Kinematics for AP Physics B and C – Multiple Choice Questions on Two Dimensional Motion

We will discuss free response questions on kinematics after discussing multiple choice questions on two dimensional motion. Here are some typical MCQ’s for practice:

(1) A particle A of mass 4m is released from rest from a point P at the top edge of a tower. Simultaneously, particles B and C of masses m and 2m respectively are projected horizontally with equal velocities. Which one of the following statements is correct? (Neglect air resistance)

(a) Particle A will reach the ground first

(b) Particles B and C will not reach the ground simultaneously

(c) Particle B will reach the ground last

(d) Statements (a), (b) and (c) are correct

(e) All the above statements are incorrect

The vertical acceleration of the particles is the same and is the gravitational acceleration g. The initial vertical component of velocity is zero for all the particles. Since the vertical displacement is the same, all the particles will reach the ground simultaneously. So the correct option is (e).

[Gravitational acceleration is independent of the mass of the particle and you should not be distracted].

(2) A particle of mass m is projected from the ground level with a velocity v making an angle 60º with the horizontal. What will be the angular momentum of the particle about the point of projection just before it hits the ground? (Neglect air resistance)

(a) 3mv³/4g

(b) 3mv³/2g

(c) mv³/2g

(d) 3mv³/4g

(e) √3 mv³/2g

The velocity of the particle when it returns to the ground level will have magnitude v and will be inclined at 60º with the horizontal as shown. Therefore, its angular momentum with respect to the point of projection A will be mv×AC where AC is drawn perpendicular to the direction (CB) of the projectile just before it hits the ground.

But AC = AB sin 60º = [(v²sin120º)/g]×sin60º

= (v²/g)(√3/2)×(√3/2) = 3v²/4g

[Note that the horizontal range (R) of a projectile is given by R = (v₀²sin 2θ₀)/g]

Therefore, angular momentum = mv×3v²/4g = 3mv³/4g

(3) The magnitude of the velocity of a projectile at the maximum height is half the magnitude of its initial velocity. If the initial velocity is u, the maximum height reached is

(a) 3u²/2g

(b) 3u²/4g

(c) 3u²/8g

(d) √3u²/2g

(e) √3u²/4g

At maximum height the projectile has horizontal velocity only . Its value is ucosθ₀ throughout where θ₀ is the angle of projection. Therefore we have ucosθ₀ = u/2 so that θ₀ = 60º

Maximum height reached, H = (u²sin²θ₀)/2g = (u²sin²60º)/2g = 3 u²/8g

(4) A stone is projected at an angle θ with the horizontal. If the initial kinetic energy of the stone is E, what is its kinetic energy at the topmost point of its trajectory? (Neglect air resistance).

(a) Zero

(b) E cosθ

(c) E/2

(d) E cos²θ

(e) E sin²θ

We have E = ½ mv₀² where v₀ is the velocity of projection of the stone (of mass m).

Since at the maximum height the projectile has horizontal velocity only and its value is v₀cosθ, the kinetic energy at the maximum height is ½ mv₀²cos²θ = E cos²θ.

(5) A bullet of mass m is fired with velocity v from an air gun at an angle of 45º to the horizontal. The magnitude of the change in momentum of the bullet on arriving at the horizontal plane passing through the point of firing is

(a) zero

(b) mv

(c) 2mv

(d) mv/√2

(e) √2 mv

The initial momentum of magnitude mv of the bullet is inclined at 45º with the horizontal and is represented by the vector AB in the adjoining figure. The momentum when the bullet reaches the horizontal plane passing through the point of projection during its return to the ground is represented by the vector AC. Its magnitude is mv itself and it is inclined at 45º with the horizontal. The change in the momentum is vector AC – vector AB. The vector –AB is represented by the vector AD so that vector (AC – AB) = vector AR. Since the magnitudes of the vectors AC and –AB are the same and equal to mv and the angle between them is 90º, the magnitude of the change in momentum of the bullet is √(m²v² + m²v²) = √2 mv.

The figure showing the vectors is given here just to make things clear. You should be able to find the answer without using the figure.

You can arrive at the answer much more easily if you remember that there is no change in the horizontal component of momentum. The vertical component of momentum which has magnitude mv sin45º gets reversed so that the change in momentum is mv sin45º – (–mv sin 45º) = 2 mv sin 45º = √2 mv.

In the next post we will discuss some more questions from this section.

You will find a useful post on AP Physics Kinematics here

AP Physics B and C – Kinematics – Answers to Multiple Choice Questions on One Dimensional Motion

Some typical multiple choice questions (for practice) on one dimensional motion were given to you in the post dated 18^th August 2008. As promised, I give below the answers with explanation along with those questions:

Questions for AP Physics B and C:

(1) The adjoining figure shows the displacement-time graph of a particle in uniformly accelerated linear motion. At the points A and B (fig.) the graph is inclined at 30º and 60º respectively with the time axis. The acceleration of the particle in ms^–2 is

(a) √2

(b) √3

(c) 3/√2

(d) 2/√3

(e) 1/√3

The slope of the displacement-time graph gives the velocity and hence the velocity of the particle at time 2 s is tan30º = 1/√3. Similarly, the velocity of the particle at time 3 s is tan 60º = √3.

Acceleration = change of velocity/time = [√3 – (1/√3)]/(3 – 2) = 2/√3.

(2) A stone released from the top of a tower falls through half the height of the tower in 3/√2 seconds. The total time taken by the stone to reach the ground is

(a) 3 s

(b) 3.3 s

(c) 3.5 s

(d) 3.8 s

(e) 4 s

If h is the height of the tower we have

h/2 = 0 + (½) g (3/√2)²

This gives h = 9g/2

If t is the total time taken to reach the ground, we have

9g/2 = 0 + (½) g t² so that t = 3 s.

(3) A particle in one dimensional motion has initial velocity of 1 ms^–1. The variation of its acceleration with time is represented by the semi circular curve shown in the adjoining figure. What is the velocity of the particle ( in ms^–1) at the end of 8 seconds?

(a) 2π

(b) 2π + 1

(c) 4π

(d) 4π + 1

(e) 1

The area under the acceleration-time graph gives the change in velocity. [This area gives the sum of the products of the acceleration and time at all instants indicated by the curve. Since acceleration = change in velocity/time, the products of acceleration and time is the change in velocity and the area under the curve indeed gives the total change in the velocity].

Therefore, the change in velocity during 4 s = Area of the semicircle.

= (½)π ×4×2 = 4π

[Since the x-plot and the y-plot are of different quantities and the scales can be chosen arbitrarily, you have to treat the shape as half ellipse and the area is (½)πab where a and b are the semi major axis and semi minor axis respectively]

Since the initial velocity is 1 ms^–1, the velocity at the end of 4 s = (4π +1) ms^–1

(4) The bob of a simple pendulum is made to move along a semicircular path of radius 1 m. If the time taken to move along the semicircle is 1 s, the average velocity of the bob is

(a) π ms^–1

(b) π/2 ms^–1

(c) 1 ms^–1

(d) 2 ms^–1

(e) zero

The displacement of the bob in 1 s is 2 m.

Therefore, the average velocity of the bob = Displacement/Time = 2 m/1 s = 2 ms^–1.

(5) A stone is projected vertically upwards from point A (fig.) at the foot of a tower. It passes the top level B of the tower after 2 s. After another 2 s it again reaches the top level of the tower while returning from the top point C of its path. What is the height of the tower?

(a) 98 m

(b) 55 m

(c) 40 m

(d) 22 m

(e) 10 m

The time required for the upward journey of the stone is equal to the time required for the downward journey.

Time required for the upward journey (from A to C) = 2 + (2/2) = 3 s.

You can imagine that a stone released from rest at the point C takes a time of 3 s for the fall from C to A so that distance CA = (½)gt² = (½)×10×3² = 45 m.

Similarly, CB = (½)×10×1² = 5 m.

Therefore, height of the tower, BA = 45 – 5 = 40 m.

Questions for AP Physics C:

(1) The displacement x of a particle moving along a straight line is given by

x = 3t² + 2t + 4 where x is in metre and t is in second.

If the particle could move for 3 seconds, what was its acceleration (in ms^–2) at t = 2 s?

(a) 15

(b) 6

(c) 4

(d) 3

(e) 2

The acceleration a = d²x/dt² = 6 ms^–2

Since this is independent of time t, the particle moves with uniform acceleration throughout the time interval. The correct option is 6 ms^–2.

(2) A particle moving along the X-direction has its displacement x related to time t as

t = α x² + β x where α and β are positive constants.

If the velocity of the particle at the instant t is v, its acceleration is

(a) α v²

(b) – α v²

(c) – 2 α v²

(d) 2α v³

(e) – 2α v³

We have t = α x² + β x. Differentiating with respect to t, we obtain

1 = 2αx dx/dt + β dx/dt

Therefore, dx/dt = 1/(2αx+ β)

Acceleration, a = d²x/dt² = – (2α dx/dt) /(2αx+ β)²

Since dx/dt = v, the above relation becomes a = – 2αv³

(3) A body falling freely under gravity covers half its total path in the last second of its fall. The total time taken by the body to cover the entire path is approximately

(a) 1.414 s

(b) 1.732 s

(c) 2.414 s

(d) 3.414 s

(e) 3.732 s

If h is the height from which the body falls and t is the total time of fall, we have

h = 0 + (½)gt² = 5t²

Also, h/2 = 0 + g(t – ½) = 10t – 5

Substituting for h from the first equation, 5t²/2 = 10t – 5

Or, t² – 4t + 2 = 0

This gives t = 2 ±√2

Since t has to be greater than 1 s, as indicated in the question (there is a last second for the fall!), the answer is (2 + √2) s = 3.414 s.

(4) The co-ordinates of a moving particle at time t are given by x = a_x t², y = a_y t² and z = a_z t² where a_x, a_y, and a_z are constants. The magnitude of its velocity at time t is

(a) 2t (a_x² + a_y² + a_z²)^1/2

(b) t (a_x² + a_y² + a_z²)^1/2

(c) 2t (a_x² + a_y² + a_z²)

(d) t (a_x² + a_y² + a_z²)

(e) 2t (a_x + a_y + a_z)/3

The components of its velocity are

v_x = dx/dt = 2a_x t

v_y = dy/dt = 2a_y t

v_z = dz/dt = 2a_z t

The magnitude of the velocity is (v_x² + v_y² + v_z²)^1/2 = 2t (a_x² + a_y² + a_z²)^1/2

You teach me baseball and I'll teach you relativity... No, we must not. You will learn about relativity faster than I learn baseball.

– Albert Einstein

AP Physics B & C- Multiple Choice Questions (for practice) on One Dimensional Motion

The essential points to be remembered in one dimensional and two dimensional motions were discussed in the post dated 14^th August 2008. As promised, I give below some multiple choice questions on one dimensional motion. I’ll give you the solution in the next post.

The following questions will be useful for AP Physics B as well as C:

(1) The adjoining figure shows the displacement-time graph of a particle in uniformly accelerated linear motion. At the points A and B (fig.) the graph is inclined at 30º and 60º respectively with the time axis. The acceleration of the particle in ms^–2 is

(a) √2

(b) √3

(c) 3/√2

(d) 2/√3

(e) 1/√3

(2) A stone released from the top of a tower falls through half the height of the tower in 3/√2 seconds. The total time taken by the stone to reach the ground is

(a) 3 s

(b) 3.3 s

(c) 3.5 s

(d) 3.8 s

(e) 4 s

(3) A particle in one dimensional motion has initial velocity of 1 ms^–1. The variation of its acceleration with time is represented by the semi circular curve shown in the adjoining figure. What is the velocity of the particle ( in ms^–1) at the end of 8 seconds?

(a) 2π

(b) 2π + 1

(c) 4π

(d) 4π + 1

(e) 1

(4) The bob of a simple pendulum is made to move along a semicircular path of radius 1 m. If the time taken to move along the semicircle is 1 s, the average velocity of the bob is

(a) π ms^–1

(b) π/2 ms^–1

(c) 1 ms^–1

(d) 2 ms^–1

(e) zero

(5) A stone is projected vertically upwards from point A (fig.) at the foot of a tower. It passes the top level B of the tower after 2 s. After another 2 s it again reaches the top level of the tower while returning from the top point C of its path. What is the height of the tower?

(a) 98 m

(b) 55 m

(c) 40 m

(d) 22 m

(e) 10 m

The following Questions are for AP Physics C aspirants:

(1) The displacement x of a particle moving along a straight line is given by

x = 3t² + 2t + 4 where x is in metre and t is in second.

If the particle could move for 3 seconds, what was its acceleration (in ms^–2) at t = 2 s?

(a) 15

(b) 6

(c) 4

(d) 3

(e) 2

(2) A particle moving along the X-direction has its displacement x related to time t as

t = α x² + β x where α and β are positive constants.

If the velocity of the particle at the instant t is v, its acceleration is

(a) α v²

(b) – α v²

(c) – 2 α v²

(d) 2α v³

(e) – 2α v³

(3) A body falling freely under gravity covers half its total path in the last second of its fall. The total time taken by the body to cover the entire path is approximately

(a) 1.414 s

(b) 1.732 s

(c) 2.414 s

(d) 3.414 s

(e) 3.732 s

(4) The co-ordinates of a moving particle at time t are given by x = a_x t², y = a_y t² and z = a_z t² where a_x, a_y, and a_z are constants. The magnitude of its velocity at time t is

(a) 2t (a_x² + a_y² + a_z²)^1/2

(b) t (a_x² + a_y² + a_z²)^1/2

(c) 2t (a_x² + a_y² + a_z²)

(d) t (a_x² + a_y² + a_z²)

(e) 2t (a_x + a_y + a_z)/3

Try to answer these questions. I’ll be back with the solution shortly.

Kinematics for AP Physics B & C- One Dimensional and Two Dimensional Motions

You will have to remember certain basic formulae for solving multiple choice questions within the permitted time. The essential things you need to remember in one dimensional uniformly accelerated motion are here:

(1) The final velocity (v) at time t of an object in one dimensional motion with uniform acceleration a is given by

v = v₀ + at where v₀ is the initial velocity (at time t = 0)

[If you use the symbol u for the initial velocity, the equation becomes v = u + at]

(2) The position x of the object at time t is related its initial position x₀ (at zero time) as

x = x₀ + v₀ t+ (½) at²

This can be rewritten in terms of the displacement s = x – x₀ as

s = v₀ t+ (½) at²

[If you use the symbol u for the initial velocity, the above equation becomes s = ut+ (½) at²]

(3) The final velocity is related to the displacement (x – x₀) as

v² = v₀² + 2a(x – x₀)

[If you use the symbol u for the initial velocity and the symbol s for the displacement, the above equation becomes v² = u² + 2as]

(4) The distance (s_n = x_n – x_n–1) traveled during the n^th second is given by

s_n = v₀ + a(n– ½)

[If you use the symbol u for the initial velocity, the above equation becomes s_n = u + a(n– ½)]

(5) (i) The slope of the displacement – time graph (obtained by plotting time on the X-axis and the displacement on the Y-axis) gives the velocity.

(ii) The slope of the velocity – time graph (obtained by plotting time on the X-axis and the velocity on the Y-axis) gives the acceleration.

(iii) The area under the velocity – time graph (obtained by plotting time on the X-axis and the velocity on the Y-axis) gives the displacement.

In two dimensional motion you have to consider mainly circular motion and projectile motion. We have already discussed the essential points required in the case of circular motion in the post dated 20^th January 2008. You can access that post as well as related posts by clicking on the label ‘circular motion’ below this post. Here are the important equations you need to remember in the case of projectile motion:

(6) In projectile motion, the gravitational force affects the vertical component of the velocity (of the projectile) only. If the projectile is launched with velocity v₀ from the origin, making an angle θ₀ with respect to the horizontal, the x-component and the y-component of the velocity v₀ are respectively

v_0x = v₀ cosθ₀ and

v_0y = v₀ sinθ₀

The x and y co-ordinates of the projectile after time t are respectively

x = (v₀ cosθ₀)t and

y = (v₀ sinθ₀)t – (½)gt²

The x and y components of the velocity of the projectile after time t are respectively

v_x = v₀ cosθ₀ and

v_y = v₀ sinθ₀ – gt

(7) Time of flight(T_f ) of the projectile is given by

T_f = (2 v₀ sinθ₀) /g

(8) Horizontal range (R) of the projectile is given by

R = (v₀²sin 2θ₀)/g

(i) The horizontal range is maximum (R_max) when the angle of projection θ₀ = 45º and

R_max = v₀²/ g

(ii) If the velocity of projection v₀ is the same, the horizontal range is the same for angles of projection θ₀ and (90º – θ₀). These directions are equally inclined to the 45º direction for maximum range.

(10) Maximum height (H) reached by the projectile is given by

H = (v₀²sin²θ₀)/2g

If the projectile is launched vertically, the maximum height reached will be v₀²/ 2g, which is half the maximum range.

(11) for any given value of the angle of projection θ₀, the horizontal range R and the maximum height H are related as

R = 4H cot θ₀

We will discuss questions from this section in the next post. Meanwhile, you may go through some useful multiple choice questions on one dimensional motion given here.
You can find useful posts on two dimensional motion here.

AP Physics B and C- Electrostatics: Answers to Free Response Practice Questions on Capacitors

In the post dated 5^th August 2008, two free response practice questions on capacitors were given to you. As promised, I give below model answers along with the questions:

Question No1 (For AP Physics B & C):

Two metal plates A and B, each of area 0.2 m² are arranged parallel to each other as shown, with a separation of 1 cm in air. A regulated direct voltage source of output V volt remains connected between the plates so that there is an electric field of 2000 N/C in the region between the plates.

(a) Determine the output voltage V of the regulated voltage source.

(b) Determine the capacitance of the parallel plate capacitor formed by the plates A and B

(c) If the separation between the plates is gradually reduced (with the voltage source remaining connected between the plates), what quantities among the following will change? Put tick (√) marks against the changing quantities:

(i) Capacitance____ (ii) Charge on the plates____ (iii) Potential difference between the plates___

Give reasons for change (if any) and for no change (if any).

(d) During the decrease in the separation between the plates, will there be a current through the wires connecting the plates to the voltage source? Justify your answer.

(e) Determine the energy of the capacitor when the separation between the plates is reduced to half the initial value.

(a) The potential difference between the plates is equal to the otput voltage V of the voltage source. Since the electric field (E) between the plates is given by E = V/d where d is the separation between the plates, we have

2000 = V/10^–2 from which V = 20 volt.

(b) Capacitance (C) of the parallel plate capacitor is given by

C = ε₀A/d where A is the area of a plate.

Therefore, C = 8.85×10^–12×0.2/10^–2 = 1.77×10^–10 farad.

[The value of ε₀ will be given in your question paper].

(c) (i) Capacitance__√__ (ii) Charge on the plates__√__ (iii) Voltage between the plates____

The capacitance is given by C = ε₀A/d where A is the area of a plate and d is the seaparation between the plates. Since the separation is decreased, the capacitance is increased.

The charge on the plates is given by Q = CV. The capacitance C is increased but the voltage V between the plates is unchanged (since the plates are connected across the fixed voltage source). The charge Q on the plates is therefore increased.

(d) During the decrease in the separation between the plates, the capacitance goes on increasing and hence more charges flow from the voltage source to the capacitor, satisfying the equation, Q = CV. So there is a charging current in the wires connecting the plates to the voltage source.

(e) The energy (U) of a capacitor of capacitance C charged to a potential difference V between its plates is given by

U = (½) CV²

Initially the capacitance of the capacitor made by the two plates is 1.77×10^–10 farad as shown in part (a) above. When the separation (d) between the plates is reduced to half the initial value, the capacitance is doubled in accordance with the expression for capacitance, C = ε₀A/d. Therefore, the energy is given by

U =(½) (2×1.77×10^–10)×20² = 7.08×10^–8 joule.

Question No.2 (For AP Physics C):

A spherical capacitor is made of two conducting spherical shells of radii r₁ and r₂ (r₂ > r₁). The inner shell has a total charge +Q and the outer shell has a total charge –Q. Now, answer the following questions:

(a) The space between the shells is filled with air of dielectric constant very nearly equal to 1. Calculate the potential difference between the shells.

(b) Determine the electric field at a point in between the shells, at distance a from the common centre of the shells.

(c) What are the values of the electric field and potential at a point outside both shells, at distance b (b > r₂) from the common centre of the shells?

Justify your answer.

(d) Determine the potential at the common centre of the shells.

(e) The air in between the shells is replaced by a material whose dielectric constant has value K₁ from r₁ to R and value K₂ from R to r₂ ((r₂ > R > r₁). Show that the capacitance of this spherical capacitor is given by

C = 4πε₀/[(1/K₂R – 1/K₁R) + (1/K₁r₁ – 1/K₂r₂)]

(a) The potential at all points inside the larger shell is the same as the potential at its surface. Therefore, the potential difference (between the shells) produced by the charge –Q on the outer shell is zero and we need consider the contribution (to the potential difference) by the charge +Q on the inner shell alone.

The potential (V₁) of the inner shell due to its charge +Q is given by

V₁ = Q/4πε₀r₁

The potential (V₂) of the outer shell due to the charge +Q on the inner shell is given by

V₂ = Q/4πε₀r₂

Therefore, potential difference V₁ – V₂ = (Q/4πε₀)[(1/ r₁) – (1/ r₂)]

(b) The electric field at points inside the outer shell due to the charge on it is zero (since there is no potential gradient because of its charge). The field (E) in between the shells at distance a is therefore due to the charge +Q on the inner shell alone and is given by

E = Q/4πε₀a²

This field is directed radially outwards.

(c) The electric potential and field at all points outside both shells (at distance b > r₂) is zero since the potential and field at points outside any charged spherical shell is produced as though the entire charge is concentrated at the centre. We have therefore positive and negative charges of equal magnitude and the net charge imagined to be concentrated at the centre is zero.

[The electric field can be proved to be zero by applying Gauss’s law, considering a Gaussian spherical surface of radius b passing through the point. The net charge enclosed by the gaussian surface is zero so that the field is zero].

(d) The potential (V) at the common centre of the shells is the sum of the potentials due to the charges on the shells and is given by

V = (1/4πε₀)[(Q/r₁) – (Q/r₂)] = (Q/4πε₀)[(1/ r₁) – (1/ r₂)]

(e) The capacitance (C) is given by

C = Q/V where V is the potential difference between the shells.

But V = V₁ – V₂ where V₁ and V₂ are respectively potentials on the inner and outer shells with the composite dielectric in between them.

If E is the electric field at a point in between the shells, at distance r from the centre, we have

V = – _r2∫^r1 Edr = _r1∫^r2 (Q/4πε₀K r²)dr where K is the dielectric constant which varies with r (K has value K₁ from r₁ to R and value K₂ from R to r₂).

Therefore, V = (Q/4πε₀) _r1 ∫^r2 dr/ Kr²

= (Q/4πε₀) [{– 1/ K₁r}^R _r1 + {– 1/ K₂r}^r2_R]

= (Q/4πε₀) [(1/K₂R – 1/K₁R) + (1/K₁r₁ – 1/K₂r₂)]

The capacitance is given by C = Q/V = 4πε₀/[(1/K₂R – 1/K₁R) + (1/K₁r₁ – 1/K₂r₂)]

[You can obtain the above expression from the expression for the capacitance of a spherical capacitor with a material of dielectric constant K, given in the post dated 14^th July 2008:

C = 4πε₀K r_a r_b /(r_b – r_a)

The radii r_a and r_b are to be replaced by r₁ and r₂. Further, C will be the series combined value of two spherical capacitors C₁ and C₂ where C₁ is made by spherical shells of radii r₁ and R, with material of dielectric constant K₁ in between and C₂ is made by spherical shells of radii R and r₂, with material of dielectric constant K₂ in between. C is given by the reciprocal relation, 1/C = 1/C₁ + 1/C₂.

Since you have to borrow the expression for the simple spherical capacitor, this method will not be as rigorous as the one we used above]