fisica1-youn-e-freedman-exercicios-resolvidos

8.58: a) The speed of the ball before and after the collision with the plate are found
from the heights. The impulse is the mass times the sum of the speeds,
( 2.00m + 1.60m) 0. 47
2
J = m(
v1 + v2)
= m(
2gy1
+ 2gy2
) = (0.040kg) 2(9.80m s
=
J
−3
b) = (0.47 N ⋅s/2.00×
10 s) 237 N.
=
∆t r
3
8.59: p F
ˆ
2
j
ˆ
2 3
2
= dt = ( αt 3) + ( βt + γt 2) j = (8.33 N s t )ˆ i + (30.0 Nt
+ 2.5 N st
) ˆ
∫ r
j
r
After 0.500 s, p = ( 1.04kg ⋅ m s)ˆ i + (15.63kg ⋅ m s) ˆj
, and the velocity is
r r
v = p m = ( 0.52m s)ˆ i + (7.82m s) ˆ. j
−3
8.60: a) J = F t = ( −380 N) (3.00×
10 s) = −1.14
N ⋅ s.
J
x
y
x
= F t =
y
(110 N) (3.00×
10
b) v = v + J m = ( m s)
v
−3
s) = 0.33 N ⋅s.
( −1.14N.s)
( 0.560 N)
20.0 +
2
(9.80 m s )
2 x 1x
x
=
= v J m
(0.33 N.s)
−4.0m s) +
((0.560 N) (9.80m s
2 y 1y
+
y
= ( =
2
))
0.05m s
1.78m s.
8.61: The total momentum of the final combination is the same as the initial momentum;
for the speed to be one-fifth of the original speed, the mass must be five times the original
mass, or 15 cars.
8.62: The momentum of the convertible must be the south component of the total
momentum, so
(800 kg ⋅ m s)cos 60.0°
v =
(1500 kg)
Similarly, the speed of the station wagon is
v
con
=
(800kg ⋅ m s) sin 60.0°
=
(2000 kg)
sw
=
2.67 m s.
3.46 m s.