Physics 215 Fall 2008 Exam 1 Version A (707459)

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1 Page 1 of 6 Physics 215 Fall 2008 Exam 1 Version A (707459) Instructions Be sure to answer every question Follow the rules shown on the screen for filling in the Scantron form Each problem is worth 10% of the exam When you are finished, check with Dr Mike or his TA to be sure you have finished the scantron correctly 1 Circular Motion [713781] Bailey D Wonderdog is chasing her tail at a constant speed of v = 14 m/s The circle she moves in has a diameter of d = 52 cm What is the magnitude and direction of her acceleration? *5*c*75385 m/s 2 in the direction of her motion *2*d*754 m/s 2 in the direction of her motion *10*a*75385 m/s 2 toward the center of the circle *7*b*754 m/s 2 toward the center of the circle *2*e*754 m/s 2 opposite to the direction of her motion The acceleration of any object going in a circle is the centripetal acceleration, which has a magnitude of a = v 2 / r = v 2 / (d/2) = [14 m/s] 2 / ([52 cm]/2) = [14 m/s] 2 / ([52 cm]/2) (100 cm / 1 m) = m/s 2 2 Constant Acceleration [711502] At t = 0, a stone is dropped from a cliff above a lake; 14 seconds later another stone is thrown downward from the same point with an initial speed of 37 m/s Both stones hit the water at the same instant Find the height of the cliff *1*d*6978 m *10*a*1613 m *3*b*961 m *5*c*1923 m *1*e*240 m Let us call T 0 the time when the first stone was thrown Then the second stone is thrown at a time T s Let us also call the time at which both stones hit the water T f The equation to use for both stones is y = y 0 + v 0y t + 1/2 a y t 2 For the first stone, we have t I = T f - T 0 and for the second, t II = T f - (T s) Thus, t II = t I - 14 s Remember this I will call it equation (1) We will need it later Now, the height of the cliff is h = y 0 - y for both stones Thus,

2 Page 2 of 6 -h = v 0yI t I + 1/2 a yi t I 2 = v 0yII t II + 1/2 a yii t II 2 For the first stone, v 0yI = 0 and a yi = -g So that, -h = - 1/2 g t I 2 This, I will call equation (2) For the second stone, v 0yII = 37 m/s and a yii = -g So that, -h = (37 m/s) t II - 1/2 g t II 2 This is equation (3) Equations (1), (2) and (3) are three equations with three unknown variables t I, t II and h You should be able to solve this for any of the three unknown variables Here is how you do it We can eliminate t II by putting equation (1) into equation (3) to get -h = (37 m/s) (t I - 14 s) - 1/2 g (t I - 14 s) 2 We can also eliminate t I be solving equation (2) for t I and putting this into the last equation Thus, we are left with -h = (37 m/s) ( (2h/g) - 14 s) - 1/2 g ( (2h/g) - 14 s) 2 Now all we need to do is solve for h After using some algebra, you will find that (2h/g) = [(37 m/s) (14 s) - 1/2 g (14 s) 2 ] / [(37 m/s) + g (14 s)] which gives us h = 1613 m 3 Vectors [771679] Find the cross product of A = 6 cos(30 o ) i and B = 5 sin(52 o )j *5*b*3000 j *10*a*2047 k *5*c*2934 i *1*e*914 *3*d*2047 The magnitude of the cross product is A X B = A B cos, where is the angle between the vectors In this case = 90 o and we have A X B = [6 cos(30 o )] [5 sin(52 o )] cos 90 o = 2047 The direction is given by the right-hand rule Point your fingers to the right and bend their tips upward Your thumb will be point toward you This is the positive z-direction (k) 4 Mathematics in Physics [711489] There are 1057 quarts in a liter and 4 quarts in a gallon A barrel equals 42 gallons How many liters are there in 55 barrrel(s)? *2*b*006 liters *4*d*087 liters *10*a*87417 liters *2*c*874 x 10 5 liters

3 Page 3 of 6 *1*e*6104 liters We begin with 55 barrels This is the same as (55 barrels)(42 gallons / 1 barrel)(4 quarts / 1 gallon) (1 liter / 1057 quarts) = liters 5 Constant Acceleration [716875] A particle moves in the xy plane with constant acceleration At t = 1 s, the particle is at x = 206 m, y = 119 m, and has velocity at time zero of v = 8 m/s i + 5 m/s j The acceleration is given by the vector a = 7 m/s 2 i + 8 m/s 2 j Find the position vector at time zero *4*c*(22345 m) *7*b*(2175 m) i + (128 m) j *7*e*(1945 m) i + (128 m) j *2*d*(25237 m) *10*a*(1945 m) i + (110 m) j The equation we use is r = r 0 + v 0 (t - t 0 ) + 1/2 a (t - t 0 ) Solving the components for r 0, we get x 0 = x - v x0 (t - t 0 ) - 1/2 a x (t - t 0 ) = (206 m) - (8 m/s) (1 s) - 1/2 (7 m/s 2 ) (1 s) 2 = 1945 m Similary, y 0 = y - v y0 (t - t 0 ) - 1/2 a y (t - t 0 ) = (119 m) - (5 m/s) (1 s) - 1/2 (8 m/s 2 ) (1 s) 2 = 110 m 6 Algebra, Trigonometry and Geometry [711497] Calculate the following with your calculator Given A=46, B=47, C=7, and D=2 What is the value of B-(2D+3/16A 2 )/C? *4*c*-5054 *10*a*-1025 *1*e*9968 *5*b*4643 *4*d*614 Remember to use the PEMDAS order (parentheses, exponents, multiplication/division, addition/subtraction) going from left to right If we had put the parentheses in explicitly, we would have had B-(((2D)+((3/16)(A 2 )))/C) = Vectors [ ]

4 Page 4 of 6 Given that, what is the unit vector that points in the same direction as? *10*a* *3*c* *3*d* *5*b* *2*e* The way to find a unit vector that has the same direction as another vector is to divide that vector by its magnitude The magnitude of is Thus, the new unit vector is 8 Method of Susbtitution [ ] Given that both of the following equations are both true Which of the answers is a possible solution for *2*d* 452 *5*b* 4 *10*a* 3 *2*e* 322 *5*c* 312 Start with the first equation and plug this into the second equation

5 Page 5 of 6 We can now use the quadratic equation to solve for 9 Algebra and Constant Acceleration [ ] The x-component of the first equation for constant acceleration, can be solved for time Which of these is the right equation? *10*a* *2*e* There is no way to know *2*d* *5*c* *7*b* If we start with and rearrange, we will find that which is a quadratic equation that has the solution 10 Preview of Calculus and Velocity [711490] Figure 2-35 shows the x-component of position of a car plotted as a function of time At which time t 0 to t 7 is the x-component of velocity zero? Figure 235 *4*e*t 7

6 Page 6 of 6 *10*a*t 2 *4*d*t 3 *4*c*t 1 *4*b*t 0 The velocity is zero when the slope is zero This only happens at t 2 Assignment Details Name (AID): Physics 215 Fall 2008 Exam 1 Version A (707459) Submissions Allowed: 100 Category: Exam Code: Locked: No Author: DeAntonio, Michael ( mdeanton@nmsuedu ) Last Saved: Sep 28, :21 PM MDT Permission: Protected Randomization: Person Which graded: Last Feedback Settings Before due date Nothing After due date Nothing

Physics 222 Spring 2009 Exam 1 Version D (808898)

Physics 222 Spring 2009 Exam 1 Version D (808898) Question 1 2 3 4 5 6 7 8 9 10 Instructions Be sure to answer every question. Follow the rules shown on the first page for filling in the Scantron form.