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printout
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have
13
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
An object having an initial momentum that
may
be
represented
by
the
vector
below
strikes an object that is initially at rest.
Which of the following sets of vectors may
represent the momenta of the two objects af
ter the collision?
Note carefully:
The original vector above
and the following vectors are all drawn to the
same length scale.
1.
correct
2.
3.
4.
5.
6.
7.
Explanation:
There is no external force for the twoobject
system, so the total momentum is a constant.
From the choices, applying the vector sum
mation for the momenta of the two objects,
we can easily identify the correct choice.
The figure below shows the sum of the
x

and
y
components of the vectors which repre
sent the correct answer.
initial momentum
The horizontal vectors add to be the same
length as the vector presented in the question.
The vertical vectors cancel, as expected, since
there is no vertical momentum.
002
10.0 points
A uranium nucleus
238
U may stay in one piece
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
2
.
45
×
10
7
m
/
s, what would be the recoil speed
of the thorium nucleus?
Correct answer: 4
.
19278
×
10
5
m
/
s.
Explanation:
Let :
v
α
= 2
.
45
×
10
7
m
/
s
,
homework 23 – Turner – (58185)
1
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hinojosa (jlh3938) – homework 23 – Turner – (58185)
2
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:
vector
P
tot
=
M
α
vectorv
α
+
M
Th
vectorv
Th
= 0
.
This means that the Thorium nucleus recoils
in the direction exactly opposite to that of the
α
particle with speed
bardbl
vectorv
Th
bardbl
=
bardbl
vectorv
α
bardbl
M
α
M
Th
=
(2
.
45
×
10
7
m
/
s) (6
.
64
×
10
−
27
kg)
3
.
88
×
10
−
25
kg
=
4
.
19278
×
10
5
m
/
s
.
003
10.0 points
A(n) 1480 kg car moving with a speed of
9
.
13 m
/
s collides with a utility pole and is
brought to rest in 0
.
344 s.
Find the magnitude of the average force
exerted on the car during the collision.
Correct answer: 39280
.
2 N.
Explanation:
The impulse equals the change in momen
tum, therefore
F
Δ
t
= Δ
p
=
m v
f
−
m v
i
.
Thus
F
=
m v
f
−
m v
i
Δ
t
=
0
−
(1480 kg) (9
.
13 m
/
s)
(0
.
344 s)
=
−
39280
.
2 N
bardbl
vector
F
bardbl
=
39280
.
2 N
,
where the negative sign indicates the direction
of the force is opposite to that of the car’s
original motion.
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 Spring '08
 Turner
 Kinetic Energy, Mass, Momentum, Work, Correct Answer, kg, Hinojosa

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