Download Phys 111 CE1 2013 Fall

Physics 111 Common Exam 1, Fall 2013, Version A
Name (Print): _______________________________
4 Digit ID:________ Section: ______
Honors Code Pledge: For ethical and fairness reasons all students are pledged to comply with the
provisions of the NJIT Academic Honor Code. You must answer the exam questions entirely by yourself.
Turn off all cell phones, pagers, or other communication devices. Use only your own calculator.
A
Instructions:
First, write your name and section number on both the Scantron card and this exam sheet.
Use the formula sheet (last exam booklet page) and no other materials.
Budget your time. There are 18 multiple choice problems.
For most, if not all, of the multiple choice problems, it will be difficult to arrive at the correct answer without showing
your work. You are strongly encouraged to do so. However, partial credit will not be awarded on the multiple choice
problems.
 Answer each question on the Scantron card using #2 pencil. Also circle your answers on question papers.
 Do not hesitate to ask for clarification of any exam question, if needed, from your proctor or Professor.
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1. Let θ be the angle that a vector makes with the + x axis and measured counter clock-wise from that axis. Vector A

o
o
has a magnitude of 20 and is at an angle θ = 30 . Vector B has magnitude 40 at an angle θ = 90 . The magnitude of the
  
vector C= A + B is:
A) 45
B) 53
C) 57
D) 62
E) 71
 


2. The dot product A B of the two vectors A
−3iˆ + 5 ˆj is:
= 3iˆ + 3 ˆj and B =
A)
B)
C)
D)
E)
9iˆ
−9iˆ
6
9
15
 


3. The cross product A × B of the two vectors A= 3iˆ + 3 ˆj and B =
−3iˆ + 5 ˆj is:
A)
B)
none of the other answers
6
C) 6kˆ
D) 24kˆ
E) 24
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

4. The angle between the vector A = iˆ + ˆj + kˆ and the vector B =−iˆ − ˆj + kˆ in degrees is closest to:
A)
B)
C)
D)
E)
-19
55
71
87
109
A
5. The position of a particle moving along the x axis is given by x = (21 + 22t − 6.0t2)m, where t is in s. What is the average velocity
during the time interval t = 1.0 s to t = 3.0 s?
A)
B)
C)
D)
E)
−6.0 m/s
−4.0 m/s
−2.0 m/s
−8.0 m/s
8.0 m/s
6. A particle moving with a constant acceleration has a velocity of 20 cm/s when its position is x = 10 cm. Its position 7.0 s later is
x = −30 cm. What is the acceleration of the particle?
A) −7.3 cm/s2
B) −8.9 cm/s2
C) −11 cm/s2
D) −15 cm/s2
E) −13 cm/s2
7. A rocket, initially at rest, is fired vertically with an upward acceleration of 10 m/s2. At an altitude of 0.50 km, the engine of the
rocket cuts off. What is the maximum altitude it achieves?
A) 1.9 km
B) 1.3 km
C) 1.6 km
D) 1.1 km
E) 2.1 km
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8. A ball thrown vertically from ground level is caught 3.0 s later by a person on a balcony which is 14 m above the ground.
Determine the initial speed of the ball.
A) 19 m/s
B) 4.7 m/s
C) 10 m/s
D) 34 m/s
E) 17 m/s
A
9. A particle starts from the origin at t = 0 with a velocity of (16 − 12 ) m/s and moves in the xy plane with a
constant acceleration of
A) 52 m/s
B) 39 m/s
C) 46 m/s
D) 33 m/s
E) 43 m/s
= (3.0 − 6.0 ) m/s2. What is the speed of the particle at t = 2.0 s?
10. At t = 0, a particle leaves the origin with a velocity of 9.0 m/s in the positive y direction and moves in the
xy plane with a constant acceleration of (2.0i − 4.0j) m/s2. At the instant the x coordinate of the particle is 15 m,
what is the speed of the particle?
A) 10 m/s
B) 16 m/s
C) 12 m/s
D) 14 m/s
E) 26 m/s

11. The position (in units of meters) of an object is given by r =
−4.00t iˆ + 6.00t 3 ˆj where t is in seconds. At t
= 2.0 s, what is the magnitude of the particle’s acceleration?
A) 0 m/s
2
B) 2.0 m/s
2
C) 17 m/s
2
D) 36 m/s
2
E) 72 m/s
2
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12. A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of
53° above the horizontal. The rock strikes the ground a horizontal distance of 25 m from the base of the
building. Assume that the ground is level and that the side of the building is vertical. How tall is the building?
A) 25.3 m
B) 29.6 m
C) 27.4 m
D) 23.6 m
E) 18.9 m
Vo
θ
H
D
13. An object moving at a constant speed requires 6.0 s to go once around a circle with a diameter of 4.0 m.
What is the magnitude of the instantaneous acceleration of the particle during this time?
A) 2.2 m/s
2
B) 2.7 m/s
2
C) 3.3 m/s
2
D) 3.8 m/s
2
E) 4.4 m/s
2
14. A block lies on a horizontal frictionless surface. A horizontal force of 100 N is applied to the block giving rise to an
2
acceleration of 3.0 m/s . Determine the mass of the block.
A) 9.8 kg
B) 33 kg
C) 10 kg
D) 300 kg
E) none of the other answers
A
15. A 50kg mass is tied to a rope and pulled upward (in the presence of gravity) with a force P such that its
acceleration is 4.0m/s. The force P is closest to
P
A) 290 N
B) 690 N
C) 490 N
D) 200 N
E) 50 N
m
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16. Two blocks are attached with a string and dragged along a horizontal frictionless surface as shown in the
figure. The mass of block 1 is 5 kg while the mass of block 2 is 25 kg. A force of 15 N pulls the blocks to the
right. What is the tension of the string between the two blocks?
A) none of the other answers
B) 0.5 N
C) 7.5 N
D) 12.5 N
E) 2.5 N
m1
m2
F
17. A crate of mass of 15kg has is placed on a horizontal frictionless surface. A person pulls the crate
horizontally across the surface with a force of F=30 N directed 30 degrees above the horizontal. What is the
magnitude of the force on the crate which is exerted by the horizontal surface?
F
A) 147 N
B) 162 N
C) 132 N
D) 121 N
E) 26 N
θ
m
18. The diagram below shows a block of mass m=1.5 kg on a frictionless horizontal surface, as seen from above. Three
forces of magnitudes F1 = 4.0 N , F2 = 5.0 N , and F3 = 8.0 N are applied to the block, initially at rest on the surface, at
angles shown on the diagram. What is the magnitude of the mass’s acceleration?
A) 0.81 m/s
2
B) 11 m/s
2
C) 8.0 m/s
2
D) 3.0 m/s
E) none of the other answers
2
A
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Physics 111 Common Exam 1 Formulas
Conversion Factors: 1 inch = 2.54 cm; 1 mi =1609.3 m; 1 cm=10-2 m; 1 mm= 10-3 m; 1 g=10-3 kg;
= 5.97 ×1024 kg ; REarth
= 6.37 ×106 m
; G 6.674 ×10−11 N m2/kg2 ; M Earth
Physical constants: g = 9.8 m/s2 =
Math: 360° = 2π radians = 1 revolution. Arc length s = rθ ; Vsphere = 4π R 3 / 3 ; Asphere = 4π R 2 ; Acircle = π R 2
−b ± b 2 − 4ac
2a




Vectors: =
A
A Ax iˆ + Ay ˆj ; Ax = A cos(θ ) ; Ay = A sin(θ ) ; =
quadratic formula to solve ax 2 + bx + c =
0 : x=
Ax 2 + Ay 2 ; tan θ =
  
Ay + By
Ax + Bx ; C=
C= A + B implies C=
y
x
   
 ˆj iˆ=
kˆ ˆj=
kˆ 0
A B = A B cos θ = Ax Bx + Ay By + Az Bz ; iˆiˆ= 1= ˆj  ˆj = kˆkˆ ; iˆ=
 
 
 
A× B =
A B sin θ ; A ×
=
B iˆ( Ay Bz − Az By ) + ˆj ( Az Bx − Ax Bz ) + kˆ( Ax By − Ay Bx )
Ay
Ax
iˆ × iˆ = ˆj × ˆj = kˆ × kˆ = 0 ; iˆ × ˆj =
kˆ ; ˆj × kˆ =
iˆ ; kˆ × iˆ =ˆj
1D and 2D motion:
∆x
∆t

∆r
=
∆t
∆v
∆t

∆v
=
∆t
vavg =
;
aavg =

vavg
;

aavg
1
x =xi + vi t + at 2
2
;
d 2x
; ∆=
x x final − xinitial ; ∆=
v v final − vinitial
dt 2

d 2r
;
dt 2
  
1
  
v= vi + at ; v 2 =
; v= vi + at
vi2 + 2a ( x − xi ) ; r =ri + vi t + at 2
2
;
dx
dt

 dx
v=
dt
v=
dv
=
dt

 dv
; =
a =
dt
; =
a
Circular motion: T = 2π R / v ; T = 2π / ω ; ac = v 2 / R




Newtons Laws: ∑ F = ma ; F12 = − F21 ; w = mg
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