fisica1-youn-e-freedman-exercicios-resolvidos

7.87: a) Eliminating β in favor of α and x β = α ) ,
0( x0
2
2
α β α x0
α α ⎡⎛
x0
⎞ ⎛ x0
⎞⎤
U ( x)
= − = − =
.
2
2 2
2 ⎢⎜
⎟ − ⎜ ⎟⎥
x x x0
x x0x
x0
⎢⎣
⎝ x ⎠ ⎝ x ⎠⎥⎦
α
U x0 ) (1 −1)
= 0 . U (x)
is positive for x < x0
and negative for x > x0
(α and β
( = x
2
0
must be taken as positive).
b)
2
2 ⎛ 2α
⎞ ⎛
⎞
0 0
( )
⎜⎛
x ⎞ ⎛ x ⎞
v x = − U =
− ⎟
⎜
.
2
⎟
⎜ ⎟ ⎜ ⎟
m ⎝ mx
0 ⎠ ⎝⎝
x ⎠ ⎝ x ⎠ ⎠
The proton moves in the positive x-direction, speeding up until it reaches a maximum
speed (see part (c)), and then slows down, although it never stops. The minus sign in the
square root in the expression for v (x)
indicates that the particle will be found only in the
region where U < 0 , that is, x > x0
.
c) The maximum speed corresponds to the maximum kinetic energy, and hence the
dU
minimum potential energy. This minimum occurs when = 0 , or
dU
dx
3
2
α ⎡ ⎛ x0
⎞ ⎛ x0
⎞ ⎤
= 3 ⎢−
2⎜
⎟ + ⎜ ⎟ ⎥ = 0,
x0
⎢⎣
⎝ x ⎠ ⎝ x ⎠ ⎥⎦
which has the solution x = 2x . α
α
0
U ( 2x0)
= − , so v = .
4x
2 2 d) The maximum speed
2mx 0
0
occurs at a point where dU = 0 , and from Eq. (7.15), the force at this point is zero. e)
dx
−
x
1
= 3x 0
, and = − = − = ( ) − ( ) −
x
)
x 2
2 α
2
2 2 α α 0 x 0
U ( 3x0 )
9 2 ; v(
x)
x
m
( U ( x1
) U ( x))
m 9 2 2
x x x
0
0 0
2
2 x0
x0
2 ( x
) − ( x
) − 2 9)
α . The particle is confined to the region where ( x)
U(
x ) 1
mx0
dx
[ ] =
U < . The