Department of Physics
Phys112: General Physics I
Final Exam: 2009-2010 (093). August 25. 2010
 Notes and textbooks are not allowed, calculators are allowed.
 Formula sheet is included.
 The exam consist of four solving problems and fifteen multiple choice questions. Answer all
questions.
 For the problem solving questions show all your work in a neat and logical manner.
 For multiple choice questions choose one answer. ( if you choose more than one answer it will
be counted zero)
Name:
ID#
Part
Value
I-MC
15
II-1
5
II-2
5
II-3
5
II-4
5
TOTAL
35
Mark
Sec #
Part I: Multiple Choices
1) If L = 350 m then
a)
b)
c)
d)
e)
L = 3.5×10-8 m
L = 3.5×10-4 m
L = 3.5×10-7 m
L = 3.5×10-5 m
L = 3.5×10-6 m
2) Two vectors and are given by: A   5iˆ  5 ˆj and B  3iˆ  6 ˆj  2kˆ , then C  A  2 B is equal to
a) C  iˆ 17 ˆj  4kˆ
b) C 11iˆ  7 ˆj  4kˆ
c) C   11iˆ  7 ˆj  4kˆ
d) C   11iˆ  7 ˆj  2kˆ
e) C   11iˆ  7 ˆj  4kˆ
3) The position of a body moving on an x axis is given by x(t) = 3t2 +8t – 5, with x in meter and t in seconds.
The average velocity between t = 1s to t = 3s is:
a) 20 m/s
b) 26 m/s
c) 35 m/s
d) 15 m/s
e) 10 m/s
4) The position vector of a particle of mass m=5.0 kg is given by: r
seconds. The displacement of the particle from t=0s to t=2s is
a) r  2iˆ  4 ˆj m
 (t  2)iˆ  (t 2 -4) ˆj , where
t is in
b) r  2iˆ  4 ˆj m
c) r   2iˆ  4 ˆj m
d) r   2iˆ  4 ˆj m
e) r  6iˆ  4 ˆj m
5) An object of mass m= 2.0 kg is rotating in circular motion of radius r= 10 m with a speed of v = 10 m/s. Its
centripetal acceleration
a) 1 m/s2
b) 5 m/s2
c) 20 m/s2
d) 2 m/s2
e) 10 m/s2
6) A horizontal force F= 42 N is applied to a 15 kg block on a rough horizontal surface. If the
coefficient of kinetic friction k = 0.20, the acceleration of the block is:
a) 29.4 m/s2
b) 12.6 m/s2
c) 2.8 m/s2.
d) 0.84 m/s2.
e) 9.8 0 m/s2.
fk
15 kg
F= 42 N
7) A force F = 26 N acts on a beam that rotate about the axis #1 as shown in figure. The torque of F with
respect to the axis of rotation is:
a)N×m
b) N×m
c) N×m
d) 27 N×m
e) N×m
F
Axis#1
120 cm
cm
60o
beam
8) An ideal spring is used to fire a 15.0-g pellet horizontally. The spring has a spring constant of 20 N/m and is
initially compressed by 7.0 cm. The kinetic energy of the pellet as it leaves the spring is:
a) zero
b) 2.5 × 10−2 J
c) 4.9 × 10−2 J
d) 9.8 × 10−2 J
e) 1.4 J
9) A disk of mass M = 20 kg and radius R = 0.40 m rolling without slipping over a horizontal surface with
angular speed ω = 20 rad/s, and the speed of the center of the disk is v = 8 m/s the kinetic energy of the disk is:
(assuming the moment of inertia of the disk is I =1.6 kg.m2):
a) 640 J
b) 96 J
c) 960 J
d) 320 J
e) 80 J
10) A cylinder of mass m = 30 kg and the moment of inertia I = 1.5 kg×m2, is rotating about a fixed axis from
the center with angular velocity ω = 8 rad/s, the angular momentum for this cylinder is equal to:
a) 12 kg×m2/s
b) 240 kg×m2/s
c) 45 kg×m2/s
d) 0.4 kg×m2/s
e) 6 kg×m2/s
11) If  = 5.5 rev/s then:
a)
b)
c)
d)
e)
= 345.4 rad/s
= 11 rad/s
= 17.27 rad/s
= 3.14 rad/s
= 34.54 rad/s
12) Two balls of mass m1= 15 kg and m2 = 8 kg are moving with velocities V1=6 m/s to the left and V2= 12 m/s
to the right. The total momentum of the system is:
a) 𝑝⃗ = (90 kg m/s) iˆ
b) 𝑝⃗ = (96 kg m/s) iˆ
c) 𝑝⃗ = (186 kg m/s) iˆ
d) 𝑝⃗ = (6 kg m/s) iˆ
e) 𝑝⃗ = (-6 kg m/s) iˆ

v2

v1
x
m2
m1
13) An object rotating with a constant angular acceleration and its initial angular velocity ωo = 4 rad/s, after
5 s the angular velocity of the object is 7 rad/s. The angular acceleration is
a) 3 rad/s2
b) 0.6 rad/s2
c) 27 rad/s2
d) 15 rad/s2
e) 2.2 rad/s2
14) A 4.0 kg block moving with initial velocity of 6 m/s makes an elastic collision with a 8 kg initially at rest.
The speed of the v2f of the 8 kg block after collision
a)
b)
c)
d)
e)
8 m/s
-4 m/s
4 m/s
3 m/s
12 m/s
4 kg
8 kg, v= 0 m/s
v1i=6 m/s
v1f
v2f
15) A wheel with rotational inertia I1=1.2 kg×m2, mounted on a vertical shaft, is rotating with angular speed ω=
4.2 rad/s. A non-rotating wheel with rotational inertia I2 = 0.4 kg×m2 is suddenly dropped onto the same shaft as
shown. The angular speed ’ of the resultant combination of the two wheels is:
a) ω’ = 3.15 rad/s
b) ω’ = 4.2 rad/s
c) ω’ = 5.0 rad/s
d) ω’ = 1.7 rad/s
e) ω’ = 3.5 rad/s
Part II: Problems solving:
1) The figure shows a block of mass m1 = 5 kg moving with a speed of 20 m/s collide with another block of
mass m2 = 15 kg which is stationary after the collision both blocks stick together and move with a velocity V.
5 kg
a) Find the speed V of both blocks after the collision.
15 kg, v= 0 m/s
20 m/s
V
b) Find kinetic energy of the system before the collision.
c) Calculate the kinetic energy of the system after collision.
2) A rotating object has an angular position is given by (t)= 0.8 + 2.5 t – 2.2 t2, where  in radians and t in
seconds.
a) Find the angular velocity of the object at t = 3s.
b) Calculate the angular acceleration of the object at t = 2s.
3) Figure shows a mass of 25 kg, falls freely from rest starting at a height of 12 m.
a) Calculate its total mechanical Energy at y= 12 m.
y
y=12 m
y=5m
y=0m
b) Calculate its potential energy at y= 5m
c) Determine its kinetic energy at y =5m
4) A rotating disk has a constant angular acceleration and an initial angular speed ωo = 3.4 rad/s. After 40 rad,
the angular speed of the disk is ω = 2 rad/s.
a) What is the angular acceleration of the disk?
b) How much time does it take to reduce the angular speed from3.4 rad/s to 2 rad/s?
Physics 101; Formulas sheet Final Exam
In one Dimension:
In two or three dimensions
dr
r
dv
v  ; vavg 
a  ; v  v0  a t
dt
t
dt
1
r  r0  v0 t  at 2
2
dx
dv
x
v
; a  ; v avg 
; aavg 
dt
dt
t
t
total dis tan ce
Speed S avg 
t
v
1
1
v v 0  at ; x  x 0 v 0t  at 2 ; x  x 0 vt  at 2
2
2
1
v 2  v 02  2a (x  x 0 ); x  x 0  (v  v 0 )t
2
Free fall: a=-g; g= 9.80m/s2

d

d
; 
;  avg 
; 
.
t
dt
t
dt
 2  02  2 (   0 ) . s = R . v = R. a =R.
Rotation
A  B  AB cos  ;
ax  0
ay   9.80 m / s 2
v x v 0 cos 
v y v 0 sin   gt
x  x 0  (v 0 cos  ) t
v2
;
Circular Motion: a 
r
1
K  I 2 ;
2
Power: Pavg 
t ;
T 
1
y  y 0  (v 0 sin  ) t  gt 2
2
2 r
. Friction: f s , max  s FN ;
v
f k  k FN
Fnet  ma.
1
K  mv 2 ; K W net . W  F  d .
2
W
dW
; P
; P F v .
t
dt
v1 f 
J  p f  pi ; J  Favg t
Disk or Cylinder: I 

v 02
R  sin 2
g
Potential energy: U = -W. U(y) = mgy. Emec = K + U. Spring: Fs = -kx ;
p  mv;
0
A  B  Ax Bx  Ay B y  Az Bz
Projectile Motion:
 net  I 

s  r  ; avg 
m1  m2
v1i ;
m1  m2
v2 f 
2m1
v1i .
m1  m2
  r F  rF  rF sin 
2
1
2
MR 2 . Sphere I  MR .
5
2
.
1
K  I 2 .
2
1
U s  kx 2 .
2
I   m i ri 2 .
0
1
 t
2
2
;
0