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We use the same procedure as in part 1
again. We already know the length of ~ D, so
p
we just need the length of ~ E :
p
j~ E j =
p
= ( pC) 2 + ( pC) 2 + ( pC) 2
x
y
z 4 q q ( pE ) 2 + ( pE ) 2 + ( pE ) 2
x
y
z
(8 :97) 2 + ( p
= 84:1858
= 9 :17528 : 1:93) 2 + (0) 2 So the dierence is given by
j~ E j j ~ Dj = 6 :8639 kg m =s:
p
p So the dierence is given by
j~ Cj j ~ Bj =
p
p 0:465923 kg m =s: Now nd
~ CD, the change between locations
p
C and D . Answer in kg m =s.
Correct answer:
Explanation: 1:36875 kg m =s. Between which two locations is the magnitude
of the change in momentum greatest?
1. Between points B and C
2. Between points D and E correct
3. Between points C and D
Explanation: cai (atc667) { homework 03 { turner { (57345) 5 Speed is just the magnitude of the velocity,
so no vectors are involved. We just need to
carry out the multiplication An electron travels at speed
j~ = 0 :998c;
vj
where 3 108 m =s m/s is the speed of light.
The electron travels in the direction given by
the unit vector
v = h0:655; 0:492; 0:573i :
^
The mass of an electron is 9
What is the value of
=s 10 kg. j~
vj
c ? 108 m =s) What is the magnitude of the electron’s momentum? Answer in kg m =s:
Correct answer: 4 :26267 1
1 −31 j~ = 0 :998 c = 0 :998(3
vj
= 2 :994 108 m =s : 10−21 kg m =s. Explanation:
Relativistically, the magnitude of momentum is given by 2 You can simplify the calculation if you notice
2
j~
vj
= (0 :998) 2. Also, remember that
that
c
is a unitless quantity. j~ = m j~ :
pj
vj
We know as well as m and j~ , so we have
vj
everything we need. We obtain Correct answer: 15 :8193.
Explanation:
This is just a plug and chug problem. You
can carry out the calculation with the more
complicated expression or you can take the
hint into account and simplify things to nd
that
=s
=s j~
vj
c
1 ~ = hpx; py; pzi :
p
2 0:998 c
c
= 15 :8193 :
1 2 What is the speed of the electron? Answer in
m/s.
Correct answer: 2 :994
Explanation: 108 m =s) The full vector momentum of the electron
would be of the form 1
1 j~ = m j~
pj
vj
= (15 :8193)(9 10−31 kg)(2 :994
= 4 :26267 10−21 kg m =s : 108 m =s. Find px in kg m =s. Remember that any vector
can be \factored" into its magnitude tim...
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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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