Pickering, Tracey – Homework 18 – Due: Mar 6 2006, noon – Inst: Drummond
1
This
printout
should
have
14
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
An object of mass
m
is moving with speed
v
0
to the right on a horizontal frictionless
surface, as shown, when it explodes into two
pieces. Subsequently, one piece of mass
24
29
m
moves with a speed
v
24
/
29
=
v
0
4
to the left.
m
v
0
before
24
29
m
v
0
4
5
29
m
v
5
/
29
after
The speed
v
f
=
k
~v
5
/
29
k
of the other piece
the object is
1.
v
f
= 6
v
0
.
2.
None of these is correct.
3.
v
f
= 5
v
0
.
4.
v
f
= 8
v
0
.
5.
v
f
= 4
v
0
.
6.
v
f
= 7
v
0
.
correct
7.
v
f
= 2
v
0
.
8.
v
f
= 9
v
0
.
9.
v
f
= 10
v
0
.
10.
v
f
= 3
v
0
.
Explanation:
The horizontal component of the momen
tum is conserved, so
0 +
m v
0
=
24
29
m v
24
/
29
+
5
29
m v
5
/
29
0 +
m v
0
=
24
29
m
‡

v
0
4
·
+
5
29
m v
5
/
29
m v
0
=

24
116
m v
0
+
5
29
m v
5
/
29
5
29
v
5
/
29
=
116
116
+
24
116
¶
v
0
5
29
v
5
/
29
=
140
116
v
0
v
5
/
29
=
140
116
29
5
v
0
k
~v
5
/
29
k
=
7
v
0
.
002
(part 1 of 1) 10 points
Bill (mass
m
) plants both feet solidly on the
ground and then jumps straight up with ve
locity
→
v
.
The earth (mass
M
) then has velocity
1.
V
Earth
= +
M
m
¶
→
v
man
.
2.
V
Earth
=

→
v
man
.
3.
V
Earth
=

r
m
M
→
v
man
.
4.
V
Earth
= +
→
v
man
.
5.
V
Earth
= +
r
m
M
→
v
man
.
6.
V
Earth
=

M
m
¶
→
v
man
.
7.
V
Earth
= +
‡
m
M
·
→
v
man
.
8.
V
Earth
=

‡
m
M
·
→
v
man
.
correct
Explanation:
The momentum is conserved. We have
m
→
v
man
+
M
→
V
Earth
= 0
So
→
V
Earth
=

‡
m
M
·
→
v
man
.
003
(part 1 of 1) 10 points
A uranium nucleus
238
U may stay in one piece
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Pickering, Tracey – Homework 18 – Due: Mar 6 2006, noon – Inst: Drummond
2
for billions of years, but sooner or later it de
cays into an
α
particle of mass 6
.
64
×
10

27
kg
and
234
Th nucleus of mass 3
.
88
×
10

25
kg,
and the decay process itself is extremely fast
(it takes about 10

20
s). Suppose the uranium
nucleus was at rest just before the decay.
If the
α
particle is emitted at a speed of
1
.
58
×
10
7
m
/
s, what would be the recoil speed
of the thorium nucleus?
Correct answer: 270392 m
/
s.
Explanation:
Let :
v
α
= 1
.
58
×
10
7
m
/
s
,
M
α
= 6
.
64
×
10

27
kg
,
and
M
Th
= 3
.
88
×
10

25
kg
.
Use momentum conservation: Before the de
cay, the Uranium nucleus had zero momentum
(it was at rest), and hence the net momentum
vector of the decay products should total to
zero:
~
P
tot
=
M
α
~v
α
+
M
Th
~v
Th
= 0
.
This means that the Thorium nucleus recoils
in the direction exactly opposite to that of the
α
particle with speed
k
~v
Th
k
=
k
~v
α
k
M
α
M
Th
=
(1
.
58
×
10
7
m
/
s) (6
.
64
×
10

27
kg)
3
.
88
×
10

25
kg
=
270392 m
/
s
.
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 Fall '08
 Turner
 Energy, Kinetic Energy, Mass, Momentum, Work, m/s, Pickering

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