# Sol-Ch10 - Exercises 1(a x C M = 35(0.13 nm/36 = 0.126 nm...

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Unformatted text preview: Exercises 1. (a) x C M = 35(0.13 nm)/36 = 0.126 nm from H; (b) x C M = (0.2 nm)(cos 52.5°)/18 = 6.76 pm from 0 along the line of symmetry. 2. x C M = (–4 – 9 +15)/10 = 0.2 m; y C M = (6 + 12 – 5)/10 = 1.3 m 3. x C M = [30(40) + 20(120) + 10(200)]/60 = 93.3 cm 4. x C M = [4M(0) – M(R/2)]/3M = –R/6 (left of center) 5. Mass of plate = σ(4R 2 ), mass of hole = – σπ(R/2) 2 = – σπR 2 /4. x C M = [4R 2 (0) – (πR 2 /4)(R/2)]/(4 – π/4) σR 2 = –0.122 R = y C M 6. x C M = [0 – ρ 4πr 3 d/3]/(R 3 – r 3 ) 4π ρ = –r 3 d/(R 3 – r 3 ) 7. Square m 1 = σL 2 at x 1 = L/2, y 1 = L/2 ; Triangle m 2 = σL 2 (3)1/2/4 at x 2 = L/2, y 2 = L +L/2(3) 1/2 x C M = L/2; y C M = 0.738 L 8. (a) x = [4(d/2) + 1(1.5d) + 1(2.5d)]/6 = d = 2cm; y = [3(d/2) + 1.5d + 2.5d +3.5d]/6 = 3d/2 = 3 cm; (b) x = [4(d/2) – 2(d/2) – 1.5d + 1.5d)]/ 8 = d/8 = 0.25 cm, y = [4(d/2) + 1.5d + 2.5d + 2(3.5d)]/8 = 13d/8 = 3.25 cm 9. x C M = [M E (0) + M m r]/(M E + M m ) = 4680 km from earth’s center 10. Canceling 4π ρ /3 : x = [0 + R 3 (3R)/(8R 3 + R 3 )]= R/3 11. (a) m 1 v 1 Δt = m 2 u 2 Δt, so (60 kg) (1.5 m) = 75d, so d = 1.2 m. Separation = 5 – (1.5 + 1.2) = 2.3 m; (b) x C M = 60(5)/135 = 2.22 m. 12. x C M = 0 = (20) (–3 × 10 –4 ) + m(0.18), thus m =33.3 g 13. (a) x C M = [(1) (L/2) + (1)(L/4) + (1)(3L/4)]/3 = L/2. y C M = [0 + 2(1/2)(0.866 L)]/3 = 0.29 L (b) x C M = [(1) (L/2) + 0 + (2) 1/2 (l/2)]/[(2 + 2 1/2 )] = 0.354 L CHAPTER 10 14. (a) 8v C M = (10i – 6 j + 8k ) + (–18i + 12 j –6k ), thus v C M = –i + 0.75 j + 0.25k m/s; (b) P = Mv C M = –8i + 6 j + 2k kg ‧ m/s 15. m collides with the box. ∑ p :–mu = 4mv 1 , so v 1 = –u/4....
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Sol-Ch10 - Exercises 1(a x C M = 35(0.13 nm/36 = 0.126 nm...

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