fisica1-youn-e-freedman-exercicios-resolvidos

8.90: a) For the x - and y - directions, respectively, and m as the common mass of a
proton,
or
mvA
1
= mvA2
cosα
+ mv2
cos β
0 = mvA2
sinα − mv B 2
sin β
v
A1
= v
A2
cosα
+ v B 2
sinα
0 = v −
A2
cos β
sin β
v B 2
.
b) After minor algebra,
v
2
A1
= v
= v
2
A2
2
A2
+ v
+ v
2
B2
2
B2
c) For a perfectly elastic collision,
+ 2v
+ 2v
v
A2
B2
v
A2
B2
( cosα
cos β − sin α sin β )
cos( α + β).
1 2 1 2 1 2 2 2 2
mv
A1
= mv
A2
+ mvB2
or v
A1 = v
A2
+ vB2.
2 2 2
Substitution into the above result gives cos( α + β)
= 0. d) The only positive angle with
π
zero cosine is (90 ).
2
°
8.91: See Problem 8.90. Puck B moves at an angle 65.0 ° ( i.e.90° − 25°
= 65°
) from the
original direction of puck A’s motion, and from conservation of momentum in the
y-direction, v = .466v
. Substituting this into the expression for conservation of
B2 0
A2
ο
ο
momentum in the x − direction, v
A2 = v A 1
(cos25.0 + 0.466cos65 ) = 13.6 m s , and so
v = 6.34
B2
m/s.
As an alternative, a coordinate system may be used with axes along the final directions
of motion (from Problem 8.90, these directions are known to be perpendicular). The
initial direction of the puck’s motion is 25.0°
from the final direction, so
v
A2 = vA
1
cos25.0°
and vB2
= vA
1
cos65.0, giving the same results.