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Unformatted text preview: baum (lmb2768) – HW06 – distler – (56295) 1 This printout should have 20 questions. Multiplechoice 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 xaxis, as shown. The 5 g object overtakes and collides with the 2 g object. Immediately after the collision, the ycomponent of the velocity of the 5 g object is 4 m / s upward. v i 5 g v i 2 g What is the ycomponent 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 ycomponent of the mo mentum must be the same before or after the collision. Since the two objects move horizon tally along the xaxis, the ycomponent of the momentum before the collision is zero. Thus, to make the ycomponent 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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 Spring '09
 IRENEPOLYCARPOU
 Physics, Mass, Momentum, kg, m/s

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