PHY 141 Midterm 1 Oct 2, 2014 Version A

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1 PHY 141 Midterm 1 Oct 2, 2014 Version A Put FULL NAME, ID#, and EXAM VERSION on the front cover of the BLUE BOOKLET! To avoid problems in grading: do all problems in order, write legibly, and show all work for full credit! Note: make sketches where appropriate and show your axis conventions. Do not forget units, number of significant digits, and check your results for consistency. Do ALL problems. This exam will last 1.3 hr. Success! 1. Projectile motion (50 points). A cannon ball is launched from a vertical cliff 35.0 m above the sea. The launching speed of the ball is 50.0 m/s and the launch angle is 36.9º with respect to the horizontal. Ignore drag. Use the local value for g=10.0 m/s 2. a. Sketch the trajectory and show both the initial velocity and the acceleration vectors. b. Calculate the horizontal range of the ball (measured from the foot of the cliff.) c. Calculate the time from the moment of launch until the ball hits the water. d. Calculate the maximum height over sea-level reached by the ball 2. The conical pendulum (50 points). A bob of mass m on a massless string of length L is rotating around the vertical such that the bob describes a perfect horizontal circle with the string held at the apex of a cone. With the bob moving through a circle, the string is observed to make an angle θ with the vertical. Ignore drag. Assume g is known. a. Sketch the free-body diagram with all the forces, the axes, and show the direction of the net force F net. b. Express the speed of rotation v in terms of the variables given in the problem. c. Calculate the period of the motion, i.e. the duration of each revolution. 3. Spring launcher (50 points). A block of mass m is held on a horizontal surface against a horizontal spring. The spring has spring constant k= N/m and is compressed by an amount d = 10.0 cm with respect to its relaxed position. The mass m = kg is released from rest and slides a total distance D before coming to a stop. The horizontal surface has a friction coefficient μ k with the block. Ignore drag and assume the spring to be massless. Use the local value for g=10.0 m/s 2. a. First, given the static friction coefficient μ s = 0.400, calculate the minimum vertical force I must apply down onto the block to keep it in place when the spring is compressed by 10.0 cm. Note: draw a freebody diagram showing all the forces on the block in this situation. b. Second, given the kinetic friction coefficient μ k = 0.300, calculate the total distance D that the block will slide when it is released from rest before coming to a stop. Make free-body diagrams of the block in its initial position (just after being released), in the position just after it loses contact with the spring, and in its final position. In the sketch, show the distances d and D involved in the problem. c. Calculate the maximum speed that the block reached during its travel above. Show all work for full credit! (Note: this does not occur at the point where the block leaves the spring!) 4. A moon is circling an unknown planet (50 points). a. The moon s period is 7.15 days and its orbital radius is km. Calculate the mass of the planet. b. Astronauts have landed on this moon while their ship remains in low orbit around it. The moon s mass is kg and its radius is 2631 km. Calculate the period of the mother ship s orbit. c. Calculate the escape velocity from this moon. (Escape velocity is the minimum initial speed required, e.g. of a bullet, to escape the gravitational attraction of the moon.) d. How fast must the moon spin before I become weightless at the equator? Calculate the speed

2 PHY141 Midterm 1 Exam Formulae: Trigonometry: sinθ b/c; cosθ a/c; tanθ b/a = slope of c sin30 = cos60 = ½ c thus: b=c sinθ, a=c cosθ, b=a tanθ, b sin60 = cos30 = ½ 3 a 2 + b 2 = c 2, hence: sin 2 θ + sin 2 θ = 1 θ sin45 = cos45 = ½ 2, tan45 =1 sinθ = cos(90 θ), sinθ = sin( θ), cosθ = cos( θ) a Components: (if θ angle with the +x-axis!) A x =A cosθ; A y =A sinθ; A = (A x,a y ) Circle: circumference= 2πR; Area=πR 2 ; π Sphere: surface= 4πR 2 ; volume=4πr 3 /3 Quadratic Equation: ax 2 + bx + c=0; constants a, b, and c Solutions: x +, x =[ b (b 2 4ac) ] /(2a) Differentiation: d/dx[ax n ]=Anx n 1 ; d/dt[cos(ωt)]= ωsin(ωt); d/dt[sin(ωt)]=ωcos(ωt); d/dx[lnx]= x 1 Prefixes: f=10 15, p=10 12, n=10 9, μ=10 6, m=10 3, k=10 3, M=10 6, G=10 9, T=10 12, P=10 15 Constants: 1 mile=1609 m, 1 L=10 3 m 3, g=9.80 m/s 2 ; M E = kg; R E = m; N A = mol 1 Vectors: A = A x i + A y j + A z k Scalar Product ( Dot Product) A B = A x B x + A y B y + A z B z = AB cosθ A,B = AB // = A // B Vector Product ( Cross Product) A B = (A y B z A z B y )i + (A z B x A x B z )j + (A x B y A y B x )k direction: right-hand rule A B = AB sinθ A,B = AB = A B B Kinematics: θ B // A the motion : position r as function of time t s = s(t) (position) velocity v; acceleration a: v ds/dt (speed v ); a dv/dt Linear motion with constant a: v = v 0 + at, r = r 0 +v 0 t + ½at 2 ; eliminating t: v 2 = v a (s s 0 ) rotation angle (radians; rotation radius R): s(=arc length)/r = (t) (angular position) angular velocity : d /dt = v/r (v speed along the circle) angular acceleration : d /dt = a // /R (// parallel to circle) Circular motion (radius R) with constant : = 0 + t ; = t + ½ t 2 eliminating t: 2 = ( 0 ) circular motion centripetal acceleration a c : a rad = a c = v 2 T /R (radially inwards) Kinetic Energy K [J Nm]: K ½ mv 2 Forces [N kgm/s 2 ] and consequences: i F i = ma = dp/dt ; F A on B = F B on A ; i τ i = dl/dt = Iα Force of Gravity between M and m, at center-to-center distance r F G = GMm/r 2 (attractive! G= Nm 2 /kg 2 ); near sea level: F G = mg( j) (downwards; g=9.80 m/s 2 ) Force of a Spring (spring constant k): F S = kx (opposes compression/stretch x) Friction: static: F f s N, kinetic: F f = k N, opposes motion; =frict n coef.; N=normal force Work done by a force F over a trajectory: W F F dx Work-Kinetic Energy relationship (from F i = ma): W tot = j W j = K K f K i Power P [W J/s]: P F dw F /dt = F v Potential Energy U of a conservative force F: U F = W F ; e.g. U G = GMm/r, U S = ½kx 2 Work by Non-Conservative forces - Total Energy: W NC. = E = E f E i K f +Σ j U fj (K i +Σ j U ij )

dt 2 x = r cos(θ) y = r sin(θ) r = x 2 + y 2 tan(θ) = y x A circle = πr 2

dt 2 x = r cos(θ) y = r sin(θ) r = x 2 + y 2 tan(θ) = y x A circle = πr 2 v = v i + at a dv dt = d2 x dt 2 A sphere = 4πr 2 x = x i + v i t + 1 2 at2 x = r cos(θ) V sphere = 4 3 πr3 v 2 = v 2 i + 2a x F = ma R = v2 sin(2θ) g y = r sin(θ) r = x 2 + y 2 tan(θ) = y x a c = v2 r