HW06-solutions - baum (lmb2768) HW06 distler (56295) 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

Page1 / 9

HW06-solutions - baum (lmb2768) HW06 distler (56295) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online