fisica1-youn-e-freedman-exercicios-resolvidos

5
8.1: a) (10,000 kg)(12.0 m s) = 1.20 × 10 kg ⋅ m s.
b) (i) Five times the speed, .0 m s.
60 (ii) 5 ( 12 .0 m s) = 26.8 m s.
8.2: See Exercise 8.3 (a); the iceboats have the same kinetic energy, so the boat with the
larger mass has the larger magnitude of momentum by a factor of ( 2m
) ( m)
= 2.
2 2 2
1 2 1 m v 1 p
8.3: a) K = mv = = .
2 2 m 2 m
b) From the result of part (a), for the same kinetic energy,
2 2
p
1
p =
2 , so the larger mass
baseball has the greater momentum; ( p ) = 0.040 0.145 0.525.
m
1
p From the result
m
bird ball
=
1m1
K
2m
1w1
K
2w
K
man
K
woman
= 450 700 =
of part (b), for the same momentum K =
2
, so K =
2
; the woman, with the
0.643
smaller weight, has the larger kinetic energy. ( ) .
2
8.4: From Eq. (8.2),
p
p
x
= mvx
y
= mvy
=
=
( 0 .420 kg)( 4.50 m s) cos 20.0°
= 1.78 kg m s
( 0 .420 kg)( 4.50 m s) sin 20.0°
= 0.646 kg m s.
8.5: The y-component of the total momentum is
( 0.145 kg)( 1.30 m s) + ( 0.0570 kg)( − 7.80 m s) = −0.256 kg ⋅ m s.
This quantity is negative, so the total momentum of the system is in the
− y -direction.
8.6: From Eq. (8.2),
y
= −( 0.145 kg)( 7.00 m s) = −1.015 kg ⋅ m s,
= ( 0.045 kg)( 9.00 m s) = 0.405 kg ⋅ m s,
x
p and
p so the total momentum has magnitude
p
2 2
= p x
+ p
y
=
.015
and is at an angle arctan ( ) = − °
2
2
( − 0.405 kg ⋅ m s) + ( −1.015 kg ⋅ m s) = 1.09 kg ⋅ m s,
− 1 68 , using the value of the arctangent function in the
+ .405
fourth quadrant ( p > 0 , p < y
0).
x