Answer to Free Response Practice Question on Kinematics in Two Dimensions for AP Physics C

A free response practice question on kinematics in two dimensions was given to you in the post dated 11^th February 2010. As promised, I give below a model answer along with the question:

A chopper at an air base is rising vertically up with a velocity of 2 ms^–1. When it is at an altitude of 50 m, a packet is thrown down with velocity 8 ms^–1 (with respect to the chopper), making an angle of 60º with the vertical. Neglecting air resistance and assuming that g = 10 ms^–2, answer the following questions

(a) Draw a diagram to show the nature of the path AB followed by the packet from the moment it leaves the chopper to the moment it hits the ground, as seen by an observer on the ground. Is the path straight, circular or parabolic?.

(b) Will the packet move up initially? Justify your answer.

(c) Calculate the time taken by the packet to reach the ground.

(d) Determine the horizontal distance traveled by the packet before hitting the ground.

(e) Determine the velocity (magnitude as well as direction) with which the packet hits the ground.

(a) The path AB of the packet is shown in the following figure. The path is parabolic.

(b) The packet will not move up initially. This is due to the fact that the vertical, downward component (8 cos 60º = 4 ms^–1) of the velocity with which the packet is thrown down is greater than the upward velocity (2 ms^–1) of the chopper.

[The situation is as though the packet is thrown down from a stationary chopper with vertical (downward) velocity component (4 ms^–1 – 2 ms^–1) = 2 ms^–1 and horizontal velocity component 8 sin 60º = 4√3 ms^–1].

(c) The time t taken by the packet to reach the ground can be found by considering its vertical motion. We have

s = ut + ½ at² where the vertical displacement s = 50 m, vertical (downward) velocity component u = 8 cos 60º – 2 = 2 ms^–1 and vertical (downward) acceleration a = g = 10 ms^–2.

Therefore, 50 = 2t + ½ ×10t².

Or, 5t² + 2t – 50 = 0

This gives t = [–2 ± √(4 – 1000)] /10 = 2.968 s.

(d) The horizontal distance R traveled by the packet before hitting the ground is the product of the above time and the horizontal velocity component v_horizontal which is equal to 8 sin 60º (= 4√3 ms^–1).

Therefore, R = 2.968×4√3 = 20.56 m.

(e) The velocity with which the packet hits the ground is the resultant of the horizontal and vertical components. The horizontal component of velocity remains unchanged (at 8sin 60º = 4√3 ms^–1) throughout the motion of the packet since the gravitational force acts vertically and cannot affect the horizontal motion. The vertical component of velocity v_vertical goes on increasing during the fall of the packet in accordance with the equation,

v_vertical = u_vertical + at where u_vertical is the initial vertical component of velocity, a is the vertical acceleration and t is the time of flight of the packet.

Here u_vertical = 8 cos 60º – 2 = 2 ms^–1, a = g = 10 ms^–2 and t = 2.968 s.

Therefore, v_vertical = 2 + 10×2.968 = 31.68 ms^–1.

The magnitude v of the velocity with which the packet hits the ground is therefore given by

v = √[(4√3)² + (31.68)²] = 32.43 ms^–1.

The direction of this velocity makes an angle θ with the horizontal (fig.) and is given by

tan θ = v_vertical/ v_horizontal = 31.68/(4√3) = 4.572

Therefore θ = 77.66º

Kinematics in Two Dimensions – Free Response Practice Question for AP Physics C

Today I’ll give you a free response practice question on Kinematics in Two Dimensions. Even though this question is meant for AP Physics C aspirants, those who prepare for AP Physics B exam also will find it useful. Here is the question:

A chopper at an air base is rising vertically up with a velocity of 2 ms^–1. When it is at an altitude of 50 m, a packet is thrown down with velocity 8 ms^–1 (with respect to the chopper), making an angle of 60º with the vertical. Neglecting air resistance and assuming that g = 10 ms^–2, answer the following questions

(a) Draw a diagram to show the nature of the path AB followed by the packet from the moment it leaves the chopper to the moment it hits the ground, as seen by an observer on the ground. Is the path straight, circular or parabolic?.

(b) Will the packet move up initially? Justify your answer.

(c) Calculate the time taken by the packet to reach the ground.

(d) Determine the horizontal distance traveled by the packet before hitting the ground.

(e) Determine the velocity (magnitude as well as direction) with which the packet hits the ground.

Try to answer this question.. You have 15 minutes at your disposal and can score up to 15 points for the right answer. Of course I’ll be back soon with a model answer for your benefit.

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

Everything that is really great and inspiring is created by the individual who can labor in freedom.
– Albert Einstein

As promised in the post dated 27^th August 2008, I give below some more multiple choice questions (with solution) on two dimensional motion. These questions will be useful for AP Physics B as well as AP Physics C aspirants.

(1) A particle is projected from the origin with velocity (2 î + 3 ĵ) ms^–1 where î and ĵ are unit vectors along the X and Y directions respectively. If the X direction is horizontal, what is the range of the projectile on the horizontal plane through the point of projection? (g = 10 ms^–2)

(a) 6.2 m

(b) 4.8 m

(c) 4.2 m

(d) 2.8 m

(e) 1.2 m

The time of flight of the projectile can be found from the vertical component of velocity (which is 3 ms^–1). This projectile will have time of flight equal to that of a particle projected vertically up with velocity 3 ms^–1. Therefore, time of flight,

T_f = 2 v_y/g = 2×3/10 = 0.6 s

[We have used the relation s =ut + ½ at² in which s = 0, u = 3 ms^–1, a = g and t = T_f]

Horizontal range = Horizontal velocity ×Time of flight = 2×0.6 = 1.2 m

(2) The horizontal rnge R and the maximum height H (both in metre) of a projectile are related to the time of flight T (in seconds) as

R = 40T and H = 15T – 1.25T²

The velocity of projection is (g = 10 ms^–2)
(a) 100 ms^–1

(b) 50 ms^–1

(c) 40 ms^–1

(d) 30 ms^–1

(e) 25 ms^–1

Projectile motion is the combination of two one dimensional motions. From the form of the expressions for R and H it is evident that the horizontal component of the velocity of projection is 40 ms^–1. [Remember that R = v_0xT where v_0x is the horizontal component of the velocity of projection]

The vertical component of the velocity of projection is 30 ms^–1 [since the maximum height, H = v_0y(T/2) – ½ g(T/2)²]

The velocity of projection is √(v_0x² + v_0y²) = √(40² + 30²) = 50 ms^–1.

(3) A body of mass 2 kg is projected from the origin with initial velocity v = (40 î + 30 ĵ) ms^–1. The only force acting on the body is a constant force F = 3 î – 4 ĵ newton. The time in which the y–component of the velocity of the body will become zero is

(a) 5 s

(b) 10 s

(c) 15 s

(d) 20 s

(e) 25 s

The y–component of acceleration is – 4/2 = – 2 ms^–1 [since the y–component of force is

– 4 N and the mass is 2 kg].

The time t in which the y–component of the velocity (30 ms^–1) will become zero is given by

0 = 30 –2t (using v = u + at)

This gives t = 15 s.

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

(1) The position co-ordinates of a particle projected from the origin are given by x = 30t and y = 40t – 5t². The vertical component of velocity after 5 seconds and the magnitude of the velocity of projection are respectively

(a) –10 ms^–1, 50 ms^–1

(b) –10 ms^–1, 40 ms^–1

(c) 10 ms^–1, 40 ms^–1

(d) –20 ms^–1, 50 ms^–1

(e) 20 ms^–1, 50 ms^–1

The vertical component (v_y) of velocity at the instant t during the flight of the projectile is given by

v_y = dy/dt = 40 – 10t

When t = 5 s we have v_y = 40 – 10×5 = –10 ms^–1

[The negative sign shows that the projectile is moving down]

The initial vertical component of velocity (v_0y) at the instant when t = 0 is 40 ms^–1. The horizontal component of velocity at the instant of projection (and throughout the motion) is 30 ms^–1 since x = 30t and dx/dt = v_x = v_0x = 30 ms^–1. [Since x = 30t, you should be able to understand that the horizontal velocity is 30 ms^–1, even without finding the value of dx/dt]

The velocity of projection has magnitude √(v_0x² + v_0y²) = √(30² + 40²) = 50 ms^–1.

(2) The displacement of a particle at time t has x-component 3 + at³ and y-component 4 + bt³. The speed of the particle at time t is

(a) 3t²√(a ² + b²)

(b) 3t²(a ² + b²)

(c) 3t√(a ² + b²)

(d) 3t(a ² + b²)

(e) t²√(a ² + b²)

The x-component of velocity (v_x) is the time derivative of the x-component of displacement: v_x = 3 at²

Similarly, y-component of velocity, v_y = 3 bt²

The magnitude (v) of the velocity at time t = √(v_x² + v_y²)

Therefore, v =√[(3 at² )² + (3 bt²)²] = 3t²√(a ² + b²)

This is the speed of the particle at time t.

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

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.