fisica1-youn-e-freedman-exercicios-resolvidos

8.63: The total momentum must be zero, and the velocity vectors must be three vectors of
the same magnitude that sum to zero, and hence must form the sides of an equilateral
triangle. One puck will move 60 ° north of east and the other will move 60°
south of east.
8.64: a) m
AvAx
+ mBvBx
+ mCvCx
= mtotvx
, therefore
v
Cx
v
Cx
( 0.100 kg)(0.50 m s) − (0.020 kg)( −1.50 m s) − (0.030 kg)( −0.50m s)cos60°
=
0.050kg
= 1.75m s
Similarly,
v
Cy
v
Cy
( 0.100kg)(0m s) − (0.020kg)(0 m s) − (0.030 kg)( −0.50m s)sin 60°
=
0.050 kg
= 0.26 m s
b)
∆K =
−
1
2
1
2
(0.100 kg)(0.5 m
s)
(0.050 kg) × [(1.75 m
2
−
s)
1
2
2
(0.020 kg)(1.50 m
s)
1
2
2
+ (0.26 m s) ] = −0.092 J
2
−
(0.030 kg)( −0.50m
s)
2
8.65: a) To throw the mass sideways, a sideways force must be exerted on the mass, and
hence a sideways force is exerted on the car. The car is given to remain on track, so some
other force (the tracks on the car) act to give a net horizontal force of zero on the car,
which continues at 5.00 m s east.
b) If the mass is thrown with backward with a speed of 5 .00 m s relative to the initial
motion of the car, the mass is at rest relative to the ground, and has zero momentum. The
( 200 kg)
speed of the car is then 5.00 m s) = 5.71m s, and the car is still moving east.
(
kg
( 175 )
c) The combined momentum of the mass and car must be same before and after the
( 200 kg)( 5.00 m s) + ( 25.0 kg)( −6.00
m s)
mass hits the car, so the speed is = 3.78 m s, with the car still
moving east.
( 225 kg )