Therefore, the third fragment must fly off toward a general
direction of (2) north of east
.
(b) Apply momentum conservation
P
o =
P
.
along the
x
axis:
(2.0 kg)(0) = (0.50 kg)(
−
2.8 m/s) + (1.3 kg)(0) + (1.2 kg)
v
x
,
so
v
x
= 1.17 m/s;
along the
y
axis: (3.0 kg)(0) = (0.50 kg)(0) + (1.3 kg)(
−
1.5 m/s) + (1.2 kg)
v
y
,
so
v
y
= 1.63 m/s.
v
=
(1.17 m/s)
2
+ (1.63 m/s)
2
= 2.0 m/s
,
θ
= tan
−
1
163
117
.
.
= 54
°
north of east
.
41.
(a) Energy conservation gives
mgh =
1
2
mv
2
v
= [2(9.80 m/s
2
)(0.10 m)]
1/2
=
1.4 m/s
(b) The balls stick together, so their speed just after the collision is 0.700 m/s, by momentum conservation.
Using energy conservation again gives
h = v
2
/2
g
= (0.700 m/s)
2
/[2(9.80 m/s
2
)] = 0.025 m =
2.5 cm
46.
m
1
= 0.010 kg,
m
2
= 3.0 kg,
v
1o
= 400 m/s,
v
2o
= 0,
v
1
= 300 m/s,
v
2
= ?
P
o
=
P
,
m
1
v
1o
+
m
2
v
2o
=
m
1
v
1
+
m
2
v
2
.
v
2
=
m
1
v
1o
+
m
2
v
2o
−
m
1
v
1
m
2
=
(0.010 kg)(400 m/s) + 0
−
(0.010 kg)(300 m/s)
3.0 kg
= 0.33 m/s
.
y
3
1
x
2

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