Free SAT II Physics - Vectors - Solutions

Physics Problems with Solutions
SAT Physics Practice Questions
Forces in Physics, tutorials and Problems with Solutions
Linear Momentum and Collisions
Vectors
Motion Examples an Problems with Solutions
Optics Tutorials, Examples and Questions with Solutions.
Projectiles in Physics
Electrostatic
Examples and Problems in Magnetism and Electromagnetism
Interactive Physics HTML5 applets

Custom Search



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



Physics Problems with Solutions -- Motion -- Forces in Physics, tutorials and Problems with Solutions --
Linear Momentum and Collisions -- Electrostatic -- Practice Tests Questions -- Vectors -- projectiles in Physics
Examples and Problems in Magnetism and Electromagnetism -- Optics Tutorials, Examples and Questions with Solutions.
-- Interactive Physics HTML5 applets


Author - e-mail


Updated: 26 September 2015

Copyright © 2012 - 2015 - All rights reserved