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Tutorial 1
GEM2507
Physical Question from Everyday Life
My Biodata
 Name : Setiawan
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1.
Would it be a good idea to measure the distance to the
closest galaxy outside our own by triangulation?
 No. Triangulation only works for distances up to about 50 pc.
 Milky Way’s diameter ~ 30kpc with thickness ~ 300 pc.
 Triangulation can only be used in the direct neighbourhood of sun
2. If we know that a star’s spectrum is blue shifted, what do we
know about its distance?
 Observation shows that all sufficiently distant celestial objects are red
shifted. The distance can be calculated from Hubble’s law v =H0xd
 If a star is blue shifted, it means it can’t be very far away from us.
 Out of billions of galaxies, only somewhat more than 100 galaxies are
blue shifted. One of them is Andromeda galaxy which is 2.57 Mly ==
788 kpc and approching us whit a speed 301 km/s
3. With the help of mathematical induction, prove that
1 1
1
1
n
 
 ... 

3 15 35
(2n  1)(2n  1) 2n  1
for any n
 The theorem is true for n = 1 because we have
1
1
1


3 2 x1  1 3
n
 Suppose that for some integer , n  1,
1
n


2n  1
k 1 2k  12k  1
 We
must show that for n+1 we have
n 1
1
n 1
 2k 12k  1  2n  1  1
k 1
 Proof :
n 1
n
1
1
1




2n  1  12n  1)  1
k 1 2k  12k  1
k 1 2k  12k  1
n
1
1

2n  12n  3
k 1 2 k  12 k  1
n
1


2n  1 2n  12n  3
2n  3n  1

2n  12n  3

2n 2  3n  1

2n  12n  3
2n  1n  1  n  1

2n  12n  3 2(n  1)  1
QED
4.
A slingshot consists of a rubber strap attached to a Y-shaped frame with
a small pouch at the center of the strap to hold a small rock or other
projectile. The rubber strap behaves much like a spring. Suppose that
for a particular slingshot a spring constant of 600 N/m is measured for
the rubber strap. The strap is pulled back approximately 40 cm (0.4m)
prior to being released

What is the potential energy of the system prior to release?

What is the maximum possible kinetic energy that can be gained by
the rock after release?

If the rock has a mass of 50 g (0.05 kg), what is its maximum
possible velocity after release?

Will the rock actually reach these maximum values of kinetic energy
and velocity? Does the rubber strap gain kinetic energy? Explain
4a) Each strap of the rubber band has a spring constant of 600 N/m.
Since there are two straps acting parallely with each other. The
total spring constant will be ktotal = 2k = 2(600N/m) = 1200 N/m
 The elastic potential energy of the spring prior to release is
Eelastic 
1
1
2
ktotal x 2  1200 N / m 0.4m   96 J
2
2
4b) EKmax  Eelastic  96J
4c) 1 m v2  96J
2
max
vmax 
296J 
 61.97m / s
0.05kg 
4d) The rock will not actually reach these maximum values of kinetic energy
and velocity because some of the elastic potential energy will be converted
to kinetic energy of the rubber strap, heat, sound etc
5. If the half-life of a radioactive substance is 12 years, how much
of the substance will have decayed after 100 years?
The substance remain after 12 years is
100 /12
1
 
2
 3.1x103 of the original amount
6. Why is it reasonable to approximate atoms as billiard balls
when discussing a gas at room temperature and one
atmosphere pressure (STP)?
•Ideal gas model assumes no interaction between the gas particles except the
elastic collision between the particles which implies a large the distance
between the gas particles.
•The size of an atom or molecule is in the order of 10-10 m and hence its
volume is (10-10)3.
•The volume of 1 mol of atoms is then equal to 6.022 x 1023 x 10-30 m3 =
6.022x10-7 m3.
•At STP, 1 mole of gas occupies a volume of 22.4 l or 2.24 x 10-2 m3.
Therefore, the volume of the atoms is only 6.022x10-7/ 2.24 x 10-2 = 0.0026%
of the space volume that the gas occupies. So, at STP, on average one gas
particle is indeed very far away from the other. Hence, the ideal gas model is a
suitable model to be used under such condition.
7. Which lattice is more efficient in packing atoms into a fixed
volume of space: simple cubic or face centered cubic?
 As can be seen from the picture, Face Centered Cubic (FCC) can pack
more atoms into the same volume of space.
 In simple cubic lattice, there are only 8 atoms situated at the corner
while in FCC lattice, in addition to these 8 atoms there are also one
atom at each face of the cube
http://feynman.phy.ulaval.ca/marleau/bravais3D_1.htm
8. We have an incompressible fluid flowing into the opening of a
canonical pipe. The radius of the opening is 1 cm and the
fluid velocity is 2 m/s. If the outflow from the pipe end is 1
m/s, what is the radius of the pipe end?
Since the fluid is incomprehensible, the amount of fluid that enters
the cone is the same as the amount of fluid which exits the cone
Q  v1 A1  v2 A2


Q  2m / s  0.01m 2  1m / 2  r2 
r2  0.01 2m
2
9. In your own words, explain the difference between hardness and
toughness. When a steel knife and a diamond are made to collide at high
speed, which of the two is more likely to break?
Toughness: the amount of energy that a material can absorb before rupturing (the
resistance to fracture of a material when stressed). It can be found by finding the
area (i.e. by taking the integral) underneath the stress-strain curve
Q9. Hardness
Definition: the resistance of
a material to permanent deformation (plastic
deformation). Hardness can be measured in various scales, depending on needs.
For example, in mineralogy, the hardness of a material is measured against the scale by
finding the hardest material that the given material can scratch, and/or the softest
material that can scratch the given material (Mohs scale).
Diamond is the hardest material in the world that can sustain high pressure without
getting deformed. However, steel knife is tougher than diamond, in the sense that it
can sustain high impact. So when the two collide at high speed, diamond will likely
to shatter first.
10. The grain size in steel depends on a number of factors
including the speed with which it is cooled. What kind of
grain size (relatively large or relatively small) should I aim for
if I want the steel to be very hard?

d
σ0 and ky are material dependent
σy = yield strength
D = average grain diameter
Adapted from Fig. 7.12, Callister 6e.
(Fig. 7.12 is from A Textbook of Materials
Technology, by Van Vlack, Pearson
Education, Inc., Upper Saddle River, NJ.)
y
 y  0
ky
B
ar
nd



n
u
bo

grain A
ai
gr
in

slip plane
a
gr

During plastic deformation, slip or dislocation
must take place across the boundary between
grain A and B in the picture
Grain boundaries are barriers to slip for two
reason
1. Since the two grains are of different
orientations, dislocation passing into grain B
will have to change its direction of motion; it
become more difficult as the crystallographic
misorientation increases
2. The atomic disorder in the grain boundary
result in discontinuity of slip planes from grain
into the other
High-angle boundaries are better in blocking
slip !
• Smaller grain size:
more barriers to slip.
The so called Hall- Petch eq. expresses this