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.

AP Physics B & C - Multiple Choice Practice Questions on One Dimensional Kinematics

"Whenever you are confronted with an opponent, conquer him with love."

– Mahatma Gandhi

Let us discuss a few interesting multiple choice practice questions on one dimensional motion. Here are some questions beneficial for AP Physics B as well as AP Physics C aspirants:

(1) The adjoining figure shows the velocity time graph of an object. Total displacement suffered by the object during the interval when it has non-zero acceleration and retardation is

(a) 80 m

(b) 70 m

(c) 60 m

(d) 40 m

(e) 30 m

The object has non-zero acceleration and retardation during the time intervals from 5 sec to 15 sec and from 20 sec to 30 sec. The total area under the velocity time graph during these intervals gives the required displacement.

Displacement from 5 sec to 15 sec = 30 m

Displacement from 20 sec to 30 sec = 40 m

Therefore, total displacement = 70 m

(2) Successive positions (x) of an object (moving from left to right) at equal time intervals are shown in the following figure:

Which one among the following position-time graphs best represents the motion of the object? 

 

Change of position is slowest in the beginning and in the end. The motion is best represented by graph (c).

[Note that graph (d) is not the answer since the change of position in the beginning and in the end is shown as fastest in it]

(3) Which one among the following velocity-time graphs best represents the motion of the object mentioned in question no.(2)? 

The graph (c) is the answer.

The following questions are meant for AP Physics C aspirants:

(4) A particle projected vertically upwards attains the maximum height h in time t. While returning from the highest point it takes an additional time t₁ to fall to the height h.2. If air resistance is negligible, how is t₁ related to t?

(a) t₁ = t/2

(b) t₁ = √(t/2)

(c) t₁ = t/√2

(d) t₁ = t/√3

(e) t₁ = t/3

For the upward motion we have

             0 – u² = – 2gh …………(i)

[We have used the equation of motion, v² – u² = 2as. The sign of the gravitational acceleration g is negative since it is opposite to the direction of the velocity of projection u]

Using the equation of motion, v = u + at we have

             0 = u – gt from which u = gt

Substituting this value of u in Eq (i), we have

             g²t² = 2gh

Therefore, t =√(2h/g) ……..(ii)

For the fall through h/2 from the highest point we have

             h/2 = 0×t₁ + ½ gt₁²

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

Therefore, t₁ =√(h/g)

Comparing this with the value of t given in Eq (ii) we obtain t₁ = t/√2.

(5) A ball projected vertically up has the same vertical displacement h at times t₁ second and t₂ second. If air resistance is negligible, the maximum height reached by the ball is

(a) gt₁t₂/2

(b) g(t₁² + t₂²)/2

(c) g(t₁ + t₂)²/4

(d) g(t₁ + t₂)²/8

(e) 2g(t₁ + t₂)²

The time taken by the ball to move from height h to the top of its trajectory and back to the height h is t₂ – t₁. Therefore, the time taken to move from height h to the top of the trajectory is (t₂ – t₁)/2.

The total time taken for the upward journey (from ground to the top most point) is evidently t₁ + t₂ – t₁)/2 = (t₁ + t₂)/2.

Time taken for the return journey (from the top most point to the ground) also is equal to (t₁ + t₂)/2. Therefore, the maximum height H (using the equation, s = ut + ½ gt²) is given by

             H = 0 + ½ g [(t₁ + t₂)/2]²

Or, H = g(t₁ + t₂)²/8

             Questions on kinematics were discussed earlier on this site. You can access them either by clicking on the label ‘kinematics’ below this post or by trying a search for                  ‘kinematics’ using the search box provided on this page.

Kinematics in One Dimension –Multiple Choice Practice Questions for AP Physics B & C

All posts related to kinematics on this blog can be accessed either by clicking on the label ‘kinematics’ below this post or by performing a search using the search box at the top of this page. Today I give you a few multiple choice practice questions with solution:

(1) A stone projected vertically up is found to be at a height h at times 1 s and 3 s. Neglect air resistance and assume that g = 10 ms^–2. The maximum height reached by the stone is

(a) 60 m

(b) 50 m

(c) 44 m

(d) 22 m

(e) 20 m

Evidently the stone reaches the same height h first during its upward trip and next during its downward trip. But you need not bother about this because everything is contained in the equation of motion, x – x₀ = v₀t + ½ at² (or, s = ut + ½ at²)

Considering the two cases given in the question, since a = – g (taking upwrd as positive and therefore downward negative), we have

h = v₀×1 – ½ ×10×1² = v₀×3 – ½ ×10×3² from which

2v₀ = 40 so that v₀ = 20 ms^–1

The naximum height h_max reached is given by by 0 = v₀² – 2gh_max

[This is the equation, v² = v₀² + 2a(x – x₀) which can also be written as v² = u² + 2as]

Therefore, h_max = v₀²/2g = 20²/(2×10) = 20 m

(2) A food packet is released from a helicopter ascending vertically up with uniform velocity of 2 ms^–1. Neglect air resistance and assume that g = 10 ms^–2. The velocity of the packet at the end of 0.2 s will be

(a) 5 ms^–1

(b) 4 ms^–1

(c) 2 ms^–1

(d) 1 ms^–1

(e) zero

The initial velocity v₀ of the packet is 2 ms^–1 and is directed upwards. The gravitational acceleration is directed downwards. The velocity v of the packet at time t is given by

v = v₀ – gt.

[Note that we have taken the upward vector as positive and hence the downward vector as negative].

Substituting for v₀, g and t, v = 2 – (10×0.2) = 0.

[After getting released, the packet moves upwards for 0.2 s with uniformly decreasing speed, momentarily comes to rest and then moves downwards with uniformly increasing speed].

(3) An astronaut on a planet having no atmosphere throws a stone of mass m vertically upwards with velocity 10 ms^–1. It reaches maximum height of 10 m. If another stone of mass 2m is thrown upwards with velocity 20 ms^–1, the maximum height reached will be

(a) 80 m

(b) 40 m

(c) 30 m

(d) 20 m

(e) 10 m

The maximum height (h_max) reached is given by

h_max = v₀²/2g

[Remember v² = v₀² + 2a(x – x₀) which can be written as v² = v₀² – 2gh_max for projection vertically apwards]

Substituting for h_max (= 10 m) and v₀ (= 10 ms^–1), the acceleration due to gravity (g) on the planet is obtained as 5 ms^–2.

For a given velocity of projection, the height reached is independent of the mass of the stone.

When the velocity of projection is changed to 20 ms^–1, the maximum height reached is 20²/(2×5) = 40 m.

[You may note that the maximum height is directly proportional to the square of the velocity of projection. If the velocity of projection is doubled, the maximum height reached must be quadrupled as in the present problem].

The following question is specifically for AP Physics C aspirants. But AP Physics B aspirants too can ‘enjoy’ it.

(4) A ball thrown vertically up from the ground moves very close to an open window of a room in the first floor of a building. A student in the room sees the ball through the window, when the ball moves up as well as when it move down. The height of the window is 2 m and the ball is in view for a total time of 0.2 s. Take g = 10 ms^–2. How high above the top of the window does the ball go (approximately)?

(a) 19 m

(b) 20 m

(c) 21 m

(d) 24 m

(e) 25 m

Since the total time for which the ball is in view is 0.2 s, the time taken by the ball to traverse the height of the window is 0.1 s. The average upward velocity (v) of the ball while moving past the window is 2/0.1 = 20 ms^–1. We may take this to be very nearly equal to the velocity of the ball at the centre of the window.

Therefore, the maximum height reached by the ball (from the centre of the window) is given by

h_max = v²/2g = 20²/(2×10) = 20 m.

[We have used the equation, v² =u² + 2as where a = – g and s = h_max].

Therefore, the ball goes to a height of nearly 19 m above the top of the window [Option (a)].

[Since the motion of the ball is accelerated, the average velocity is not exactly equal to the velocity at the centre of the window. The answer we obtain is an approximate one.

If you wnt to calculate the exact height, you will proceed as follows:

If t is the time taken by the ball to fall from the topmost point of the trajectory to the top of the window, the maximum height h_max (measured from the top of the window) is given by

h_max = 0 + ½ gt² = 5 t² ………(i)

Also, (h_max + 2) = 0 + ½ g (t +0.1)² = 5t² + 5×0.01 + 5×2×0.1t

Or, h_max + 2 = 5 t² + 0.05 + t ……….(ii)

Subtracting Eq (i) from Eq (ii) we obtain t = 1.95 s.

Therefore, h_max = 5 t² = 5×(1.95)² = 19.0125 m.