fisica1-youn-e-freedman-exercicios-resolvidos

8.40: a) In the notation of Example 8.10, with the large marble (originally moving to the
right) denoted as A ,(3.00)
v A 2
+ (1.00) vB2
= 0.200 m s. The relative velocity has switched
direction, so v
A2 − vB2
= −0.600 m s.
Adding these eliminates v
B2
to give
( 4.00) v
A2 = −0.400 m s, or v
A2
= −0.100 m s, with the minus sign indicating a final
velocity to the left. This may be substituted into either of the two relations to obtain
v
B2
= 0.500 m s; or, the second of the above relations may be multiplied by 3.00 and
subtracted from the first to give ( 4.00) v = B2 2.00 m s, the same result.
b) ∆P
= −0.009 kg ⋅ m s, ∆P
= 0.009 kg ⋅ m s
A
−4 −4
c) ∆K
A
= −4.5×
10 , ∆KB
= 4.5×
10 .
Because the collision is elastic, the numbers have the same magnitude.
B
8.41: Algebraically, v
B2
= 20 m s. This substitution and the cancellation of common
factors and units allow the equations in α and β to be reduced to
2 = cosα
+
0 = sin α −
1.8 cos β
1.8 sin β.
Solving for cos α and sin α , squaring and adding gives
2
2
( 2 − 1.8 cos β ) + ( 1.8.sin β) = 1.
1 .2
Minor algebra leads to cos β = , or β = 26.57°
. Substitution of this result into the first
4
of the above relations gives cos α = , andα
= 36.87°
.
5
1.8
v 1 u 2 u 1
8.42: a) Using Eq. 8.24),
A −
= +
= . b) The kinetic energy is proportional to the
(
V 1u 2 u 3
K A
1
square of the speed, so =
K 9
c) The magnitude of the speed is reduced by a factor of
1
after each collision, so after collisions, the speed is ( )
3
, consider
⎛ 1 ⎞
⎜ ⎟
⎝ 3⎠
3
=
1
59,000
= 59,000
ln(3) = ln(59,000)
1
of its original value. To find
ln(59,000)
= = 10.
ln(3)
to the nearest integer. Of course, using the logarithm in any base gives the same result.
or
3