Homework 03 - Bautista Aldo Homework 3 Due 4:00 am Inst...

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Bautista, Aldo – Homework 3 – Due: Sep 20 2005, 4:00 am – Inst: Maxim Tsoi 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points Consider the setup of a gun aimed at a target (such as a monkey) as shown in the figure. The target is to be dropped from the point A at t = 0, the same moment as the gun is fired. The bullet hits the target at a point P, which is at the same horizontal level as the gun. Let the initial speed of the bullet be v 0 , let the angle between the vector v 0 and the horizontal ( x -) direction be θ , and OP= L and AP= h . The gravitational acceleration is g . Denote the time taken to hit the target by T . The acceleration of gravity is 9 . 8 m / s 2 . A P v θ 0 O This time is given by 1. T = p g h . 2. T = 2 h g . 3. T = s h g . 4. T = s 2 h g . correct 5. T = s h 2 g . 6. T = 3 p g h . 7. T = 2 p g h . Explanation: Basic Concepts: Constant acceleration: x - x 0 = v 0 t + 1 2 at 2 (1) v = v 0 + at (2) Solution: The time T is the time it takes for the monkey to fall from A to P, a distance h . For any object falling a distance h from rest (neglecting air friction), we have from (1): 0 - h = 0 + 1 2 ( - g ) T 2 or T = s 2 h g 002 (part 2 of 3) 10 points Find the initial speed v 0 (magnitude of the vector ~ v 0 ) which allows the projectile to meet the target at location P. ( Hint: T defined in part 1 is also the time taken for the bullet to travel, following the projectile trajectory, from O to P). Let the distance OP be L = 2 . 91 m, the angle θ = 46 . 1 , and the time T = 0 . 785577 s. (Given g = 9 . 8 m / s 2 ). Correct answer: 5 . 34219 m / s. Explanation: We divide v 0 into components: v x = v 0 cos θ ( * ) v y = v 0 sin θ The x -velocity is unchanged (there is no hor- izontal acceleration) so we simply use (1) where a = 0 and the bullet is traversing a distance L: L - 0 = v 0 x T + 0 or v 0 x = L/T Now we use ( * ) to find the speed v 0 : v 0 = v 0 x cos θ = L T cos θ With the given values we find v 0 = 2 . 91 m 0 . 785577 s cos(46 . 1 ) m / s = 5 . 34219 m / s 003 (part 3 of 3) 10 points
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Bautista, Aldo – Homework 3 – Due: Sep 20 2005, 4:00 am – Inst: Maxim Tsoi 2 Now the same setup is to take place at some planet where the gravitational acceleration is g 0 = g/ 4. Keep v 0 , θ and h to be the same as before. Find the new height, i.e. the y -coordinate of the new point of collision. ( Hint: you should convince yourself that for this new case, the time taken for the bullet to travel from O to the new point of collision P’ should still be T ). 1. y = h 4 2. y = h 2 3. y = h 3 4. y = 2 h 5. y = 3 h 4 correct 6. y = 2 h 3 7. y = h 5 8. y = 3 h 5 9. y = h 10. y = 3 h Explanation: If g 0 = g/ 4, we would expect the bullet to hit its target at a higher point, since the target will not fall quite as fast. However, as noted in the hint, the x -motion is still unaccelerated so T will still be the same. So the only difference is that the event, while taking the same amount of time to transpire, will occur at a height y P 0 instead of y P = 0.
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