fisica1-youn-e-freedman-exercicios-resolvidos

8.11: a) With t = 0 1
,
,
t
2
7 2
J
x
= ∫ Fx
dt = (0.80 × 10 N s) t
2
− (2.00 × 10
0
9
N
2
s ) t
3
2
which is 18.8 kg ⋅ m s , and so the impulse delivered between t=0 and
−3
t = 2.50 × 10 s is (18.8 kg m s)î. b)
2
⋅
2 −3
J = −(0.145 kg) (9.80 m s ) (2.50×
10 s), and the impulse is
y
( −3.55×
10
−3
(7.52 × 10 3 N)î.
r r r
d) p p + j
2
= 1
kg ⋅ m s) ĵ
J
c) x
=7.52× 10 3 N, so the average force is
t 2
= −(0.145kg)(40.0ˆ
i + 5.0 ˆ)m/s j + (18.8ˆ i − 3.55×
10
= ( 13.0 kg.m/s)ˆ i − (0.73 kg.m/s) ˆj.
ˆ)
− 3 j
The velocity is the momentum divided by the mass, or (89.7 m/s)
i ˆ − (5.0m/s) ˆj.
8.12: The change in the ball’s momentum in the x-direction (taken to be
positive to the right) is
o
( 0.145 kg) ( −(65.0 m s) cos 30 − 50.0m/s) = −15.41kg
⋅ m/s, so the x-
component of the average force is
−15.41kg
⋅ m/s
3
= −8.81×
10 N,
− 3
1.75×
10 s
and the y-component of the force is
( 3
0.145kg)(65.0 m/s) sin 30°
= 2.7 × 10
− 3
(1.75×
10 s)
N.
t
B
8.13: a) = ∫ 2
3 3
J Fdt = A(
t2
− t1)
+ ( t2
− t1
),
t1
3
3
p J A B 3
or J = At2 + ( B / 3) t2
if t1
= 0.
b) v = = = t2 + t 2
.
m m m 3m