Department of Natural Sciences
Clayton State University
Department of Natural Sciences
January 18, 2006
Physics 1111 – Quiz 1
Name _SOLUTION____________________________________
1. The base SI unit of time is
a. Hour.
b. Minute.
c. Second.
d. Millisecond.
2. Given that a = 1.50 m and b = 3.00 m, find:
a. The length of side c.
a2 + b2 = c2
c = (a2 + b2)1/2
c = ((1.50 m)2 + (3.00 m)2)1/2 = 3.35 m
b. The cosine of angle 

cos() = b/c = (3.00 m) / (3.35 m) = 0.894

c. The tangent of angle 

tan() = a/b = (1.50 m) / (3.00 m) = 0.500
d. Angle 

 = tan-1(0.500) = 26.6o

Department of Natural Sciences
Clayton State University
Department of Natural Sciences
January 25, 2006
Physics 1111 – Quiz 2.a
Name ___SOLUTION__________________________________
An airplane travels 280 m down the runway before taking off. It starts from rest and
moves with constant acceleration. The speed of the airplane is 70.0 m/s as it takes off.
a. What is its acceleration?
V2 = V02+ 2 a (x - x0)
V2 = 2 a (x - x0)
a = V2 / (2 (x - x0))
a = 8.75 m/s2
b. How long did it take to reach 70.0 m/s speed?
V = V0+ a t
t = (V - V0)/ a
t = (70.0 m/s)/ (8.75 m/s2) = 8.00 s
c. What is the average speed of the airplane as it travels down the runway?
Vav = (x - x0) / t = (280 m) / (8.00 s) = 35.0 m/s
Department of Natural Sciences
Clayton State University
Department of Natural Sciences
January 25, 2006
Physics 1111 – Quiz 2
Name __SOLUTION___________________________________
An airplane travels 280 m down the runway before taking off. If it starts from rest,
moves with constant acceleration, and becomes airborne in 8.00 s,
a. What is its speed, in m/s, when it takes off?
x - x0 = ½ (V + V0) t
x - x0 = ½ V t
V = 2 ( x - x0 )/ t
V = 2 (280 m) / (8.00 s) = 70.0 m/s
b. What is its acceleration?
V = V0+ a t
a = (V - V0)/ t
a = (70.0 m/s) / (8.00 s) = 8.75 m/s2
c. What is the average velocity of the airplane during these 8 seconds?
Vav = (x - x0) / t = (280 m) / (8.00 s) = 35.0 m/s
Department of Natural Sciences
Clayton State University
February 1, 2006
Physics 1111 – Quiz 3
Name ____SOLUTION_________________________________
Deep inside an ancient physics text you discover two vectors:
A: 450.0 m @ 140.0 o
B: 290.0 m @ -13 .0 o
Not content with these hoary relics, you are asked to find a new vector R = A + B.
Note: Find the magnitude and direction of vector R.
Ax = A cos(A) = (450.0 m) cos(140.0o) = - 345 m
Ay = A sin(A) = (450.0 m) sin(140.0o) = 289 m
Bx = B cos(B) = (290.0 m) cos(-13.0o) = 283 m
By = B sin(B) = (290.0 m) sin( -13.0o) = - 65.2 m
Rx = Ax + Bx = (- 345 m) + (283 m) = - 62.0 m
Ry = Ay + By = (289 m) + (-65.2 m) = 224 m
R = ( Rx2 + Ry2)1/2 = ( (-62.0 m)2 + (224 m)2)1/2 = 232 m
R = tan-1 (Ry/Rx) = tan-1 ((224 m)/(-62.0 m)) = -74.5o+ 180o = 106o
Department of Natural Sciences
Clayton State University
March 1, 2006
Physics 1111 – Quiz 5
Name ___SOLUTION__________________________________
1. An 88-g arrow is fired from a bow whose string exerts an average force of 110 N
on the arrow over a distance of 78 cm. What is the speed of the arrow as it leaves the
bow?
W = F x cos( = (110 N)(0.780 m)cos(0o) = 85.8 J
W = ½ m V2
V2 = 2 W/ m
V = (2 W/ m)1/2
V = (2 ()85.8 J/ (0.0880 kg))1/2 = 44.2 m/s
2. A 7.0-kg monkey swings from one branch to another 1.2 m higher. What is the
change in potential energy?
y = yf – yi = 1.20 m

U = m g yf – m g yi = m g (y) = (7.00 kg) (9.81 m/s2)(1.20 m) = 82.4 J
Department of Natural Sciences
Clayton State University
March 15, 2006
Physics 1111 – Quiz 6
Name __SOLUTION___________________________________
1.
A skier traveling 12.0 m s reaches the foot of a steady upward 18.0º incline and glides
12.2 m up along this slope before coming to rest. What was the average magnitude
of friction force acting on the skier?
Wnc = Ef - Ei
Ei = Ki + Ugi = ½ m Vi2
(Ugi = 0)
Ef = Kf + Ugf = m g yf
(Kf = 0)
Where yf = d sin () = (12.2 m) sin (18.0º) = 3.77 m
Wnc = Wfriction = fk cos(180o)s = - fk s
m) = 5.88 N
fk = - Wnc / s = - (-9.00 J)/(1.53
Department of Natural Sciences
Clayton State University
March 15, 2006
Physics 1111 – Quiz 6a
Name __SOLUTION___________________________________
1.
A skier traveling 12.0 m s reaches the foot of a steady upward 18.0º incline and glides
12.2 m up along this slope before coming to rest. What was the average magnitude
of friction force acting on the skier? Mass of the skier is 70.0 kg.
Wnc = Ef - Ei
Ei = Ki + Ugi = ½ m Vi2
(Ugi = 0)
Ef = Kf + Ugf = m g yf
(Kf = 0)
Where yf = d sin () = (12.2 m) sin (18.0º) = 3.77 m
Wnc = fk d cos(180o) = - fk d
- fk d = m g yf - ½ m Vi2
fk = (m g yf - ½ m Vi2) / (-d) = 201 N
Department of Natural Sciences
Clayton State University
March 22, 2006
Physics 1111 – Quiz 7
Name __SOLUTION___________________________________
1. A 15.0-kg object moving in the  x direction at 5.5 m s collides head-on with a 10.0kg object moving in the  x direction at 4.0 m s .
a. Find the final velocity of each mass if the objects stick together.
m1 V1i + m2 V2i = (m1 + m2 ) V
(m1 V1i + m2 V2i ) / (m1 + m2 ) = V
V = ((15.0 kg) (5.50 m/s) + (10.0 kg)(-4.00 m/s)) / (15.0 kg + 10.0 kg) = + 1.70 m/s
V = 1.70 m/s in the positive x direction
b. How much energy was lost during this collision?
KEi = KE1i + KE2i = ½ m1 V1i2 + ½ m2 V2i2 = ½ (15.0 kg)(5.50 m/s)2 + ½ (10.0 kg)(4.00 m/s)2 = 307 J
KEf = ½ (m1 + m2 )Vf2 = ½ (15.0 kg + 10.0 kg)(1.70 m/s)2 = 36.1 J

KE = KEf - KEi = 36.1 J – 307 J = - 271 J
Department of Natural Sciences
Clayton State University
April 5, 2006
Physics 1111 – Quiz 8
Name ____SOLUTION_________________________________
1. A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant
angular acceleration of 0.600 rad/s2. After the flywheel has turned through 60.0o,
a. What is the magnitude of the centripetal acceleration of the point on the rim of
the flywheel?
)
0.600 rad/s2)
1.12 rad/s
acp = r = (1.12 rad/s) (0.300 m) = 0.376 m/s2
b. What is the magnitude of the tangential acceleration of the same point?
at = r = (0.600 rad/s2) (0.300 m) = 0.180 m/s2
c. What is the magnitude of the total acceleration of the point?
atot = (acp2 + at2)1/2 = 0.417 m/s2
Department of Natural Sciences
Clayton State University
April 12, 2006
Physics 1111 – Quiz 9
Name __SOLUTION___________________________________
1. Two objects attract each other gravitationally. If the distance between their centers is
cut in half, the gravitational force
a. Is cut to one fourth.
b. Is cut in half.
c. Doubles.
d. Quadruples.
(Fg ~ 1/r2)
2. As a rocket moves away from the Earth's surface, the rocket's weight
a. Increases.
b. Decreases. (Fg ~ 1/r2)
c. Remains the same.
d. Depends on how fast it is moving.
3. Let the average orbital radius of a planet be R. Let the orbital period be T. What
quantity is constant for all planets orbiting the Sun?
a. T/R
b. T/R2
c. T2/R3
(T2 = KS R3)
d. T3/R2
Clayton College & State University
Department of Natural Sciences
April 19, 2006
Physics 1111 – Quiz 10
Name _____SOLUTION______________________________
A uniform meter stick with a mass of 180 g is supported horizontally by two vertical
strings, one at the 0-cm mark and the other at the 90-cm mark. What is the tension in the
string (a) at 0 cm? (b) at 90 cm?
Translational equilibrium requirements:
TLx = TL cos (90o) = 0
TLy = TL sin (90o) = TL
TRx = TR cos (90o) = 0
TRy = TR sin (90o) = TR
wx = (mg) cos (-90o) = 0
wy = (mg) sin (-90o) = - mg = - (0.180 kg)(9.81 m/s2) = 1.77 N
Fx = 0: 0 = 0

Fy = 0:
TL + TR – 1.77 N = 0
TL + TR = 1.77 N
Rotational equilibrium requirements:

 TL = 0

 TR = + TR (0.900 m)

 w = - (1.77 N)(0.500 m) = -0.885 N-m

= 0
TR (0.900 m) – 0.885 N-m = 0
TR = 0.983 N
TL= 1.77 N – 0.983 N = 0.786 N