{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW 8-solutions

HW 8-solutions - kuruvila(lk5992 HW 8 opyrchal(11113 This...

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

View Full Document Right Arrow Icon
kuruvila (lk5992) – HW 8 – opyrchal – (11113) 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(n) 70 . 8 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 31 . 8 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0 . 556 kg camera in her hand and decides to get back to the shuttle by throwing the camera at a speed of 12 m / s in the direction away from the shuttle. How long will it take for her to reach the shuttle? Answer in minutes. Correct answer: 5 . 62411 min. Explanation: Let : M = 70 . 8 kg , m = 0 . 556 kg , d = 31 . 8 m , and v = 12 m / s . Because of conservation of linear momentum, we have 0 = M V + m ( - v ) m v = M V where V is the velocity of the astronaut and it has a direction toward the shuttle. V = m v M = (0 . 556 kg) (12 m / s) 70 . 8 kg = 0 . 0942373 m / s . And the time it takes for her to reach the shuttle t = d V = 31 . 8 m 0 . 0942373 m / s · 1 min 60 s = 5 . 62411 min . 002 10.0 points A child bounces a 49 g superball on the side- walk. The velocity change of the superball is from 28 m / s downward to 17 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: 1764 N. Explanation: Let : m = 49 g = 0 . 049 kg , v u = 17 m / s , v d = 28 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 = (49 g) (17 m / s + 28 m / s) 0 . 00125 s = 1764 N . 003 (part 1 of 2) 10.0 points A(n) 20 g object moving to the right at 24 cm / s overtakes and collides elastically with a 33 g object moving in the same direction at 11 cm / s. Find the velocity of the slower object after the collision. Correct answer: 7 . 81132 cm / s. Explanation: Basic Concepts: Momentum conserva- tion gives m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f . Solution: For head-on elastic collisions, we know that v 1 i - v 2 i = - ( v 1 f - v 2 f ) .
Background image of page 1

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

View Full Document Right Arrow Icon
kuruvila (lk5992) – HW 8 – opyrchal – (11113) 2 For the relative velocities, we have v 1 - v 2 = v 2 f - v 1 f v 2 f = v 1 - v 2 + v 1 f . Momentum is conserved, so m 1 v 1 + m 2 v 2 = m 1 v 1 f + m 2 v 2 f m 1 v 1 + m 2 v 2 = m 1 v 1 f + m 2 ( v 1 - v 2 + v 1 f ) m 1 v 1 + m 2 v 2 - m 2 ( v 1 - v 2 ) = v 1 f ( m 1 + m 2 ) v 1 f = m 1 v 1 + m 2 v 2 - m 2 { v 1 - v 2 } m 1 + m 2 = 1 (20 g) + (33 g) × bracketleftBig (20 g)(24 cm / s) + (33 g)(11 cm / s) - (33 g) { (24 cm / s) - (11 cm / s) } bracketrightBig = 7 . 81132 cm / s . 004 (part 2 of 2) 10.0 points Find the velocity of faster object after the collision.
Background 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 ]}