HW06-solutions

# HW06-solutions - baum(lmb2768 – HW06 – distler...

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Unformatted text preview: baum (lmb2768) – HW06 – distler – (56295) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two objects of mass 5 g and 2 g , respectively, move parallel to the x-axis, as shown. The 5 g object overtakes and collides with the 2 g object. Immediately after the collision, the y-component of the velocity of the 5 g object is 4 m / s upward. v i 5 g v i 2 g What is the y-component of the velocity v of the 2 g object immediately after the collision? 1. v y = 5 . 33333 m / s downward 2. v y = 10 . 5 m / s downward 3. v y = 8 m / s downward 4. v y = 5 m/s upward 5. v y = 22 . 5 m / s downward 6. v y = 10 m / s downward correct 7. v y = 7 m/s upward 8. v y = 4 . 5 m / s downward Explanation: Let : m 1 = 5 g , m 2 = 2 g , and v 1 y = 4 m / s . Momentum p is conserved for collision m 1 v 1 = m 2 v 2 . In particular, the y-component of the mo- mentum must be the same before or after the collision. Since the two objects move horizon- tally along the x-axis, the y-component of the momentum before the collision is zero. Thus, to make the y-component of the momentum after the collision zero, the y- component of the velocity is 0 = m 1 v 1 y + m 2 v 2 y v 2 y =- m 1 m 2 v 1 y =- (5 g) (2 g) (4 m / s) =- 10 m / s , and the direction is downward. keywords: 002 (part 1 of 4) 10.0 points A 29 kg gun is standing on a frictionless sur- face. The gun fires a 57 g bullet with a muzzle velocity of 315 m / s. The positive direction is that of the bullet. Calculate the momentum of the bullet im- mediately after the gun was fired. Correct answer: 17 . 955 kg · m / s. Explanation: Let : m b = 57 g and v b = 315 m / s . The momentum of the bullet is p b = m b v b = (57 g) (315 m / s) = 17 . 955 kg · m / s . 003 (part 2 of 4) 10.0 points Calculate the momentum of the gun immedi- ately after the gun was fired. Correct answer:- 17 . 955 kg · m / s. Explanation: baum (lmb2768) – HW06 – distler – (56295) 2 By conservation of momentum, p bo + p go = p bf + p gf 0 + 0 = p b + p g p g =- p b =- 17 . 955 kg · m / s . 004 (part 3 of 4) 10.0 points Calculate the kinetic energy of the bullet im- mediately after the gun was fired. Correct answer: 2827 . 91 J. Explanation: The kinetic energy of the bullet is K b = 1 2 m b v 2 b = 1 2 (57 g) (315 m / s) 2 = 2827 . 91 J . 005 (part 4 of 4) 10.0 points Calculate the kinetic energy of the gun imme- diately after the gun was fired. Correct answer: 5 . 55831 J. Explanation: Let : m g = 29 kg . The recoil velocity of the gun comes from its momentum: p g = m g v g v g = p g m g Thus the kinetic energy of the gun is K g = 1 2 m g v 2 g = p 2 g 2 m g = (- 17 . 955 kg · m / s) 2 2 (29 kg) = 5 . 55831 J ....
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HW06-solutions - baum(lmb2768 – HW06 – distler...

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