MasteringPhysics Homework #7
10.5. Model:
This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not
change as the ball rises and falls.
Visualize:
The figure shows a ball’s beforeandafter pictorial representation for the three situations in parts (a), (b) and (c).
Solve:
The quantity
K
+
U
g
is the same during free fall:
K
f
+
U
gf
=
K
i
+
U
gi
.
We have
(a)
1
2
mv
1
2
+
mgy
1
=
1
2
mv
0
2
+
mgy
0
⇒
y
1
=
v
0
2
−
v
1
2
( )
g
=
[(10 m/s)
2
−
(0 m/s)
2
]/(2
×
9.8 m / s
2
)
=
5.10 m
5.1 m is therefore the maximum height of the ball above the window. This is 25.1 m above the ground.
(b)
1
2
mv
2
2
+
mgy
2
=
1
2
mv
0
2
+
mgy
0
Since
y
2
=
y
0
=
0,
we get for the magnitudes
v
2
=
v
0
=
10 m / s.
(c)
1
2
mv
3
2
+
mgy
3
=
1
2
mv
0
2
+
mgy
0
⇒
v
3
2
+
2
gy
3
=
v
0
2
+
2
gy
0
⇒
v
3
2
=
v
0
2
+
2
g
(
y
0
−
y
3
)
⇒
v
3
2
=
(10 m / s)
2
+
2(9.8 m / s
2
)[0 m
−
(
−
20 m)]
=
492 m
2
/ s
2
This means the magnitude of
v
3
is equal to 22 m/s.
Assess:
Note that the ball’s speed as it passes the window on its way down is the same as the speed with which
it was tossed up, but in the opposite direction.
pendulum’s beforeandafter pictorial representation for the two situations described in parts (a) and (b).
Solve:
(a)
The quantity
K
+
U
g
is the same at the lowest point of the trajectory as it was at the highest point.
Thus,
K
1
+
U
g1
=
K
0
+
U
g0
means
1
2
mv
1
2
+
mgy
1
=
1
2
mv
0
2
+
mgy
0
⇒
v
1
2
+
2
gy
1
=
v
0
2
+
2
gy
0
⇒
v
1
2
+
2
g
(0 m)
=
2
+
2
gy
0
⇒
v
1
=
2
gy
0
From the pictorial representation, we find that
y
0
=
L
−
L
cos30
°
.
Thus,
v
1
=
2
gL
(1
−
cos30
°
)
=
2(9.8 m/s
2
)(0.75 m)(1
−
cos30
°
)
=
1.403 m/s
The speed at the lowest point is 1.40 m/s.
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View Full Document(b)
Since the quantity
K
+
U
g
does not change,
K
2
+
U
g
2
=
K
1
+
U
g
1
. We have
1
2
mv
2
2
+
mgy
2
=
1
2
mv
1
2
+
mgy
1
⇒
y
2
=
v
1
2
−
v
2
2
( )
g
⇒
y
2
=
[(1.403 m/s)
2
−
(0 m/s)
2
]/(2
×
9.8 m/s
2
)
=
0.100 m
Since
y
2
=
L
−
L
cos
θ
,
we obtain
cos
=
L
−
y
2
L
=
(0.75 m)
−
(0.10 m)
(0.75 m)
=
0.8667
⇒
=
cos
−
1
(0.8667)
=
30
°
That is, the pendulum swings to the other side by
30
°
.
Assess:
The swing angle is the same on either side of the rest position. This result is a consequence of the fact that
the sum of the kinetic and gravitational potential energy does not change. This is shown as well in the energy bar
chart in the figure.
10.16.
Model:
Assume an ideal spring that obeys Hooke’s law.
Visualize:
Solve: (a)
The spring force on the 2.0 kg mass is
F
sp
=
−
k
Δ
y
.
Notice that
Δ
y
is negative, so
F
sp
is positive.
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 Spring '09
 Mylander
 Kinetic Energy, Momentum, VIX, lab frame, s′

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