Download Physics Honors Exam 1: Problems on Acceleration, Forces, and Motion and more Exams Mechanics in PDF only on Docsity! Physics 218 Honors Exam 1 1. An astronaut (mass 80 kg) is tethered to a spacecraft (mass 105 kg) by a cord. Both are traveling initially in uniform motion. The astronaut pulls on the cord with a force of 100 N. Calculate a) the acceleration of the astronaut; b) the acceleration of the spacecraft; c) the tension in the cord. A = acceleration of spacecraft a = acceleration of astronaut M = mass of spacecraft M = mass of astronaut T = tension in cord = 100 N Newton #2: T = MA A = T/M = (100 N)/(105 kg) = 10-3 m/s2 towards the astronaut T = ma a = T/m = (100 N)/(80 kg) = 1.125 m/s2 towards the spacecraft 2. A clever physics student (mass 60 kg) decides that she can descend from a height of 10 m above ground by grasping a rope attached to a pulley, mounted from a beam above her, and jumping horizontally off the building. The other end of the rope is attached to a sandbag of mass 50 kg sitting on the ground. What is the velocity with which the student lands on the ground? M = mass of sandbag M = mass of student T = tension in rope Take up to be + for force and acceleration Newton #2 for student: +T-mg = -ma Newton #2 for sandbag: +T โ Mg = +Ma Note: if the acceleration of the student is down with value -a, the acceleration of the sandbag will be up with the value +a. Eliminate T by subtracting the two equations: (M+m)a = (m-M)g ( ) 2/89.08.9 110 10 smg Mm Mma == + โ = Now calculate the velocity when she hits ground. Her equations of motion are 2 0 2 1 atyy โ= v=-at t = -v/a 2)( 2 1100 a vaโ= smv /2.489.0202 =โ
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= 5. An exciting amusement park ride consists of a car that travels with the acceleration shown in the graph. The car is at rest at t=0. How far does the car travel in the time from t=0 to t=20 s? What is its maximum velocity? What is the average velocity of the car during the first 16 seconds of motion? 0 5 10 15 20 25 30 0 5 10 15 20 25 time (s) ac ce le ra tio n (m /s2 ) T<6 s v = 0 6<t<16 s: โซโซโซ ==โ== 10 0 2 60 20 25 10 25)6( 10 25)()( ttdtdttdttatv tt v(16s) = 125 m/s t>16 s: v = 125 m/s This is the maximum velocity. distance traveled in first 20 s: mdtdttdttvx 916)1620(125)616( 60 25125)6( 20 25)( 3 20 0 20 16 16 6 2 =โ+โ=+โ== โซ โซโซ Average velocity on the interval 0<t<16 s: sm s xsxv /26 16 )616( 60 25 16 )0()16( 3 = โ = โ = 6. A stone is thrown vertically upward. On its way up, it passes point A with speed v0, and it passes point B (3.00 m higher than point A) with speed 0.70 v0. Calculate the speed v0 and the maximum height reached by the stone above point B. Choose y-axis vertical, origin at point A, t=0 when stone passes point A. Equations of motion: gtvtv gttvty โ= โ= 0 2 0 )( 2 1)( Now the stone has velocity 0.7 v0 when it passes point B, y = 3 m: smv g v g v gttv gvt gtvv BB B B /7.10 255.0) 2 3.03.0( 2 13 /3.0 7.0 0 2 0 2 0 2 2 0 0 00 = =โ=โ= = โ= Maximum height above B: v(t) = 0: t = v0/g = 1.09 s mssmssmy 88.5)09.1)(/8.9( 2 1)09.1)(/7.10( 22 =โ= So max height above point B is 2.88 m.
