Multiple Choice Practice Questions on Vectors for AP Physics B and C

“Iron rusts from disuse, stagnant water loses its purity and in cold weather becomes frozen; so does inaction sap the vigors of the mind.”

– Leonardo da Vinci

If you have a concrete idea about manipulation of vectors you will be at an advantage in problem solving in many situations. Today we shall discuss a few multiple choice practice questions on vectors.

(1) In the following figure two vectors P and Q are shown. One of the vectors shown in options (a), (b), (c), (d) and (e) represents their resultant. Pick out the correct one. 

The correct option is (b).

You can verify this from the following figure in which a parallelogram is constructed with the vectors P and Q as adjacent sides. The diagonal of the parallelogram indeed represents the resultant R of the vectors P and Q.

(2) A small toy car is in uniform circular motion with speed 2 ms^–1 along horizontal ground. When the acceleration of the car is directed towards east its velocity is directed towards south. If the magnitude of its acceleration is 4 ms^–2, the circular path followed by the car is

(a) clockwise with radius 1 m

(b) anticlockwise with radius 1 m

(c) clockwise with radius 2 m

(d) anticlockwise with radius 2 m

(e) clockwise with radius 4 m

The path of the car is shown in the adjoining figure in which the normal acceleration (centripetal acceleration) vector is indicated eastwards and the velocity vector is indicated southwards. Evidently the path is anticlockwise. The normal acceleration ‘a’ is given by

             a = v²/r where ‘v’ is the speed and ‘r’ is the radius of the circular path.

Therefore, r = v²/a = 2²/4 = 1 m.

The correct option is (b). 

(3) A motor boat can cross a 60 m wide river in a minimum time of 10 s when the water is still. What will be the minimum time required by the boat to cross the river when the water in the river flows steadily at a speed of 1 ms^–1?

(a) 9.86 s

(b) 10.14 s

(c) 8.57 s

(d) 12 s

(e) 10 s

The minimum time required will be 10 s itself. In still water the boat will have to move in a direction at right angles to the bank in order to reach the other bank in minimum time. When the river is flowing, the boat will reach the opposite bank in the same minimum time if the engine drives the boat at right angles to the bank. In this case the velocity component of the boat at right angles to the river bank will be unaffected by the flow of the river. But, because of the flow of the river the boat will be carried downstream through 10 m by the time it reaches the other bank.

The following questions are meant for AP Physics C aspirants:

 

(4) A and B shown in the figure are electric field vectors. The component of vector A along the direction of vector B is

(a) (A × B)/ | A |

(b) (A × B)/ | B |

(c) (A . B)/ | A |

(d) (A . B)/ | B |

(e) | (A × B) |

If the angle between A and B is θ, the component of A along the direction of B is A cosθ. We have

             A . B = AB cosθ where A and B are the magnitudes of A and B respectively.

Therefore, A cosθ = (A . B)/B = (A . B)/ | B |

(5) Specific charge of proton is 9.6×10⁷ C/kg. A proton having velocity (4 i + 6 j)  m/s enters a magnetic field of flux density (j + 2 k) tesla where i, j and k are unit vectors along the x, y and z directions respectively. The acceleration produced in the proton in m/s² is

(a) 3.84×10⁸ (3i – 2j + k)

(b) 3.84×10⁸ (6i – 4j + 2k)

(c) 9.6×10⁷(6i – 4j)

(d) 1.92×10⁸ (3i – 2j + k)

(e) 1.92×10⁸ (3i + k)

The magnetic force F on the proton is given by

             F = e(v×B) where  e is the charge on the proton, v is its velocity and B is the magnetic flux density.

The acceleration a of the proton is given by

             a = e(v×B)/m where m is the mass of the proton.

Or, a = (e/m) (v×B)

Substituting for the specific charge (e/m), v and B, we have

             a = 9.6×10⁷(4 i + 6 j) × (j + 2 k)

Or, a = 9.6×10⁷(4 k – 8 j + 12 i)

This gives a = 3.84×10⁸ (3i – 2j + k), as given in option (a).

[Remember that i × j = k, i × k = – j, j × j = 0, j × k = i]

You will find a couple of multiple choice questions (with solution) on vectors here.

AP Physics B & C - Multiple Choice Practice Questions on Motion in Two Dimensions including Projectile Motion

"If man is not rising upward to be an angel, he is sinking downward to be a devil. He cannot stop at the beast."

– Samuel Taylor Coleridge

Let us discuss a few questions (MCQ) involving motion in two dimensions. Questions in common for AP Physics B as well as AP Physics C aspirants are discussed first. Questions specifically meant for AP Physics C aspirants are discussed next.

(1) The adjoining figure represents the path of an ant on a horizontal floor between the instants t₁ and t₂. The velocities of the ant at positions A an B at the instants t₁ and t₂ are represented by the vectors V₁ and V₂ respectively. Which one among the following vectors best represents the net acceleration of the ant during the interval t₂ – t₁?

The acceleration a is given by

a = (V₂ – V₁)/(t₂ – t₁)

The direction of the vector V₂ – V₁ is given correctly in option (b).

[To find V₂ – V₁ you have to add the vector V₂ to the vector –V₁ using the parallelogram law (for convenience). On reversing the direction of V₁ and using it along with V₂ to form the parallelogram, you get the diagonal which points along the vector shown in option (b)].

The correct option is indeed (b).

(2) The path of a particle projected from point A with velocity v at an angle θ with respect to the horizontal is shown in the adjoining figure. Which one among the following graphs best represents the vertical component (v_y) of the velocity of the particle as a function of time t? (Assume that air resistance is negligible)

Since the motion of the particle is under gravity, the vertical component of velocity decreases uniformly as the particle moves up. At maximum height the vertical component of velocity momentarily becomes zero. Then the particle moves down and the vertical component of velocity goes on increasing uniformly until it hits the ground. The correct graph is (d).

[Note that the upward direction of the vertical component of velocity is taken as positive and the downward direction is taken as negative].

(3) In the above question, if g represents the gravitational acceleration, the values of acceleration of the particle at points A, B and C of the trajectory are

(a) greater than g, zero and less than g respectively

(b) greater than g, equal to g and less than g respectively

(c) less than g, zero and greater than g respectively

(d) less than g, equal to g and greater than g respectively

(e) same and equal to g

The gravitational force is the same everywhere along the trajectory and hence the acceleration of the particle is the same (equal to g).

The correct option is (e).

The following questions are specifically meant for AP Physics C aspirants:

(4) A particle starts from the origin at time t = 0 with velocity 2 ĵ and moves in the x-y plane with a constant acceleration of 2 î + 4 ĵ where î and ĵ are unit vectors along the x-direction and y irection respectively. What will be the x-coordinate of the particle when its y-coordinate becomes 12 m?

(a) 4 m

(b) 6 m

(c) 6.8 m

(d) 8 m

(e) 12 m

The position vector r_t of the particle at the instant t is given by

r_t = v₀t + ½ at² where v₀ is the initial velocity and a is the constant acceleration.

We have v₀ = 2 ĵ and a = 2 î + 4 ĵ

Therefore, r_t = 2 ĵ t + (½) (2 î + 4 ĵ)t² = t² î + (2t + 2t²) ĵ

The above equation shows that the x-coordinate of the particle at time t is t² and the y-coordinate is (2t + 2t²)

The time t at which the y-coordinate becomes 12 metre is given by

2t + 2t² = 12

Or, 2t² + 2t – 12 = 0

This gives t = [–2 ±√(4 + 96)]/4 = 2 seconds, ignoring the negative time.

Since the x-coordinate of the particle is t², its value when the y-coordinate becomes 12 m (at time 2 seconds) is 4 m [Option (a)].

(5) What is the speed of the particle in the above question at the time t = 1 s?

(a) 2 ms^–1

(b) √(10) ms^–1

(c) √(20) ms^–1

(d) √(40) ms^–1

(e) 8 ms^–1

The position vector r_t of the particle at time t, as shown above is given by

r_t = t² î + (2t + 2t²) ĵ

The velocity v_t of the particle at time t is given by

v_t = dr_t/dt = 2t î + (2 + 4t) ĵ

When the time t = 1 s, the velocity of the particle is 2 î + 6 ĵ.

The speed of the particle at time 1 s is the magnitude of the above velocity and is equal to √(2² + 6²) = √(40) m

You will find a useful post in this section here.

Multiple Choice Practice Questions on Vectors (for AP Physics B & C)

There is no substitute for hard work.

– Thomas A. Edison

AP Physics C aspirants are expected to have a fairly good idea regarding vector methods in Physics. AP Physics B aspirants also should have some idea of vectors. Today we will discuss some multiple choice practice questions on vectors.

(1) The magnetic flux density B at a point P is 0.5 tesla. What is the maximum number of components into which the vector B can be resolved?

(a) 1

(b) 2

(c) 3

(d) 6

(e) infinite

The resultant of all possible components should make the vector B of the given direction and of magnitude 0.5 tesla. We can imagine an infinite number of components in various directions to combine and produce the field B and hence the correct option is (e).

(2) The force F acting on a particle has its line of action (direction) lying in the XY plane and is inclined at an angle θ with the x-axis. What is the z-component of the force F?

(a) zero

(b) F cos θ

(c) F sin θ

(d) F/cos θ

(e) F/sin θ

Since the direction of the vector F is lying in the XY plane it has no z-component. The correct option is (a).

(3) The resultant of two forces F₁ and F₂ of the same magnitude F has magnitude F itself. The angle between F₁ and F₂ is

(a) 30º

(b) 45º

(c) 60º

(d) 90º

(e) 120º

The magnitude F of the resultant force is given by

F = √(F₁² + F₂² + 2 F₁ F₂ cos θ) where θ is the angle between the forces F₁ and F₂.

Since the magnitudes F₁ and F₂ of the two forces are the same and equal to F, we have

F = √(F² + F² + 2 F F cos θ)

Squaring, F² = F² + F² + 2 F² cos θ

This gives cos θ = – ½ so that θ = 120º

[The adjoining figure will be useful to understand how the two forces produce the resultant satisfying the given conditions].

(4) The resultant of two forces is 40 N and the smaller force, which has magnitude 30 N, is normal to the resultant. The larger force is

(a) 40 N

(b) 45 N

(c) 50 N

(d) 60 N

(e) 80 N

The forces and their resultant are shown in the adjoining figure.

Since the resultant R of magnitude 40 N is perpendicular to the force F₁ of 30 N, we have (from the right angled triangle),

F₂ = √(30² + 40²) = 50 N.

(5) Rain drops are falling vertically with a speed of 10 ms^–1. A boy holding an umbrella runs southward with a speed of 5 ms^–1. What is the direction in which he should hold his umbrella so that he will not get drenched?

(a) At an angle of tan^–1(1/2) with the vertical, towards north

(b) At an angle of tan^–1(1/2) with the vertical, towards south

(c) At an angle of tan^–1(2) with the vertical, towards south

(d) At an angle of 30º with the vertical, towards north

(e) At an angle of 30º with the vertical, towards south

In the figure the velocities of the boy and rain drops are represented respectively by the vectors OA and OB.

The relative velocity of the rain drops with respect to the boy = (velocity of rain drops) – (velocity of boy).

The quantity on the right hand side is the vector sum of the velocity of rain drops and the negative of the velocity (reversed velocity shown by vector OC) of the boy. This is given by the vector OD which makes an angle θ with the vertical and is given by

tan θ = 5/10 = ½.

Therefore, the boy should hold his umbrella at an angle of tan^–1(1/2) with the vertical, towards south.

The following multiple choice questions are specifically meant for AP Physics C aspirants:

(6) Suppose that the ground coincides with the XZ plane of a right handed Cartesian coordinate system and i, j, k are unit vectors along x, y, z directions respectively. An inclined plane of inclination 60º is placed on the ground and a force F = – 20 j N is applied at a point on the inclined plane. What is the component of this force parallel to the inclined plane?

(a) 20/√3 N

(b) 10 N

(c) 10√3 N

(d) 5√3 N

(e) Zero

The force F of magnitude 20 N is directed vertically downwards (along the negative y-direction).

With reference to the figure, you can easily see that the component of F parallel to the inclined plane is 20 sin 60º = 10√3 N.

(7) The magnitude of the area of a parallelogram formed by the vectors A = i + 2 j + 2 k metre and B = 2 i – 2 j + k metre as adjacent sides is (i, j, k are unit vectors)

(a) √78 m²

(b) 9 m²

(c) √91 m²

(d) 11 m²

(e) 12 m²

The magnitude of the area of the parallelogram is equal to the magnitude of the vector product (cross product) of the vectors A and B.

[Remember that area is a vector].

Area = A × B = (i + 2 j + 2 k) × (2 i – 2 j + k)

Or, Area = – 2 k – j – 4 k + 2i + 4 j + 4 i

= 6 i + 3 j – 6 k

The magnitude of area vector = √(6² + 3² + 6²) = √81 = 9 m²

You will find similar useful questions (with solution) on vectors here.