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DEPARTMENT OF NATURAL SCIENCES
PHYS 1111, Exam 3
Version 2
Total Weight: 100 points
Section 2
November 27, 2012
1.
Check your examination for completeness prior to starting. There are a total of ten (10)
problems on seven (7) pages.
2.
Authorized references include your calculator with calculator handbook, and the Reference
Data Pamphlet (provided by your instructor).
3.
You will have 75 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and four (4) calculation problems. Work five (5) multiple
choice problems and all calculation problems. Show all work; partial credit will be given for correct
work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
5:00 p.m.
6:15 p.m
PROBLEM
POINTS
1-6
25
7
15
8
20
9
15
10
25
TOTAL
100
PERCENTAGE
CREDIT
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR
PARTIAL CREDIT.
1. As seen from above, a car rounds the curved path shown below while moving at a constant speed
counterclockwise. Which vector best represents the acceleration of the car?
a. A.
b. B.
c. C.
d. D.
2. Andy and Charlie are riding on a merry-go-round. Andy rides on a horse at the outer rim
of the circular platform, five times as far from the center of the circular platform as
Charlie, who rides on an inner horse. When the merry-go-round is rotating at a constant
angular speed, which of the following best describes Andy's angular speed?
a. One-fifth of Charlie's.
b. The same as Charlie's
c. Impossible to determine.
d. Five times Charlie's
3. When the merry-go-round is rotating at a constant angular speed, describe Andy's tangential
speed.
a. Five times Charlie's.
b. One-fifth of Charlie's.
c. The same as Charlie's.
d. Impossible to determine.
4. What is the SI unit of momentum?
a.
N∙m.
b. N/s.
c. N∙s.
d.
N/m.
5. The area under the curve on a Force versus time (F vs. t) graph represents
a. Momentum.
b. Work.
c. Kinetic energy.
d. Impulse.
6. A 57.0-g tennis ball is traveling straight at a player at 19.0 m/s. The player volleys the ball
straight back at 21.0 m/s. If the ball remains in contact with the racket for 0.060 0 s, what is
the magnitude of the average force on the ball?
a. 38 N.
b. 56.05 N.
c. 1.9 N.
d. 19.95 N.
PHYS 1111 Exam 2, Version 2
Spring 2012
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7. The magnitude of the net force exerted in the x direction on a 2.45-kg particle varies in time as
shown in the figure below.
a.
Find the impulse of the force over the 5.00-s time interval.
I  12.0 Ns
b. Find the final velocity the particle attains if it is originally at rest.
I  mV f  mVi
I  mV f
Vf 
I
 4.90 m / s
m
8. A 10.1-g bullet is fired into a stationary block of wood having mass m = 5.05 kg. The bullet
imbeds into the block. The speed of the bullet-plus-wood combination immediately after the
collision is 0.596 m/s.
a.
What was the original speed of the bullet?
m1V1i  m2V2i  (m1  m2 )V f
m1V1i  (m1  m2 )V f
V1i  (m1  m2 )V f / m1
PHYS 1111 Exam 2, Version 2
Spring 2012
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V1i  300 m / s
b.
How much energy was lost during collision?
KEi 
1
1
2
2
m1V1i  m2V2i  455 J
2
2
KE f 
1
2
(m1  m2 )V f  0.898 J
2
KE  KE f  KEi  454 J
9. An electric fan is turned off, and its angular velocity decreases uniformly from 480 rev/min to
210 rev/min in a time interval of 4.45 s.
a. Find the angular acceleration in rev/s2.
480 rev / min  8.00 rev / s
210 rev / min  3.50 rev / s
  0  t
  (  0 ) / t  1.01 rev / s 2
b. Find the number of revolutions made by the motor in the time interval of length 4.45 s.
   0  0 t  1 2 t 2  25.6 rev
10. A uniform beam of mass m = 10.0 kg and length l = 2.00 m is hung from two cables, one at the
end of the beam and the other 1.25 m of the way to the other end as shown below. A box of 20.0
kg mass stands at a distance of 0.75 m from the same end. Determine the magnitudes of the forces
the cable exerts on the beam.
PHYS 1111 Exam 2, Version 2
Spring 2012
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(25)
TLx  TL cos(90.0 o )  0
TLy  TL sin(90.0 o )  TL
TRx  TR cos(90.0 o )  0
TRy  TR sin(90.0 o )  TR
wBEAMx  mBEAM g cos(90.0 o )  0
wBEAMy  mBEAM g sin(90.0 o )  98.0 N
PHYS 1111 Exam 2, Version 2
Spring 2012
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wBOXx  mBOX g cos(90.0 o )  0
wBOXy  mBOX g sin(90.0 o )  196.0 N
Fx  0
Fy  0
TL  TR  98.0 N  196.0 N  0
TL  TR  294 N
Choosing the center of rotation at the left end of the beam:
 TL  0
 WBEAM  (98.0 N )(1.00 m)
 WBOX  (196.0 N )(0.750 m)
 TR  TR (1.25 m)
  0
 (98.0 N )(1.00 m)  (196.0 N )(0.750 m)  TR (1.25 m)  0
TR  196 N
TL  98.0 N
PHYS 1111 Exam 2, Version 2
Spring 2012
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