Free SAT II Physics - Vectors - Solutions
Solutions to the Sat Physics subject questions on vectors, with detailed explanations.

  1. 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.
  2. 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.
  3. 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 | )
  4. 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
  5. 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.
  6. 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
  7. 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
  8. Two forces F1 and F2 are used to pull an object. The angle between 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)
  9. 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)
  10. 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 )

    component 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

  1. C
  2. A
  3. E
  4. C
  5. B
  6. C
  7. D
  8. A
  9. B
  10. D