Answers  Two Dimensional Motion


1.

You find the resultant of two vectors
graphically by (a) the parallelogram method or (b) the polygon
method. In the parallelogram method, complete the parallelogram,
as shown by the dashed lines in Fig. for #1a below. The resultant
s is the diagonal. In the polygon method, move the
tail of one of the vectors, for example s_{1},
to the head of s_{2}. The resultant s
goes from the tail of s_{2} to the head of
s_{1} (Fig. for #1b below in the statement
of the problem).



2.

(a) For vectors along a straight line,
you can assign a positive sign for a vector to the right and
a negative sign for a vector to the left. Letting 8 m be to
the right and 6 m to the left, 8 m  6 m = 2 m. This is also
shown in a vector diagram in Fig. for #2a below. Because the
two vectors lie along the same straight line, I have drawn
the vector for 6 m below the vector for +8 m. Remember two
vectors are equal if they have the same magnitude and direction
even if they do not start at the same point.
(b) Now let both 8 m and 6 m be to the right: 8 m + 6 m =
14 m. This is also shown in a vector diagram in Fig. for #2b
below.
(c) Note that (8^{2 }+ 6^{2})^{1/2}
= 10. If the vectors are at a right angle then the resultant
vector has a magnitude of 10 m, as shown in a vector diagram
in Fig. for #2c below.



3.

First find the components of A.
A_{x} = A cos 37^{o} = 10 cm (0.8)
= 8 cm
A_{y} = A sin 37^{o} = 10 cm (0.6)
= 6 cm
We are given that B_{x} = 8 cm and B_{y} =
 2 cm.
For C = A + 1/2 B,
C_{x} = A_{x} + B_{x}/2
= (8 + 4) cm = 12 cm and C_{y }= A_{y} + B_{y}/2
= (6  1) cm = 5 cm.
C = (12^{2 }+ 5^{2})^{1/2}
cm = 13 cm.
tan Θ = C_{y}/C_{x}
= 5/12 = 0.42. Θ
= 26^{0}.
Figure for #3 is shown below.



4.

The vector r_{1} is totally
in the +Ydirection.
The component of r_{1} along the Xaxis, r_{1x
}= r_{2} cos 90^{0 }= (7.00 m)0 = 0.
The component of r_{1} along the Yaxis, r_{1y
}= r_{1} sin 90^{0 }= 7.00 m (1) = 7.00
m. The component of r_{2} along the Xaxis,
r_{2x }= r_{2} cos 45^{0 }= 7.07 m
(0.707)
= 5.00 m.
The component of r_{2} along the Yaxis, r_{2y
}= r_{2} sin 45^{0 }= 7.07 m (0.707)
= 5.00 m.
The component of r along the Xaxis, r_{x }=
r_{1x }+ r_{2x} = 0 + 5.00 m = 5.00 m.
The component of r along the Yaxis, r_{y }=
r_{1y }+ r_{2y} = 7.00 m + 5.00 m = 12.0 m.
(a) r = r_{x}i + r_{y}j
= (5.00i + 12.0j) m.
(b) The magnitude of r = r = (r_{x}^{2}
+ r_{y}^{2})^{1/2 }= (5.00^{2}
+ 12.0^{2})^{1/2 }m^{ } = 13.0 m.
tan Θ= 12/5. Θ=
67.4^{0}.
See Fig. 2 below.



5.

For motion in the horizontal direction
along the Xaxis,
t = x/v_{x }= 60 ft/(132 ft/s) =
0.454 s.
For motion in the vertical direction along the Yaxis,
the vertical drop y = 1/2 at^{2}
= 1/2 (32 ft/s^{2})(0.454 s)^{2} = 3.3 ft.


6.

(a) For horizontal motion of the rock,
x
= 24 m/s t. (Equation
1)
For horizontal motion
of the car,
x
= 90 m 1/2(4.0 m/s^{2})t^{2 }(Equation
2)
Substituting Eq. 1 into
Eq. 2:
24
m/s t = 90 m 1/2(4.0 m/s^{2})t^{2}
or
t^{2 }+ 12s t  45s^{2} = 0
(t  3s)(t + 15s) = 0 or
t = 3 s.
(b) Distance fallen by the rock = y = 1/2 at^{2} =
1/2(10 m/s^{2})(9s^{2}) = 45 m.


7.

For the vertical motion of the ball,
y = 1.0 m = 1/2(10 m/s^{2})t^{2} and t = 0.45
s. For the horizontal motion, x = v_{x}t = (4.0 m/s)(0.45
s) = 1.8 m.


8.

The expressions for the quantities t,
y_{max}, v_{y}(t). and x will all remain the
same. This is exactly what the principle of superposition
is telling us. The object is subjected to two different influences,
an initial velocity and a continuing acceleration due to gravity,
and the object responds to each without altering its response
to the other. The fact that the object has an initial horizontal
velocity does not alter its vertical motion or the time it
takes the object to hit the ground. The fact that the object
is subjected to a vertical acceleration does not change its
constant horizontal velocity motion. As a result you may separate
a twodimensional problem of a projectile into two onedimensional
problems.


9.

(a) At the initial position i, v_{ix
}= v_{o} cos 37^{0} = 25 m/s (0.80) =
20 m/s and
v_{iy } = v_{o} sin 37^{0} = 25 m/s
(0.60) = 15 m/s.
(b) At any time, the horizontal component of the velocity
v_{x} = v_{ix} + a_{x}t =
v_{ix }+ (0)t = 20 m/s. There is no acceleration in
the xdirection and the object continues to move with a constant
horizontal velocity. At the highest point, the vertical component
of velocity is zero. It does not continue to move upward.
It momentarily stops moving upward and then moves down. It
does continue, however, to have the same constant horizontal
velocity. Notice that this velocity at h is tangent to the
path, as it should be.
(c) At f, the horizontal velocity still equals 20 m/s. It
takes 3.0 s. to move 60 m (see x position of object just before
it hits the ground in Fig. 3 below) at a constant horizontal
velocity of 20 m/s. At t = 3.0 s, the vertical velocity v_{fy}(3.0s)
= v_{iy }+ a_{y}t = 15m/s + (10 m/s^{2})(3.0
s) = 15 m/s. For this symmetrical situation, v_{fy }=
v_{iy}. The final velocity is 25 m/s at an angle
of 37^{0}. Notice also for this case where y_{i
}= y_{f} = 0, the time to go half the distance
horizontally equals onehalf the total time of flight.
(d) The velocity at h and f are shown in Fig. 3 below.
(e) The acceleration is a constant. It is vertically downward
and we took it to be 10 m/s^{2}. The acceleration
vectors are shown at x = 20.0, 30.0, and 40.0 m in Fig. 3
below.



10.

Divide this problem into two onedimensional
problems:
X 
Y 
x_{o} =
0

y_{o }=
20 m

(a)v_{ox }= 25 m/s (cos
37^{0})
= 25 m/s (0.80) = 20 m/s

(b)v_{oy }= 25 m/s (sin
37^{0})
= 25 m/s (0.60) = 15 m/s

a_{x } = 0

a_{y } = 10 m/s^{2}

v_{x} = v_{ox}
+ a_{x}t = v_{ox }= 20 m/s

v_{y} = v_{oy }+
a_{y}t = 15 m/s  10 m/s^{2}t

At highest point when y = y_{max,}

v_{y}(y_{max})
= 0

(c) In general,

v_{y}^{2}(y) =
v_{o}^{2} + 2a_{y}(y  y_{o})

For this case,

= (15 m/s)^{2} 20 m/s^{2}(y
 20m)

At highest point with y = y_{max,}

0 = 225 m^{2}/s^{2}
20 m/s^{2}(y_{max } 20m)
y_{max} = 31.25 m

(d) In general,

y(t) = y_{o} + v_{oy}t
+ 1/2 a_{y}t^{2}

For this case,

y(t) = 20 m + 15 m/s t  5 m/s^{2}t^{2}

When the object returns to earth,

0 = 20 m + 15 m/s t  5 m/s^{2}t^{2}
or
t^{2}  3 s t  4 s^{2} t^{2}
= 0 or factoring

(e) x(t) = x_{o} + v_{ox}t
+ 1/2 a_{x}t^{2}
x(t) = 0 + 20 m/s t +
0

(t  4 s)(t + 1 s) = 0 and t =
4 s

x(4s) = (20 m/s)(4 s) = 80 m

v_{y}(t) = v_{oy }+
a_{y}t

(f) For example, for


v_{x}(4s) = 20 m/s = constant

v_{y}(4s) =15 m/s 10 m/s^{2}(4s)
= 25 m/s

v(4s) = {(20 m/s)^{2} +
(25 m/s)^{2}}^{1/2 }=^{ }32.0
m/s. tan Θ = 25/20; Θ = 51.3^{0}

(a), (b), (f), and (g) are drawn on Fig.
4 below.



11.

To catch up to the ball, the horizontal
velocity v_{b} of the boy must equal the initial horizontal
velocity of the ball:
v_{b} = v_{ox} = v_{o
}cos Θ.
cos Θ = v_{b}/v_{o}
= (20 m/3s)/20 m/s = 1/3.
Θ =
70.5^{o}. v_{oy }= v_{o
}sin Θ = 20 m/s sin
70.5^{o }= 18.9 m/s.
For motion in the ydirection, v_{y}(y) = v_{oy}^{2}
+ 2 a_{y}(y  y_{o}).
For the highest point, v_{y }= 0 and 0 = (18.9 m/s)^{2}
 20 m/s^{2}(y  y_{o}).
The ball rises (y  y_{o}) = (18.9)^{2}m/20
= 17.8 m.


12.

X 
Y 
x_{o} = 0

y_{o } = 0

v_{ox }= v_{o }cos
Θ

v_{oy }= v_{o }sin
Θ

a_{x} = 0

a_{y} = 10 m/s^{2}

x_{max }= 180 m for which
y = 0


x(t) = v_{ox}t

y(t) = v_{oy}t + 1/2 a_{y}t^{2}

For x = 180 m = (v_{o }cos
Θ)t (Equation
1)

y = 0 = v_{o} sin Θ
t  1/2 a_{y}t^{2
}t = 2v_{o} sin Θ/a_{y} (Equation
2)

Substituting Eq. 2 into Eq. 1:
180 m = (v_{o }cos )(2v_{o} sin Θ/a_{y})
= (2v_{o}^{2} sin cos Θ/a_{y}),
or
v_{o}^{2} = 90 m a_{y}/sin Θ
cos Θ (Equation
3)
When the ball clears the tree,
x = 30 m = (v_{o }cos Θ)t
and y = 15 m = v_{o}
sin Θ t  1/2 a_{y}t^{2}
(Equation
4)
t = 30 m/(v_{o }cos Θ)
(Equation
5)
Substituting Eq. 5 into Eq. 4:
15 m = (v_{o} sin Θ)(30
m/v_{o }cos Θ)
 1/2 a_{y}(30 m/v_{o }cos Θ)^{2},
or
15 m = 30 m tan  1/2 a_{y}(30 m/v_{o }cos
)^{2 }(Equation6)
Substituting Eq. 3 into Eq. 6:
15 m = 30 m tan Θ  1/2
a_{y} (30 m/cos Θ)^{2}(sin
Θ cos Θ/90
m a_{y})
15 m = 30 m tan Θ  5
m tan Θ= 25 tan Θ.
tan Θ = 15/25. Θ=
31^{0}.
v_{o}^{2} = 90 m a_{y}/sin Θ
cos Θ = 180 m a_{y}/sin
2Θ
= 180 m(10 m/s^{2})/0.88
v_{o}= 45 m/s.


13.

(a) and (b). The velocity v and
the acceleration a at positions A, B, and C for an
object moving with uniform circular motion are shown in Fig.
6 below. The magnitude of the velocity is a constant and always
tangent to the circle at the point where you wish to find
it. The centripetal acceleration is constant in magnitude
and always in toward the center of the circle. The centripetal
acceleration a = v^{2}/r = (2.0 m/s)^{2}/0.4
m = 10 m/s^{2}.
(c) While the magnitudes of the velocity and acceleration
are constants, the directions are not. Thus neither v
nor a is constant.
(d) A change in direction of velocity results in an acceleration
just as much as a change in the magnitude of the velocity.
You could use this device to see how many "g's"
an object could withstand. For the radius and speed in this
problem, the object is subjected to approximately 1g since
10 m/s^{2}≈1 g.



14.

(a) The period = distance moved in one
complete rotation/velocity
= 2 πr/v
= 2 π(0.4 m)/2.0 m/s =
0.4 π s.
(b) The frequency = 1/period = 2.5/ π
s^{1
} (c) From (a) v = 2 πr/T.
The centripetal acceleration a = v^{2}/r = (2 πr/T)^{2}/r
= 4 π^{2}r/T^{2}
= 4 π^{2}rf^{2}
= (d)

