fisica1-youn-e-freedman-exercicios-resolvidos

8.17: The change in velocity is the negative of the change in Gretzky’s
momentum, divided by the defender’s mass, or
v
m
A
B2
= vB
1
−
A2
A
mB
= −5.00
m s −
= 4.66 m s.
( v − v ) 1
756 N
(1.50m s
900 N
−13.0 m s)
Positive velocities are in Gretzky’s original direction of motion, so the
defender has changed direction.
b) 1 2 2 1 2
K
( ) (
2
2
− K1
= mA
v
A2
− v
A1
+ mB
vB2
− vB
1)
2
2
2
2
1 ⎡(756 N)((1.50 m/s) − (13.0 m/s) )
=
2 ⎢
2
2(9.80 m/s )
⎣ + (900 N)((4.66 m/s) − ( −5.00 m/s)
= −6.58 kJ.
2
⎤
⎥
)
⎦
8.18: Take the direction of the bullet’s motion to be the positive direction. The total
momentum of the bullet, rifle, and gas must be zero, so
and
gas
( 0.00720 kg)(601m/s −1.85 m/s) + (2.80 kg)( −1.85 m/s) + p
gas
= 0,
p = 0.866 kg ⋅ m s.
Note that the speed of the bullet is found by subtracting
the speed of the rifle from the speed of the bullet relative to the rifle.
v
=
3.00 kg
8.19: a) See Exercise 8.21; ( )(0.800 m s) 3.60 m s.
A
1.00 kg
2
2
b) (1 2) (1.00 kg) (3.60 m/s) + (1/ 2)(3.00 kg)(1.200 m/s) = 8.64 J.
=
8.20: In the absence of friction, the horizontal component of the hat-plus-adversary
system is conserved, and the recoil speed is
(4.50 kg)(22.0 m s) cos 36.9°
= 0.66 m s.
(120 kg)