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820
Chapter 8
EXECUTE:
///
AE
AB
BE
=+
vvv
!!!
/
5.00 m/s
v
/
5.00 m/s
v
=−
(minus since the mass is moving west relative to the car). This gives
/
0;
v
=
the mass is at rest
relative to the earth after it is thrown backwards from the car.
As in part (a),
()
12
2
.
A
A
xB
B
x
mm
vm
v m
v
+=
+
Now
2
0,
Ax
v
=
so
.
B
B
x
v
21
200 kg
5.00 m/s
5.71 m/s.
175 kg
Bx
B
vv
m
⎛⎞
+
==
=
⎜⎟
⎝⎠
The final velocity of the car is 5.71 m/s, east.
EVALUATE:
The thrower exerts a force in the
direction
x
−
so the mass exerts a force on him in the
direction
x
+
and he and the car speed up.
(c) SET UP:
Let
A
be the 25.0 kg mass and
B
be the car (mass
200 kg).
B
m
=
Figure 8.70b
x
P
is conserved so
( )
11
2
AAx
BBx
A
B
x
mv
m
m v
+
.
EXECUTE:
2
AA
BB
A
B
x
m
−+
=
+
.
( )( ) ( )( )
2
200 kg 5.00 m/s
25.0 kg 6.00 m/s
3.78 m/s.
200 kg
25.0 kg
x
v
−
−
=
++
The final velocity of the car is 3.78 m/s, east.
EVALUATE:
The mass has negative
x
p
so reduces the total
x
P
of the system and the car slows down.
8.71.
IDENTIFY:
The horizontal component of the momentum of the sand plus railroad system is conserved.
SET UP:
As the sand leaks out it retains its horizontal velocity of 15.0 m/s.
EXECUTE:
The horizontal component of the momentum of the sand doesn&t change when it leaks out so the
speed of the railroad car doesn&t change; it remains 15.0 m/s. In Exercise 8.27 the rain is falling vertically and
initially has no horizontal component of momentum. Its momentum changes as it lands in the freight car.
Therefore, in order to conserve the horizontal momentum of the system the freight car must slow down.
EVALUATE:
The horizontal momentum of the sand does change when it strikes the ground, due to the force that
is external to the system of sand plus railroad car.
8.72.
IDENTIFY:
Kinetic energy is
2
1
2
K
mv
=
and the magnitude of the momentum is
p
mv
=
. The force and the time
t
it acts are related to the change in momentum whereas the force and distance
d
it acts are related to the change in
kinetic energy.
SET UP:
Assume the net forces are constant and let the forces and the motion be along the
x
axis. The impulse
momentum theorem then says
Ft
p
=Δ
and the workenergy theorem says
Fd
K
= Δ
.
EXECUTE:
(a)
24
1
N
2
(840 kg)(9.0 m/s)
3.40 10 J
K
×
.
1
P
2
(1620 kg)(5.0 m/s)
2.02 10 J
K
×
. The Nash has
the greater kinetic energy and
N
P
1.68
K
K
=
.
(b)
3
N
(840 kg)(9.0 m/s)
7.56 10 kg m/s
p
×
⋅
.
3
P
(1620 kg)(5.0 m/s)
8.10 10 kg m/s
p
×
⋅
. The Packard has
the greater magnitude of momentum and
N
P
0.933
p
p
=
.
(c)
Since the cars stop the magnitude of the change in momentum equals the initial momentum. Since
PN
p
p
>
,
FF
>
and
NN
PP
0.933
Fp
.
(d)
Since the cars stop the magnitude of the change in kinetic energy equals the initial kinetic energy. Since
NP
K
K
>
,
>
and
1.68
FK
.
EVALUATE:
If the stopping forces were the same, the Packard would have a larger stopping time but would
travel a shorter distance while stopping. This consistent with it having a smaller initial speed.
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821
8.73.
IDENTIFY:
Use the impulsemomentum theorem to relate the average force on the bullets to their rate of change
in momentum. By Newton&s third law, the average force the weapon exerts on the bullets is equal in magnitude
and opposite in direction to the recoil force the bullets exert on the weapon.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Mass

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