Kuwait University Physics Department Physics 121

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Kuwait University
Physics Department
Physics 121
Final Exam
Summer Semester 2011-2012
July 30, 2012 (11.00 am – 1.00 pm)
Name:……………………………………
Student No. ……………………..
Instructors: Drs. Yahya Kokaj, Fikry El-Akkad
Atmospheric Pressure = 1.013 x105 Pa
Density of water = 103 kg/m3
Density of mercury = 13.6 x 103 kg/m3
Instructions to the students:
1.
2.
3.
4.
Solve all the problems. Points for each problem are indicated.
The solution should be given explicitly for each problem.
Take g = 10 m/s2 except otherwise mentioned.
Check the name of your instructor.
NOTE: IT IS STRICTLY FORBIDDEN TO BRING ANY MOBILE COMMUNICATION
DEVICES (MOBILE PHONES, PAGERS, ETC.), INTO THE EXAMINATION HALL.
For use by Instructors only
Problem
Marks
1
2
3
4
5
6
7
8
9
10
Total
Answer the following questions
1- The figure shows a system of two rods of lengths l 1 and l 2 and masses m 1 = 10 kg and m 2 = 20 kg .
Find the distance between the center of mass of this system and the origin O.
[4 points]
M xCM = m1 x1 + m2 x 2
xCM =
10(0 ) + 20 (4 )
= 2.67 m
10 + 20
l1= 6m
M y CM = m1 y1 + m2 y 2
y CM =
10 (3) + 20 (0 )
=1 m
10 + 20
2
2
rcm = xcm
+ y cm
=
(2.67 )2 + 12
O
l2= 8m
= 2.8 m
2- A stone is thrown vertically upward from the top of a 24 m high building as shown in the figure. It strikes
the ground with a speed of 26 m/s. Find the Initial speed of the stone.
[3 points]
vi
v = v − 2 g∆y
2
2
0
(26)2 = v02 − 2 (10)(− 24)
vo =14 m / s
24 m
3. A 2-kg mass moves with a speed of 4 m/s on a frictionless surface. It strikes a spring of force constant
k= 2200 N/m. Find the maximum compression of the spring.
[4 points]
Ei = E f
m
1 2 1 2
mv = k x
2
2
2(4 )
x=
= 0.12 m
2200
2
4. A disk of radius 2.5 m rotates with constant angular acceleration of 1200 rev/min2. At the moment when
its angular speed is 2.0 rad/s, Find
a- The tangential acceleration of a point at the rim of the disk
αT = αr =
[2 points]
1200 (2π )
2.5 = 5.2 m / s 2
2
(60)
b- The radial acceleration of a point at the rim of the disk
α R = ω 2 r = (2 )2 2.5 = 10 m / s 2
c- The total linear acceleration.
α t = α T2 + α R2 =
(5.23)2 + (10)2
[2points]
[1points]
=11.3 m / s 2
5.
A 6-kg block moves on a rough surface (µ k =0.30) under the action of a force F =60 N as shown in the
figure.
a. Find the normal force on the block
[3 points]
F
ΣFy = N + F sin θ − mg = 0
N = 6 (10) − 60 sin 30 = 30 N
30 ̊
b. Find the frictional force
[2 points]
f k = µk N
f k = 0.3 (30) = 9 N
6. A uniform rod of mass m = 4 kg and length 42 m is pivoted at point “A” while two forces are acting on
it as shown in the figure. A block of mass M is hanged on the rod at a distance x from point “A” so that
the system becomes in a state of equilibrium.
F1= 35 N
a. Find the value of M
F2= 60 N
[2 points]
ΣFy = F1 + F2 − mg − Mg = 0
M =
35 + 60 − 4(10 )
= 5.5 kg
10
b. Find the value of x
A
12m
28m
[3points]
Σiˆ = F1 r1 + F2 r2 − mgr − Mg x = 0
x=
35 (12 ) + 60 (40) − 4(10)21
= 36 m
5.5 (10)
7. A point source emits a spherical wave. At a distance of 10 m from the source, the wave intensity is 84
W/m2. At what distance from the source will the intensity be 8.0 W/m2.
I1  r2 
= 
I 2  r1 
r2 = r1
[3points]
2
I1
84
= 10
= 32m
I2
8
8. A loud music passage produces a sound level of 70 dB. What is the ratio between the intensity of this
sound and that of a person speaking normally whose sound level is 40 dB?
β1 − β 2 =10 log
I1
I2
70 − 40 = 10 log
I1
=10 3
I2
I1
I2
[3points]
9. A garden hose of area A = 12 cm2 is used to fill a round swimming pool 8.0 m in diameter. How long
(in hours) will it take to fill the pool to a depth of 3.8 m if water issues from the hose at a speed v = 4.0
m/s?
[4 points]
v
( )
πr h
V
=
t
t
2
3.14 (4 ) 3.8
−4
12 ×10 (4 ) =
t
4
t = 3.98 ×10 s = 11 h
Av =
A
2
3.8 m
10- A U-Shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully
poured into the left arm of the U-shaped tube until the vertical height of the water column is 27.2 cm.
Calculate the vertical distance h (in cm) of the mercury ( see figure)
[4 points]
ρ ω ghω = ρ m gh
h=
10 3 (272 )
= 2 cm
13.6 × 10 3
Water
h
27.2 cm
Mercury
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