HW 7 solution_pdf - ghei(mg42382 HW 7 opyrchal(11130 This...

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

View Full Document Right Arrow Icon
ghei (mg42382) – HW 7 – opyrchal – (11130) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A child bounces a 45 g superball on the side- walk. The velocity change of the superball is from 23 m / s downward to 14 m / s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk? Correct answer: 1332 N. Explanation: Let : m = 45 g = 0 . 045 kg , v u = 14 m / s , v d = 23 m / s , and Δ t = 0 . 00125 s Choose the upward direction as positive. The impulse is I = F Δ t = Δ P = m v u - m ( - v d ) = m ( v u + v d ) F = m ( v u + v d ) Δ t = (45 g) (14 m / s + 23 m / s) 0 . 00125 s = 1332 N . 002 10.0 points A 86 . 4 kg ice skater, moving at 9 . 1 m / s, crashes into a stationary skater of equal mass. After the collision, the two skaters move as a unit at 4 . 55 m / s. Suppose the average force a skater can experience without breaking a bone is 4632 N. If the impact time is 0 . 103 s, what is the magnitude of the average force each skater experiences? Correct answer: 3816 . 7 N. Explanation: Let : m = 86 . 4 kg , v i = 9 . 1 m / s , v f = 4 . 55 m / s , F = 4632 N , and Δ t = 0 . 103 s . From the impulse-momentum equation, F av Δ t = Δ p = m ( v f - v i ) . F av = m ( v f - v i ) Δ t = (86 . 4 kg) (4 . 55 m / s - 9 . 1 m / s) 0 . 103 s = - 3816 . 7 N , which has a magnitude of 3816 . 7 N . The average force on skater 2 has the same magni- tude but opposite direction by Newton’s third law. This force is not great enough to break a bone. 003 10.0 points A(n) 810 N man stands in the middle of a frozen pond of radius 7 . 9 m. He is un- able to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1 . 4 kg physics textbook horizontally toward the north shore, at a speed of 7 m / s. The acceleration of gravity is 9 . 81 m / s 2 . How long does it take him to reach the south shore? Correct answer: 66 . 5606 s. Explanation: Let : W m = 810 N , r = 7 . 9 m , m b = 1 . 4 kg , and v b = 7 m / s . The mass of the man is m m = W m g .
Image of page 1

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

View Full Document Right Arrow Icon
ghei (mg42382) – HW 7 – opyrchal – (11130) 2 From conservation of momentum, m m v m + m b v b = m m v m + m b v b 0 = m m v m + m b v b v m = - m b m m v b = - g m b W m v b = - ( 9 . 81 m / s 2 ) (1 . 4 kg) 810 N (7 m / s) = - 0 . 118689 m / s .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern