Chapter 9
Page 1
CHAPTER 9  Linear Momentum and Collisions
1.
We find the force on the expelled gases from
F
= ?
p
/?
t
= (?
m
/?
t
)
v
= (1200 kg/s)(50,000 m/s) = 6.0
×
10
7
N.
An equal, but opposite, force will be exerted on the rocket:
6.0
×
10
7
N, up
.
2.
For the momentum
p
= 4.8
t
2
i
– 8.0
j
– 8.9
t
k
, we find the force from
F
= d
p
/dt =
9.6
t
i
– 8.9
k
, in SI units
.
3. (
a
)
p
=
mv
= (0.030 kg)(12 m/s) =
0.36 kg
·
m/s
.
(
b
) The force, opposite the direction of the velocity, changes the momentum:
F
= ?
p
/?
t
;
–
2
.
0
×
10
–2
N = (
p
2
– 0.36 kg
·
m/s)/(12 s), which gives
p
2
=
0.12 kg
·
m/s
.
4.
The change in momentum is
?
p
=
p
2
–
p
1
=
mv
j
–
mv
i
= (0.145 kg)(30 m/s)
j
– (0.145 kg)(30 m/s)
i
=
– (4.35 kg
·
m/s)
i
+ (4.35 kg
·
m/s)
j
.
5.
The force changes the momentum:
F
= (26 N)
i
– (12 N/s
2
)
t
2
j
= d
p
/d
t
.
Because
F
is a variable force, we integrate to find the momentum change:
d
p
=
F
d
t
;
∆
p
=(
2
6
N)
i
– (12 N/s
2
)
t
2
j
dt
1.0 s
2.0 s
2
6 N)
t
i
– (4.0 N/s
2
)
t
3
j
1.0 s
=
(26 N
⋅
s)
i
– (28 N
⋅
s)
j
.
6. If
M
is the initial mass of the rocket and
m
2
is the mass of the
expelled gases, the final mass of the rocket is
m
1
=
M
–
m
2
.
Because the gas is expelled perpendicular to the rocket in the
rocket’s frame, it will still have the initial forward velocity,
so the velocity of the rocket in the original direction will not
change.
We find the
y
component of the rocket’s velocity after
firing from
v
1
⊥
=
v
0
tan
θ
= (120 m/s) tan 23.0° = 50.9 m/s.
Using the coordinate system shown, for momentum
conservation in the
y
direction we have
0 + 0 =
m
1
v
1
⊥
–
m
2
v
2
⊥
,
or
(
M
–
m
2
)
v
1
⊥
=
m
2
v
2
⊥
;
(4200 kg –
m
2
)(50.9 m/s) =
m
2
(2200 m/s), which gives
m
2
=
95 kg
.
v
0
Before
v
2
v
1
After
x
y
gas
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Page 2
7. (
a
) We choose downward as positive.
For the fall we have
y
=
y
0
+
v
0
t
1
+
!
at
1
2
;
h
= 0 + 0 +
!
gt
1
2
, which gives
t
1
= (2
h
/
g
)
1/2
.
To reach the same height on the rebound, the upward motion must be a reversal of the downward
motion.
Thus the time to rise will be the same, so the total time is
t
t
otal
= 2
t
1
= 2(2
h
/
g
)
1/2
=
(8
h
/
g
)
1/2
.
(
b
) We find the speed from
v
=
v
0
+
at
1
= 0 +
g
(2
h
/
g
)
1/2
=
(2
gh
)
1/2
.
(
c
) To reach the same height on the rebound, the upward speed at the floor must be the same as the
speed striking the floor.
Thus the change in momentum is
?
p
=
m
(–
v
) –
mv
= – 2
m
(2
gh
)
1/2
=
=
– (8
m
2
gh
)
1/2
(up)
.
(
d
) For the average force on the ball we have
F
= ?
p
/?
t
= – (8
m
2
gh
)
1/2
/(8
h
/
g
)
1/2
= –
mg
(up).
Thus the average force on the floor is
mg
(down), a surprising result
.
8.
We convert the speed:
(100 km/h)/(3.6 ks/h) = 27.8 m/s.
In a time ?
t
all of the air within a distance
v
?
t
will strike the building; so the mass being brought to rest is
?
m
=
ρ
Av
?
t
.
Thus the average force on the air is
F
= ?
p
/?
t
= (?
m
/?
t
) ?
v
=
Av
(0 –
v
) = –
Av
2
= – (1.3 kg/m
3
)(40 m)(60 m)(27.8 m/s)
2
= – 2.4
×
10
6
N.
The average force on the building is the reaction to this:
2.4
×
10
6
N
.
9.
For the onedimensional motion, we take the direction of the first car for the positive direction.
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 Spring '09
 Fielding
 Physics, Force, Kinetic Energy, Momentum, kg, m/s

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