fisica1-youn-e-freedman-exercicios-resolvidos

8.78: The speed of the block after the bullet has passed through (but before the block has
begun to rise; this assumes a large force applied over a short time, a situation
characteristic of bullets) is
V
=
2gy
=
2
2(9.80 m/s )(0.45×
10
−2
m) = 0.297 m/s.
The final speed v of the bullet is then
p mv0
− MV M
v = = = v0
− V
m m m
1.00 kg
= 450 m/s −
(0.297 m/s) = 390.6 m/s,
−3
5.00×
10 kg
or 390 m/s to two figures.
8.79: a) Using the notation of Eq. ( 8.24),
K
0
− K
2
1
= mv
2
1
= mv
2
= K
= K
0
0
2
2
1 2
− mvA
2
2
⎛
⎞
⎜ ⎛ m − M ⎞
1−
⎟
⎜ ⎟
⎝ ⎝ m + M ⎠ ⎠
2
⎛ ( m + M ) − ( m − M )
⎜
2
⎝ ( m + M )
⎛ 4mM
⎜
⎝ ( m + M )
b) Of the many ways to do this calculation, the most direct way is to differentiate the
expression of part (a) with respect to M and set equal to zero;
d ⎛ M ⎞
0 = (4mK0)
⎜ ⎟,
or
2
dM ⎝ ( m + M ) ⎠
1 2M
0 = −
2
3
( m + M ) ( m + M )
0 = ( m + M ) − 2M
m = M.
c) From Eq.(8.24), with m
A
= mB
= m, vA
= 0; the neutron has lost all of its kinetic
energy.
2
⎞
⎟ .
⎠
2
⎞
⎟
⎠