Physics 430 Classical Mechanics Exam 1 2010 Oct 05 Name Instructions No books notes or cheat sheet allowed You may use a calculator but no other electronic devices during the exam Please turn your cell phone off Please note that the NJIT honor code applies to this exam as it does to all activities related to this course Each part of each question is worth 5 points as noted for a total of 80 points You may use additional sheets of paper if you need more room to work Clearly label the portion of each additional sheet with the problem number Show all work Right answers with no work will be marked wrong Be careful with notation e g vectors underlined unit vectors with hats etc 1 Given the two vectors a 2 1 1 and b 0 0 5 a 5 points Find the dot product and use it to find the angle between the two vectors Start with definition a b ab cos a b ax bx a y by az bz 2 0 1 0 1 5 5 a 22 1 2 1 2 6 and b 02 02 52 5 cos 5 a b 0 4082 so 114 1 ab 5 6 b 5 points Find the cross product and use its magnitude to find the same angle between the two vectors If you get a different answer from a state why and say which is the correct angle Start with definition a b ab sin a b a y bz az by x az bx ax bz y ax by a y bz z 1 5 1 0 x 1 0 2 5 y 2 0 1 0 z 5x 10y a b 5 2 10 2 125 5 5 a b 0 9129 so 65 9 ab 5 6 This is different from part a because of 180 ambiguity sin The correct angle is 114 1 as in part a 2 A particle s potential energy is U r x y z y2 a 5 points What is the vector force on the particle F U x y z U r y z x x 2 y y xz y z x or F y z x 2 y x 1 Physics 430 Classical Mechanics Exam 1 2010 Oct 05 2 b 5 points What is the work W Fidr done against the force you derived in 1 part a in going from the point 0 0 0 to the point 2 1 0 along a straight line path as shown in the Figure 1 Show your work Hint Write down the equation for the line y in terms of x then find dy in terms of dx You can then convert the integral into one over a single variable x y 2 1 0 z 0 0 0 x Figure 1 Equation of the line is y 12 x so dy 12 dx 2 1 F dr Fx dx Fy dy ydx x 2 12 x 12 dx 2 2 1 2 1 0 2 xdx 12 0 2 2 x 2 1 0 3 The equation of motion for a vertically falling body under linear air resistance drag force f v bv is mv mg bv a 5 points Use this equation to derive the expressions for the terminal velocity Explain your reasoning The body accelerates until the drag force equals the gravity force When that occurs the body stops accelerating and v 0 The velocity at that time is v vter Then mg bvter 0 so vter mg linear case b b 5 points What is the corresponding equation for a horizontally moving body under linear air resistance Write it in separated form terms involving dv on one side and dt on the other Write the quantity m b as time constant dv b v dt m In separated form this is dv b 1 dt dt v m mv bv 2 Physics 430 Classical Mechanics Exam 1 2010 Oct 05 c 5 points Solve your equation derived in part b for v t then integrate to get the equation for x t for the initial conditions x 0 0 v 0 vo What is the maximum distance the body will reach after infinite time Integrating both sides dv 1 v t so 1 v ln t v t vo e t vo Integrating again vdt x t v e t t o 0 vo 1 e t At t x vo 4 An ion rocket engine has an exhaust velocity of 29 000 m s almost 65 000 mph but very low thrust mvex 950 mN milli Newtons If a small satellite has a payload non fuel part of the rocket of 100 kg and carries 50 kg of fuel in the form of Xenon gas to supply the ions what is its maximum speed when its fuel is exhausted What is the rate of fuel consumption m in kg s How long will the rocket take to reach maximum speed if the rate of fuel consumption is constant Recall that the rocket equation is v vo vex ln mo m 5 points The initial mass is payload fuel so mo 150 kg while the final mass is m 100 kg v vex ln mo m so v 29000 ln 150 100 11 758 m s The rate of fuel consumption is 950 mN 0 950 N m 3 3 10 5 kg s 29000 m s vex Since 50 kg are lost at the above rate it takes 50 kg t 1 5 10 6 s 17 6 days 5 3 3 10 kg s 5 a 5 points The element of volume in Cartesian coordinates is dV dxdydz Give the corresponding expression in cylindrical coordinates z The element of volume in cylindrical coordinates is dV d d dz 3 Physics 430 Classical Mechanics Exam 1 2010 Oct 05 b 5 points By direct integration V dV z over the appropriate limits find the volume of the one quarter cylinder of radius R and height h shown in Fig 1 Show all steps including the limits on each of the three integrals needed Show that your volume is 1 4th of a complete cylinder R V dV d 0 2 0 R h d dz 0 y R h R2h 2 2 4 2 x R2h V 1 4 2 Vcyl R h 4 Fig 1 Slice of a cylinder of radius R height h c 5 points Using the definition of center of mass 1 R rdV M write down the expression for the x component of the CM vector call it X in terms of integrals over d d and dz assuming the density we are using for density to avoid confusion with the coordinate is constant and the total mass is M Indicate the appropriate limits for each coordinate Do not do the integration yet you ll do it in part d Using x cos then xdV 2 d cos d dz so R X 2d 2 h 0 0 M 0 R 2 h 4 2 d d cos 0 0 dz R 2 h …