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 - Newton’s Laws - Multiple Choice Practice Questions

"I am a friend of Plato, I am a friend of Aristotle, but truth is my greater friend."

 – Sir Isaac Newton

 
Let us discuss a few multiple choice practice questions involving Newton’s laws of motion.

(1) Coefficient of static friction between a horizontal conveyor belt and a 3 kg packet placed on it is 0.2. What is the maximum permissible acceleration of the conveyor belt so that the packet will not slip?

(a) 0.2 ms^–2

(b) 0.6 ms^–2

(c) 2 ms^–2

(d) 3 ms^–2

(e) 6 ms^–2

It is the frictional force between the packet and the conveyor belt that keeps the packet stationary with respect to the belt when the belt moves. The maximum value of this frictional force is μ_smg where μ_s is the coefficient of static friction, m is the mass of the packet and g is the acceleration due to gravity.

If the maximum permissible acceleration of the belt is a_max, we have

            ma_max = μ_smg

Therefore, a_max = μ_sg = 0.2×10 = 2 ms^–2

[Note that the mass of the packet is given just to serve the purpose of a distraction]

(2) An empty open box of mass M (Fig.) is sliding along a smooth horizontal surface with constant velocity ‘v₁’. An object of mass M/4 is dropped vertically into the box and the box continues to move forward with velocity v₂. After some time another object of mass 3M/4 is dropped vertically into the box and then the box moves forward with velocity v₃. Which one among the following is correct?

(a) v₁ = v₂ = v₃

(b) v₂ =  v₁/4

(c) v₂ = 3 v₁/4

(d) v₃ =  v₁/4

(e) v₃ =  v₁/2

The objects dropped vertically into the box will not contribute any horizontal momentum to the box. But they can change the velocity of the box obeying the law of conservation of momentum. The momentum of the system throughout its motion is equal to the initial momentum Mv₁ of the box and we have

Mv₁ = [M + (M/4) + (3M/4)]v₃   

Therefore v₃ =  v₁/2

Therefore, the correct option is (e).

[We have Mv₁ = [M + (M/4)]v₂ and this yields v₂ = 4v₁/5. But we don’t have this answer in the given options]

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

(3) A bullet of mass m voving horizontally with speed v strikes a wooden block of mass M which rests on a horizontal surface. After the collision the block and the bullet move together and come to rest after moving through a istanc d. The coefficient of kinetic friction between the block and the horizontal surface is

(a) m²v²/[2gd(M + m)²]

(b) mv²/[2gd(M + m)]

(c) m²/[2(M + m)²]

(d) (M + m)v²/2gdM

(e) 2m²v²/[gd(M + m)²]

The common velocity (of the bullet and wooden block) after the impact (v_f) is given by the law of conservation of momentum:

mv + M×0 = (M + m)v_f

Therefore, v_f = mv/(M + m)

The kinetic energy possessed by the the bullet and wooden block, just after the impact, is utilized in doing work against friction while moving through the distance d.

Therefore we have

            ½ (M + m)v_f² = μ_kRd where μ_k is the coefficient of kinetic friction and R is the normal force which is the weight of the bullet and the wooden block.

Substituting for v_f  and R, we have

            ½ (M + m) [mv/(M + m)]² = μ_k(M + m)gd

This gives μ_k = m²v²/[2gd(M + m)²]

(4) A balloon with its payload has mass 40 kg. It is found to descend with acceleration equal to g/4 where g is the acceleration due to gravity. Keeping the volume of the balloon unchanged, the payload is reduced by M kg. The balloon is then found to ascend with acceleration of the same magnitude g/4. What is the value of M? (Ignore viscous drag force)

(a) 4 kg

(b) 6 kg

(c) 8 kg

(d) 16 kg

(e) 20 kg

If the acceleration due to gravity is g we have

            (40g – F)/40 = g/4 where F is the upthrust on the balloon.

Or, g – (F/40) = g/4  from which F = 30g   

If the total mass of the balloon while ascending is x we have

             (30g – xg)/x = g/4

Or, (30/x) – 1 = ¼

Or, 30/x = 5/4

Therefore x = 24 kg

The reduction (M) in the mass of the payload is given by

             M = 40 kg – 24 kg = 16 kg