# 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 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 ) 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 C A E C B C D A B D