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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
Fall Term 2010
Problem Set 8: Continuous Mass Transfer Solutions
Problem 1
A rocket has a dry mass (empty of fuel)
m
r
,0
=
2
!
10
7
kg
, and initially carries
fuel with mass
m
f
,0
=
5
!
10
7
kg
. The fuel is ejected at a speed
u
=
2.0
!
10
3
m
"
s
-1
relative
to the rocket. What is the final speed of the rocket after all the fuel has burned?
Solution
The initial mass of the rocket included the fuel is
m
r
,
i
=
m
r
,0
+
m
f
,0
=
2
!
10
7
kg
+
5
!
10
7
kg
=
7
!
10
7
kg
(1)
The ratio of the initial mass of the rocket (including the mass of the fuel) to the final dry
mass of the rocket (empty of fuel) is
R
=
m
r
,
i
m
r
,
d
=
7
!
10
7
kg
2
!
10
7
kg
=
3.5
(0.0.2)
The final speed of the rocket is then
v
r
,
f
=
u
ln
R
=
(2.0
!
10
3
m
"
s
-1
)ln3.5
=
2.5
!
10
3
m
"
s
-1
.
(0.0.3)

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Problem 2:
An ice skater of mass
m
is holding a bag of sand of mass
m
s
that is leaking
sand at a constant rate
b
. The ice skater is pushed with a constant force of magnitude
F
on a frictionless ice surface. At the instant that all the sand has leaked out, what is the
speed of the skater?
Solution:
Choose the positive x-direction to point in the direction that the ice skater is
moving. Let’s take as our system the amount of sand of mass
s
m
!
that leaves the bag
during the time interval
[ ,
]
t
t
t
+
!
, and the ice skater with bag of remaining sand.
At the beginning of the interval the ice skater with bag sand is moving with speed
v
so
the x-component of the momentum at time
t
is given by
p
x
(
t
)
=
(
!
m
s
+
m
(
t
))
v
)
,
(0.0.4)
where
m
(
t
)
is the mass of the car and sand in it at time
t
. Denote by
m
0
=
m
1
+
m
s
where
m
is the mass of the skater and
m
s
,0
is the mass of the sand
in the bag at
0
t
=
,
and
( )
s
m
t
bt
=
is the mass of the sand that has left the car
at time
t
since
m
s
(
t
)
=
dm
s
dt
dt
0
t
!
=
bdt
0
t
!
=
bt
.
(0.0.5)
Thus
m
(
t
)
=
m
0
!
bt
=
m
1
+
m
s
,0
!
bt
.
(0.0.6)
During the interval
[ ,
]
t
t
t
+
!
, the small amount of sand of mass
s
m
!
leaves the bag with
the speed of the skater at the end of the interval
v
v
+
!
.
So the x-component of the
momentum at time
t
t
+
!
is given by
p
x
(
t
+
!
t
)
=
(
!
m
s
+
m
(
t
))(
v
+
!
v
)
.
(0.0.7)
Throughout the interval a constant force
F
is applied to the skater so

- 3 -
0
(
)
( )
lim
x
x
t
p
t
t
p
t
F
t
! "
+
! #
=
!
.
(0.0.8)
From our analysis above Eq. (0.0.8) becomes
F
=
lim
!
t
"
0
(
m
(
t
)
+
!
m
s
)(
v
+
!
v
)
#
(
m
(
t
)
+
!
m
s
)
v
!
t
.
(0.0.9)
Eq. (0.0.9) simplifies to
F
=
lim
!
t
"
0
m
(
t
)
!
v
!
t
+
lim
!
t
"
0
!
m
s
!
v
!
t
.
(0.0.10)
The second term vanishes when we take the
0
t
! "
because it is of second order in the
infinitesimal quantities (in this case
s
m
v
!
!
) and so when dividing by
t
!
the quantity is
of first order and hence vanishes since both
0
s
m
!
"
and
0
v
! "
. So Eq. (0.0.10)
becomes
F
=
lim
!
t
"
0
m
(
t
)
!
v
!
t
.
(0.0.11)
We now use the definition of the derivative:
0
lim
t
v
dv
t
dt
! "
!
=
!
(0.0.12)
in Eq. (0.0.11) which then becomes the differential equation
F
=
m
(
t
)
dv
dt
.
(0.0.13)
Using Eq. (0.0.6) we have
F
=
(
m
1
+
m
s
,0
!
bt
)
dv
dt
.
(0.0.14)
(b) We can integrate this equation through the separation of variable technique. Rewrite
Eq. (0.0.14) as
dv
=
Fdt
(
m
1
+
m
s
,0
!
bt
)
.
(0.0.15)

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