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Ch3HW2 – florin – (56950)
1
This printout should have 16 questions.
Multiplechoice questions may continue
on the next column or page – find all
choices before answering.
And we use the magnitude to help us
find the unit vector:
ˆr
21
=
~r
21
~r
21

001 (part 1 of 7) 4.0 points
In outer space, far from other objects,
block 1 of mass 40 kg is at position
~r
1
= h6, 8, 0O m,
and block 2 of mass 1000 kg is at position
~r
2
= h19, 12, 0O m.
Since the z components of position are
zero, the (vector) gravitational force acting
on block 2 due to block 1 will be of the form
~
21
21
21
F
gr
= hF
gr
,x
, F
gr
,y
, 0O N.
Find F
gr
21
,x
. It helps to make a sketch of the sit
uation. Use G = 6.67 × 10
−
11
m
3
· kg
−
1
· s
−
2
.
Correct answer:
−1.37839 × 10
−
8
N.
Explanation:
The formula for the gravitational force
between two objects is
~
Gm
1
m
2
F
gr
= −
~r
2
ˆr.
We know the masses of the blocks, so
we just need to find the magnitude of the
distance
between
them
and
the
unit
vector pointing from block 1 to block 2.
To find the vector pointing from block 1 to
block 2 (let’s call it ~r
21
), we subtract the
position of block 1 from that of block 2:
~r
21
= ~r
2
− ~r
1
= h19, 12, 0O m
− h6, 8, 0O m =
h13, 4, 0O m .
We need the length of this vector and the
unit vector pointing in the same direction.
The length (or magnitude) is given by
~r
21
=
q
(13 m)
2
+ (4 m)
2
₃
₃
13.6015 m .
₃
₃
=
h
13,
4,
0O
m
13.6015 m
= h0.955779, 0.294086, 0O .
Now we have everything we need to calcu
~
late the components of F
gr
. Let’s go
ahead
and
find
all
three
components
using vector algebra.
~
21
Gm
1
m
2
21
F
gr
= −
~r
21

2
ˆr
=
−
(
G)(40 kg)(1000 kg)
(13.6015 m)
2
× h0.955779, 0.294086, 0O
=
h−1.37839 × 10
−
8
, −4.24119 × 10
−
9
, 0O N
where G = 6.67 × 10
−
11
m
3
· kg
−
1
· s
−
2
.
002 (part 2 of 7) 4.0 points
Find F
gr
21
,y
.
Correct answer:
−4.24119 × 10
−
9
N.
Explanation:
See the explanation for part 1.
003 (part 3 of 7) 4.0 points
At 4.6 s after noon both blocks were at
rest at the positions given above. At 4.7 s
after noon, what is the (vector) momentum
of
block
2?
Note
that
the
momentum
vector will also be of the form
~p
2
f
= hp
2
f,x
, p
2
f,y
, 0O kg · m/s,
so start by finding p
2
f,x
.
Correct answer:
−1.37839 × 10
−
9
N.
Explanation:
We already know the force on block 2.
Using the Momentum Principle, defining
block 2 as our system, we can write
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2
2
2
~
~p
f
=
~p
i
+
F
net
t
~
=
F
net
(
t
f
−
t
i
)
.
According to this formula, the
x
component will be
p
2
f,x
=
₃
−
1
.
37839
×
10
−
8
N (0
.
1 s)
=
−
1
.
37839
×
10
−
9
kg
·
m
/
s
.
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This note was uploaded on 04/04/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner

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