ◦ from the initial direction of motion of the 3 . 8 kg object. 6 . 8 m / s 1 . 4 m / s 3 . 6 kg 3 . 8 kg 61 . 7788 ◦ What is the speed of the two objects after the collision if they remain stuck together? Correct answer: 7 . 38242 m / s. Explanation: Let : m 1 = 3 . 8 kg , m 2 = 3 . 6 kg , m f = m 1 + m 2 = 7 . 4 kg , v 1 = 6 . 8 m / s , v 2 = 10 . 4 m / s , p 1 = m 1 v 1 = 25 . 84 kg m / s , p 2 = m 2 v 2 = 37 . 44 kg m / s , 2 p 1 p 2 = 2 m 1 v 1 m 2 v 2 = 1934 . 9 kg 2 m 2 / s 2 , p x = p 1 + p 2 cos θ = 43 . 5445 kg m / s , p y = p 2 sin θ = 32 . 9895 kg m / s , θ = 61 . 7788 ◦ , and π- θ = 118 . 221 ◦ ....
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Physics 303k homework 8 - ang(nza67 – HW#8 –...

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