Download Physics 215 Fall 2008 Makeup Exam D (759376)

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Physics 215 Fall 2008 Makeup Exam D (759376)
Instructions
Be sure to answer every question. Follow the rules shown on the first page for filling in the Scantron form. Each problem is
worth 10% of the exam. When you are finished, check with Dr. Mike or his TA to be sure you have finished the scantron
correctly.
1. Work/Energy and Impulse/Momentum [504242]
A package of mass 36 kg is released from rest at a warehouse loading dock and slides down a 3 m high frictionless chute to a
waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 72 kg ,
from the bottom of the chute. Suppose the packages stick together. How much kinetic energy is lost during the collision?
*03*d*353.16 J
*05*b*264.87 J
*10*a*706.32 J
*00*e*None of the other answers are correct.
*04*c*941.76 J
Solution or Explanation
The first box starts with a gravitational potential energy of mgh which is converted to a kinetic energy (energy conservation) 1/2
(2gh). Thus, it has a momentum of m (2gh). The other box has no initial momentum.
m v2. So that is has a speed of v =
The final momentum of both boxes is the same but can also be written as 3m vf. Thus vf = (2gh) / 3 and the kinetic energy of
the boxes is 3/2 m (2gh) / 9. The difference is mgh - (mgh / 3) = 2/3 (mgh).
2. Kinematics [504239]
As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of
your window is 1.3 m. Take down to be the positive direction, so that downward velocities are positive and the acceleration due
to gravity is the positive quantity. Using a special device, you are able to measure the speed of the pot as it passes the bottom
of your window to be 20 m/s. What is the value of t?
*10*a*0.06 seconds
*00*d*2.10 seconds
*05*b*1.52 seconds
*05*c*0.00 seconds
*00*e*None of the other answers are correct.
Solution or Explanation
The flower pot has a displacement of - 1.3 m j in this time. Its acceleration is a constant -g j. The final speed of the pot is -20
m/s j. Use these values in the constant accleration equation
vfy2 = viy2 + ∆y ay
to get vfy. And then use this value in the equation
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vfy2 - viy2 = ay ∆t
to find the time.
3. Rotation [504237]
An 3 m diameter, 30 kg solid spehere is rolling without slipping on a flat frictionless surface with an angular velocity of 32 rad/s.
What is the ratio of its translational kinetic energy over its rotational kinetic energy? (The moment of inertia of a solid sphere is
2/5 m R2.)
*00*e*None of the other answers is correct.
*10*a*2.5 J
*07*d*0.625 J
*05*b*2.25 J
*05*c*2 J
Solution or Explanation
The translational kinetic energy is
1/2 m v2.
The rotational kinetic energy is
1/2 I ω 2 = 1/2 (2/5 m R2) (v/R)2 = 1/5 m v2.
The ratio of these is 2.5.
4. Force [504234]
For which of the following forces is the direction always parallel to the surface of contact?
*05*d*Normal Force
*02*c*All of these forces always have that direction.
*02*b*Gravity
*10*a*Friction
*00*e*None of these forces always have that direction.
Solution or Explanation
Only friction has this direction. It also opposes the motion of the object or its possibility of motion.
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5. Force and Systems [504241]
An external force that points to the right is exterted on a two particle system. What can you say about the motion of the centerof-mass of the system?
*10*a*It could be moving to the left, moving to the right, or be instantaneously at rest.
*01*e*It must be at rest.
*01*c*It must be moving to the left.
*06*b*It could be moving to the left or to the right, but it cannot be at rest.
*03*d*It must be moving to the right.
Solution or Explanation
The force acting on the system only changes the accleration of the box. If the box was moving to the left initially, it could now
be moving to the left, to the right or momentarily at rest.
6. Rotation [504243]
A man lifts a 5 kg board of uniform density from one end. The board has a length 5.3 m. The force with which the man lifts is
constant and has a magnitude of 200 N. Assume that he lifts straight up. What is the magnitude of the total torque acting on the
board?
*02*d*0.00 Nm
*04*c*1060.00 Nm
*03*b*129.98 Nm
*00*e*None of the other answers are correct.
*10*a*930.02 Nm
Solution or Explanation
The center-of-mass of the board is L/2 and gravity acts there pointing downward. The force from Dr. Mike acts at L pointing
upward. Thus the torques are - L/2 mg k and L F k. The total torque is then - L/2 mg k + L F k.
7. Impulse/Momentum [504235]
A 500 g cart is released from rest 5.0 m from the bottom of a frictionless, 30o ramp. The cart rolls down the ramp and bounces
off a rubber block at the bottom. The figure shows the force during the collision. What is the impulse on the cart during the
collision?
*07*b*3.50 kg m/s
*10*a*2.67 kg m/s
*01*d*0.83 kg m/s
*00*e*None of the other answers is correct.
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*03*c*0 kg m/s
Solution or Explanation
The impulse is given by the integral of force over time. In this case, it is the area under the curve in the figure. Since it is a
triangle, the area is A = 1/2 base x height. Don't forget to convert ms to s.
8. Force [504233]
Bailey D. Wonderdog is standing in an elevator that is accelerating upward at a rate of 15 m/s2. Her mass is about 30 kg. What
is the magnitude of the normal force on her paws?
*07*b*155.70 N
*10*a*744.30 N
*01*d*294.30 N
*05*e*588.60 N
*02*c*450.00 N
Solution or Explanation
Two forces act on Bailey, gravity and the normal force acting on her paws. Gravity has a magnitude of mg = 294.3 N and points
downward. The normal force points upward and we do not know its magnitude. The acceleration has a magnitude of 15 m/s2
and points upward. The sum of both forces is
N j - mg j = ma j.
Thus,
N = ma + mg.
9. Work/Energy [504236]
Which of the following has the larger kinetic energy?
*10*a*A 2000 kg car moving at at 20 m/s.
*01*d*A 50 kg dog running on a frictionless surface at 20 m/s.
*01*e*All of the other answers have the same kinetic energy.
*05*b*A 0.25 kg bullet fired at 200 m/s.
*01*c*A 0.01 kg marble ball rolling at 10 m/s.
Solution or Explanation
A 2000 kg car moving at 20 m/s has a kinetic energy of 4. 105 J.
A 0.25 kg bullet moving at 200 m/s has a kinetic energy of 5000 J.
A 50 kg dog running at 20 m/s has a kinetic energy of 100 J.
A 0.01 kg car rolling at 10 m/s has a kinetic energy of 0.5 J.
10. Mathematics [488212]
Estimate to within an order of magnitude the number of students Dr. Mike will teach in his career at NMSU. In order to get the
correct answer, you should assume he will teach for about 25 years and use a reasonable estimate of the number of student he
teaches each year.
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*05*c*105
*01*d*106
*10*a*103
*01*b*101
*00*e*109
Solution or Explanation
To order-of-magnitude, Dr. Mike (or any teacher) teaches between 10 and 100 students a year. If he teaches for 25 years, this
would be 10 years to order of magnitude. Thus, to order-of magnitude, he will teach between 1000 and 10,00 students in his
career. (By the way, he has already taught close to 2000 students in only 7 years.)
Assignment Details