# CSULA- PHYSICS 211

Document Sample

```					CSULA- PHYSICS 211                    Summer’12                    Solution: Material to Study                        Chapter 3

OBJECTIVE QUESTIONS

OQ-3 (7e- Q3)
                                                                                    
The vector  2D1 will be twice as long as D1 and in the opposite direction, namely northeast. Adding D 2 , which is about equally long
and southwest, we get a sum that is still longer and due east, choice (a).

OQ-8 (7e-None)
8m/s, choice (c).

OQ-10 (7e-Q8)
Take the difference of the coordinates of the ends of the vector. Final first means head end first: 1  (2) = 3 cm, answer (a).

CONCEPTUAL QUESTIONS
CQ-2 (7e-Q10)
Vectors A and B are perpendicular to each other.

CQ-3 (7e-Q12)
Addition of a vector to a scalar is not defined. Think of numbers of apples and of clouds.

CQ-4 (7e-Q11)
No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction, not magnitude.

PROBLEMS

P6 (7e-P5)
y
We have r  x 2  y 2 and   tan 1   .
x
 y 
(a)      The radius for this new point is:          x 2  y 2  x 2  y 2  r        and its angle is    tan 1    180  .
 x 
(b)       (2x)2  (2y)2  2r      This point is in the third quadrant if  x , y  is in the first quadrant or in the fourth quadrant if  x , y 

is in the second quadrant. It is at an angle of 180  .

(c)       (3x)2  (3y)2  3r      This point is in the fourth quadrant if  x , y  is in the first quadrant or in the third quadrant

if  x , y  is in the second quadrant. It is at an angle of  .

P7 (7e-P7)

x
tan 35.0 
100 m
x   100 m  tan 35.0  70.0 m

P15 (7e-P15)
A x  25.0                                                                                     Ay
A y  40.0                                                        We observe that: tan              .
Ax
A  Ax  Ay 
2    2
 25.02   40.02  47.2 units
 Ay        40.0
So:            tan 1      tan       tan 1  1.60  58.0 .
   
Ax        25.0

The diagram shows that the angle from the +x axis can be found by subtracting from 180°:   180 58  122 .
P23 (7e-P25)
(a)       A  B   3ˆ  2ˆj    ˆ  4ˆj  
i              i             2ˆ  6ˆ
i    j                            (b)            A  B   3ˆ  2ˆj    iˆ  4ˆj  
i                                ˆ
4i  2ˆ
j

(c)       A  B  22  62  6.32                                                        (d)          A  B  42  22  4.47

 A  B  tan 1     71.6  288                              A B  tan 1    26.6
6                                                               2
(e)                                                            AND                         
 2                                                            4

P29 (7e-P31)
(a)
F  F1  F2
ˆ                                          ˆ
F  120cos 60.0  i  120sin  60.0  ˆ  80.0cos  75.0  i  80.0sin  75.0  ˆ
j                                           j
F  60.0i  104ˆ  20.7ˆ  77.3ˆ  39.3i  181ˆ N
ˆ      j       i       j       ˆ
     j                                                                (b)        F3  F       39.3iˆ  181ˆj  N
F  39.32  1812  185 N

  tan 1 
181 
       77.8
 39.3 

P45 (7e-P40)
(a)                                 ˆ
E   17.0 cm  cos27.0i  17.0 cm  sin 27.0ˆ
j                         E      15.1iˆ  7.72ˆj  cm
(b)                                        ˆ
F    17.0 cm  sin 27.0i   17.0 cm  cos27.0ˆ
j                  F       7.72ˆi  15.1ˆj  cm      Note that we do not need to

explicitly identify the angle with the positive x axis.

(c)                                    ˆ
G    17.0 cm  sin 27.0i   17.0 cm  cos27.0ˆ
j                      G       7.72iˆ  15.1ˆj  cm                               y

27.0° 27.0°

F        G
E
27.0°
x

P57 (7e-P47)
Let  represent the angle between the directions of A and B . Since A and B have the same magnitudes, A , B , and R  A  B
        
form an isosceles triangle in which the angles are 180   , , and . The magnitude of R is then R  2A cos   . This can be seen
 
2        2                                       2
from applying the law of cosines to the isosceles triangle and using the fact that B  A .
Again, A , – B , and D  A  B form an isosceles triangle with apex angle . Applying the law of cosines and the identity

                                                     
1  cos   2sin 2   gives the magnitude of D as
                                         D  2A sin   .   The problem requires that R  100D .
 2                                                2

Thus, 2A cos    200A sin    . This gives tan     0.010 and   1.15 .
                                                                                                               R                   B
 2              2
 
 2                                                    /2

A
A          

D             –B

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 0 posted: 9/29/2012 language: Latin pages: 2
How are you planning on using Docstoc?