Momentum, Impulse, and Collisions
8-35
(b)
For the wagon,
24
1
1
2
(300 kg)(10.12 m/s)
1.54 10 J
K
==
×
. Assume that the two heroes drop from a small
height, so their kinetic energy just before the wagon can be neglected compared to
1
K
of the wagon.
1
2
2
(435 kg)(6.98 m/s)
1.06 10 J
K
×
. The kinetic energy of the system decreases by
3
12
4.8 10 J
KK
−=×
.
EVALUATE:
The wagon slows down when the two heroes drop into it. The mass that is moving horizontally
increases, so the speed decreases to maintain the same horizontal momentum. In the collision the vertical
momentum is not conserved, because of the net external force due to the ground.
8.110.
IDENTIFY:
Gravity gives a downward external force of magnitude
mg
. The impulse of this force equals the
change in momentum of the rocket.
SET UP:
Let +
y
be upward. Consider an infinitesimal time interval
dt
. In Example 8.15,
ex
2400 m/s
v
=
and
0
120 s
dm
m
dt
=−
. In Example 8.16,
0
/4
mm
=
after
90 s
t
=
.
EXECUTE:
(a)
The impulse-momentum theorem gives
ex
()
(
)
(
)
(
)
mgdt
m
dm v
dv
dm v v
mv
−=
+
+
+
−
−
. This
simplifies to
ex
mgdt
mdv v dm
+
and
ex
dv
dm
mv
m
g
dt
dt
−
.
(b)
ex
dv
v dm
ag
dt
m dt
−
−
.
(c)
At
0
t
=
,
22
ex
0
1
(2400 m/s)
9.80 m/s
10.2 m/s
120 s
vd
m
md
t
⎛⎞
− =−
−
−
=
⎜⎟
⎝⎠
.
(d)
ex
v
dv
dm
gdt
m
−
. Integrating gives
0
0e
x
ln
m
vv
v
g
t
m
−
=+
−
.
0
0
v
=
and
2
(2400 m/s)ln4 (9.80 m/s )(90 s)
2445 m/s
v
−
=
.
EVALUATE:
Both the initial acceleration in Example 8.15 and the final speed of the rocket in Example 8.16 are
reduced by the presence of gravity.
8.111.
IDENTIFY
and
SET UP:
Apply Eq. 8.40 to the single-stage rocket and to each stage of the two-stage rocket.
(a) EXECUTE:
x
0
ln
/
;
vv v
0
0
v
=
so
( )
ex
0
ln
/
=
The total initial mass of the rocket is
0
12,000 kg 1000 kg
13,000 kg.
m
=
Of this, 9000 kg
700 kg
9700 kg
+=
is fuel, so the mass
m
left after all the fuel is burned is 13,000 kg 9700 kg
3300 kg.
−
=
( )
ex
ex
ln 13,000 kg/3300 kg
1.37
v
.
(b)
First stage:
ex
0
ln
/
=
0
13,000 kg
m
=
The first stage has 9000 kg of fuel, so the mass left after the first stage fuel has burned is
13,000 kg
9000 kg
4000 kg.
( )
ex
ex
ln 13,000 kg/4000 kg
1.18
v
.
(c)
Second stage:
0
1000 kg,
m
=
1000 kg
700 kg
300 kg
m
=
.
( ) ( )
0
ex
0
ex
ex
ex
ln
/
1.18
ln 1000 kg/300 kg
2.38
v v
v
=
+
=
.
(d)
7.00 km/s
v
=
( )
ex
/2.38
7.00 km/s /2.38 2.94 km/s
=
. q
EVALUATE:
The two-stage rocket achieves a greater final speed because it jetisons the left-over mass of the first
stage before the second-state fires and this reduces the final
m
and increases
0
/.
8.112.
IDENTIFY:
During an interval where the mass is constant the speed of the rocket is constant. During an interval
where the mass is changing at a constant rate, the equations of Section 8.6 apply.
SET UP:
For 0
90 s
t
≤≤
,
0
120 s
dm
m
dt
. From Example 8.15,
ex
2400 m/s
v
=
.