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Recitation ClassWork 5 solutions
Physics 203
1.
Chapter 7 problem 3
REASONING
The impulse that the roof of the car applies to the hailstones can be found from the impulse
momentum theorem, Equation 7.4. Two forces act on the hailstones, the average force
F
exerted by the
roof, and the weight of the hailstones. Since it is assumed that
F
is much greater than the weight of the
hailstones, the net average force
(
)
Σ
F
is equal to
F
.
SOLUTION
From Equation 7.4, the impulse that the roof applies to the hailstones is:
( )
Impulse
Final
Initial
momentum
momentum
tm
m
m
∆=
−
=
f
0
f0
F
v
v
vv
Solving for
F
(with
up
taken to be the positive direction) gives
[ ]
(
)
(0.060 kg/s) ( 15 m/s)
( 15 m/s) = +1.8 N
m
t
=
−
=
+
−−
∆
F
This is the average force exerted on the hailstones by the roof of the car. The positive sign indicates that
this force points upward. From Newton's third law, the average force exerted by the hailstones on the roof
is equal in magnitude and opposite in direction to this force. Therefore,
Force on roof
=
1.8 N
−
The negative sign indicates that this force
points downward .
2.
Problem 8
REASONING
We will apply the impulse momentum theorem as given in Equation 7.4 to solve this problem.
From this theorem we know that, for a given change in momentum, greater forces are associated with
shorter time intervals. Therefore, we expect that the force in the stifflegged case will be greater than in the
kneesbent case.
SOLUTION
a. Assuming that upward is the positive direction, we find from the impulsemomentum
theorem that
( )( ) ( )( )
5
3
75 kg 0 m/s
75 kg
6.4 m/s
2.4 10 N
2.0 10 s
mm
t
−
−
Σ=
=
=+ ×
∆
×
F
b.
Again using the impulsemomentum theorem, we find that
( )( ) ( )( )
3
75 kg 0 m/s
75 kg
6.4 m/s
4.8 10 N
0.10 s
t
−
=
∆
F
c.
The net average force acting on the man is
Ground
FF
W
, where
F
Ground
is the average upward
force exerted on the man by the ground and
W
is the downwardacting weight of the man. It follows, then,
that
=
Σ−
Ground
F
FW
. Since the weight is
W
=
−
mg
, we have
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( )
( )
( )
5
25
3
23
2.4 10 N
75 kg 9.80 m/s
4.8 10 N
75 kg 9.80 m/s
5.5 10 N
−
=
Σ−
=
+×
−
−
=
−
=
=
−
−
=
Ground
Ground
Stiff
legged
F
F W
Knees bent
F
F W
3.
Problem 12
REASONING AND SOLUTION
According to the impulsemomentum theorem (Equation 7.4)
( )
()
tm
Σ ∆=
−
f0
F
vv
(1)
Conservation of mechanical energy can be used to relate the velocities to the heights. If the floor is used to
define the zero level for the heights, we have
mgh
mv
0
1
2
2
=
B
where
h
0
is the height of the ball when it is dropped and
v
B
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This note was uploaded on 06/22/2009 for the course PHYS 1101 taught by Professor Richardson, b during the Fall '08 term at Cornell University (Engineering School).
 Fall '08
 RICHARDSON, B
 Physics, Force, Momentum, Work

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