276_Physics ProblemsTechnical Physics - 278*P9.63 Linear...

278
Linear Momentum and Collisions
*P9.63 (a) Each object swings down according to
mgR mv =
1
2
1
2
MgR Mv =
1
2
1
2
v gR
1
2 =
The collision: − + = + + mv Mv m M v
1 1 2
a f
v
M m
M m
v
2 1
=
−
+
Swinging up:
1
2
1 35
2
2
M m v M m gR + = + − °
a f a f a f
cos
v gR
2
2 1 35 = − ° cos
a f
2 1 35 2
0 425 0 425
1 425 0 575
0 403
gR M m M m gR
M m M m
m M
m
M
− ° + = −
+ = −
=
=
cos
. .
. .
.
a fa f a f
(b) No change is required if the force is different. The nature of the forces within the system of
colliding objects does not affect the total momentum of the system. With strong magnetic
attraction, the heavier object will be moving somewhat faster and the lighter object faster
still. Their extra kinetic energy will all be immediately converted into extra internal energy
when the objects latch together. Momentum conservation guarantees that none of the extra
kinetic energy remains after the objects join to make them swing higher.
P9.64 (a) Use conservation of the horizontal component of
momentum for the system of the shell, the cannon,
and the carriage, from just before to just after the
cannon firing.
p p
xf xi
= : m v m v
shell shell cannon recoil
cos . 45 0 0 °+ =
200 125 45 0 5 000 0
a fa f b g
cos . °+ = v
recoil
or v
recoil
m s = −