AP Physics Kinematics Practice Questions (Multiple Choice)

“Being ignorant is not so much a shame, as being unwilling to learn.”

– Benjamin Franklin

Many questions (with solution) involving kinematics in one dimension and two dimensions have been posted on this site earlier. You may click on the label ‘kinematics’ below this post to access them. After obtaining the first result, you will have to click on the ‘older posts’ tab to access all the posts in this section. Alternatively you may try a search for ‘kinematics’ using the search box provided on this page.

Today we shall discuss a few more multiple choice practice questions on kinematics:

(1) A car travels from station A and to station B separated by a distance of d km. The average speeds of the car while covering the first and second halves of the distance are v₁ and v₂ respectively. What is the average speed of the car for the entire trip from station A to station B?

(a) √(v₁v₂)

(b) (v₁v₂)/(v₁+v₂)

(c) (v₁+v₂)/2

(d) (v₁v₂)/(v₁–v₂)

(e) (2v₁v₂)/(v₁+v₂)

The times taken for covering the first and second halves of the trip are d/2v₁ and d/2v₂ respectively.

Therefore, the total time taken to cover the entire istance d is (d/2v₁ + d/2v₂) = d[(1/2v₁) + (1/2v₂)] = d[(v₁+v₂)/2v₁v₂]

The average speed v for he entire trip is is given by

             v = d/d[(v₁+v₂)/2v₁v₂] = (2v₁v₂)/(v₁+v₂)

(2) An object has acceleration. Then

(a) its speed must be decreasing

(b) its speed must be increasing

(c) its speed must be decreasing or increasing

(d) its direction must be changing

(e) its speed or direction must be changing

An object moving with varying speed has acceleration. But this does not mean that all accelerated objects must move with a varying speed. For instance, an object in uniform circular motion has constant speed, even though it has a centripetal acceleration. Its direction of motion changes continuously and it is the change in direction that makes it an accelerated object.

For an object to be in accelerated motion, it is enough that its speed or direction of motion changes. Therefore, the correct option is (e).

(3) A bullet is fired from a gun in a direction inclined at angle θ with respect to the horizontal ground. Which one among the following graphs represents the plot of the vertical velocity v of the bullet against time t between the instant of firing and the instant just before the bullet hits the ground? (Take the upward direction as positive).

 

At the instant of firing, the bullet has the highest vertical velocity. When the bullet rises up, its vertical velocity goes on decreasing linearly (because of gravity) and at the highest point of its trajectory the vertical velocity becomes zero. The bullet then starts falling down with linearly increasing speed. In other words, the vertical velocity of the bullet becomes negative and its magnitude goes on increasing until it hits the ground.

The vertical velocity of the bullet as a function of time is therefore correctly represented by  graph (b).

(4) The adjoining figure shows forces F₁ and F₂ with their lines of action in the XY plane and acting on a particle.P.

If F₁ = a₁ î + b₁ ĵ

and F₂ = a₂ î + b₂ ĵ where î and ĵ are unit vectors in the x-direction and y-direction respectively, which one among the following statements is correct?

(a) a₁, b₁, a₂, and b₂ are positive.

(b) a₁ and  b₁ are negative where as a₂ and b₂ are positive.

(c) a₁ is negative where as b₁, a₂, and b₂ are positive.

(d) a₁, b₁ and a₂ are positive where as b₂ is positive.

(e) a₁, b₁, a₂, and b₂ are negative.

Imagine the rectangular components of F₁ and F₂. You can easily see that the x-component of F₁ is along the negative x-direction while the y-component is along the positive y-direction. The x-component of F₂ is along the positive x-direction while the y-component is along the positive y-direction.

This means that a₁ is negative where as b₁, a₂, and b₂ are positive [Option (c)].

(5) A small object at the foot of a smooth inclined plane AB (Fig.) is projected up along the inclined plane with an initial speed v. The object returns before reaching the top of the inclined plane and after reaching the foot of the plane, it moves further along a smooth horizontal surface AC. Which one among the following graphs represents the variation of the speed v of the object against time t? 

of the object therefore gets decreased uniformly and becomes zero when the object reaches its highest position on the incline. Then the object retraces its path with uniformly increasing speed until it reaches the foot of the incline. Then it moves along the horizontal surface AC with uniform speed.

The above facts are correctly represented by the  graph (d).

AP Physics C - Multiple Choice Practice Questions on One Dimensional Motion

“Maturity is often more absurd than youth and very frequently is most unjust to youth.”

– Thomas A. Edison

Today’s post covers a few multiple choice practice questions related to one dimensional kinematics meant for AP Physics C aspirants.

(1) Water drops from a leaking overhead tank falls on the ground 5 m below, at regular intervals, the 11^th  drop just beginning to fall when the first drop strikes the ground. What will be the height of the 9^th drop when the first drop strikes the ground? (Acceleration due to gravity = 10 ms^–2)

(a) 4.9 m

(b) 4.8 m

(c) 4.6 m

(d) 4.4 m

(e) 4.2 m

The time t taken by a drop to fall through 5 m is given by the relevant equation of motion,

            5 = 0 + ½ gt²

Since g = 10 ms^–2 the above equation gives t = 1 second

When the first drop strikes the ground after falling freely for one second, the 9^th  drop has fallen freely for one-fifth of one second, as is evident from the adjoining figure (remembering that the drops fall at regular intervals of time). Therefore, the 9^th drop has fallen through a distance s given by

            s = 0 + ½ g(0.2)² = (½)×10×0.04 = 0.2 m

Therefore the height of the 9^th drop when the first drop strikes the ground is 5 m – 0.2 m = 4.8 m.

(2) An object projected vertically upwards with initial velocity u attains maximum height in 5 s. The ratio of the distance traveled by the object in the 1^st second and the 6^th second is (Acceleration due to gravity = 10 ms^–2)

(a) 6 : 1

(b) 8 : 1

(c) 9 : 1

(d) 10: 1

(e) 11 : 1

The velocity of projection (u) of the object is given by the equation of uniformly accelerated linear motion,

             0 = u – gt

[We have used the equation, v = u + at with usual notations].

Substituting for g and t we have

             0 = u – 10×5

Therefore u = 50 ms^–1

The distance s₁ traveled by the object in the 1^st second is given by

             s₁ = (50×1) – (½ ×10×1²) = 45 m

[We have used the equation, s = ut + ½ at² with usual notations]

At the end of 5 seconds the object is at the highest point of its trajectory where its velocity is zero. Therefore, the distance s₆ traveled by the object during the next one second (6^th second) is given by

             s₆ = 0 + ½ ×10×1² = 5 m.

[We have used the equation of motion, s = ut + ½ gt² with usual notations]

The ratio of the distance traveled by the object in the 1^st second and the 6^th second is

             s₁ : s₆ = 45 : 5 = 9 : 1

(3) A ball projected vertically upwards with initial velocity u reaches maximum height h in t sec. What is the total time taken by the ball (from the instant of projection) to reach a height h/4 while returning?

(a) 1.75t

(b) 1.65t

(c) 1.5t

(d) 1.4t

(e) 1.3t

From the equation of motion, v² = u² + 2as with usual notations, we have for the upward journey

             0 = u² – 2gh

Therefore h = u²/2g

From the equation of motion, v = u + at with usual notations, we have for the upward journey

             0 = u – gt so that u = gt

Substituting for u in the expression for h we have

             h = gt²/2 …………. (i)

We now use the equation of motion, s = ut + ½ gt² for the free fall of the ball from the maximum height h to the height h/4.

Since distance of fall is 3h/4) we have

             3h/4 = 0 + ½ gt₁² where t₁ is the time of fall from the maximum height h to the height h/4.

Substituting for h from equation (i) we have

             gt²/8 = gt₁²/2

This gives t₁ = t/2

The total time time taken by the ball (from the instant of projection) to reach a height h/4 while returning is t + t₁ = t + (t/₂) = 3t/2 = 1.5 t.

(4) The velocity-time graph of an object moving along the x-direction is shown in the figure. What is the displacement of the object when it moves with the maximum acceleration?

(a) 12 m

(b) 8 m

(c) 6 m

(d) 4.8 m

(e) 4.2 m

The object has maximum acceleration from 10 sec to 12 sec since the slope of the velocity-time graph is maximum during this interval. The displacement is given by the area under the velocity-time graph for the interval from 10 sec to 12 sec which is 8 m.

[The displacement of the object when it moves with the maximum acceleration can be calculated using the equation,  s = ut + ½ gt² as well:

As is evident from the velocity-time graph, the maximum acceleration a is given by

             a = Change of velocity/Time = (5 – 3)/(12 – 10) = 1 ms^–2

Since the initial velocity u = 3 ms^–1 and the time interval t = 2 s, we have

             s = (3×2) + (½)×1×2² = 8 m].

(5) A particle moving along the x-axis is initially at the origin with velocity 2 ms^–1. If the acceleration a of the particle is given by a = 6t, the position of the particle after 4 seconds is

(a) 24 m

(b) 48 m

(c) 72 m

(d) 96 m

(e) 120 m

The velocity v of the particle is given by

             v = ∫adt = ∫6t dt = 3t² + C where C is the constant of integration which we can find from the initial condition.

Initially (at time t = 0) the particle has velocity 2 ms^–1. Therefore from the above expression for velocity we have

             C = 2 ms^–1

Thus the expression for velocity becomes

             v = 3t² + 2

The position x of the particle is given by

             x = ∫v dt = ∫(3t² + 2)dt = t³ + 2t + C’ where C’ is the constant of integration in this case.

Initially (at t = 0) since the particle is at the origin (where x = 0) we obtain (from the above expression for v)

             C’ = 0

Therefore, the expression for the position x becomes

             x = t³ + 2t

The position x’ at t = 4 seconds is given by

             x’ = 4³ + (2×4) = 72 m

You can find a few more useful questions (with solution) in this section here.