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Unformatted text preview: min (mm64262) – Homework 5 – lee – (GEDB00864) 1 This printout should have 12 questions. Multiplechoice 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.
 Spring '11
 LEE

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