fisica1-youn-e-freedman-exercicios-resolvidos

8.99: The information is not sufficient to use conservation of energy. Denote the emitted
neutron that moves in the + y - direction by the subscript 1 and the emitted neutron that
moves in the –y-direction by the subscript 2. Using conservation of momentum in the x-
and y-directions, neglecting the common factor of the mass of neutron,
v
0
= (2v
0 = (2v
0
0
3)cos10° + v
3)sin 10° + v
1
1
cos 45° + v
sin 45° − v
2
2
cos 30°
sin 30°
.
With sin 45° = cos 45°
, these two relations may be subtracted to eliminate v 1,
and
rearrangement gives
v
0
( 1−
(2 3) cos10° + (2 3)sin 10°
) = v2(cos 30° + sin 30°
),
3
3
from which v
2
= 1.01×
10 m s or 1.0×
10 m s to two figures. Substitution of this into
either of the momentum relations gives v = 221m
1
s. All that is known is that there is no
z - component of momentum, and so only the ratio of the speeds can be determined. The
ratio is the inverse of the ratio of the masses, so v = ( 1.5 v .
Kr
)
Ba
mA
8.100: a) With block B initially at rest, vcm
=
m m
v
A + B A1
. b) Since there is no net external
force, the center of mass moves with constant velocity, and so a frame that moves with
the center of mass is an inertial reference frame. c) The velocities have only x -
components, and the x -components are
mB
m A
u = v − v = v u = −v
= u . Then, P m u + m u 0 .
A1 A1
cm
A1, B1
cm
−
m A + mB
m A + mB
A1
cm
=
A A1
B B1
=
d) Since there is zero momentum in the center-of-mass frame before the collision, there
can be no momentum after the collision; the momentum of each block after the collision
must be reversed in direction. The only way to conserve kinetic energy is if the
momentum of each has the same magnitude so in the center-of-mass frame, the blocks
change direction but have the same speeds. Symbolically, uA2
= −uA
1, uB2
= −uB
1. e) The
velocities all have only x –components; these components are
0.200
0.400
u .00 m s = 2.00m s, u = − 6.00 m s = −4.00 m s, u = −2.00m s, u 4.
A1 =
0.600
6
B1
0.600
A2
B2
=
1
v
A2 = + 2.00
m s, vB2
= 8.00 m and Equation (8.24) predicts v
A2 +
3
v
A1
and s.
4
(8.25) predicts v
B
= u
1
, which are in agreement with the above.
2 3 A
= and Eq.