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Unformatted text preview: Husain, Zeena – Homework 3 – Due: Sep 16 2003, 4:00 am – Inst: H L Berk 1 This printout should have 20 questions, check that it is complete. Multiplechoice questions may continue on the next column or page: find all choices before making your selection. 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 , let the angle between the vector v 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 . A P v θ O This time is given by: 1. T = s 2 h g correct 2. T = p gh 3. T = 2 h g 4. T = s h 2 g 5. T = s h g 6. T = 2 p gh 7. T = 3 p gh Explanation: Basic Concepts: Constant acceleration: x x = v t + 1 2 at 2 (1) v = v + 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): 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 (magnitude of the vector ~ v ) 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 = 1 . 5 m, the angle θ = 34 . 2 ◦ , and the time T = 0 . 456115 s. (Given g = 9 . 8 m / s 2 ). Correct answer: 3 . 97621 m / s. Explanation: We divide v into components: v x = v cos θ ( * ) v y = v sin θ The xvelocity 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 x T + 0 or v x = L/T Now we use ( * ) to find the speed v : v = v x cos θ = L T cos θ With the given values we find v = 1 . 5 m . 456115 scos(34 . 2 ◦ ) m / s = 3 . 97621 m / s 003 (part 3 of 3) 10 points Now the same setup is to take place at some Husain, Zeena – Homework 3 – Due: Sep 16 2003, 4:00 am – Inst: H L Berk 2 planet where the gravitational acceleration is g = g/ 4. Keep v , θ and h to be the same as before. Find the new height, i.e. the ycoordinate 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 = 3 h 4 correct 2. y = h 3 3. y = h 2 4. y = h 4 5. y = 2 h 3 6. y = h 7. y = 2 h 8. y = h 5 9. y = 3 h 10. y = 3 h 5 Explanation: If g = 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 xmotion is still unaccelerated so T will still be the same. So the only difference is that the event, while taking the...
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This note was uploaded on 01/27/2009 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
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