# HW 9 - johnson(mjj622 – HW 9 – Coker –(56625 1 This...

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Unformatted text preview: johnson (mjj622) – HW 9 – Coker – (56625) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This assignment covers Chapter 10 mate- rial: momentum, the center of mass, system momentum and kinetic energy, generalized 2nd Law. 001 10.0 points What is the momentum of a two-particle sys- tem composed of a 1000 kg car moving east at 90 m / s and a second 900 kg car moving west at 35 m / s? Let east be the positive direction. Correct answer: 58500 kg · m / s. Explanation: Let : m 1 = 1000 kg , v 1 = 90 m / s , m 2 = 900 kg , and v 2 = 35 m / s . Thus p = m 1 v 1 + m 2 v 2 = (1000 kg)(90 m / s) + (900 kg)( − 35 m / s) = 58500 kg · m / s . 002 10.0 points Bill (mass m ) plants both feet solidly on the ground and then jumps straight up with ve- locity → v . The earth (mass M ) then has velocity 1. V Earth = − radicalbigg m M → v man . 2. V Earth = + parenleftBig m M parenrightBig → v man . 3. V Earth = + radicalbigg m M → v man . 4. V Earth = − → v man . 5. V Earth = + parenleftbigg M m parenrightbigg → v man . 6. V Earth = − parenleftBig m M parenrightBig → v man . correct 7. V Earth = − parenleftbigg M m parenrightbigg → v man . 8. V Earth = + → v man . Explanation: The momentum is conserved. We have m → v man + M → V Earth = 0 So → V Earth = − parenleftBig m M parenrightBig → v man . 003 10.0 points The mass of the Earth is 5 . 98 × 10 24 kg . A 7 . 64 kg bowling ball initially at rest is dropped from a height of 2 . 84 m. The acceleration of gravity is 9 . 8 m / s 2 . What is the speed of the Earth coming up to meet the ball just before the ball hits the ground? Correct answer: 9 . 5319 × 10- 24 m / s. Explanation: The speed of the ball just before impact is v = radicalbig 2 g h = 7 . 46083 m / s . From conservation of momentum Δ p = 0, or M e V + m b v = 0 . Therefore bardbl vector V bardbl = v m b M e = (7 . 46083 m / s) (7 . 64 kg) (5 . 98 × 10 24 kg) = 9 . 5319 × 10- 24 m / s . 004 10.0 points Three point mass particles are located in a plane: 4 . 27 kg located at the origin, 7 . 8995 kg at [(5 . 77 cm) , (11 . 54 cm)], and 2 . 23321 kg at [(12 . 117 cm) , (0 cm)]. johnson (mjj622) – HW 9 – Coker – (56625) 2 How far is the center of mass of the three particles from the origin? Correct answer: 8 . 09307 cm. Explanation: Basic Concepts: Center of mass vectorr CM = ∑ i m i vectorr i ∑ i m i . Solution: The center of mass ( x CM , y CM ) is defined as x CM = ∑ i m i x i ∑ i m i = m 1 x 1 + m 2 x 2 m + m 1 + m 2 = (7 . 8995 kg) (5 . 77 cm) + (2 . 23321 kg) (12 . 117 cm) (4 . 27 kg) + (7 . 8995 kg) + (2 . 23321 kg) = 5 . 04349 cm y CM = ∑ i m i y i ∑ i m i = m 1 y 1 + m 2 y 2 m + m 1 + m 2 = (7 . 8995 kg) (11 . 54 cm) + (2 . 23321 kg) (0 cm) (4 . 27 kg) + (7 . 8995 kg) + (2 . 23321 kg) = 6 . 32938 cm and r CM = radicalBig x 2 CM + y 2 CM = radicalBig (5 . 04349 cm) 2 + (6 . 32938 cm) 2 = 8...
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HW 9 - johnson(mjj622 – HW 9 – Coker –(56625 1 This...

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