fisica1-youn-e-freedman-exercicios-resolvidos

8.37: Let +y be north and +x be south. Let v
S1 and vA1
be the speeds of Sam and of
Abigail before the collision. m 80.0 kg, m = 50.0 kg, v
2
= 6.00 m s, v
2
= 9.00 m s.
S
=
A
S
A
P
x
is constant gives
m v = m v cos 37.0
v
v
S1
m
A1
b) K
K
S S1
2
= 9.67 m s (Sam)
P is constant gives
y
v
A A1
= 2.26 m s (Abigail)
1
=
=
1
2
∆K
= K
= m v
1
2
m v
2
S S 2
S S 2
m v
2
S S1
2
S S 2
− K
+
1
sin 37.0° − m v
+
1
2
1
2
m
m
v
v
2
A A2
= −640 J
° + m v
2
A A1
A A2
A A2
= 4101J
= 3465 J
cos 23.0°
sin 23.0°
8.38: (a) At maximum compression of the spring, v = v = . Momentum conservation
gives ( 2.00 kg)(2.00 m s) = (12.0 kg) V
2 10
V
V = 0.333 m s
1 2 1
2
Energy conservation : m
2
v0
= ( m2
+ m10
) V + U
2 2
1
2 1
(2.00 kg)(2.00 m s) = (12.0 kg)(0.333 m
2
2
U
spr
= 3.33 J
(b) The collision is elastic and Eqs. (8.24) and (8.25) may be used:
v
= −1.33 m s, v10
2
= +
0.67 m s
spr
2
s ) +Uspr
8.39: In the notation of Example 8.10, with the smaller glider denoted as A, conservation
of momentum gives ( 1.50) v
A2 + (3.00) vB2
= −5.40 m s. The relative velocity has
switched direction, so v A 2
− v B 2
= −3.00 m s. Multiplying the second of these relations
by (3.00) and adding to the first gives ( 4.50) v
A2 = −14.4 m s, or v
A2
= −3.20 m s,
with
the minus sign indicating a velocity to the left. This may be substituted into either relation
to obtain v
B2 = −0.20 m s; or, multiplying the second relation by (1.50) and subtracting
from the first gives ( 4.50) v
2
= −0.90 m s, which is the same result.
B