Solutions to the Sat Physics subject questions on vectors, with detailed explanations.
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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.
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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.
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A andB are vectors with angle θ between them such that 0 < θ < 90°. If |A |, |B | and |A +B | are the magnitudes of vectorsA ,B andA +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 + 2A ·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 | )
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Which of the following is a unit vector in the same direction as vectorA = 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 vectorA = 3 i - 4 j is given by
u =A / |A | = ( 3 i - 4 j ) / √(3 2 + (-4) 2)
= (3/5) i - (4/5)j
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IfA andB are vectors, thenA ·(A ×B ) =
A) |A | 3
B) 0
C) |A | 2
D) 1
E) |A |
Solution - Explanations
The cross productA ×B gives a vector perpendicular to both vectorsA andB and therefore the scalar product between vectorsA andA ×B , which is perpendicular to A, is equal to zero.
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VectorU has a magnitude of 3 and points Northward. VectorV has a magnitude of 7 and points Eastward. |U +V | is the magnitude of vectorU +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 ofU +V are given as follows
U +V = 3j + 7 i
magnitude is now calculated
|U +V | = √(7 2 + 3 2) = √ 58
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Given vectorsU = 2 i + 2j andV = 2i - 2j. What angle does the vectorU -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. HenceU -V makes 90° with the positive x-axis
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Two forcesF1 andF2 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
LetR 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)
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Two forcesF1 = 3i + bj andF2 = 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
LetR 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 minimizeR the magnitude
144 + (b + 12)2 is a quadratic expression and it has a minimum value at b = -12 (position of vertex)
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Find m so that the vectorsA = 5 i - 10 j andB = 2 m i + (1 / 2) j are parallel.
A) 5
B) - 40
C) 8
D) - 1 / 8
E) - 5
Solution - Explanations
For vectorsA andB to be parallel, there must be a real K so thatA = KB
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