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. 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} as in Fig. 1b below of the
statement of the problem.


2.

 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 above. 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.
 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 above.
 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 above.


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^{o}.
Fig. for #3 is shown above.


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.
 r = r_{x}i + r_{y}j
= (5.00i + 12.0j) m.
 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}.


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.

 For horizontal motion of the rock,
x = 24 m/s t (Equation
1)
For horizontal motion of the car,
90 m  x = 1/2(4.0 m/s^{2})t^{2}
(Equation
2)
Substituting Eq. 1 into Eq. 2:
90 m  24 m/s t = 1/2(4.0 m/s^{2})t^{2}
or
t^{2 }+ 12s t  45s^{2} t^{2}
= 0 or
(t  3s)(t + 15s) = 0 or
t = 3 s.
 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 you. 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.

 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.
 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.
 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 above) 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.
 The velocity at h and f are shown in Fig. 3 above.
 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 above.


10.

Divide this problem into two onedimensional problems:

X 
Y 

x_{o}
= 0 
y_{o
}= 20 m 

v_{ox
}= 25 m/s (cos 37^{o}) 
v_{oy
}= 25 m/s (sin 37^{o}) 

=
25 m/s (0.80) = 20 m/s 
=
25 m/s (0.06) = 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 
(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 } 20 m) = 11.25 m
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
(t  4 s)(t + 1 s) = 0 and
t = 4 s 
(e) 
x(t) =
x_{o} + v_{ox}t + 1/2 a_{x}t^{2}
x(t) = 0 + 20 m/s t + 0
x(4s) = (20 m/s)(4 s) = 80 m



v_{x}
= v_{ox} 
v_{y}(t)
= v_{oy }+ a_{y}t 
(f) 
v_{x}(4s)
= 20 m/s 
v_{y}(4s)
=15 m/s 10 m/s^{2}(4s)
= 25 m/s 
(g) 
v(4s) =
{(20 m/s)^{2} + (25 m/s)^{2}}^{1/2
}=^{ }32.0 m/s.
tan Θ= 25/20
Θ= 51.3^{o}

Velocity vectors are drawn on Fig. 4
above.


11.

(a) r_{o } = 0i + 20 m j =
20 m j
(b) v_{o} = (20 i + 15 j)
m/s
(c) a = 10 m/s^{2 }j
(d) v(t) = 20 m/s i + (15 m/s  10 m/s^{2}
t)j
(e) v(t) = (20 m/s i +15 m/s j)
 (10 m/s^{2} t)j
v(t) = v_{o}
+ at
(f) r(t) = (20 m/s i)t + {20 m
+15 m/s t + 1/2 (10m/s^{2})t^{2}}j
(g) = 20 m j + {(20 i + 15 j)
m/s}t + 1/2(10 m/s^{2})j t^{2
} =
r_{o }+ v_{o }t+ 1/2 at^{2}
The figure above shows how to find r(1s), r(2s),
r(3s) and r(4s) using r(t) = r_{o
}+_{ }v_{o }t_{ }+ 1/2 at^{2}.
To find r(3s) start out at the origin and go up
20 m, add a vector 37^{o} above the Xaxis of v_{o
}x 3s = 25 m/s x 3 s = 75 m. Finally add a vector down
= 1/2 (10 m/s^{2})(3s)^{2} = 45 m and you arrive
at the head of r(3s).


12.

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.


13.

X 
Y 
x_{o}
= 0 
y_{o
} = 0 
v_{ox
}= v_{o }cosΘ 
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 Θ)(2v_{o}
sin Θ/a_{y}) = (2v_{o}^{2}
sin Θ 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 Θ  1/2
a_{y}(30 m/v_{o }cos Θ)^{2}
(Equation
6)
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 . Since
sin 2Θ= 2 sin Θcos
Θ,
v_{o}^{2} = 180 m a_{y}/sin 2Θ=
180 m(10 m/s^{2})/0.88.
v_{o} = 45 m/s.


14.

When, and only when, a projectile starts
at the ground and ends up at the ground, the time to reach the
highest point equals the time to go from the highest point to
the ground. Since they all go up the same vertical height, the
time involved must be the same. In more detail,
 To go from the highest point take y_{o }= y_{max}.
At that point, the vertical velocity is zero, so y
= y_{o} + 1/2 at^{2}, where a is the acceleration
of the object. When the object hits the ground, y = 0 =
y_{max} + 1/2 at^{2}. Since y_{max
}and a are the same for all objects, the time for the
object to fall or to be in flight is the same.
 At the highest point, the vertical velocity of the object
is zero.
0 = v_{y }= v_{oy}+ a_{y}t
and the time is the same for all three projectiles
so v_{oy }= v_{o }sin Θ
= a_{y}/t = constant. Thus v_{o }= constant/sin
Θ or v_{o}
is inversely proportional to sin Θ.
The larger the angle, the larger the sine of the angle and
the smaller the initial velocity v_{o}. The initial
velocity is greatest for the smallest angle Θ_{3}.
 The horizontal distance x = (v_{o} cos Θ)t.
As the angle decreases the cosine increases and v_{o}
increases, so the range is the greatest for the smallest
angle Θ_{3}.


15.

(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 above. 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.


16.

(a) The period = distance moved in one complete rotation/velocity
T = 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
(d) The centripetal acceleration a = v^{2}/r = (2 πr/T)^{2}/r
= 4 π^{2}r/T^{2}
= 4 π^{2}rf^{2}
= (d)


17.

 By definition of an angle in radians,
ΔΘ= Δs/r
(Equation
1)
 Dividing both sides of Eq. 1 by Δt,
ΔΘ/ Δt=
(1/r) Δs/ Δt
^{lim}^{Δ}^{t
}^{0}
ω= dΘ/dt
= (1/r) ds/dt = 1/r (v)
By definition, the velocity v = rate of change of position
= ds/dt.
ω = v/r.
 Curling the fingers of your right hand counterclockwise,
your thumb points out of the page.
 Since the magnitude of the angular velocity = v/r = a
constant and the direction of ω
for counterclockwise motion is out of the page, uniform
circular motion is a case of constant angular acceleration.
Note: use "out of the page" rather than up because
up is in the plane of the page.


18.

 r_{x} = r cos ωt
r_{y }= r sin ωt
r(t) = r(cos ωt
i + sin ωt j)
 v(t) = dr/dt =
v(t) = ωr(sin
ωt i + cos ωt
j)
 v_{x} = ωr
sin ωt
v_{y} = ωr cos
ωt
v = (v_{x}^{2} + v_{y}^{2})^{1/2
} = ωr(sin^{2}ωt
+ cos^{2} ωt)^{1/2}^{
}= ωr
 a(t) = dv/dt
a(t)= ω^{2}r(cos
ωt i + sin ωt
j) = ω^{2}r(t)
 a_{x} = ω^{2}r
cos ωt. a_{y }=
ω^{2}r sin
ωt
a = (a_{x}^{2} + a_{y}^{2})^{1/2
}= ω^{2}r(cos^{2}^{
}ωt + sin^{2}
ωt)^{1/2} =
ω^{2}r
Since ω = v/r, a = (v/r)^{2}
r = v^{2}/r


19.

 Since v_{1} is perpendicular to r_{1}
and v_{2} is perpendicular to r_{2},
the two angles Θ are
equal. Since the two triangles are isosceles (r_{1 }=
r_{2} = r and v_{1 }= v_{2} = v),
they are similar and
Δv/v = Δr/r or
Δv = Δr(v/r)
and
Δv/ Δt
= ( Δr/ Δt)
(v/r).
Taking the limit of this equation,
Δv/ Δt
= ( Δr/ Δt)
(v/r) or
^{lim}^{Δ}^{t
}^{0
}
dv/dt = (v)(v/r) since dr/dt = v.
^{lim}^{Δ}^{t
}^{0}
Since dv/dt = a, a = v^{2}/r.
 As Δt goes to 0, Θ goes
to 0, and Δv is
perpendicular to v.
Since a = Δv/ Δt,
a is perpendicular to v or parallel to r
into the center of the circle. For this reason it is
called a centripetal acceleration.

