CONFIDENTIAL
UNIVERSITI TUN HUSSEIN ONN MALAYSIA
FINAL EXAMINATION
SEMESTER II
SESSION 2013/2014
COURSE NAME
PHYSICS TECHNOLOGY
COURSE CODE
BWM 12603
PROGRAM
BND
EXAMINATION DATE
JUNE 2014
DURATION
3 HOURS
INSTRUCTION
ANSWER ALL QUESTIONS IN
PART A. ANSWER ONLY 3
QUESTIONS IN PART B
THIS QUESTION PAPER CONSISTS OF NINE (9) PAGES
BWM 12603
PART A: Answer all questions.
Q1 a) A brother and sister try to communicate with a string tied between two tin cans
as shown in Figure Q1 a). If the string is 9.5 m long, has a mass of 55 g and is
pulled taut with a tension of 7.5 N, how much time does it take for a wave to
travel from one end of the string to the other? Express your answer using two
significant figures.
(4 Marks)
b) During an earthquake, the floor of an apartment building is measured to
oscillate in approximately simple harmonic motion with a period of 1.90 s and
an amplitude of 8.35 cm.
i) Determine the maximum speed of the wave acting to the floor during
this motion.
ii) Determine the maximum acceleration of the wave acting to the floor
during this motion.
(7 marks)
c) The wavelength and period of a wave are 15 cm and 0.03 s respectively.
i)
Calculate the phase difference between two points of the wave
separated by a distance of 7 cm.
ii) Calculate the velocity of this wave
iii) How much is the change in phase for a particle after 0.04 s
(9 marks)
2
a) It takes 487.5 J to heat 25 grams of copper from 25 C to 75 C. What is the
specific heat in Joules/gC?.
(4 marks)
b)
A 1.0 kg Aluminium block has an initial temperature of 10.0C. What will the
final temperature of the Aluminium block be if 3.0 x 104 J of heat is added?
(4 marks)
c) A 0.50 kg block of metal with an initial temperature of 54.5 is dropped into a
container holding 1.1 kg of water at 20.0. If the final temperature of
the block water system is 21.4, what is the specific heat of
the metal? Assume the container can be ignored, and that no
heat is exchanged with the surroundings.
(7 marks)
PART B : Please answer 3 questions only.
Q3 a) State the Archimedes principle.
BWM 12603
A block of wood floats in a container of water, sealed and transported to the
moon. How would the same block of wood float in the same sealed container
of water on the surface of the moon?
(4 marks)
b)
A huge rising balloon has a volume of 2300 m and is filled with hot air with a
density of 0.92 kg m. The cold air surrounding the balloon has a density of
1.29 kg m. How much load can the balloon carry?
(6 marks)
c) In a hydraulic car lift, the input piston has a radius of r 1 = 0.0120 m and a
negligible weight. The output plunger has a radius of r 2 = 0.150 m. The
combined weight of the car and the plunger is F 2 = 20 500 N. The lift uses
hydraulic oil that has a density of 8.00 102 kg/m3. What input force F 1 is
needed to support the car and the output plunger when the bottom surfaces of
the piston and plunger are at
(i) the same level and
(ii) the levels with h = 1.10 m?
(10 marks)
Q4 a) Suppose your 60.0-L (15.9-gal) steel gasoline tank is full of gas, so both the
tank and the gasoline have a temperature of 15.0C. How much gasoline has
spilled by the time they warm to 35.0C?
(6 marks)
b) As a result of a temperature rise of 32 C a bar with a crack at its centre
buckles upward, as shown in Figure Q4 (b). If the fixed distance between the
ends of the bar is 3.77 m and the coefficient of linear expansion of the bar is
2.5x10-5 K-1, find the rise at the centre.
(8 marks)
c) A hole with a diameter of 0.85 cm is drilled into a steel plate. At 30.0C, the
hole exactly accommodates an aluminum rod of the same diameter. What is the
spacing between the plate and the rod when they are cooled to 0.0C?
(6 marks)
Q5 a) The velocity of a body of mass m that has been acted upon by a force F for a
time t is given by Ft=mv. Show that this equation is dimensionally correct.
(6 marks)
BWM 12603
b) An airplane flies due south at 175 km/hwith respect to the air. There is a wind
blowing at 85km/h to the east relative tothe ground. What are the planes speed
and direction with respect to the ground?
(4 marks)
c) Lilian and Chrissy are going to sleepover night in their tree house and are using
some ropes to pull up a box containing their pillows and blankets,which have a
total mass of 3.20 kg. The girls stand on different branches, as shown in Figure
5 (c) , and pull at the angles and with the forces indicated. Find the x- and ycomponents of the net force on the box.
(10 marks)
Q6 a) A bicycle rider pushes a bicycle that has amass of 13 kg up a steep hill. The
incline is 25 andthe road is 275 m long, as shownin Figure. The rider pushes
the bikeparallel to the road with a force of 25 N.
i. How much work does the rider do on the bike?
ii.
How much work is done by the force ofgravity on the bike?
(7 marks)
b) A 2.00-g bullet, moving at 538 m/s,strikes a 0.250-kg piece of wood at reston a
frictionless table. The bullet sticks inthe wood, and the combined mass
movesslowly down the table.
i)
ii)
iii)
iv)
v)
Draw energy bar graphs and momentum vectors for the collision.
Find the speed of the system after the collision.
Find the kinetic energy of the system before the collision.
Find the kinetic energy of the system after the collision.
What percentage of the systems original kinetic energy was lost?
(13 marks)
END OF QUESTION
FINAL EXAMINATION
SEMESTER / SESSION
: SEM I / 2012/2013
PROGRAMME
BWM 12603
COURSE : PHYSICS TECHNOLOGY
COURSE CODE
: BWM 12603
Figure Q1 (a)
FINAL EXAMINATION
SEMESTER / SESSION : SEM II / 2012/2013
COURSE : PHYSICS TECHNOLOGY
Figure Q1 (a)
PROGRAMME
COURSE CODE
:
: BWM 12603
BWM 12603
Figure Q5 (b)
Figure 5 (c)
FINAL EXAMINATION
SEMESTER / SESSION : SEM II / 2012/2013
COURSE : PHYSICS TECHNOLOGY
PROGRAMME
COURSE CODE
:
: BWM 12603
BWM 12603
Figure Q6 (a
FORMULA
SEMESTER / SESSION : SEM II / 2012/2013
COURSE : FIZIK TEKNOLOGI
Gravity acceleration,
PROGRAMME
COURSE CODE
1 feet = 12 in
1 feet =30.48cm=0.3048 m
:
: BWM 12603
P mv
BWM 12603
g = 9.81 m/s2
1 mi = 1.609 km
W = Fs = Fs cos
1
mv 2
2
Eu
E J E k Eu
U mgh
s ut
a 2 x
T 2
v A2 x 2
1
1
mv 2 m 2 ( A 2 x 2 )
2
2
d
dt
d
dt
1 2
at
2
ac
v2
2r
r
v 2 u 2 2as
a r 4 2
F ma
o t
W mg
1
o t t 2
2
fk = k N
v r
v u at
1
1
mv 22 mv12 (mgh2 mgh1 )
2
2
s r
a r
R x2 R y2
Ry
Rx
Wn = K
Ek
1
m 2 A 2
2
tan 1
K = - U
1 2 1
kx m 2 x 2
2
2
fs = s N
2 o2 2
BWM 12603
BWM 12603
Schema for Final BWM 12603
Q
Answer
Total
Q1
a) v = d/t
d/t = sqrt (F / )
9.5/t = sqrt (7.5 N / (0.055 kg / 9.5 m))
9.5/t = sqrt (7.5 / 0.005789)
9.5/t = sqrt (1296)
9.5/t = 36
t = 0.26 seconds
F1
S1
A1
U1
b) i)T is related to angular motion by:
7
F1
T = 2/
: angular velocity. and amplitude A as the radius to
convert to linear units (m/s)
T = 2/
T = 2
= 2/T
= 6.28318 / 1.90
= 3.30694 radians/s
F1
S1
Convert to linear units (m/s) by multiplying by the amplitude (which
needs to be converted from cm to m):
A1
U1
m/s = * r
m/s = 3.30694 * 0.0835
v = 0.276m/s
ii)
Maximum acceleration is given by:
F1
a=A
a = 3.306942*0.0835
a = 10.936*0.0835
A1
10
BWM 12603
a = 0.913m/s2
c) The wavelength and period of a wave are 15 cm and 0.03 s
respectively.
i)
Calculate the phase difference between two points of the
wave
separated by a distance of 7 cm.
ii) Calculate the velocity of this wave
iii) How much is the change in phase for a particle after 0.04 s
(i)
= 0.07 m
=
9
F1
S1
A1
(0.07)
= 2.932 rad
(ii)
v = f
F1
S1
A1
=
=
= 5 ms-1
= 0.04 s
(iii)
F1
S1
A1
=
= 2.67 rad
Q2
a) Use the formula
11
BWM 12603
F1
S1
A1
U1
Q = mcT
487.5 J = (25 g)c(75 C - 25 C)
487.5 J = (25 g)c(50 C)
c = 487.5 J/(25g)(50 C)
c = 0.39 J/gC
The specific heat of copper is 0.39 J/gC
F1
S1
A1
U1
F1
7
F1
S1
S2
A1
U1
Q3
12
BWM 12603
(a)
Archimedes Principle stated that the upward buoyant force exerted
on a body immersed in a fluid is equal to the weight of the fluid the
body displaces.
The wood will float the same as the density of the wood would not
be affected by the moon gravity or the changes in weight for water
and wood does not change the mass or density.
Presure; given by balloon:
(b)
Pbaloon= gh should be equal to Pressure by cold air
Pcold air = cold air gh
M1
4
M1
M1
M1
F1
F1
S1
A1
A2
Pbaloon= gh
= 0.92 kg m x 9.81ms2 x h
Pcold air = cold air gh
= 1.29 kg m x9.81 ms2 x h
Pcold air - Pbaloon = (12.6549 9.0252) kgm-1 h = (m/V)gh.
m/2300m3 = 9.81 ( 3.6297) kgm-1
The load = m= 851 kg
i)
F1
S1
S1
A1
(c)
ii) P2 = P1 + gh
P2 =F2/(pr22) and P1 =F1/(pr12):
F2
F
12 gh
2
r2 r1
Q4
10
F1
F1
S1
A1
A2
a) Use the equation for volume expansion of the steel tank:
Vs=sVsT.
13
6
F1
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The increase in volume of the gasoline:
F1
Vgas=gasVgasT.
Find the difference in volume to determine the amount spilled as
F1
S1
A1
U1
Vspill=VgasVs.
Vspill = (gass)VT
= [(95035)106/C](60.0L)(20.0C)
= 1.10L
T = 32 C
Lo = 3.77/2 m = 1.885 m
-5
-1
= 2.510 K
M1
F1
F1
S1
S1
Linear expansion
L = Lo + L = Lo + Lo T
From Pythagoras theorem
L2 = Lo2 + h2
h2 = L2 Lo2
= (Lo + Lo T)2 Lo2
= 2 Lo2T + 2 Lo2 T2
h = (2 T) Lo,
neglecting very small terms
-5
h = {(2)(2.510 )(32)} (1.885) m
h = 0.075 m
S1
A1
U1
The aluminum shrinks more than the steel. Let L be the diameter
of the rod.
6
F1
S1
A1
= (25 10-6C-1)(0.85 cm) (0.0C - 30.0C)
=-6.3810-4 cm
For the steel, the diameter of the hole shrinks by
S1
-6
-1
= (1210 C )(0.85 cm)(0.0C - 30.0C)
= - 3.0610-4 cm
The spacing between the rod and the hole will be
A1
A1
-4
= 1.6 10 cm
Q5
a)
[m][v]
[Ft] = [ma]
[mv] =
14
F1
F1
BWM 12603
[F][t] = [m][a][t]
= kg [ ms-1 ]
= [m][v/t][t]
= kgms-1 (or MLT-1)
= kg[ms-1/s][s]
= kgms-1 (or MLT-1)
S1
S1
A1
A1
[t] = s or T
[m] = kg or M
[v] = ms-1 or LT-1
[F][t] = kgms-1 or MLT-1 = [m]
[v]
b)
F1
A1
F1
S1
A1
VR =
= 190 km/h
VR = 190 km/h, 64 south of east
F1
c) FA on box, x =FA on box cos
= (20.4 N)(cos 120)
= -10.2 N
FA on box, y =FA on box sin
= (20.4 N)(sin 120)
=17.7 N
10
F1
F1
FC on box, x =FC on box cos
= (17.7 N)(cos 55)
= 10.2 N
FC on box, y = FC on box sin
= (17.7 N)(sin 55)
= 14.5 N
Fg, x = 0.0
Fg, y = - mg
= - (3.20 kg)(9.80 m/s2)
= - 31.4 N
Fnet on box, x = FA on box, x + FC on box, x + Fg, x
= - 10.2 N + 10.2 N + 0.0 N
= 0.0 N
Fnet on box, y = FA on box, y + FC on box, y + Fg, y
= 17.7 N + 14.5 N - 31.4 N
= 0.8 N
15
F1
S1
S1
A1
S1
A1
BWM 12603
The net force is 0.8 N in the upward direction.
Q6
a) i) How much work does the rider do on the bike?
Force and displacement are in the same direction.
W = Fd
= (25 N)(275 m)
= 6.9 103 J
ii) How much work is done by the force ofgravity on the bike?
The force is downward ( -90), andthe displacement is 25
above thehorizontal or115 from the force.
W = Fd cos
= mgd cos
= (13 kg)(9.80 m/s2)(275 m)(cos 115)
= -1.5 104 J
i.
7
F1
S1
A1
F1
S1
A1
U1
Draw energy bar graphs and momentum vectors for the
collision.
13
D1
ii.
Find the speed of the system after the collision.
From the conservation of momentum,
mv =(m + M)V
so V =
F1
S1
A1
=
F1
= 4.27 m/s
iii.
Find the kinetic energy of the system before the collision.
K=
S1
A1
16
BWM 12603
=
= 289 J
iv.
Find the kinetic energy of the system after the collision.
A1
Kf =
F1
S1
A1
= 2.30 J
v.
F1
S1
What percentage of the systems original kinetic energy
was lost?
%K lost =
=
= 99.3%
17
