fisica1-youn-e-freedman-exercicios-resolvidos

8.25: m + m )(3.00 m s) = m (4.00 m s) + m (2.25 m s), so
(
Kim Ken
Kim
Ken
m
m
Ken
(3.00 m s) − (2.25 m s)
=
(4.00 m s) − (3.00 m s)
Kim
=
and Kim weighs ( 0.750)(700 N) = 525 N.
0.750,
4
8.26: The original momentum is (24,000 kg)(4.00 m s) = 9.60 × 10 kg ⋅ m s, the
final mass is 24 ,000 kg + 3000 kg = 27,000 kg, and so the final speed is
4
9.60×
10 kg ⋅ m s
4
2.70 × 10 kg
= 3.56 m s.
8.27: Denote the final speeds as v
A
and vB
and the initial speed of puck A as v
0
, and
omit the common mass. Then, the condition for conservation of momentum is
v = v A
cos 30.0°
+ v
0 B
0 = v
sin 30.0°
−
A
v B
o
cos 45.0
o
sin 45.0 .
The 45 .0°
angle simplifies the algebra, in that sin 45 .0° = cos 45.0°
, and so the
vB
terms cancel when the equations are added, giving
v A
v
=
cos 30.0° + sin 30.0°
0
=
29.3m s
From the second equation, v = v A
= 20.7 m
B
s. b) Again neglecting the common mass,
2
K
K
2 2
2
(1 2)( vA
+ vB
) (29.3 m s) + (20.7 m s)
=
=
2
2
(1 2) v
(40.0 m s)
2
2
=
1
so 19.6% of the original energy is dissipated.
0
0.804,