Phys 263/Amath 261
Assignment 5
SOLUTIONS
1. First, consider the center of mass momentum, which reacts to two forces: the tension
on the rope (by the hand), and gravity:
˙
P
=
T

Mg.
(1)
Now, the mass of the suspended rope is the mass density times the length
x
:
M
=
μx
(2)
so that the momentum of the system is
P
=
Mv
0
=
μxv
0
(3)
and
˙
P
=
μ
˙
xv
0
=
μv
2
0
(4)
since the speed
v
0
is a constant. Equating these expressions gives
μv
2
0
=
T

Mg
=
T

μxg
(5)
which at
x
=
a
gives
T
=
μ
(
v
2
0
+
ag
)
(6)
2.
(a) The total angle between the two particles in the LAB frame is given by Θ =
ψ
+
ζ
,
where
ζ
is the angle of the emerging particle. From
cos Θ =
(
m
1

m
2
)
v
2
2
m
1
v
1
(7)
since
m
1
=
m
2
, we get cos Θ = 0 or Θ =
π/
2. Since
ψ
=
π/
4, this means
ζ
=
π
4
= 45
◦
(8)
(b) To relate LAB and CM angles, use the formula derived in class
tan
ψ
=
sin
θ
cos
θ
+ (
m
1
/m
2
)
=
sin
θ
cos
θ
+ 1
(9)
and since tan
π/
4 = 1
sin
θ
= cos
θ
+ 1
(10)
which is true for
θ
=
π/
2, plus larger angles like
π
, etc. The smallest angle is the
CM angle of the first ball,
θ
=
π
2
= 90
◦
(11)
1
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(c) We can write conservation of momentum in this case as:
x
component
mu
1
=
mv
1
cos(
π/
4) +
mv
2
cos(
π/
4)
(12)
u
1
=
v
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 Spring '10
 RogerMelko
 Mass, Sin, Fundamental physics concepts, θ ϕ

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