fisica1-youn-e-freedman-exercicios-resolvidos

8.47: a) With x = 1
0 in Eq. (8.28), kg.
m
1
= m2
(( x2
/ xcm
) −1)
= (0.10 kg)((8.0 m)/(2.0 m) −1)
= 0.30
b) P = M v (0 40 kg)(5.0 m s) ˆ
cm = .
i = (2.0 kg ⋅ m s) iˆ.
c) In Eq. (8.32),
r r r
v = so v = P /(0.30 kg) = (6.7 m s)ˆ. 0
2
,
1
i
8.48: As in Example 8.15, the center of mass remains at rest, so there is zero net
momentum, and the magnitudes of the speeds are related by m
1v1
= m2v2
, or
v = m / m ) v = (60.0 kg / 90.0 kg)(0.70 m s) 0.47 m s.
2
(
1 2 1
=
8.49: See Exercise 8.47(a); with y = 0 1
, Eq. (8.28) gives m = (( / ) −1)
1
m2
y2
ycm
=
( 0.50 kg)((6.0 m) /(2.4 m) −1)
= 0.75 kg, so the total mass of the system is 1.25 kg.
r r
d
3
b) a v (1.50 m s ) tiˆ
cm
=
.
dt cm
=
r
3
c) F = m a
r = (1.25 kg) (1.50m s ) (3.0 s)ˆ i = (5.63 N)ˆ i.
cm
z
= , so Fz
8.50: p 0 = 0.
The x -component of force is
F
x
dp
=
dt
x
= ( −1.50 N s) t.
F
y
dp
= dt
y
= 0.25 N
8.51: a) From Eq. (8.38), F = ( 1600 m s)(0.0500 kg s) = 80.0 N. b) The absence of
atmosphere would not prevent the rocket from operating. The rocket could be steered by
ejecting the fuel in a direction with a component perpendicular to the rocket’s velocity,
and braked by ejecting in a direction parallel (as opposed to antiparallel) to the rocket’s
velocity.