TWO DIMENSIONAL MOTION

MPHIRISENI NORMAN KHWANDA

17 March 2015

PRE-CONCEPT QUESTIONS

1. A projectile is fired into the air, and it follows the

parabolic path as shown in the drawing

There is no air resistance. At any instant, the projectile has a velocity 𝒗 and

an acceleration 𝒂.

Which one or more the drawings on the right could not represent the

directions for 𝒗 and 𝒂 at any point on the trajectory?

A: Diagram 1 and 2

B: Diagram 3 and 4

C: Diagram 1 and 3

D: Diagram 2 and 4

The gravitational acceleration cannot be towards the horizontal

direction and again cannot face up.

THE CONVERTIBLE CAR

2. Suppose you are driving in a convertible with the top down. The car is moving to the right with constant velocity as the diagram illustrates. You point a rifle upwards and fire it. In the absence of air resistance, where would the bullet land?

A: Behind you (or the car)

B: Ahead of the you (car)

C: In the barrel of the riffle (back to where it was originally fired)

The bullet moves with the same velocity horizontally as the car it only accelerates up and down due to gravity. The vertical and horizontal motions are independent of each other

THE CASE OF TWO BALLS

3. Ball 1 is thrown into the air and it follows the trajectory for a projectile motion shown in the drawing. At the instant Ball 1 is at the top, Ball 2 is dropped from rest from the same height, which ball reaches the ground first?

A: Ball 1 reaches the ground first since it is moving while Ball 2 is stationery

B: Ball 2 reaches the ground first because it has a shorter distance to travel

C: Both balls reach the ground at the same time

D: There is not enough information to tell which ball reaches the ground first

Explanation:

Both have zero velocity vertically and same height and under the gravitational acceleration g. hence they will reach at the same time

THE PATH TAKEN BY THE BALL4. The diagram represents a ball moving at a constant velocity of 50 m/s towards the right. Complete the path that the ball would take after reaching the edge B.

Explanation:

The ball will continue to move with the same velocity of 50 m/s horizontally but at the same time the force of gravity will be pulling the ball down immediately after leaving the edge, hence the parabolic path

MOTION IN TWO DIMENSIONS

Summary:

The motion is under the influence of gravity which always points towards the centre of the earth.

The vertical and horizontal motions are independent from one another

The motion horizontally remains constant (acceleration is zero)

Air resistance is ignored

The maximum velocity during free-fall depends on the initial velocity and initial position (height)

Same height, same initial velocity implies same final velocity

EQUATIONS OF KINEMATICS HORIZONTALLY (ALONG THE X)

tavv xoxx tvvxx xoxo 21

)(222

oxoxx xxavv

2

21 tatvxx xoxo

3.2 EQUATIONS OF KINEMATICS VERTICALLY (ALONG THE Y)

gtvv oyy

2

21 gttvyy oyo

tvvyy yoyo 21

)(222

ooyy yygvv

THE COMBINATION:

SIMULTANEOUS MOTION ALONG THE 𝒙 AND 𝒚

The x part of the motion occurs exactly as it would if the

y part did not occur at all, and vice versa.

ASSUMPTIONS

The motion along the 𝒙 and the 𝒚are independent of each other.

The acceleration horizontally is zero

The acceleration vertically is constant and due to gravity

The effects of air resistance is ignored.

THE FINAL VELOCITY OF AN OBJECT

Note:

The magnitude of velocity along the 𝒙 doesn’t change and is given by 𝒗𝒙 = 𝒗𝒐𝒙because the acceleration 𝒂𝒙 = 𝟎.

The magnitude of the final velocity is given by 𝒗 = (𝒗𝒙)𝟐+(𝒗𝒚)

𝟐 and the

direction is given by 𝜽 = 𝒕𝒂𝒏−𝟏(𝒗𝒚

𝒗𝒙)

The magnitude of velocity along the y changes because of the acceleration 𝒂 =− 𝒈 𝐢𝐟 𝐮𝐩 > 𝟎

Note:

The magnitude of velocity along the 𝒙doesn’t change and is given by

𝒗𝒐𝒙 = 𝑣𝑜𝑐𝑜𝑠𝜃0 because the acceleration

𝒂𝒙 = 𝟎.Hence:

𝒗𝒙 = 𝒗𝒐𝒙 + 𝒂𝒙𝒕 = 𝑣𝑜𝑐𝑜𝑠𝜃0

Note:

The velocity along the y

changes due to the gravitational

acceleration g

𝒗𝒐𝒚 = 𝑣𝑜𝑠𝑖𝑛𝜃0because the acceleration 𝒂𝒚 =

𝒈 = −𝟗. 𝟖𝐦

𝒔𝟐𝐢𝐟 𝐮𝐩 > 𝟎

Hence the final velocity along

y is

𝒗𝒚 = 𝒗𝒐𝒚 + 𝒈𝒕 = 𝒗𝒐𝒔𝒊𝒏𝜽𝟎 + 𝒈𝒕

THE MAXIMUM HEIGHT H AND THE RANGE R

(1) Maximum Height H:

Data: set 𝑎𝑛𝑑 𝒚𝟎 = 𝟎𝒎 𝑎𝑛𝑑 𝑦 = 𝐻Then 𝒗𝒐𝒙 = 𝑣𝑜𝑐𝑜𝑠𝜃0and 𝒗𝒐𝒚 = 𝑣𝑜𝑠𝑖𝑛𝜃0From the diagram , at maximum height

𝑦 = 𝐻 𝒗𝒚 = 𝟎𝒎/𝒔

Hence the maximum height can be

found using the equation:

𝑣𝑦2 = 𝑣𝑜𝑦

2 + 2𝑔(𝑦 − 𝑦0)

𝑦 − 𝑦0 =𝑣𝑦

2−𝑣𝑜𝑦2

2𝑔

Substituting:

𝐻 − 0 =02−(𝑣𝑜𝑠𝑖𝑛𝜃0)

2

2(−𝑔)=

𝑣02𝑠𝑖𝑛2𝜃0

2𝑔

The maximum height H 𝐻 =𝑣0

2𝑠𝑖𝑛2𝜃0

2𝑔

(2)The time to reach the maximum

height 𝑣𝑦 = 𝑣𝑜𝑦 + 𝑔𝑡

𝑡 =−𝑣𝑜𝑦

𝑔=−𝑣𝑜𝑠𝑖𝑛𝜃0

−𝑔=𝑣𝑜𝑠𝑖𝑛𝜃0

𝑔

(3):Range R: Horizontal distance

Data set: 𝒙𝟎 = 𝟎𝒎 and 𝒙 = 𝑹The range R can be given by

:𝑥 − 𝑥0 = 𝑣𝑜𝑥𝑡 +1

2𝒂𝒙𝑡

2 = 𝑣𝑜𝑥𝑡 since 𝒂𝒙 = 𝟎

Substituting

R−0 = 𝑣𝑜𝑥𝑡, hence R= 𝑣𝑜𝑥𝑡 = 𝑣𝑜𝑐𝑜𝑠𝜃0t

But due to symmetry 𝒕𝒊𝒎𝒆 𝒖𝒑 =𝒕𝒊𝒎𝒆 𝒅𝒐𝒘𝒏

Hence the total time is 𝑡 =2𝑣𝑜𝑠𝑖𝑛𝜃0

𝑔

The range R= 𝑣𝑜𝑐𝑜𝑠𝜃02𝑣𝑜𝑠𝑖𝑛𝜃0

𝑔=2𝑣0

2𝑐𝑜𝑠𝜃0𝑠𝑖𝑛𝜃0

𝑔

Or 𝑹 =𝒗𝟎

𝟐𝒔𝒊𝒏𝟐𝜽𝟎

𝒈

because 𝑠𝑖𝑛2𝜃0 = 2𝑐𝑜𝑠𝜃0𝑠𝑖𝑛𝜃0

YOUR TURN

A football was shot at an angle of 37° above the horizontal with the speed of 20 𝑚/𝑠. Calculate

(a) The maximum height.

(b) The time the ball travel before it hits the ground

(c) How far did the ball travel just before hitting the ground?[The Range]

(d) The magnitude and direction of velocity at maximum height

(e) The magnitude and direction of acceleration at maximum height.

Question: A football is kicked at an angle 𝜽𝟎 = 𝟑𝟕° with the velocity of 20.0 m/s.

Direction: Assume up to be positive and For the sake of referencing, set 𝑥0 = 0𝑚 𝑎𝑛𝑑 𝑦0 =0𝑚

Initial velocities: 𝑣0𝑥 = 20𝑐𝑜𝑠37° = 16𝑚/𝑠 and 𝑣0𝑦 = 20𝑠𝑖𝑛37° = 12 𝑚/𝑠 .

At maximum height, 𝑣𝑦 = 0𝑚/𝑠 and 𝑣𝑥 = 𝑣0𝑥 = 20𝑠𝑖𝑛37° = 12 𝑚/𝑠

(a) For maximum height: 𝑣𝑦2 = 𝑣𝑜𝑦

2 + 2𝑔(𝑦 − 𝑦0) hence 𝑦 − 𝑦0 =𝑣𝑦

2−𝑣𝑜𝑦2

2𝑔

substituting we get 𝑦 − 0 =0 2− 12 2

2 −9.8= 7.35 𝑚. Hence maximum height 𝑦 = 7.35 𝑚

(b) The total time the ball is in air is given by:

𝑦 − 𝑦0 = 𝑣0𝑦𝑡 +1

2𝑔𝑡2

0−0 = 12 𝑡 + 1

2(−9.8)𝑡2

0 = 𝑡(12 − 4.9𝑡)

Solving we get 𝑡 = 0 𝑠 𝑎𝑛𝑑 𝑡 = 2.45 𝑠

𝒕 = 𝟎 𝒔 corresponds to the initial position

and 𝒕 = 𝟐. 𝟒𝟓 𝒔 is the total time of travel

of the ball.

(c) Range:The total distance travelled horizontally

is the range R given by 𝑥 − 𝑥0 = 𝑣0𝑥𝑡𝑥 − 0 = 16 2.45 = 39.2 𝑚

So that 𝑥 = 39.2 𝑚

Motion in two dimensions

7. A shell is fired with a horizontal velocity in the

positive x direction from the top of an 80-m high cliff.

The shell strikes the ground 1330 m from the base of

the cliff. The drawing is not to scale.Calculate the initial speed of the shell.

A) 4.0 m/s

B) 9.8 m/s

C) 82 m/s

D) 170 m/s

E) 330 m/s

The answer is E

8. A shell is fired with a horizontal velocity in the

positive x direction from the top of an 80-m high cliff.

The shell strikes the ground 1330 m from the base of

the cliff. The drawing is not to scale.

What is the magnitude of the velocity as

it hits the ground?

A) 4.0 m/s

B) 9.8 m/s

C) 82 m/s

D) 170 m/s

E) 330 m/s

The answer is E

9. A shell is fired with a horizontal velocity in the

positive x direction from the top of an 80-m high cliff.

The shell strikes the ground 1330 m from the base of

the cliff. The drawing is not to scale.

What is the magnitude of the acceleration of the

shell just before it strikes the ground?

The answer is E

A) 4.0 m/s2

B) 9.8 m/s2

C) 82 m/s2

D) 170 m/s2

E) 330 m/s2

10. A tennis ball is thrown upward at an angle from point A. It

follows a parabolic trajectory and hits the ground at point D. At

the instant shown, the ball is at point B. Point C represents the

highest position of the ball above the ground. While in flight,

how do the x and y components of the velocity vector of the ball

compare at the points B and C?

A) The velocity components are non-zero at B and zero at C.

B) The x components are the same; the y component at C is zero m/s.

C) The x components are the same; the y component has a larger magnitude at C than at B.

D) The x component is larger at C than at B; the y component at B points up while at C, it points

downward.

E) The x component is larger at B than at C; the y component at B points down while at C, it points

upward.

The answer is B

11. A tennis ball is thrown upward at an angle from point A. It

follows a parabolic trajectory and hits the ground at point D. At

the instant shown, the ball is at point B. Point C represents the

highest position of the ball above the ground. While in flight,

how do the x and y components of the velocity vector of the ball

compare at the points A and D?

The answer is E

A) The velocity components are non-zero at A and are zero m/s at D.

B) The velocity components are the same in magnitude and direction at both points.

C) The velocity components have the same magnitudes at both points, but their directions are

reversed.

D) The velocity components have the same magnitudes at both points, but the directions of the x

components are reversed.

E) The velocity components have the same magnitudes at both points, but the directions of the y

components are reversed.

12.A football is kicked with a speed of 18 m/s at an

angle of 65° to the horizontal.

12.1What are the respective horizontal

and vertical components of the initial

velocity of the football?

A) 7.6 m/s, 16 m/s

B) 16 m/s, 7.6 m/s

C) 8.4 m/s, 13 m/s

D) 13 m/s, 8.4 m/s

E) 9 m/s, 9 m/s

The answer is

A

12.2 How long is the football in the air?

A) 1.1 s

B) 1.6 s

C) 2.0 s

D) 3.3 s

E) 4.0 s

The answer is

D

12.3 How far does the football travel

horizontally before it hits the ground?

A) 18 m

B) 25 m

C) 36 m

D) 48 m

E) 72 m The answer is

B

Chapter 2 Problem 58:

Two stones are thrown simultaneously, one straight up from the

base of the cliff and the other straight down from the top of the

cliff. The height of the cliff is 6.00m. The stones are thrown with

the same speed of 9.00 m/s. Find the location(above the base of

the cliff) of the point where the stones cross paths.

Reasoning:

The stone that is thrown upward loses

speed on the way up. The stone that is

thrown downward gains speed on the way

down. The initial velocity v0 is known for both stones, as

is the acceleration a due to gravity. In addition,

we know that at the crossing point the stones are

at the same place at the same time t.

210 2

y v t at The equation to be used will be

Applying the equation of motion we get

The distances above and below the

crossing point must add to equal the height

of the cliff H, and hence

This equation can be solved for t to show that the

travel time to the crossing point is

Substituting this result into the expression

for y up gives

Thus, the crossing is located at a distance of

2.46 m above the base of the cliff

ASSIGNMENT/TUTORIAL QUESTIONS