cutnell7e_ISM_ch03 - CHAPTER 3 KINEMATICS IN TWO DIMENSIONS...

CHAPTER 3 KINEMATICS IN TWO DIMENSIONS
CONCEPTUAL QUESTIONS
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1. REASONING AND SOLUTION In addition to the x and y axes, the z axis would be
required to completely describe motion in three dimensions. Motion along the z axis can be
described in terms of the kinematic variables z , a
z
, v
z
,
v
0 z
, and t . In analogy with the
equations of kinematics for the x and y components, the following equations would be
necessary to describe motion along the z axis:
v v a t
z z z
= +
0
z v t a t
z z
= +
0
1
2
2
z v v t
z z
= + 1
2 0
( )
v v a z
z z z
2 2 = +
0
2
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2. SSM REASONING AND SOLUTION An object thrown upward at an angle θ will
follow the trajectory shown below. Its acceleration is that due to gravity, and, therefore,
always points downward. The acceleration is denoted by a
y
in the figure. In general, the
velocity of the object has two components, v
x
and v
y
. Since a
x
= 0, v
x
always equals its
initial value. The magnitude of the y component of the velocity, v
y
, decreases as the object
rises, drops to zero at the highest point, and then increases as the object falls downward.
a. Since v
y
= 0 when the object is at its highest point, the velocity of the object points only
in the x direction. As suggested in the figure below, the acceleration will be perpendicular to
the velocity when the object is at its highest point and v
y
= 0.
θ
θ – v
0y
v
0x
v
f
a
y
a
y
v
0y
a
y
v
0x
v
0x
b. In order for the velocity and acceleration to be parallel, the x component of the velocity
would have to drop to zero. However, v
x
always remains equal to its initial value; therefore,
the velocity and the acceleration can never be parallel.
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96 KINEMATICS IN TWO DIMENSIONS
3. REASONING AND SOLUTION As long as air resistance is negligible, the acceleration of
a projectile is constant and equal to the acceleration due to gravity. The acceleration of the
projectile, therefore, is the same at every point in its trajectory, and can never be zero.
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4. REASONING AND SOLUTION If a baseball were pitched on the moon, it would still fall
downwards as it travels toward the batter. However the acceleration due to gravity on the
moon is roughly 6 times less than that on earth. Thus, in the time it takes to reach the batter,
the ball will not fall as far vertically on the moon as it does on earth. Therefore, the pitcher's
mound on the moon would be at a lower height than it is on earth.
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5. REASONING AND SOLUTION The figure below shows the ball’s trajectory. The
velocity of the ball (along with its x and y components) are indicated at three positions. As
long as air resistance is neglected, we know that a
x
= 0 and a
y
is the acceleration due to
gravity. Since a
x
= 0, the x component of the velocity remains the same and is given by v
0 x
.
The initial y component of the velocity is v
0 y
and decreases as the ball approaches the
highest point, where v
y
= 0. The magnitude of the y component of the velocity then increases
as the ball falls downward. Just before the ball strikes the ground, the y velocity component
is equal in magnitude to v
0 y
. The speed is the magnitude of the velocity.