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9.1.
Model:
Model the car and the baseball as particles.
Solve:
(a)
The momentum
pm
v
=
=
()
(
)
1500 kg 10 m/s
=×
15 10
4
.
kg m/s.
(b)
The momentum
v
=
=
( )
0.2 kg
40 m/s
=
8.0 kg m/s.
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Model:
Model the bicycle and its rider as a particle. Also model the car as a particle.
Solve:
From the definition of momentum,
pp
car
bicycle
=
⇒=
⇒
=
mv
m
v
v
m
m
v
car
car
bicycle
bicycle
bicycle
car
bicycle
car
=
()
=
1500 kg
100 kg
5.0 m/s
m/s
75 0
.
Assess:
This is a very high speed (
≈
168 mph). This problem shows the importance of mass in comparing two
momenta.
9.3.
Visualize:
Please refer to Figure Ex9.3.
Solve:
The impulse is defined in Equation 9.6 as
JF
t
d
t
xx
t
=
() =
∫
i
f
t
area under the
F
x
(
t
) curve between
t
i
and
t
f
⇒
()
⇒=
6 0
8
1500
1
2
. N s =
ms
N
max
max
FF
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Model:
Model the rubber ball and the steel ball as particles, and their interaction as a collision.
Solve:
When objects collide, Newton’s third law tells us that the forces between the objects are equal and
opposite. The time of contact is the same for each object. Therefore, the impulse on each will be the same.
9.5.
Model:
The particle is subjected to an impulsive force.
Visualize:
Please refer to Figure Ex9.5.
Solve:
Using Equation 9.5, the impulse is the area under the curve. From 0 s to 2 ms the impulse is
Fdt
=−
()
×
−
∫
1
2
3
21
0
500 N
s
0.5 N s
From 2 ms to 8 ms the impulse is
Fdt
=+
−
∫
1
2
2000 N 8 ms
2 ms
6.0 N s
From 8 ms to 10 ms the impulse is
Fdt
−
∫
1
2
500 N 10 ms
8 ms
0.5 N s
Thus, from 0 s to 10 ms the impulse is
−+ −
05 60 05
.
.
.
N s
=
5.0 N s.
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Model:
Model the object as a particle and the interaction as a collision.
Visualize:
Please refer to Figure Ex9.6.
Solve:
The object is initially moving to the right (positive momentum) and ends up moving to the left (negative
momentum). Using the impulsemomentum theorem
ppJ
xx
x
fi
=+
,
−=
++
26
kg m/s
J
x
⇒=
−
=
−
J
x
8 kg m/s
8 N s
Since
JFt
x
=
avg
∆
, we have
Ft
avg
8 N s
∆=−
−
=−
F
avg
8 N s
10 ms
800 N
The force is
r
F
=
()
800 N, left .
9.7.
Model:
Model the object as a particle and the interaction with the force as a collision.
Visualize:
Please refer to Figure Ex9.7.
Solve:
(a)
Using the equations
p
f
x
=
p
i
x
+
J
x
JF
t
d
t
xx
t
t
=
∫
()
i
f
=
area under force curve
⇒=
+
v
x
f
1.0 m/s
2.0 kg
1
(area under the force curve)
=
+
=
2.0 kg
1.0 N s
m/s
1
15
.
(b)
Likewise,
v
x
f
2.0 kg
=
+
1
(area under the force curve)
=
+
−
=
2.0 kg
1.0 N s
1
05
.
Assess:
For an object with positive velocity, a negative impulse slows an object, whereas a positive impulse
increases an object’s speed. The opposite is true for an object with negative velocity.
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Model:
Use the particle model for the sled, the model of kinetic friction, and the impulsemomentum
theorem.
Visualize:
Note that the force of kinetic friction
f
k
imparts a negative impulse to the sled.
Solve:
Using
∆
pJ
xx
=
, we have
p
p
F t dt
f
dt
f
t
x
t
t
t
t
fi
k
k
i
f
i
f
−
=
=−
∫
∫
()
∆ ⇒
−
mv
mv
n t
mg t
k
k
µµ
∆∆
We have used the model of kinetic friction
f
k
=
µ
k
n
, where
k
is the coefficient of kinetic friction and
n
is the
normal (contact) force by the surface. The force of kinetic friction is independent of time and was therefore taken
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This note was uploaded on 04/02/2008 for the course PHYS 131133 taught by Professor All during the Spring '08 term at Cal Poly.
 Spring '08
 ALL
 Momentum

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