PHYSICS DEPARTMENT
PHY2004, Stanton Exam #2 October 24, 1996
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NAME
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PLEASE SIGN THE SHEET AT THE TOP. THIS INDICATES THAT YOU HAVE NOT
DISCUSSED THIS TEST WITH ANYONE AND THAT YOU NEITHER HAVE GIVEN OR
RECEIVED HELP FROM ANYONE OTHER THAN THE INSTRUCTOR/PROCTOR.
DIRECTIONS
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(1) Code your test number on your green answer sheet (use 76-80 for the
5-digit number). Code your name and student number on your answer
sheet. Darken circles completely (errors can occur if too light).
(2) Print your name on this sheet and sign it also.
(3) For most questions, one of the answers listed will be considered
correct. Sometimes none of the listed answers is close enough to
be considered correct. In that case the correct answer will be
no answer.
(4) Do all scratch work on this exam to the right of the questions, and
anywhere else on this exam. At the end of the test, this exam
printout is to be turned in. No credit will be given without both
the answer sheet and printout, with the scratch work which most
questions demand of anyone.
(5) Work the questions in any order. Incorrect answers are not taken
into account in any way; you may guess at answers you don't know if
you feel that a correct answer is listed. Guessing on all
questions will most likely result in failure.
(6) Black the circle of your intended answer completely, using a number
2 pencil on the answer sheet. Do not make any stray marks or the
answer sheet may not read properly.
(7) As an aid to the examiner (and yourself) in the case of poorly
marked answer sheets, please circle your selected answer on the
examination sheet.
(8) Good luck!!!!
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>>>>>>>> BEFORE YOU FINISH <<<<<<<<
Fold the computer print out so your name is on top, include any figure
sheet inside the print out. Hand in the green answer sheet separately.
If none of the answers are correct, please leave the answer sheet
blank. It is not our intention to omit the right answer, but in
case of a mistake please leave the answer sheet blank.
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g = 32.2ft/s**2 = 9.8 m/s**2
1 hp = 746 watts
* * Problems marked with a * (* *) are more (most) difficult. * *
- An 80 kg skier, skies down a slope of 30 degrees (see figure 1). How
much work is done (in Joules) BY GRAVITY when the skier is at a height
100 m below the originial height?
- 7.84 x 104
- -7.84 x 104
- 3.92 x 104
- -3.92 x 104
- 8.0 x 103
- A fireman jumps down a 20 foot fire pole. Neglecting all friction (both
with the pole and with the air) what is the speed, in miles per hour, of
the fireman as he reaches the bottom of the slide?
- 24.5
- 35.9
- 19.8
- 13.5
- 17.3
- In figure 2 is shown a wire along which a 50-g bead slides. If friction
forces can be ignored, and the bead starts from rest at A, how fast will
it be going at point B in m/s?
- 4.4
- 3.1
- 7.0
- 4.9
- 3.5
- A crane is operated by a 100 hp motor. If it lifts its load at the
rate of 0.1 m/s, then what is the maximum mass (in kg) that the crane can
lift? Assume the crane is 100% efficient.
- 7.6 x 104
- 7.5 x 105
- 102
- 1.0 x 105
- 746
- A simple machine requires an input force of 80 N acting over a distance
of 40 m to raise a 3500 N weight by 30 cm. What is the % efficiency of
this machine?
- 33
- 44
- 100
- 55
- 25
- A pulley system consists of two sets of two pulleys as shown in figure
3. If the system has an efficiency of 85%, Then how much weight W, in lbs.,
can be lifted with a 150 lb input force?
- 510
- 600
- 638
- 255
- 383
- What is the IMA of the lever system shown in figure 4?
- 0.5
- 2
- 3
- 0.33
- 1.5
- A certain hydraulic press has an output piston of 10.0 cm in diameter.
The input piston has a diameter of 0.5 cm. How large a force (in N) must
be supplied to the input piston to provide an output force of 30,000 N?
Assume the press is ideal.
- 75
- 1500
- 150
- 750
- 600
- How large an impulse (in N-s) does a 20 g bullet moving at 400 m/s
exert on a tree as it strikes and comes to rest in the tree?
- 8
- 8000
- 20
- 20,000
- 1600
- * A 2.5 kg ball is moving at 5.0 m/s in the +y direction. It strikes
a stationary 6.0 kg ball and bounces off so that it is going 0.75 m/s in
the +x direction. What is the x-component of the 6.0 kg ball after the
collision (in m/s)?
- -0.31
- 0.31
- 2.1
- -2.1
- 3.2
- An 80 kg man on ice skates is moving at 6.5 m/s when he runs squarely
into the back of a 60 kg women on ice skates standing at rest. He holds
onto her and they move off together. What is their speed (in m/s) after
the collision?
- 3.7
- 2.8
- 4.9
- 8.7
- 3.3
- * * Particle A has mass 1 kg and is initially moving with a velocity
of 2 m/s toward particle B which has mass 2 kg and is initially at rest.
If they collide head on (i.e. in one dimension) and the collision is perfectly
elastic, then what is the velocity (in m/s) of particle B after the collision?
- 1.3
- 1.0
- -0.7
- 2.7
- 0.7
- * What is the angular speed, in RADIANS/s of the Earth going around
the Sun?
- 2.0 x 10-7
- 3.2 x 10-8
- 7.2 x 10-4
- 7.3 x 10-5
- 4.8 x 10-6
- After the power is turned off to a motor whose shaft speed is 5000
revs/min, the motor takes 10.0 s to stop. Through how many REVOLUTIONS
did the shaft turn while the motor came to a rest? Assume uniform acceleration.
- 417
- 833
- 208
- 25,000
- 500
- What is the maximum angular velocity (in RAD/s) that a 2 kg mass on
a 1.5 m rope can be swung if the rope can support a tension of 350 N before
breaking?
- 10.8
- 5.4
- 67.9
- 16.2
- 117
- If a bicycle is travelling at 40 km/h, how fast are its 70 cm diameter
wheels turning in REVS/MIN?
- 303
- 5.1
- 152
- 48
- 952
- A pendulum (shown in figure 5) has a WEIGHT of 98 N and length of 10
m. What is the moment of inertia of the pendulum about its fixed point
in (kg-m2)? (Assume the rope has no mass.)
- 1000
- 9800
- 980
- 100
- 4900
- What is the moment of inertia for rotation in-plane about the center
sphere for the masses shown in figure 6 in units of (kg-m2)?
Take m = 2.0 kg and r = 1.0 m. The black spheres have mass 2m and the white
spheres have mass m.
- 24
- 12
- 20
- 10
- 16
- What is the rotational kinetic energy (in Joules) for a wheel that
is rotating at 5.0 REVS/s if I for the wheel is 2.7 kg-m2?
- 1332
- 34
- 2665
- 68
- 46
- A figure skater is spinning at the rate of 3 REVS/sec. He pulls in
his arms so that his new radius of gyration is 1/2 his original radius
of gyration. What is his new angular velocity in REVS/sec?
- 12
- 6
- 0.5
- 0.75
- 1.5