Free SAT II Physics - Vectors - Solutions
Solutions to the Sat Physics subject questions on vectors, with detailed explanations.
Which of the following is represented by a vector?
I) velocity II) speed III) displacement IV) distance V) force VI) acceleration
A) I only
B) III and IV only
C) I , III, V and VI
D) III, V and VI only
E) None
Solution - Explanations
In physics velocity, displacement, force and acceleration are defined as vector quantities.
Which of the following is not represented by a vector?
I) mass II) electric field III) magnetic field IV) current V) voltage VI) work
A)I , IV, V and VI
B) II and III
C) All
D) I , IV and V only
E) I and IV only
Solution - Explanations
In physics mass, current, voltage and work are defined as scalar quantities and therefore are NOT represented by vectors.
A and B are vectors with angle θ between them such that 0 < θ < 90°. If | A |, | B | and | A + B | are the magnitudes of vectors A, B and A + B respectively, which of the following is true?
A) | A + B | > | A | + | B |
B) | A + B | = | A | + | B |
C) | A + B | = √(| A |2 + | B |2)
D) | A + B | < √(| A |2 + | B |2)
E) | A + B | < | A | + | B |
Solution - Explanations
Start with the following identity
(A + B)·(A + B) = (A + B)·(A + B)
use scalar product to rewrite the above as follows
| A + B | 2 = | A | 2 + | B | 2 + 2 A · B
rewrite as
| A + B | 2 = | A | 2 + | B | 2 + 2 |A| |B | cos (θ)
Since 0 < θ <90°, we can write
0 < cos (θ) < 1
multiply all terms of inequality by 2 |A| |B | to obtain
0 < 2 |A| |B | cos (θ) < 2 |A| |B |
add | A | 2 + | B | 2 to all terms of above inequality to obtain
| A | 2 + | B | 2 <
| A | 2 + | B | 2 + 2 |A| |B | cos (θ) <
| A | 2 + | B | 2 + 2 |A| |B |
the above inequality can now be written as
| A | 2 + | B | 2 <
| A + B | 2 <
( | A | + | B | ) 2
all terms of the above inequality are positive, we can write the following inequality taking the square root as follows
√[ | A | 2 + | B | 2 ] <
| A + B | <
( | A | + | B | )
Which of the following is a unit vector in the same direction as vector A = 3 i - 4 j, where i and j are the unit vector along the x and y axis respectively?
A) (3/5) i + (4/5) j
B) (9/5) i - (16/5) j
C) (3/5) i - (4/5) j
D) - (3/5) i + (4/5) j
E) (3/25) i - (4/25) j
Solution - Explanations
The unit vector u in the same direction as vector A = 3 i - 4 j is given by
u = A / | A | = ( 3 i - 4 j ) / √(3 2 + (-4) 2)
= (3/5) i - (4/5)j
If A and B are vectors, then A ·( A × B ) =
A) | A | 3
B) 0
C) | A | 2
D) 1
E) | A |
Solution - Explanations
The cross product A × B gives a vector perpendicular to both vectors A and B and therefore the scalar product between vectors A and A × B , which is perpendicular to A, is equal to zero.
Vector U has a magnitude of 3 and points Northward. Vector V has a magnitude of 7 and points Eastward. | U + V | is the magnitude of vector U + V . Which of the following is true?
A) | U + V | > 10
B) | U + V | = 10
C) | U + V | = √58
D) | U + V | = √10
E) | U + V | = 4
Solution - Explanations
The x and y axes are directed Eastward and Northward respectively and therefore the components of U + V are given as follows
U + V = 3j + 7 i
magnitude is now calculated
| U + V | = √(7 2 + 3 2) = √ 58
Given vectors U = 2 i + 2j and V = 2i - 2j. What angle does the vector U - V make with the positive x - axis?
A) 0 °
B) 45 °
C) -90 °
D) 90 °
E) -45 °
Solution - Explanations
U - V = 2 i + 2j - (2i - 2j) = 4j
U - V is proportional to the unit vector j which makes 90° with the positive x-axis. Hence U - V makes 90° with the positive x-axis
Two forces F1 and F2 are used to pull an object. The angle between the two forces is θ. For what value of θ is the magnitude of the resultant force equal to √(|F1|2 + |F2|2)
A) 90 °
B) 135 °
C) 45 °
D) 180 °
E) 0 °
Solution - Explanations
Let R be the resultant and write
R = F1 + F2
use scalar product to write
R R = (F1 + F2)
· (F1 + F2)
expand the terms on the right
|R|2 = |F1|2 + |F2|2 + |F1| |F2| cos (θ)
For θ = 90°, cos (θ) = 0
and
|R|2 = |F1|2 + |F2|2
taking the square root gives
|R| = √ (|F1|2 + |F2|2)
Two forces F1 = 3i + bj and F2 = 9i + 12j act on the same object.(i and j are the unit vectors along the positive x-and y- axes). For what value of b will the magnitude of the resultant force be minimum?
A) 0
B) - 12
C) 9
D) - 10
E) 4
Solution - Explanations
Let R be the resultant force
R = F1 + F2 = 3i + bj + 9i + 12j = 12i + (b + 12)j
calculate magnitude
R = √( 144 + (b + 12)2 )
the quantity under the radical is positive and therefore a value of b that minimizes 144 + (b + 12)2 will also minimize R the magnitude
144 + (b + 12)2 is a quadratic expression and it has a minimum value at b = -12 (position of vertex)
Find m so that the vectors A = 5 i - 10 j and B = 2 m i + (1 / 2) j are parallel.
A) 5
B) - 40
C) 8
D) - 1 / 8
E) - 5
Solution - Explanations
For vectors A and B to be parallel, there must be a real K so that A = K B
A = 5 i - 10 j = K ( B = 2 m i + (1 / 2) j )
components are equal, hence
5 = K (2 m)
-10 = K (1 / 2)
Solve the second equation fo K: K = -20
substitute K by -20 in the equation 5 = K (2 m)
5 = -20 (2 m)
solve for m
m = - 5 / 40 = - 1 / 8
Answers to the Above questions
- C
- A
- E
- C
- B
- C
- D
- A
- B
- D