Physics 430, Classical Mechanics Eam, 00 Oct 05 Name Instructions: No books, notes, or cheat sheet allowed. You may use a calculator, but no other electronic devices during the eam. Please turn your cell phone off. Please note that the NJIT honor code applies to this eam, 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. ight answers with no work will be marked wrong. Be careful with notation (e.g. vectors underlined, unit vectors with hats, etc.).. Given the two vectors a = (,, ) 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: ab = abcosθ ab = ab + ab + ab = 0 0 5= 5 y y z z a= + + = = + + = ab 5 cosθ = = = 0.408 so θ=4. ab 5 6 ( ) ( ) 6 and b 0 0 5 5 (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 = absinθ a b= ( ab ab) ˆ + ( ab ab) yˆ + ( ab ab) zˆ y z z y z z y y z = ( 5 0) ˆ + ( 0 5) yˆ + ( 0 0) zˆ = 5ˆ 0yˆ ( 5) ( 0) 5 a b = + = a b 5 5 sinθ = = = 0.99 so θ=65.9 ab 5 6 This is different from part (a), because of 80 ambiguity. The correct angle is 4. as in part (a).. A particle s potential energy is U(r) = (y + z) y. (a) (5 points) What is the (vector) force on the particle? F= U = ˆ + yˆ + zˆ U( r) = ( y+ z) ˆ + ( y) yˆ + zˆ y z or F = ( y+ z, y, ) [ ] - -
Physics 430, Classical Mechanics Eam, 00 Oct 05 (b) (5 points) What is the work W = Fdr i done against the force you derived in part (a) in going from the point (0,0,0) to the point (,,0) along a straight line path as shown in the Figure? Show your work. [Hint: Write down the equation for the line, y in terms of, then find dy in terms of. You can then convert the integral into one over a single variable.] y (,,0) z (0,0,0) Equation of the line is y =, so dy =. y 0 0 [ ] F dr = ( F + F dy) = y + ( ) = 0 = = Figure. 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 epressions for the terminal velocity. Eplain 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 mg bvter =0 so vter = (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 τ (time constant). dv b mv = bv = v. dt m In separated form, this is: dv b = dt = dt. v m τ - -
Physics 430, Classical Mechanics Eam, 00 Oct 05 (c) (5 points) Solve your equation derived in part (b), for v(t), then integrate to get the equation for (t), for the initial conditions (0) = 0, v(0) = v o. What is the maimum distance the body will reach after infinite time? Integrating both sides: dv = t v τ so v t/ τ ln = t v( t) = voe. vo τ Integrating again: t'/ τ t t/ τ 0 oτ vdt = () t = v τ( e ) = v ( e ). At t, = vτ. o o 4. An ion rocket engine has an ehaust velocity of 9,000 m/s (almost 65,000 mph), but very low thrust mv e = 950 mn (milli-newtons). If a small satellite has a payload (non-fuel part of the rocket) of 00 kg, and carries 50 kg of fuel in the form of Xenon gas to supply the ions, what is its maimum speed when its fuel is ehausted? What is the rate of fuel consumption m in kg/s? How long will the rocket take to reach maimum speed, if the rate of fuel consumption is constant? ecall that the rocket equation is v v o = v e ln(m o /m). (5 points) The initial mass is payload + fuel, so mo = 50 kg, while the final mass is m = 00 kg. v= ve ln( mo / m) so v = 9000 ln(50 /00) =, 758 m/s. The rate of fuel consumption is 950 mn 0.950 N m 3.3 0 5 = = = kg/s. ve 9000 m/s Since 50 kg are lost at the above rate, it takes 50 kg 6 Δ t = =.5 0 s 7.6 days. 5 3.3 0 kg/s 5. (a) (5 points) The element of volume in Cartesian coordinates is dv = dydz. Give the corresponding epression in cylindrical coordinates ρ, φ, z. The element of volume in cylindrical coordinates is dv = ρ dρdφdz. - 3 -
Physics 430, Classical Mechanics Eam, 00 Oct 05 (b) (5 points) By direct integration, V = dv, over the appropriate limits, find the volume of the one-quarter cylinder of radius and height h, shown in Fig.. Show all steps, including the limits on each of the three integrals needed. Show that your volume is /4 th of a complete cylinder. z V V cyl π / V = dv = ρdρ dφ dz 0 0 0 π π = h= 4 π h 4 = = π h 4 h h Fig.. Slice of a cylinder of radius, height h. y (c) (5 points) Using the definition of center of mass = dv M σ r write down the epression for the -component of the CM vector (call it X), in terms of integrals over dρ, dφ, and dz, assuming the density σ (we are using σ for density to avoid confusion with the coordinate ρ) 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 = ρcos φ, then dv = ρ d ρcos φdφdz, so σ M 4 = π h π / h X = ρ dρ cosφdφ dz 0 0 0 π / h cos ρ ρ 0 φ φ 0 0 (d) (5 points) Do the integration to show X d d dz 4 =. 3π 3 3 4 π / 4 4 φ φ 0 π π π X = h cos d = h=. h 3 h 3 3-4 -
Physics 430, Classical Mechanics Eam, 00 Oct 05 6. The plot below schematically shows the graph of the one-dimensional potential energy, U() = 4 3 3 6 J, acting without friction on an object of mass 0 kg, where is in m. (a) (5 points) Find all equilibrium points n of the object under potential energy U(), mathematically, and determine which are stable and which are unstable. Plot these equilibrium points on the graph and label their locations on the ais. energy Turning points E = 3/4 J -½ E = - J U() Equilibrium points The equilibria occur where the slope of the potential energy curve is zero, i.e. du 6 6 0. This has roots m, m. = = =+ To find if they are stable or unstable, check the curvature (nd derivative) at each point: du du = = 3 = 4() 6 = 8 > 0 stable = 4( ) 6 = 8 < 0 unstable (b) (5 points) Let the total energy of the object moving in the vicinity of the stable equilibrium point be E = J. Show that = 0 is a turning point of the object. Sketch the total energy and both turning points on the graph [do not calculate the second turning point]. What is the force on the object at = 0, in N? The turning points occur where the potential energy equals the total energy, E = U( ). = = = The force at this location is found from: du F = U = ˆ = ( + 6+ 6) ˆ = 6 ˆ N at = 0. 3 3 J 4 3 6 J. Thus 4 3 6 0, which clearly has a root at 0. - 5 -
Physics 430, Classical Mechanics Eam, 00 Oct 05 (c) (5 points) For the case of E = J, what is the speed (m/s) of the object when it is at the = 0 turning point? What is its speed when it is at the equilibrium point? At a turning point, the kinetic energy is zero, so the object's speed is zero there. At the equilibrium point, =, we have E = T + U( = ) : T = E U = = + + + = + + + = The speed is determi T v = = m/s. m 3 ( ) 4 3 6 J 4 3 6 J 5 J. ned from T = mv, where m = 0 kg, so (d) (5 points) What is the maimum total energy the object could have and still be bound, i.e. remain always within a finite distance of the stable equilibrium point? Sketch and label this total energy on the graph. Looking at the graph, the left turning point ceases to eist when the total energy eceeds the U( = ) for the unstable turning point, i.e. E = U( = ) = 4( ) 3( ) 6( ) = + 3 = J. 3 3 3 4 4-6 -