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II
. (16 points) (
This is the same situation as problem I.
) Two balls are placed at opposite ends of the inclined
plane of length
L
= 2
.
0 m shown. At time
t
= 0 s ball #1 at position
x
= 0 m (top of plane) is released. At
the same time ball #2 is given a kick up the plane so its initial speed is 4.0 m/s.
At what time do the two balls meet? (
On Earth.
)
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Since both balls undergo constant acceleration, the constant accelera
tion kinematics equations can be used.
x
1
=
x
10
+
v
10
Δ
t
+
1
2
a
1
(Δ
t
)
2
and
x
2
=
x
20
+
v
20
Δ
t
+
1
2
a
2
(Δ
t
)
2
When the balls meet, they are in the same position at the same time.
x
1
=
x
2
⇒
x
10
+
v
10
Δ
t
+
1
2
a
1
(Δ
t
)
2
=
x
20
+
v
20
Δ
t
+
1
2
a
2
(Δ
t
)
2
Although the accelerations are not known, they are the same (
a
1
=
a
2
), so the
1
2
a
(Δ
t
)
2
terms can be
subtracted from each side.
x
10
+
v
10
Δ
t
=
x
20
+
v
20
Δ
t
The initial position of ball #1 is zero, and the initial velocity of ball #1 is zero. The initial position of ball
#2 is
L
. Note that the initial velocity of ball #2 is 4.0 m/s, as its initial speed is 4.0 m/s in the negative
direction.
0 + 0 =
L
+
v
20
Δ
t
⇒
Δ
t
=

L
v
20
=

2
.
0 m

4
.
0 m
/
s
= 0
.
50 s
1. (6 points) Which direction is ball #2 travelling at the instant the balls meet?
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The direction of travel for ball #2 would be determined by the sign of its velocity at the time Δ
t
found
above.
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 Spring '08
 UZER
 Acceleration, Velocity, ball, Solutions Page

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