Unformatted text preview: = m v:
First, however, we need to convert our units.
It’s easy to see that
155 g = 0 :155 kg :
And if we look up the conversion factors, we
have
Following the trajectory of the planet as it
goes along its orbit, it’s clear that at point
C the momentum vector will point along the
positive z direction (downward in the diagram), and that at point
D the momentum
vector will point along the positive x direction (to the right). Therefore the two vectors
will be as follows: 95 mph If we subtract these two vectors ~ D
p
the resultant vector will be of the form
~D
p ~ C;
p ~ C = h+ pD; 0; pCi :
p
x
z Since
p
~ C = ~ D = 15000 kg m =s;
p
this vector will point at a 45
the upper right. ◦ 1 hr
= 42 :4687 m =s:
3600 s Then we can write
p = mv
= (0 :155 kg) (42 :4687 m =s)
= 6 :58265 kg m =s: ~ C = h0; 0; + pCi ; and
p
z
~ D = h+ pD; 0; 0i :
p
x 1609:34 m
1 mile The following gure shows a portion of the
trajectory of a ball traveling through the air.
Arrows indicate its momentum at several locations.
~C
p
~D
p
~E
p
D
C
~B
p
E
B angle toward
is: At various locations, the ball’s momentum cai (atc667) { homework 03 { turner { (57345)
~B =
p
~C =
p
~D =
p
~E =
p h3:03; 2:83; 0i kg m =s
h3:55; 0:97; 0i kg m =s
h2:24; 0:57; 0i kg m =s
h8:97; 1:93; 0i kg m =s You will calculate the change in the magnitude of momentum between each pair of adjacent locations. Begin by calculating
j~ BCj,
p
the change between locations B and C . Answer in kg m =s.
Correct answer: 0:465923 kg m =s. Explanation:
To nd the magnitude of the change in momentum between points B and C , we rst nd
the magnitudes of the momentum vectors corresponding to these two locations, and then
we subtract them, always subtracting the earlier magnitude from the later. The magnitude
of ~ B is given by
p
j~ Bj =
p
= q q ( pB) 2 + ( pB) 2 + ( pB) 2
x
y
z
(3 :03) 2 + (2 :83) 2 + (0) 2 p
= 17:1898
= 4 :14606 kg m =s :
And for ~ C,
p
j~ Cj =
p
= q q p We use the same procedure as in part 1.
We already know the length of ~ C, so we just
p
need the length of ~ D:
p
j~ Dj =
p
= (3 :55) 2 + (0 :97) 2 + (0) 2 = 13:5434
= 3 :68014 kg m =s : q q ( pD) 2 + ( pD) 2 + ( pD) 2
x
y
z
(2 :24) 2 + ( 0:57) 2 + (0) 2 p
= 5:3425
= 2 :31138 kg m =s :
So the dierence is given by
j~ Dj j ~ Cj =
p
p 1:36875 kg m =s: Now nd
~ DE, the change between locations
p
D and E . Answer in kg m =s.
Correct answer: 6 :8639 kg m =s.
Explanatio...
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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.
 Spring '08
 Turner
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