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Physics 1311
Spring 2010
PS #2 Solutions
Page 1 of 9
Physics 1311
Problem Set #2 Solutions
Problem 1:
You slide a coin across the foor, and observe that it slows down and eventually
stops. A sensitive thermometer shows that its temperature increased. What can we conclude?
As your textbook explains in detail, interactions cause changes. In this example, both the
temperature and the motion of the coin changed. (Its speed changed even if its direction
remained the same.) Both of these changes indicate that the coin interacted with another
object or set of objects.
Problem 2:
In which situations is it reasonable to use the approximate Formula For the momen
tum oF an object, instead oF the Full relativistically correct Formula?
The criterion for determining whether we can approximate the momentum as
±
p
=
m
±
v
is how
close
γ
is to being 1. This in turn depends on how fast the object is moving compared to
the speed of light. Well, clearly for highway travel and jet planes, the speed is negligible, so
the approximation works. A neutron traveling at about 3000 m/s is still traveling at 0
.
001%
the speed of light, which is pretty darn slow. What about a vacuum tube electron traveling
at 0
.
02
c
? That sounds pretty fast, but
γ
=1
/
√
1

0
.
02
2
.
0002, so it’s still a very good
approximation to use Newtonian momentum. In fact,
γ <
1
.
01 even for speeds up to 14%
of light speed. So it’s really only the outerspace proton that requires the full relativistic
momentum formula in this case.
Problem 3:
An object oF mass
1
.
34
mg is traveling at a speed oF
0
.
480
c
. (a) Calculate
the Newtonian momentum oF the object. (b) Calculate the correct relativistic momentum oF
the object. (c) Using the technique oF Taylor expansion, ±nd an approximate expression For the
relativistic Factor
γ
that is correct to order
v
2
. (d) Calculate the percentage error (including
sign) between the momentum calculated using your approximate expression For
γ
, and the actual
momentum oF the object. (Just enter a number, without the percent sign.)
(a) The Newtonian momentum is simply
p
=
mv
= (1
.
34
×
10

6
kg)(0
.
480)(3
.
00
×
10
8
m/s) = 192
.
96 kg m/s
.
=
193 kg m/s
.
(b) The relativistic momentum has an extra factor
γ
compared with the Newtonian mo
mentum. For the given numbers,
γ
=
1
±
1

v
2
/c
2
=
1
√
1

0
.
480
2
.
140
.
Using the result of part (a), we compute the relativistic momentum to be
p
=
γmv
= (1
.
140)(192
.
96) = 219
.
96 kg m/s
.
=
220 kg m/s
.
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Spring 2010
PS #2 Solutions
Page 2 of 9
(c) There are several ways to compute the smallspeed Taylor series (polynomial) expan
sion for the relativistic factor
γ
. The most straightforward way is to write
γ
(
v
)
.
=
γ
(0) +
γ
±
(0)
v
+
γ
±
(0)
2
v
2
+
. . . ,
for which we need the Frst and second derivatives of
γ
with respect to
v
. Using the
chain rule, we obtain:
γ
(
v
)=
1
±
1

v
2
/c
2
γ
±
(
v
v
c
2
²
1

v
2
c
2
³
3
/
2
γ
±
(
v
1
c
2
²
1

v
2
c
2
³
3
/
2
+
3
v
2
c
4
²
1

v
2
c
2
³
5
/
2
.
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 Spring '10
 Wiegert
 Physics

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