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Unformatted text preview: use the
Pythagorean theorem to nd the magnitude
of a vector, but in this case, our vectors have
only one component each, so nding the magnitudes is easy. For ~ i ,
p
j~ i j = j(0 :057 kg) h 50; 0; 0i m =sj
p
= jh2:85; 0; 0i kg m =sj
= 2 :85 kg m =s : We are looking down on the orbit from above
the north poles of the Sun and Mars, with + x
to the right and + z down the page.
Notice that there is no y component of
momentum at either position, so the change
in momentum vector will be of the form
~ = h ~ x; 0; ~ zi :
p
p
p
Find Correct answer: ~ f = j(0 :057 kg) h 42; 0; 0i m =sj
p
= jh 2:394; 0; 0i kg m =sj
= 2 :394 kg m =s : 9: 6 1027 kg m =s. ~ x = ~B ~ A
p
px px
= m~ x m ~ x
vB
vA
= m ~ x ~x
vB vA
= m ( 15000 m =s 0)
= (6 :4 1023 kg) ( 15000 m =s)
= 9:6 1027 kg m =s : So the change in magnitude is
j ~ij =
p 0:456 kg m =s: The \red planet," Mars, has a mass of 6 :4
1023 kg and travels in a nearly circular orbit
around the Sun, as shown in the following
gure.
~B
v
C ~ x. Answer in kg m =s.
p Explanation:
To nd
~ x, we take the dierence between
p
the initial and nal momenta of the planet: And for ~ f ,
p ~f
p 2 B Sun Find ~ z. Answer in kg m =s.
p Correct answer: 9 :6 1027 kg m =s. Explanation:
We nd
~ z by the method of part 1:
p
~A
v
A +x D
+z
When Mars is at location A , its velocity is ~ z = ~B ~A
p
pz pz
= m~ z m ~ z
vB
vA
= m ~z ~z
vB vA
= m [0 ( 15000) m =s]
= (6 :4 1023 kg) [ ( 15000) m =s]
= 9 :6 1027 kg m =s : ~ A = h0; 0; 15000i m =s:
v
When it reaches location
locity is B , the planet’s ve ~ B = h 15000; 0; 0i m =s:
v Consider the momentum of Mars at positions
C and D on the diagram above. If you found
the change in momentum vector between
C
and D , which direction would it point? cai (atc667) { homework 03 { turner { (57345) 3 A baseball has a mass of about 155 g. What is
the magnitude of the momentum of a baseball
thrown at a speed of 95 mph? Answer in
kg m =s. (Note that you need to convert mass
to kilograms and miles per hour to meters per
second.) 1. Correct answer: 6 :58265 kg m =s.
Explanation:
Since the question asks for the magnitude
of momentum, we don’t need to worry about
direction or vectors. The magnitude of momentum is given by
p...
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This note was uploaded on 02/11/2014 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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