텍사스5회솔루션

텍사스5회솔루션

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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 10.0points 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 13 . 1 m / s 13 . 1 m / s 52 . 4 52 . 4 If the ball is in contact with the wall for 0 . 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 = m v f - m v 0 = m ( - v ) - m v = - 2 m v = - 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 0 . 161 s = 221 . 418 N . 002 10.0points A block of mass M is hanging from a string of length . A bullet of mass m traveling hor- izontally with speed v 0 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 0 m Find the string length in simple terms of m , M , g and v 0 . 1. None of these
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min (mm64262) – Homework 5 – lee – (GEDB00864) 2 2. = v 2 0 2 g radicalbigg 1 + M m 3. = v 2 0 2 g parenleftbigg 1 - M m parenrightbigg 2 4. = v 2 0 g parenleftbigg 1 + M m parenrightbigg 2 5. = v 2 0 2 g parenleftbigg 1 + M m parenrightbigg 2 correct 6. = v 2 0 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 m v 0 + 0 = ( M + m ) v v = m v 0 M + m . The block rises a height of 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 m v 0 M + m parenrightbigg 2 · 1 m 2 1 m 2 h = v 2 0 2 g parenleftbigg 1 + M m parenrightbigg 2 . 003 10.0points 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. 7 . 7 m 7 m How far is the child from the pier when she reaches the far end of the boat? Assume there is no friction between boat and water.
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