Ch 3:  2d Motion & Vectors
Updated 8/19/07

Home

Answers to Ch 3

3.1 Introduction to Vectors, pg 82

Section 3.1 Review Q, pg 85: 
1a] vector   [b] scalar   [c] scalar   [d] vector   [e] scalar
2] 126 m at 10^o above the horizontal
3] 204 km/h at 75^o north of east
4] 89 km/h at 54o north of east
5] zero

Ch 3 Review Q, pg 108: 
1] A scalar represents the magnitude of a physical quantity.
2] No, for the sum to be zero the vectors must be equal in size and opposite in direction.
3] Speed is the magnitude of velocity.
4] 30 m/s east
5] No, because the scalar has no direction.
6a] 5.00 units at 53.1^o below the positive x-axis.
 b] 5.00 units at 53.1^o above the positive x-axis.
 c] 8.54 units at 53.1^o below the positive x-axis.
 d] 5.00 units at 127^o clockwise from the positive x-axis.
7a] 5.20 m at 60.0^o above the positive x-axis
 b] 3.00 m at 30.0^o below the positive x-axis
 c] 3.00 m at 150^o counterclockwise from the positive x-axis
 d] 5.20 m at 60.0^o below the positive x-axis
8] 7.9 m at 4.3^o north of west
9] 15.3 m at 58.4^o south of east
10] when the vectors point in the same direction.
11] Velocity & Acceleration in same direction: 
      A car moving straight and speeding up,
    Velocity & Acceleration in opposite directions: 
      A car moving straight and slowing down.
12] 55; it cannot be 85 for the maximum the resultant can be is
        55+25 = 80
        it cannot be 20 for the minimum value the resultant can be is
        55-25 = 30
13] Yes, the distance from the tail of the first vector to the ehad of the last vector is zero.

Top

3.2 Vector Operations, pg 86

Practice A, Finding Resultant Magnitude and Direction, pg 89
[1a] 23 km   [b] 17 km East
[2] 45.6 m at 9.5^o East of North
[3] 15.7 m at 22^o to the side of downfield
[4] 1.8 m at 49^o below the horizontal

Practice B, Resolving Vectors, pg 92
[1] 95 km/h     [2] 44 km/h     [3] 21 m/s, 5.7 m/s     [4] 0 m, 5 m

Practice C, Adding Vectors Algebraically, pg 94
[1] 49 m at 7.3^o to the right of downfield
[2] 7.5 km at 26^o above the horizontal
[3] 13.0 m at 57^o North of East
[4] 171 km at 34^o East of North

Section 3.2 Review Q, pg 94: 
1a] x-axis:  forward and backward on sidewalk
    y-axis:  left and right on sidewalk
 b] x-axis:  forward and backward on rope
    y-axis:  up and down
 c] x-axis:  horizontal at water level
    y-axis:  up and down
2a] 5.8 m/s at 59^o downriver from its intended path
 b] 6.1 m/s at 9.5o form the direction the wave is traveling
3a] 7.07 km north, 7.07 km east
 b] 1.6 m/s² horizontal, 1.1 m/s² vertical
4] Nonperpendicular vectors need to be resolved into components because the Pythagorean theorem and the tangent function can be applied only to right triangles.

Ch 3 Review Q, Vector Operations, pg 109: 
14] Yes, the second component could be nonzero.
    Ex:  a 30 displacement on the x-axis has a zero y-axis component
15] No, the hypotenuse is always greater than the legs.
16] Vector addition is combining vectors to find a resultant vector.
17] To add two vectors that are not perpendicular or parallel
    a] Resolve both vectors into their vector x & y components,
    b] Add the x components to find Rx
    c] Add the y components to find Ry
    c] Use the Pythagorean Theorem to find the resultant, R.
18] They are equal and opposite.
19] If the vector is oriented at 45^o from the axes.
20] They form a closed triangle when laid head to tail.
21a] 5 blocks at 53o North of East    b]13 blocks
22] 42.7 yards
23] Four possible answers:
    61.8 m at 76.0^o S of E (or S of W)
    25.0 m at 53.1^o S of E (or S of W)
24] 108 m, -19.1 m
25] 2.81 km east, 1.31 km north
26] 240 m at 57.2^o south of west

Top

3.3 Projectile Motion, pg 95

Practice D, Projectiles Launched Horizontally, pg 99
[1] 0.66 m/s     [2] 4.9 m/s     [3] 7.6 m/s     [4] 5.6 m

Practice E, Projectiles Launched at an Angle, pg 101
[1] Yes, Δy = -2.3 m    [2] 35.1 m    [3] 2.0 s; 4.8 m    [4] 6.2 m/s

Section 3.3 Review Q, pg 101: 
1] a, b, d
2] 5.05 s; 454 m
3a] 192 m   [b] 69.4 m/s at 64.4^o below the horizontal
4]

Ch 3 Review Q, Projectile Motion, pg 109: 
27] Both hit at the same time.
28] Yes, neglecting air resistance.
29] No, neglecting air resistance.
30] The vertical components of each velocity vector will be the same,
    but the thrown ball will also have a horizontal component of
    velocity.  As a result, the thrown ball will have a greater
    speed.
31] 45.1 m/s
32] 3.3 s; 36 m/s
33] 11 m
34a] 2.77x105 m     [b] 284 s = 4 min 44 sec
35a] clears the goal by 1 m     [b] falling
36] 4.11 m
37] 80 m; 210 m

Top

3.4 Relative Motion, pg 102

Practice F, Relative Velocity, pg 105
[1] 0 m/s
[2] 14.5 m/s in the direction that the aircraft carrier is moving     [3] 3.90 m/s at 40^o north of east
[4] 26.4 m/s at 2.17^o east of north

Section 3.4 Review Q, pg 105: 
1] 10 m/s in the opposite direction
2] 1.51 m/s at 5.7^o north of east
3a] south with a speed equal to the train’s speed
 b] moves north
 c] appears to fall straight down
 d] moves in a parabola

Ch 3 Review Q, Relative Motion, pg 110: 
38] Displacement and velocity depend on the frame of reference in
    which they are measured.
39] A frame of reference is the coordinate system used to describe
    the motion.
40] Earth
41a] 70 m/s East     [b] 20 m/s
42a] To the passenger, the ball appears to move in a straight line. 
     To an outside observer, the ball moves along a parabolic
     trajectory.
  b] The passenger would see the ball move backward, while the
     stationary observer would see no change from part A
43a] 10.1 m/s at 8.53^o East of North.     [b] 48.8 m
44a] 14.1 North of West                   [b] 199 km/h
45] 7.5 min
46a] 23.2^o upstream from straight across
  b] 8.72 m/s across the river.

Top

Ch 3 Mixed Review, pg 111

     47a] 41.7 m/s     [b] 3.81 s    
 [c] Vyf = -13.5 m/s, Vxf = 34.2 m/s, Vf = 36.7 m/s
48a] 15 m/s       [b] 15 m/s
49] 10.5 m/s
50a] 22.2 s       [b] 200 s = 3 min 20 s
51a] 2.66 m/s     [b] 0.64 m
52] 22.5 s
53] 157 m
54] 70 s = 1 min 10 s
56]
57]
58]
59]
60]
61]
62]

Ch 3 Standardized Test Prep, pg 114

     1] B   [2] H  [3] A  [4] J  [5] C  [6] G  [7] B  [8] J [9] D  [10] H 11] B  [12] H      
13] They are perpendicular     14] 31.5 m horizontal, 26.4 m vertical
15] 29.4 m/s                   16] 10.8 m

Top

Solutions Ch 3a Vector Test Practice

1    State the six vector rules.
a] Vectors can be drawn with an arrow, which has magnitude (length) and direction (angle).
b] Vectors can be moved if you do not change their size or direction.
c] To add vectors move them head to tail.
d] The order of addition does not matter.
e] The net vector or resultant is drawn from the tail of the first arrow to the head of the last. 
f] Any vector can be resolved into two perpendicular components

2    a] Define concurrent vectors.
Concurrent vectors act on the same object at the same time.
b] Diagram three concurrent vectors.
c] Diagram the addition of the concurrent vectors in [b].

3    Diagram and verbally compare vector addition with vector resolution.
Resolution:  Given the Resultant, R & θ
             Find the components Rx & Ry.
Addition:    Given the Components Ry & Rx
             Find the Resultant, R & θ

4    a] Any vector can be resolved into two perpendicular components
b] Draw the vector diagram for a box on a ramp.
c] Resolve a 100 lb force at 30^o into components.  [P1] 45^o [P2] 55^o

5    a] Diagram & label three diagrams for the addition of a
   6 N and a 8 N vector separated by 0^o, 90^o, and 180^o 
b] As the angle between 2 vectors increases their sum decreases.

6    a] State the law of similar triangles.
If two triangles are similar then the ratio of their corresponding sides are equal.
b] Diagram and label it.

7    a] Diagram a 9-12-15 right triangle and label the angles 1 & 2.
b] Use above diagram to       DEFINE   | ANGLE 1 | ANGLE 2
   fill in the ratios   SIN(θ) =  Opp |   12    |    9
      /|   in this table.          Hyo |   15    |   15
     /2|                
 15 /  |                 COS(θ) =  Adj |    9    |   12
   /   | 12                       Hyp |   15    |   15
  /    |                 
 /     |                 TAN(θ) =  Opp |   12    |    9
/1     |                           Adj |    9    |   12
   9
b] Translate into a sentence:  Sin^-1(Ry/R), Cos^-1(Rx/R), Tan^-1(Ry/Rx)
Sin^-1(Ry/R) = θ = The angle whose Sine = Ry/R,
Cos^-1(Rx/R) = θ = The angle whose Cosine = Rx/R
Tan^-1(Ry/Rx) = θ = The angle whose Tangent = Ry/Rx
Ry = Y component of the Resultant
Rx = X component of the Resultant
R = The Resultant

8    SKETCH & MATHEMATICALLY RESOLVE A 140 N FORCE INTO ITS COMPONENTS IF IT IS ACTING AT 30^o WITH THE HORIZONTAL. 
a] SKETCH & LABEL THE PROBLEM   
b] SHOW EQ & SUBSTITUTION, & ANSWERS.

9    a] Define equilibrium
The state in which the vector sum of the forces on an object is zero.
Objects that are at rest or moving in a straight line at constant speed are in equilibrium.
b] Define equilibrant.
The one force that will put all the other forces in equilibrium.
c] Draw and label the equilibrant diagram.

10   AT 8 m/s A BOAT HEADS ACROSS A 200 m WIDE RIVER FLOWING AT 6 m/s.  FIND:  a] THE MAGNITUDE & DIRECTION OF THE BOAT'S NET VELOCITY.
Velocity = Speed with direction
         _______
Speed = √8² + 6² = 10 m/s
Direction = θ = Angle with shoreline = Tan^-1(8/6) = 57^o 
       b] HOW LONG IT TAKES TO REACH THE OTHER SIDE.
The time to reach the other side depends on the boat’s speed and is independent of the river’s speed.  The river could go at 1,000 m/s and it would not make any difference.
D = Boat Speed x time ---> Time = 200 m/8 m/s = 25 s
       c] HOW FAR DOWN STREAM IT IS WHEN IT REACHES THE OTHER SIDE.
D = River Speed x Time = 6 m/s x 25 s = 150 m down stream.

11   a] Diagram the vectors below.
b] Show sample calculations and solve for R, E, Ry, and Rx.
Vector| Size |Angle| Fy = F Sin(θ)         | Fx = F Cos(θ)          |
  F1  | 50 N |  20 | F1y = 50Sin(20) = 17.1| F1x = 50Cos(20) = +47.0|
  F2  | 50 N |  45 | F2y = 50Sin(45) = 35.4| F2x = 50Cos(45) = +35.4|
  F3  | 50 N | 120 | F3y = 50Sin(120)= 43.3| F3x = 50Cos(120)= -25.0|
  R   | 112 N|  θ  | Ry = F1y+F2y+F3y= 95.8| Rx = F1x+F2x+F3x = 57.4|
  E   | 112 N|  Φ  | Ey  =  -Ry = -95.8    | Ex  = -Rx = --57.4      |
                                 
R = √Ry² + Rx² = √95.8² + 57.4² = 112

θ = The angle whose Tan = Ry/Rx = Tan^-1(95.8/57.4) = 59.1^o
Φ = 180^o + θ = 180 + 59.1 = 239^o 
c] Find the acceleration for a 15 kg object.
   Fnet = ma
   Fnet = vector sum of all vectors = R
   112 N, 57.1^o = 15 kg x Acceleration
   112/15, 57.1^o = Acceleration
   7.47 m/s², 57.1^o = Acceleration

Top

Solutions Ch 3b Projectile Motion Test Practice

1 mi = 5,280 ft = 1609 m

1    a] Define projectile. 
Any object moving through air or space under the influence of gravity only; No lift (wings), no thrust (engine), no drag (air friction).
b] Give 3 examples of projectiles.
A thrown rock or ball, a bullet, a satellite (Moon, planet, man made)

2    a] When does a = 9.8 m/s²?
   9.8 is the vertical acceleration when an object is falling.
b] Translate 4 m/s into a sentence.
   An object changes its distance (displacement) by 4m every second.
c] Translate 4 m/s² into a sentence.
   An object changes its speed (velocity) by 4m/s every second

3    State the assumptions and translate into a sentence:
a] D = S t:  An object is moving at constant speed.

b] b] ΔX = (Vf + Vi)Δt:  An object is moving with constant
               2         acceleration.  Δt = acceleration time.

4A   A STONE IS THROWN HORIZONTALLY AT 20 ft/s FROM A 100 ft HIGH CLIFF.
a] Draw & label the diagram.
b] How long does it take to hit the ground?
c] Find the final vertical velocity, Vvf.
Vertical Analysis            a = Vf - Vi        H = (Vf - Vi)t
Vi = 0 ft/s                         t                   2
Vf = ?                                          100 = (32t - 0)t
a = 32 ft/s²                 32 = Vf - 0                   2
t = ?                                t          100 = 16t²
H = 100 ft                                      ________
                             Vf = 32 t          √(100/16) = t
                             Vf = 32(2.5)<-------     2.50 = t
                             Vf = 80 ft
d] How far from the cliff does the stone strike the ground? 
Horizontal Analysis
a = 0 (No air friction)      D = Vh t
Vi = Vf = Vh = 20 ft/s       D = 20(2.5)
t = 2.50 sec                 D = 50 ft
D = ?

4B   A STONE THROWN HORIZONTALLY AT 25 ft/s LANDS 75 ft FROM THE CLIFF. 
a] Draw & label the diagram.
b] How long does it take to hit the ground?
   Horizontal Analysis
a = 0 (No air friction)      D = Vh t
Vi = Vf = Vh = 25 ft/s       75 = 25t
t = ?                        75/25 = t
D = 75 ft                    3.00 = t
c] Find the final vertical velocity, Vvf.
d] How high is the cliff? 
Vertical Analysis            a = Vf - Vi        H = (Vf - Vi)t
Vi = 0 ft/s                         t                   2
Vf = ?                                          H = (96 - 0)3.00
a = 32 ft/s²                 32 = Vf - 0                2
t = 3.00 s                           3          H = (48)3
H = ?                                          
                             Vf = 32(3)         H = 144 ft
                             Vf = 96 ft/s      
d] Find the landing speed in mi/h              _________
   Landing Speed = Size of Landing Velocity = √Vf² + Vh² 
                    _________
   Landing Speed = √96² + 25² = 99.2 ft/s

   Convert to mi/h:     99.2 ft |   1 mi  | 3600 s | = 67.6 mi/h
                             s  | 5280 ft |  1 hr  |

4C   A car runs off a 80 m cliff and lands 40 m from the bottom.
a] Draw & label the diagram.
b] Find the cars horizontal velocity in mi/h
Vertical Analysis            a = Vf - Vi        H = (Vf - Vi)t
Vi = 0 ft/s                         t                   2
Vf = ?                                          80 = (10t - 0)t
a = 10 ft/s²                 10 = Vf - 0                  2
t = ?                                t          80 = 5t²
H = 80 ft                                       ______
                             Vf = 10 t          √(80/5) = t
                                                 __
                                                √16 = t
                             Vf = 10(4) ft<------4 s = t
                             Vf = 40 ft/s
Horizontal Analysis
a = 0 (No air friction)          D  = Vh t
Vi = Vf = Vh = ? ft/s          40   = Vh(4)
t = 4 sec                      40/4 = Vh
D = 40 m                     10 m/s = Vh
d] Find the landing speed in mi/h              _________
   Landing Speed = Size of Landing Velocity = √Vf² + Vh² 
                    _________
   Landing Speed = √40² + 10² = 41.2 m/s

   Convert to mi/h:     41.2 m |  1 mi  | 3600 s | = 92.3 mi/h
                             s | 1609 m |  1 hr  |

5    a] Prove that at the same elevation, Vf = -Vi for an object launched vertically
   ΔX = (Vf + Vi)t      same elevation means ΔX = 0
            2
    0 = (Vf + Vi)t      Since t does not equal zero
            2           Vf + Vi = 0
                        Vf = - Vi
b] Draw the vector diagram for a ball hit at
   Vvi = 12 m/s, Vh = 6 m/s, g = 4 m/s².
   t | Vv                                                         
   0 | 12 m/s                                                     
   1 |  8      (Since g = 4m/s² = it looses 4m/s every second
   2 |  4                                                         
   3 |  0 (Top)
   4 | -4                                                         
   5 | -8                                                         
   6 | -12     (Since Vf = -Vi this must be the BOTTOM)

6    A BALL HIT AT A VELOCITY OF 70 m/s & AN ELEVATION OF 60^o.
a] Draw & label the diagram (include landing speed).
VERTICAL ANALYSIS--FIND
b] THE INITIAL VERTICAL VELOCITY.              
   Vvi = Vm Sin(θ) = 70 Sin(60) = 60.6 m/s
c] THE TIME TO REACH THE PINICAL OR TOP OF THE ARC.
   Half Arch Analysis        a = Vf - Vi 
   Vi = 60.6 m/s                    t
   Vf = 0                   
   a = - 9.8 m/s²            -9.8 = 0 - 60.6 
   t_up = ?                            t_up
                             -9.8 t = 60.6
                             t_up = -60.6/-9.8 = 6.19 s
d] THE MAXIMUM HEIGHT IT REACHES. 
   Hmax = (Vf + Vi)t = (0 + 60.6)6.19 = 187 m
              2           2
HORIZONTAL ANALYSIS--FIND
e] HORIZONTAL VELOCITY.
   Vh = Vmcos(θ) = 70 Cos(60) = 35 m/s
f] THE TIME IT IS IN THE AIR.
   Time in air = time up + time down = 2 t_up = 2(6.19) = 12.4 s
g] THE HORIZONTAL DISTANCE IT TRAVELS.         
   D = Vh t = 35(12.4) = 433 m
Landing Speed: 
Since Vh is constant and Vvf = - Vvi the Landing Velocity has the same components as the launch velocity.  Therefore they must be equal in size.
Launch Velocity  = 70 m/s at 60^o above the horizontal.
Landing Velocity = 70 m/s at 60^o below the horizontal.

7*   a] DERIVE THE EQUATION FOR THE HORIZONTAL DISTANCE AS A FUNCTION OF
MUZZLE VELOCITY AND ANGLE.  [Hint:  Sin(2θ) = 2 Cos(θ)Sin(θ)]
Vertical Analysis           
   Half Arch Analysis        a = Vf - Vi 
   Vi = Vm Sin(θ)                   t
   Vf = 0                   
   a = -g                    -g = 0 - Vm Sin(θ) 
   t_up = ?                            t_up
                            
                             t_up = -Vm Sin(θ)/-g = Vm Sin(θ)/g
Horizontal Analysis
   Vh = VmCos(θ)
   Time in air = 2 t_up = 2 Vm Sin(θ)/g
   D = Vh t
     = VmCos(θ) 2 Vm Sin(θ)/g
Simplify:  D = Vm² 2Cos(θ)Sin(θ)
                g
Substitute Sin(2θ) for 2 Cos(θ)Sin(θ) yields:  D = Vm² Sin(2θ)
                                                  g

b] PROVE THE MAXIMUM DISTANCE IS OBTAINED AT A 45^o ANGLE
   D is a maximum when Sin(2θ) is a maximum.
   The maximum value of Sin(2θ) is one, and it occurs when 2θ = 90^o
   θ = 90/2 = 45^o 
   Thus Dmax = Vm²/g Sin(2●45) = Vm²(1)/g = Vm²/g

8*   Your cannon has a muzzle velocity of 300 m/s, the target is 8 km away.               a] Find the range.
                 Range = Dmax = Vm²/g = 300²/9.8 = 9184 m
              b] Draw the Diagram.
                 Include:  cannon, θ = ?
                        One arch for the target (8 km = 8000 m)
                        One arch for the range 9184 m
              c] Find the angle needed to hit the target. 
              D = 8000 m
              D = Vm² Sin(2θ)
                   g
              8000 = 9184 Sin(2θ)
              8000/9184 = Sin(2θ)
              2θ = the angle whose Sine = 8000/9184
              2θ = Sin^-1(8000/9184)
              2θ = 60.6^o
              θ = 60.6/2 = 30.3^o
              d] Use Sin(2θ) = Sin(2(90-θ)), to find the 2nd angle.
                θ2 = 90 - θ1 = 90 - 30.3 = 59.7^o

9*   The farthest you can throw a baseball is 125 ft. 
a] How fast did you throw it? (mi/h)  
   Dmax = Vm²/g
    125 = Vm²/32
   125(32)  = Vm²
   √125(32) = Vm             63.2 ft | 3600 s |  1 mi   | = 43 mi/h
   63.2ft/s = Vm                   s |  1 hr  | 5280 ft
b] State your 2 assumptions.
   i]  No air resistance.
   ii] The ball was thrown at a 45^o angle

10*  A ship within 3000 m of an island's 1800 m high mountain peak and fires a projectile at an enemy ship 700 m on the other side of the peak. The projectile has an initial velocity of 250 m/s at 50^o.
a] How close to the enemy ship does the projectile land.
b] How close (vertically) does the projectile come to the peak?

Top