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Physics 111 PRACTICE Final Exam, Spring 2013, Version B
STUDENTS: PLEASE NOTE that this practice exam is only a small selection of possible
problems which could be on the final. While all of these problems would be ‘fair game’
for the final exam, there will be problems on the final exam which are NOT represented
in this practice exam. Please use other practice exams, common exams from this year,
online HW problem sets, and class examples as other sources for your final exam
preparation.
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.
Instructions:
A
 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 25 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.
____________________________________________________________________________________
1. An automobile traveling along a straight road increases its speed from 30.0 m/s to 50.0 m/s in a distance of 180 m. If the
a.
b.
c.
d.
e.
acceleration is constant, how much time elapses while the auto moves this distance?
6.00 s
4.50 s
3.60 s
4.00 s
9.00 s
2. If F = 5.0 N, what is the magnitude of the force exerted by block 2 on block 1?
a.
b.
c.
d.
e.
17 N
19 N
21 N
23 N
5.0 N
3. The position of a particle as it moves along the x axis is given by x = 15e−2t m, where t is in s. What is the acceleration of the
a.
b.
c.
d.
e.
particle at t = 1.0 s?
22 m/s
60 m/s
8.1 m/s
15 m/s
35 m/s
Page 1 of 9
4. A rock is thrown downward from an unknown height above the ground with an initial speed of 10 m/s. It strikes the ground 3.0 s
a.
b.
c.
d.
e.
later. Determine the initial height of the rock above the ground.
44 m
14 m
74 m
30 m
60 m
5. A ball is thrown by a baseball player with an initial velocity of 20m/s directed at an angle of 40 degrees with respect
to the horizontal. What is the EARLIEST time that the ball reaches a height of 5m above its launch point?
A) 0.37 s
B) 0.48 s
C) 2.1 s
D) 0.25 s
E) The ball never reaches that height.
A
6. If F = 40 N and M = 1.5 kg, what is the tension in the string connecting M and 2M? Assume that all surfaces are frictionless.
a.
b.
c.
d.
e.
13 N
23 N
36 N
15 N
28 N
7. A block is pushed across a horizontal surface by the force shown. If the coefficient of kinetic friction between the block and the
surface is 0.30, F = 20 N, θ = 30°, and M = 3.0 kg, what is the magnitude of the acceleration of the block?
a.
b.
c.
d.
e.
2.8 m/s2
2.3 m/s2
1.8 m/s2
3.3 m/s2
5.4 m/s2
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8. A 0.50 kg mass attached to the end of a string swings in a vertical circle (radius = 2.0 m). When the mass is at the highest point of
a.
b.
c.
d.
e.
the circle the speed of the mass is 8.0 m/s. What is the magnitude of the force of the string on the mass at this position?
21 N
11 N
16 N
26 N
36 N
9. As a 1.0-kg object moves from point A to point B, it is acted upon by a single conservative force which does −40 J of work during
a.
b.
c.
d.
e.
this motion. At point A the speed of the particle is 6.0 m/s and the potential energy associated with the force is +50 J. What is the
potential energy at point B?
+72 J
+10 J
+90 J
+28 J
+68 J
10. A certain pendulum consists of a 1.5-kg mass swinging at the end of a string (length = 2.0 m). At the lowest point in the swing the
a.
b.
c.
d.
e.
tension in the string is equal to 20 N. To what maximum height above this lowest point will the mass rise during its oscillation?
77 cm
50 cm
63 cm
36 cm
95 cm
11. A 10-kg block on a horizontal frictionless surface is attached to a light spring (force constant = 1.2 kN/m). The block is initially at
rest at its equilibrium position when a force (magnitude P) acting parallel to the surface is applied to the block, as shown. When
the block is 8.0 cm from the equilibrium position, it has a speed of 0.80 m/s. How much work is done on the block by the force P
as the block moves the 8.0 cm?
a.
b.
c.
d.
e.
8.3 J
6.4 J
7.0 J
7.7 J
3.9 J
12. If the scalar product of two vectors,
a.
b.
c.
d.
e.
and
, is equal to −3.5, if
, and the angle between the two vectors when they are
drawn starting from the same point is equal to 130°, what is the magnitude of
2.1
2.5
2.3
2.7
3.1
?
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13. The block shown is released from rest when the spring is stretched a distance d. If k = 50 N/m, m = 0.50 kg, d = 10 cm, and the
coefficient of kinetic friction between the block and the horizontal surface is equal to 0.25, determine the speed of the block when
it first passes through the position for which the spring is unstretched.
a.
b.
c.
d.
e.
92 cm/s
61 cm/s
71 cm/s
82 cm/s
53 cm/s
14. A 2.0-kg block is projected down a plane that makes an angle of 20° with the horizontal with an initial kinetic energy of 2.0 J. If
a.
b.
c.
d.
e.
the coefficient of kinetic friction between the block and plane is 0.40, how far will the block slide down the plane before coming
to rest?
3.0 m
1.8 m
0.30 m
1.0 m
1.3 m
15. A 2.0-kg object moving 3.0 m/s strikes a 1.0-kg object initially at rest. Immediately after the collision, the 2.0-kg object has a
a.
b.
c.
d.
e.
velocity of 1.5 m/s directed 30° from its initial direction of motion. What is the x component of the velocity of the 1.0-kg object
just after the collision?
3.7 m/s
3.4 m/s
1.5 m/s
2.4 m/s
4.1 m/s
16. At t = 0, a wheel rotating about a fixed axis at a constant angular acceleration of −0.40 rad/s2 has an angular velocity of 1.5 rad/s
a.
b.
c.
d.
e.
and an angular position of 2.3 rad. What is the angular position of the wheel at t = 2.0 s?
4.9 rad
4.7 rad
4.5 rad
4.3 rad
4.1 rad
Page 4 of 9
17. Particles (mass of each = 0.40 kg) are placed at the 60-cm and 100-cm marks of a meter stick of negligible mass. This rigid body
a.
b.
c.
d.
e.
is free to rotate about a frictionless pivot at the 0-cm end. The body is released from rest in the horizontal position. What is the
magnitude of the initial linear acceleration of the end of the body opposite the pivot?
15 m/s2
9.8 m/s2
5.8 m/s2
12 m/s2
4.7 m/s2
18. A uniform rod of mass M = 1.2 kg and length L = 0.80 m, lying on a frictionless horizontal plane, is free to pivot about a vertical
axis through one end, as shown. The moment of inertia of the rod about this axis is given by (1/3)ML2. If a force (F = 5.0 N, θ =
40°) acts as shown, what is the resulting angular acceleration about the pivot point?
a.
b.
c.
d.
e.
16 rad/s2
12 rad/s2
14 rad/s2
10 rad/s2
33 rad/s2
19. A cylindrical shell rolls without slipping down an incline as shown in the figure. The linear acceleration of its center of mass is
a.
b.
c.
d.
e.
(5/7)g sin θ
(1/2)g sin θ
(3/5)g sin θ
(2/3)g sin θ
(4/5)g sin θ
20. If M = 0.50 kg, L = 1.2 m, and the mass of each connecting rod shown is negligible, what is the moment of inertia about an axis
perpendicular to the paper through the center of mass? Treat the mass as particles.
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a.
b.
c.
d.
e.
3.7 kg⋅m2
2.8 kg⋅m2
3.2 kg⋅m2
2.3 kg⋅m2
3.9 kg⋅m2
21. When an object is effectively isolated from external torques, like an ice skater twirling on the tip of one skate, the angular
a.
b.
c.
d.
e.
momentum of the object
can be increased by shifting mass out away from the axis of rotation.
can be decreased by shifting mass out away from the axis of rotation.
can be increased by shifting mass in toward the axis of rotation.
can be decreased by shifting mass in toward the axis of rotation.
cannot be changed except by friction at the point of contact.
22. A uniform 100-lb beam is held in a vertical position by a pin at its lower end and a cable at its upper end. A horizontal force
(magnitude P) acts as shown in the figure. If P = 75 lb, what is the tension in the cable?
a.
b.
c.
d.
e.
54 lb
69 lb
47 lb
61 lb
75 lb
23. A horizontal uniform 1.20-N meter stick is held up by two vertical strings, one at the 20-cm mark and the other at the 60-cm
a.
b.
c.
d.
e.
mark. What is the tension in the string at the 60-cm mark?
0.30 N
0.45 N
0.60 N
0.90 N
1.0 N
24. What is the gravitational force on a 20-kg satellite circling the Earth (radius = 6.4 × 106 m, mass = 6.0 × 1024 kg) with a period of
a.
b.
c.
d.
e.
5.0 h?
88 N
55 N
36 N
98 N
18 N
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25. An object is released from rest when it is a height h above the surface of a planet of mass M and radius R. What is the speed of the
a.
b.
c.
d.
e.
object just before striking the surface of the planet? Neglect any air resistance. Let h = 4.0 × 106 m, R = 5.0 × 106 m, and M = 4.0
× 1024 kg.
7.8 km/s
3.5 km/s
5.4 km/s
6.9 km/s
4.8 km/s
FORMULAS – Final Exam
Conversion Factors: 1 inch = 2.54 cm; 1 mi =1609.3 m; 1 cm=10-2; 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:
vavg

vavg
∆x
=
∆t

∆r
=
∆t
;
;
∆v
aavg =
∆t

∆v

aavg =
∆t
1
x =xi + vi t + at 2
2
;
;
;
dx
v=
dt

 dx
v=
dt
dv
; =
a =
dt

 dv
; =
a =
dt
d 2x
dt 2

d 2r
dt 2
1
  
v= vi + at ; v 2 =
vi2 + 2a ( x − xi ) ; r =ri + vi t + at 2
2
  
; v= vi + at
Circular motion: T = 2π R / v ; T = 2π / ω ; ac = v 2 / R




Newtons Laws: ∑ F = ma ; F12 = − F21
Friction:
Energies:
𝑓𝑠 ≤ 𝜇𝑠 𝑁 ;
1
𝐾 = 2 𝑚𝑣 2 ;
𝑓𝑘 = 𝜇𝑘 𝑁
𝑈𝑔 = 𝑚𝑔𝑦 ;
 
 
1
𝑈𝑠 = 2 𝑘𝑥 2 ; W =
− ∫ F dr =
− F ∆r
 
Etotal =K + U g + U S ; ∆Emech = ∆K + ∆U g + ∆U s = − f=
W
P dW
=
/ dt F v ∆K =
sd
;
;


 

Momentum and Impulse: p = mv ; I = ∫ Fdt = ∆p




Center of mass: rcm = ∑ mi ri / ∑ mi ; vcm = ∑ mi vi / ∑ mi
i
i
i
i


Collisions: p = const and E≠ const (inelastic) or p = const and E= const (elastic)
a=
vt2 / =
r ω 2r
Rotational motion: ω = 2π / T ; ω = dθ / dt ; α = d ω / dt ; vt = rω ; at = rα a=
c
r
2
atot
= ar2 + at2 ; vcm = rω (rolling, no slipping) ; acm = rα
ω
= ωo + α t ; θ f =θi + ωot + α t 2 / 2 ; ω 2f =
ωi2 + 2α (θ f − θi )
Page 7 of 9
I point = MR 2 ; I hoop = MR 2 ; I disk = MR 2 / 2 ; I sphere = 2 MR 2 / 5 ; I shell = 2 MR 2 / 3 ; I rod ( center ) = ML2 /12
  


  
I I cm + Mh 2 ; τ = r × F ; ∑τ = Iα ; L= r × p ; L = I ω
I rod ( end ) = ML2 / 3 ; I = ∑ mi ri 2 ; =
i
K K rot + K cm ; ∆K + ∆U = 0 ; W= τ ∆θ ; Pinst = τω
Energy: K rot = I ω / 2 ; =

Gm1m2
4π 2 3
2
2
ˆ
r
U
=
−
Gm
m
/
r
Gravitation: Fg = −
;
;
;
g
(
r
)
=
GM
/
r
T
a
=
12
1 2
r2
GM
2
Page 8 of 9