텍사스5회솔루션

텍사스5회솔루션

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Unformatted text preview: min (mm64262) – Homework 5 – lee – (GEDB00864) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 2 . 23 kg steel ball strikes a massive wall at 13 . 1 m / s at an angle of 52 . 4 ◦ with the plane of the wall. It bounces off with the same speed and angle. x y 1 3 . 1 m / s 1 3 . 1 m / s 52 . 4 ◦ 52 . 4 ◦ If the ball is in contact with the wall for . 161 s , what is the magnitude of average force exerted on the ball by the wall? Correct answer: 221 . 418 N. Explanation: Let : m = 2 . 23 kg , v = 13 . 1 m / s , θ = 52 . 4 ◦ , and t = 0 . 161 s . x y vectorv f m vectorv i m θ θ F Δ t = Δ p. Only the component of the ball’s velocity perpendicular to the wall will change. This velocity component before hitting the wall is v ⊥ = v cos θ = (13 . 1 m / s) cos 52 . 4 ◦ = 7 . 9929 m / s . After hitting the wall, this component is- 7 . 9929 m / s, because the rebound angle is also 52 . 4 ◦ . The change in momentum during contact with the wall is therefore Δ p = mv f- mv = m (- v ⊥ )- mv ⊥ =- 2 mv ⊥ =- 2 (2 . 23 kg) (7 . 9929 m / s) =- 35 . 6483 kg · m / s , so the average force on ball is F = vextendsingle vextendsingle vextendsingle vextendsingle Δ p t vextendsingle vextendsingle vextendsingle vextendsingle = 35 . 6483 kg · m / s . 161 s = 221 . 418 N . 002 10.0 points A block of mass M is hanging from a string of length ℓ . A bullet of mass m traveling hor- izontally with speed v strikes the block and embeds itself inside it. The system of block and bullet swings until the string is precisely horizontal, at which point the system is mo- mentarily at rest. v M M + m v m Find the string length in simple terms of m , M , g and v . 1. None of these min (mm64262) – Homework 5 – lee – (GEDB00864) 2 2. ℓ = v 2 2 g radicalbigg 1 + M m 3. ℓ = v 2 2 g parenleftbigg 1- M m parenrightbigg 2 4. ℓ = v 2 g parenleftbigg 1 + M m parenrightbigg 2 5. ℓ = v 2 2 g parenleftbigg 1 + M m parenrightbigg 2 correct 6. ℓ = v 2 2 g parenleftBig 1 + m M parenrightBig 2 Explanation: Momentum is conserved during the colli- sion, so considering the horizontal compo- nents, p ix = p fx mv + 0 = ( M + m ) v v = mv M + m . The block rises a height of ℓ during the upward swing, and gravity is the only force doing work, so the total energy E = K + U g is conserved: E 1 = E 2 1 2 ( m + M ) v 2 + 0 = 0 + ( m + M ) g h g h = 1 2 v 2 = 1 2 parenleftbigg mv M + m parenrightbigg 2 · 1 m 2 1 m 2 h = v 2 2 g parenleftbigg 1 + M m parenrightbigg 2 . 003 10.0 points A 62 kg boat that is 7 m in length is initially 7 . 7 m from the pier. A 38 kg child stands at the end of the boat closest to the pier. The child then notices a turtle on a rock at the far end of the boat and proceeds to walk to the far end of the boat to observe the turtle....
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This note was uploaded on 05/24/2011 for the course ENGINEERIN 1 taught by Professor Lee during the Spring '11 term at Sungkyunkwan.

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텍사스5회솔루션

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