fisica1-youn-e-freedman-exercicios-resolvidos

8.43: a) In Eq. (8.24), let mA = m and mB
= M.
Solving for M gives
v − vA
M = m
v + v
7
7
In this case, v = 1.50 × 10 m s, and v
A
= −1.20×
10 m s, with the minus sign indicating
1.50
+ 1.20
a rebound. Then, M m = 9m.
Either Eq. (8.25) may be used to find
vB
vB
= 3.00
= v ×
5
= (1.50×
10
7
10
6
=
1.50 + ) −1.20)
m s,
or Eq. (8.23), which gives
7
m s) + ( −1.20×
10 m s), the same result.
A
8.44: From Eq. (8.28),
x
y
cm
cm
(0.30 kg) ( 0.
20 m) + (0.40 kg)(0.10 m) + (0.20 kg)( −0.30 m)
=
= + 0.044 m,
(0.90 kg)
(0.30 kg)(0.30 m) + (0.40 kg)( −0.40 m) + (0.20 kg)(0.60 m)
=
= 0.056 m.
(0.90 kg)
8.45: Measured from the center of the sun,
11
( 8
30
27
1.99×
10 kg)(0) + (1.90×
10 kg)(7.78×
10
30
27
1.99×
10 kg + 1.90×
10 kg
The center of mass of the system lies outside the sun.
m)
= 7.42 × 10
m.
8.46: a) Measured from the rear car, the position of the center of mass is, from Eq. (8.28),
(1800 kg)(40.0 m)
= 24.0 m, which is 16.0 m behind the leading car.
(1200 kg + 1800 kg)
b) (1200 kg)(12.0 m
c) From Eq.(8.30),
v
cm
d) (1200 kg
s) + (1800 kg)(20.0 m s) = 5.04×
10
kg ⋅ m s.
(1200 kg)(12.0 m s) + (1800 kg)(20.0 m s)
=
= 16.8 m s.
(1200 kg + 1800 kg)
4
+ 1800 kg)(16.8 m s) = 5.04×
10
kg ⋅ m s.
4