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The spring 2009 final exam for the introductory physics course, phys 101. The exam includes multiple choice questions and problems covering various topics in physics such as torque, projectiles, pendulums, and circular motion. Students are required to provide symbolic justifications for their answers and may qualify for partial credit if their answers are incorrect but well-justified.
Typology: Exams
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Name_____________________________________________________ (please print)
Instructions and Rules:
Useful constants:
Magnitude of gravitational acceleration at the Earth’s surface: g 9.80 m/s^2
Question 4: (4 points) Two cannons launch two identical projectiles vertically upward simultaneously. First projectile has mass m and is launched with a force F. The second projectile has mass m and is launched with a force F/. What is the ratio between the accelerations of the two projectiles after 10 seconds in the air? a) 3 b) g c) 1 d) 1/
Solution: When the projectiles are in the air, the only force acting onto them is the gravitational force, so their acceleration is g, and the ratio is 1.
Question 5: (4 points) A simple pendulum consists of a small ball of mass m suspended by light string of length L. If the light string is replaced by one of length 4 L , the frequency of the pendulum changes by what multiplicative factor? a) 2 b) 1 c) 4 d) 1/ e) 1/
Solution: The frequency of a simple pendulum isf = (^21) ¼
p g=L, so the frequency changes in a
factor of 1/2.
Question 6: (4 points) Bob, of mass m , drops from a tree limb at the same time that Esther, also of mass m, begins her descent down a frictionless slide. If they both start at the same height above the ground, which of the following is true about their kinetic energies as they reach the ground? a) The answer depends on the shape of the slide. b) Esther's kinetic energy is greater than Bob's. c) Bob's kinetic energy is greater than Esther's. d) They have the same kinetic energy.
Solution: Since the slide is frictionless, energy is conserved. Both have the same initial potential energy, so when they reach the ground both have the same kinetic energy.
Question 7: (4 points) A skier leaves the end of a horizontal ski jump at 25 m/s and falls 4.5 m before landing. Neglecting friction, how far horizontally does the skier travel in the air before landing? a) 12 m b) 24 m c) 4.8 m d) 20 m e) 9.8 m
Solution: If h is the vertical height, from y = y 0 + v 0 yt ¡ 1 = 2 gt^2 , we have 0 = h ¡ 1 = 2 g t^2 f, so
tf =
p 2 h=g =
p 2 £ 4 :50 m=(9:80m=s^2 ) = 9:6 s. Then ¢x = v 0 tf = 25:0 m=s £ 0 :96 s = 24 m.
Question 8: (4 points) Object A has a mass m and momentum p. Object B has mass 2m and momentum 2p. How do the magnitudes of their kinetic energy compare? a) KEA > KEB b) Not enough information given c) KEA < KEB d) KEA = KEB
Solution: The kinetic energy can be written as KE = p^2 =(2m), so KEB = (2p)^2 =(4m) = 2 £ p^2 =(2m) = 2 £ KEA and (^) KEA < KEB.
Question 9: (4 points) You lift a book with your hand such that it moves up at a constant speed. What is the net work done on the book while it is moving? a) 0 J b) Fhand ¢ h c) mg ¢ h
Solution: Since the book is moving at constant speed, the net force on it zero, so the net work is zero.
h
F hand
mg
Problem 1: (20 points) A car moves along a straight road. The position-versus-time ( x t ) graph is shown below. (NB: On the x t -representation, the graph in the intervals A-B and B-C are parabolas and in the interval C-D is a straight line. The tangent at point B is horizontal.)
a) (6 points) Make a qualitative plot of this motion in a v-t graph (velocity versus time) and in an a-t graph (acceleration versus time). Use the frames provided above.
Solution: Showed in the frames.
b) (4 points) Compute the acceleration in the intervals A-B.
¡1m = v 0 (1s) + 12 a(1 s)^2. Substituting the expression for v 0 we already found, we have
¡1m = (¡a(1s)(1s) + 12 a(1s)^2 or^ ¡1m = ¡^12 a(1 s)^2. Thena^ = 2 m=s^2
c) (2 points) Compute the acceleration in the intervals C-D.
Solution: The acceleration is zero since the velocity is constant (the x-tgraph is a straight line).
d) (3 points) Compute the velocity of the car at A.
e) (3 points) Compute the velocity of the car at D.
f) (1 point) In which intervals the velocity is positive? Circle the correct answer(s).
(A,B) (B,C) (C,D)
g) (1 point) In which intervals the car is accelerating? Circle the correct answer(s).
(A,B) (B,C) (C,D)
c) (4 Points) Calculate the speed of the bullet-block system at the top of the track (position B) by using conservation of mechanical energy.
Solution: Immediately after the collision, the mechanical energy is EA = 12 (m + M)v^2 A. The
mechanical energy at point B is (^) EB = 12 (m + M)v^2 B + (m + M) g (2R). Since energy is
conserved, EA = EBandvB =
p v A^2 ¡ 4 g R = 5:46 m=s.
d) (4 Points) Calculate the normal force exerted by the track to the block when the bullet-block system reaches the top of the track (position B).
Solution: At point B, the normal is vertical pointing down, soP Fy = ¡n ¡ (m + M )g. By Newton’s 2nd^ Law P Fy = (m + M )ay. But ay = ¡ac = ¡v^2 B=R. Then
n = (m + M)(v^2 B=R ¡ g) = 26:6 N
e) (4 Points) Calculate the angular momentum L of the system at the top of the track (position B). To this end you can assume that the block-bullet system can be considered as a point-like object.
Solution: The angular momentum is L = I! = ((m + M) R^2 ) £ (vB=R) = (m + M) R vBso L = 56:8 kg m^2 =s.
Problem 3: (20 points) A small hermetically sealed wood box of mass m = 0:500kgand volume VBox = 6 £ 10 ¡^4 m^3 is floating on the surface of a lake (½ (^) w = 1000 kg=m^3 ).
a) (4 Points) Draw a free body diagram and represent the forces acting on the box. Find the magnitude of the buoyant force.
Solution: Since the box is in equilibrium,
F = 0. Only vertical forces are relevant, so
Fy = FB ¡ W = 0, FB = W = mg = 4:90N
b) (4 Points) Find the volume of the submerged part of the box.
Solution: The buoyant force has a magnitude equal to the weight of the displaced fluid, so FB = Vsubm ½w g^. Thus^ Vsubm =^ m=½w = 5:^00 £^10 ¡^4 m^3.
Now we place inside the box a small chunk of iron of mass (^) mI = 0:400 kg, and we hermetically seal the box again. The box starts to sink until it reaches the bottom of the lake, 3 meters deep.
c) (3 Points) What is the net force acting on the on the box when is entirely submerged in the lake (but not at the bottom yet)?
Solution: Since the box is entirely submerged, the buoyant force is FB = ½w VBox g = 5: 88 N. The weight is
Fy = FB ¡ W = 5:88 N ¡ 8 :82 N = ¡ 2 :94 N.
