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Unformatted text preview: es its
unit vector; for example, ~ = j~ a .
a
aj ^
Correct answer: 2 :79205 10−21 kg m =s. Explanation:
Since we already know the magnitude of the
momentum from part 3, we just need to multiply by the x component of the unit vector
pointing in the direction of the momentum.
But since, relativistically,
~ = m~
p
v; cai (atc667) { homework 03 { turner { (57345)
meaning momentum and velocity point in the
same direction, we can simply use the appropriate component of the given velocity unit
vector: Here, punit;x refers to the x component of
the unit momentum vector, and similarly for
the unit velocity vector.
Find py in kg m =s.
2:09723 10−21 kg m =s. Explanation:
We follow the same procedure as in part 4: Correct answer: 4. d~
r
dt
d
h 3 + 4 t; 3t 2; 2t
=
dt
= h4; 6t; 2 21t 2i : ~=
v 7t 3i Clearly, vx(0) = 4.
Find vy(0) :
Correct answer: 0. py = j~ punit;y
pj
= j~ vunit;y
pj
= (4 :26267 10−21 kg m =s)( 0:492)
= 2:09723 10−21 kg m =s : Find pz in kg m =s.
Correct answer: Find the instantaneous velocity at time
t=
0. Start by nding the
x component, vx(0). Explanation:
To obtain the velocity from the position, we
just take the derivative with respect to time.
To take the derivative of a vector, we take the
derivative of each component: px = j~ punit;x
pj
= j~ vunit;x
pj
= (4 :26267 10−21 kg m =s)(0 :655)
= 2 :79205 10−21 kg m =s : Correct answer: 6 2:44251 vy(0) = 6(0) = 0 : Find vz(0) :
10−21 kg m =s. Explanation:
Same procedure again: Correct answer: 2.
Explanation:
Taking the z component from above: pz = j~ punit;z
pj
= j~ vunit;z
pj
= (4 :26267 10−21 kg m =s)( 0:573)
= 2:44251 10−21 kg m =s : Suppose the position of an object at time
given by
~ = h3 + 4 t; 3t 2; 2t
r Explanation:
We just take the y component from above: 7t 3i : t is vz(0) = 2 21(0) 2 = 2 : Now nd the instantaneous acceleration at
time 0. Start with a x(0).
Correct answer: 0.
Explanation:
To nd the acceleration from the velocity,
we take the derivative again: cai (atc667) { homework 03 { turner { (57345)
d~
v
dt
d
=
h 4; 6t; 2 21t 2i
dt
= h0; 6; 42ti : 7 What is the magnitude of the correct relativistic momentum of the proton?
Your answer must be within
5.0%
−17
Correct answer: 1 :12144 10
kg m =s. ~=
a Explanation:
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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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