Tutorial 1_Physics 132_Summer 2022

Please answer the following questions. Show all work and write neatly. Answers without

justifying work will receive no credit. Partial credits will be given as appropriate. Do not

leave any problem blank. The point values of problems are indicated in parentheses.

1.

(1.5 marks) Find the X and Y components of the following:

A. 35 m at 57Âº from the x-axis.

B. 12 m at 34 Âº S of W

C. 8 m South

D. 20 m 275 Âº from the x-axis

2.

(1.5 marks) Find the resultant vector (magnitude and direction) given the following information:

A. Ax = 5.7, Ay = 3.4

B. Bx = -10, By = -3

C. Cx = 12, Cy = -20

D. -30i + 27j

E. 48i â€“ 12 j

3. (1.5 marks) What are x and y components?

A. What are Supermanâ€™s x and y components?

B. What are the x and y components of the planeâ€™s velocity?

40 m/s

C. How fast is the plane traveling? What is the angle Î¸?

80 m/s

120 m/s

4.

(2 marks) Observe the following vectors:

A = 15

60 Âº

28Âº

B = 10

D = 50

C = 25

Construct an X-Y chart for all vectors:

X

A

Y

B

C

D

A. Find the resultant (magnitude and direction) of the following:

A+B+C+D

A-C

D + 2C

5.

(0.5 marks) Determine the A+B+C for the following:

A = 40

15Âº

B = 90

C = 60

2ND SEMESTER

Deanship of Preparatory Year

and Supporting Studies

ENGINEERING TRACK

PHYSICS 132

WEEK 10 Lecture 9

Chapter 4

Motion in Two Dimensions

2021-2022

Learning Objectives:

Understand

projectile motion in the earthâ€™s gravity.

Identify

the terms and properties of a projectile, such as acceleration

due to gravity, range, maximum height, and trajectory

Analyze

the location and velocity of a projectile in 2D at different

points of its trajectory.

Solve

problems and exercises for projectiles by using the basic

kinematics equations of motion in 2D.

Apply

principles of vector addition to determine relative velocity.

Explain

the significance of the observer in the measurement of

velocity.

2

4-3 Projectile Motion (P. 171)

A projectile is an objectâ€™s motion, moving in two dimensions

(vertical and horizontal) under the influence of Earthâ€™s gravity; its

path is a parabola.

3

Guess now!

Which statements are NOT valid for a projectile motion ?

(a)The projectile has the same x- velocity (vx) at any point on its path.

(b)The acceleration of the projectile is positive and decreasing when the

projectile is moving upwards, zero at the top, and increasingly negative

as the projectile descends.

(c) The acceleration of the projectile is a constant negative value.

(d)The y-velocity (vy) of the projectile is zero at the highest point of its path.

(e) The velocity at the highest point is zero (vy , vx).

4

Projectiles with Angle: Finding components

ð’—ðŸŽ

ð’—ðŸŽð’š

Î¸

ð’—ðŸŽð’™

ð‘£!” = ð‘£! cos Ó¨

ð‘£!# = ð‘£! sin Ó¨

horizontal velocity (Direction: +x)

vertical velocity (Direction: +y or â€“y)

5

Ex.1: An object is fired from the ground at 100 m/s at an angle of 30 Â° with

the horizontal. Calculate the horizontal and vertical components of the

initial velocity.

Solution :

ð‘£!” = ð‘£’ cos ðœƒ = 100Ã—ð‘ð‘œð‘ 30Â° = 87 m/s

ð‘£!# = ð‘£’ sin ðœƒ = 100Ã—ð‘ ð‘–ð‘›30Â° = 50 m/s

ð’—ðŸŽ

ð’—ðŸŽð’š = ðŸ“ðŸŽ ð’Ž/ð’”

Î˜= 30 Â°

ð’—ðŸŽð’™ = ðŸ–ðŸ•ð’Ž/ð’”

6

4-3 Projectile Motion (P. 171)

horizontal velocity (Direction: +x)

vertical velocity

(Direction: -y)

vertical velocity (Direction: +y)

7

Analyzing Projectile motion P.59

The speed in the ð’™-direction (Horizontal) is constant; in the ð’š-direction

(vertical) the object moves with constant acceleration ð’ˆ.

q This photograph shows two balls that start to

fall at the same time.

q The one on the right has an initial speed in the

ð‘¥-direction.

q It can be seen that vertical positions of the

two balls are identical at identical times, while

the horizontal position of the yellow ball

increases linearly.

8

Analyzing Projectile motion

q Projectile motion is motion with constant acceleration in 2D, where the

acceleration is g and is downward.

v(! = v! cos Î¸!

v)! = v! sinÎ¸!

q In projectile motion, The horizontal motion and the vertical

motion are independent from each other.

q Thus, we can divide the problem into two separate and easy 1D

problems (x and y)

9

Analyzing Projectile motion

Horizontal Motion (in +x axis)

(no acceleration in x-axis)

ð’—ðŸŽ

v( = v!(

v( = v! cos(Î¸! )

Î¸

x âˆ’ x! = v!( ð‘¡

Vertical Motion (in y axis)

constant all the time

x âˆ’ x! = v! cos(Î¸! )ð‘¡

(constant acceleration in -y)

ð’‚ð’š = âˆ’ð

1 *

we substitute v!” and ay in y âˆ’ y! = v!# ð‘¡ + ð‘Žð‘¡

2

v!) = v! sin(Î¸! )

1 *

y âˆ’ y! = v! sin Î¸! ð‘¡ âˆ’ ð‘”ð‘¡

2

v) = v! sin Î¸! âˆ’ gt

ð‘£#* = v! sin Î¸!

* âˆ’ 2g(y âˆ’ y )

!

10

Ex.2: An object is fired from the ground at 100 m/s at an angle of 30 Â° with

the horizontal

i.

ii.

Calculate the horizontal and vertical components of the initial velocity

After 2.0 seconds, how far has the object traveled in the horizontal

direction?

iii. How high is the object at this point?

Solution :

ii)

i) ð‘£'” = ð‘£’ cos ðœƒ = 100Ã—ð‘ð‘œð‘ 30Â° = 87 m/s

ð‘£’# = ð‘£’ sin ðœƒ = 100Ã—ð‘ ð‘–ð‘›30Â° = 50 m/s

ð›¥ð‘¥

ð‘£'” =

ð›¥ð‘¡

ð›¥ð‘¥ = ð‘£'” ð‘¡ = 87Ã—2.0

= 174ð‘š

1

iii) âˆ†ð‘¦ = ð‘£’# ð‘¡ âˆ’ ð‘”ð‘¡ *

2

1

âˆ†ð‘¦ = 50Ã—2 âˆ’ 9.8Ã— 2 * = 80.4 ð‘š

2

11

Ex.3: A ball rolls off a tabletop with an initial velocity of 3 m/s. If the tabletop is

1.25 m above the floor:

A)How long does it take for the ball to hit the floor?

B)How far does the ball travel horizontally?

a) Ball rolls off in +x direction, so ð‘£'” = 3 m/s . But ð‘£’# = 0 , means it is a free

fall

1 *

1

y=ð‘¦! + ð‘£’# tâˆ’ ð‘”ð‘¡

ð‘£’# = 0

âˆ’ 1.25=0âˆ’ 9,8ð‘¡ *

ð‘¡ = 0.5 ð‘

2

2

b) Î”ð‘¥ = ð‘£'” ð‘¡

Î”ð‘¥ = 3ð‘¥0,5 = 1.5ð‘š

ð‘£'” = 3 m/s

ð‘¦! = 0

â„Ž = 1,5ð‘š

Answers: a) ð‘¡ = 0.5 ð‘

, b) âˆ†ð‘¥ = 1.5 ð‘š

ð‘¦+ = âˆ’1.25ð‘š

12

Â§ To use the following formulae , the final

height should be the same as the lunch

height.

The horizontal range of a projectile is the distance along the

horizontal plane it would travel, before reaching the same vertical

position as it started from.

13

Projectile motion: The horizontal Range

Ex.4: Which angle makes the horizontal range is maximum?

sin(2ðœƒ, ) = 1 ð‘†ð‘œ 2ðœƒ, = 90 Â°

ðœƒ, = 45 Â°

14

Ex.5: Range of a cannon ball.

Suppose one of Napoleonâ€™s cannons had a muzzle speed, ð‘£! , of 60 m/s. At

what angle should it have been aimed (ignore air resistance) to strike a target

320 m away?

Solution :

ð‘£,* sin(2ðœƒ, )

ð‘…=

g

ð‘…g (320)(9.8)

sin(2ðœƒ, ) = * =

= 0.871

*

(60)

ð‘£,

2ðœƒ, = sin-. 0.871 = 60.6Â° ð‘œð‘Ÿ 119.4Â°

ðœƒ, = 30.3Â° ð‘œð‘Ÿ 59.7Â°

15

4-5 Relative Motion in One and Two Dimensions (P. 189)

Relative Motion in 1D: Take the example of the person sitting in a train moving

east. If we choose east as the positive direction and Earth as the reference frame,

then we can write the velocity of the train with respect to the Earth as ð’— #$ . Letâ€™s

now say the person gets up out of her seat and walks toward the back of the train

at a velocity with respect to the train as ð’—%# We can add the two velocity vectors

to find the velocity of the person with respect to Earth. This relative velocity is

written as

ð’—!” = ð’—!# + ð’— #”

16

Section 3.1: Position, displacement, and average velocity | PP.106

Ex.6: While you are sitting on a train travelling at 80 km/h, someone

walks down the aisle at a speed of 5 km/h. What is the vPE?

Solution: vPE: Speed of Person with respect to the Earth ?

Â§ This 5 km/h is the speed of the person with respect to the train as frame

of reference.

ð’— =ð’— +ð’—

!”

!#

#”

speed of the Person with respect to the Train + speed of the Train with respect to the Earth

ð’—%$ = 5 km/h + 80 km/h = 85 km/h

It is important to specify the frame of reference when starting a speed.

17

Ex.7: In the figure; the velocity of the river flow relative to Earth is 5 km/h

due east, and the velocity of the boat relative to the water is 10 km/h due

north. What is the velocity of the boat relative to Earth?

Solution :

velocity of the river flow relative to Earth is ð’—/0 = 5 ðš¤Ì‚ ð‘˜ð‘š/â„Ž

velocity of the boat relative to the water is ð’—1/ = 10 ðš¥Ì‚ ð‘˜ð‘š/â„Ž

velocity of the boat relative to Earth is ð’—10 = ? ?

ð’—10 = ð’—/0 + ð’—1/ = ( 5 ðš¤Ì‚ + 10 ðš¥Ì‚ ) ð‘˜ð‘š/â„Ž

ð‘£10 = ð‘£/0 * + ð‘£1/ * =

= 11.2 ð‘˜ð‘š/â„Ž

(5)* +(10)*

3

ðœƒ = tan-. 22$% = tan-. .!

= 26.56Â°

&$

18

Ex.8: A bullet is fired horizontally with an initial velocity of 900 m/s at a

target located 150 m from the rifle.

A)How much time is required for the bullet to reach the target?

B)How far does the bullet fall in this time?

a) Î”ð‘¥ = ð‘£'” ð‘¡

150 = 900 ð‘¥ ð‘¡

.

ð‘¡ = 0.167ð‘

.

b) y=y0+Vot– ð‘”ð‘¡ * = 0+0– 9,8(0.167)* = -0.139 m (minus because target is

*

*

below y0)

Answers: a) ð‘¡ = 0.167 ð‘

, b) âˆ†ð‘¦ = 0.14 ð‘š

19

Summary

Â§ If an object is launched with

the horizontal, 2 factors would

affect projectile motion:

Â§ Angle ðœƒ0

Â§ Initial velocity

20

Friday, March 18, 2022

â€¢ https://phet.colorado.edu/en/simulation/legacy/projectile-motion

Physics Group – Basic Sciences Department

Simulations (Animations)

21

2ND SEMESTER

Deanship of Preparatory Year

and Supporting Studies

ENGINEERING TRACK

PHYSICS 132

Week 10 Tutorial 8

2021-2022

Learning Objectives:

A. Identify position vectors in 2D.

B. Use the unit vectors in 2D to present vector quantities.

C. Understand the rules of vector addition and subtraction in 2D

using analytical methods.

D. Apply analytical methods to determine vertical and horizontal

component vectors.

E. Apply analytical methods to determine the magnitude and

direction of a resultant vector.

F. Determine the dot and cross products of two vectors.

2

Ex.1: Two vectors are given: ð‘Žâƒ— = 4ðš¤Ì‚ + ðš¥Ì‚ , ð‘ = âˆ’12ðš¤Ì‚ + 5ðš¥Ì‚

a) ð‘Žâƒ— + ð‘ = ?

b) What is the magnitude of ð‘Žâƒ— + ð‘ ?

c) What is the direction of ð‘Žâƒ— + ð‘ ?

Solution

a) ð‘Žâƒ— + ð‘ = âˆ’12 + 4 ðš¤Ì‚ + 5 + 1 ðš¥Ì‚ = âˆ’8ðš¤Ì‚ + 6ðš¥Ì‚

b) |ð‘Ž + ð‘|! = (âˆ’8)!+(6)!

ð‘Ž + ð‘ = 64 + 36 = 10

$

“#

c) ðœƒ = ð‘¡ð‘Žð‘› “% = 128Â°

3

Ex.2 (Page74): Find the magnitude of the vector ð‘ª that satisfies the

#>

>

equation 2ð‘¨ âˆ’ 6ð‘© + 3ð‘ª = ðŸðš¥Ì‚ , where ð‘¨ = ðš¤Ì‚ âˆ’ 2ð‘˜ and ð‘© = âˆ’ðš¥Ì‚ + ! ð‘˜.

Solution

2ð‘¨ âˆ’ 6ð‘© + 3ð‘ª = ðŸðš¥Ì‚

3ð‘ª = ðŸðš¥Ì‚ âˆ’ 2ð‘¨ + 6ð‘©

!

!

ð‘ª = & ðš¥Ì‚ âˆ’ & ð‘¨ + 2ð‘©

!

!

> + 2(âˆ’ðš¥Ì‚ + # ð‘˜)

>

ð‘ª = ðš¥Ì‚ âˆ’ (ðš¤Ì‚ âˆ’ 2ð‘˜)

&

&

!

!

‘>

ð‘ª = & ðš¥Ì‚ âˆ’ & ðš¤Ì‚ + & ð‘˜ âˆ’ 2ðš¥Ì‚ + ð‘˜>

!

‘

(>

ð‘ª = âˆ’ & ðš¤Ì‚ + & ðš¥ Ì‚ + & ð‘˜

!

4

Ex.3: Four vectors, ð‘¨, ð‘© , ð‘ª, and ð‘«, are shown in the figure. The sum of

these four vectors is a vector having magnitude and direction is:

A. 4.0 cm, along + ð‘¥ âˆ’ ð‘Žð‘¥ð‘–ð‘

B. 4.0 cm, along âˆ’ ð‘¦ âˆ’ ð‘Žð‘¥ð‘–ð‘

C. 4.0 cm, 45Â° above + ð‘¥ âˆ’ ð‘Žð‘¥ð‘–ð‘

D. 4.0 cm, 45Â° below + ð‘¥ âˆ’ ð‘Žð‘¥ð‘–ð‘

5

Ex.4: Vector ð´âƒ— has x and y components of 8 cm and 15 cm respectively,

vector ðµ has x and y components of 14 cm and -6 cm respectively .

If ð‘¨ âˆ’ ð‘© + ðŸ‘ð‘ª = 0, what are the components of ð‘ª ?

A)ð¶* = 2,

ð¶+ = âˆ’7

ðµ)ð¶* = âˆ’2,

ð¶+ = 7

C)ð¶* = âˆ’2,

ð¶+ = âˆ’7

D)C, = âˆ’7,

C- = 2

x

y

A 8

15

-B -14

6

+3C 3Cx 3Cy

0

0

8-14+ 3Cx=0

15+6+3Cy=0

Cx= 2

Cy= -7

6

Ex.5:

ðš¤Ì‚

ðš¥Ì‚ ð‘˜*

ð‘Žð‘¥ð‘

âƒ— = ð‘âƒ— = 3 âˆ’4 0

âˆ’2 0 3

(“#”)

=ðš¤(âˆ’1)

Ì‚

(-12-0) +

(“#%)

ðš¥(âˆ’1)

Ì‚

(9-0)+

ð‘˜( âˆ’1 “#& 0 âˆ’ 8

=ðš¤Ì‚ (-12)-ðš¥Ì‚ (9)+ð‘˜( (-8)= -12ðš¤-9

Ì‚ ðš¥-8

Ì‚ ð‘˜(

ð‘âƒ— = ð‘âƒ—! + ð‘âƒ—” + ð‘âƒ—#

7

Active Learning:

=3, 4 , cos 60 = 6

8

9

2ND SEMESTER

Deanship of Preparatory Year

and Supporting Studies

ENGINEERING TRACK

PHYSICS 132

WEEK 9 â€“ TUTORIAL 8

2020-2021

Learning Objectives:

A. Understand projectile motion in the earthâ€™s gravity.

B. Identify the terms and properties of a projectile, such as

acceleration due to gravity, range, maximum height, and

trajectory

C. Analyze the location and velocity of a projectile in 2D at

different points of its trajectory.

D. Solve problems and exercises for projectiles by using the

basic kinematics equations of motion in 2D.

E. Apply principles of vector addition to determine relative

velocity.

2

Physics Group – Basic Sciences Department

Tuesday, August 2, 2022

4-3 Projectile Motion

3

1

1

âˆ†ð‘¦ = ð‘£ð‘¦0 ð‘¡ + ð‘Žð‘¦ ð‘¡ 2 = âˆ’ð‘” ð‘¡ 2

2

2

1

âˆ†ð‘¥ = ð‘£ð‘¥0 ð‘¡ + ð‘Žð‘¥ ð‘¡ 2 = ð‘£ð‘¥0 ð‘¡ + 0

2

â‡›t=

2âˆ†ð‘¦

=

âˆ’ð‘”

â‡› ð‘£ð‘¥0 =

2 Ã— (âˆ’50.0)

= 3.19ð‘

âˆ’9.80

Tuesday, August 2, 2022

Solution :

Physics Group – Basic Sciences Department

Example: A movie stunt driver on a motorcycle speeds horizontally off a 50.0 m

high cliff. How fast must the motorcycle leave the cliff top to land on level

ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air

resistance.

âˆ†ð‘¥ 90.0

=

= 28.2 ð‘šÎ¤ð‘

ð‘¡

3.19

4

Solution :

ð‘£ð‘¥0 = ð‘£0 ð‘ð‘œð‘ ðœƒ = 30 ð‘ð‘œð‘ 45Â° = 21.2 ð‘š/ð‘

ð‘£ð‘¦0 = ð‘£0 ð‘ ð‘–ð‘›ðœƒ = 30 ð‘ ð‘–ð‘›45Â° = 21.2 ð‘š/ð‘

1

(a) âˆ†ð‘¦ = ð‘£ð‘¦0 ð‘¡ âˆ’ ð‘”ð‘¡ 2

2

1

+10 = 21.2ð‘¡ âˆ’ 9.8 ð‘¡ 2 Öœ

2

4.9ð‘¡ 2 âˆ’ 21.2ð‘¡ + 10 = 0

(b) ð‘£ð‘¦ = ð‘£ð‘¦0 âˆ’ ð‘”ð‘¡ = 21.2 âˆ’ 9.8 3.79 = âˆ’15.9 ð‘š/ð‘

ð‘£=

ð‘£ð‘¥2 + ð‘£ð‘¦2 =

ðœƒð‘£ = tanâˆ’1

ð‘£ð‘¦

ð‘£ð‘¥

(21.2)2 +(âˆ’15.9)2 = 26.5 ð‘š/ð‘

= tanâˆ’1 âˆ’15.9

= âˆ’36.9Â° = 323.1Â°

21.2

Öœ ð‘¡1 = 0.54ð‘ , ð‘¡2 = 3.79ð‘

Solution :

Tuesday, August 2, 2022

Example: A kicked football leaves the ground at an angle ðœ½ðŸŽ = ðŸ‘ðŸ•. ðŸŽÂ° with a

velocity of 20.0 m/s . Calculate (a) the maximum height, (b) the time of travel

before the football hits the ground, and (c) how far it hits the ground. Assume the

ball leaves the foot at ground level, and ignore air resistance and rotation of the

ball.

2

2

= ð‘£ð‘¦ð‘œ

âˆ’ 2ð‘”âˆ†ð‘¦

(a) ð‘£ð‘¦ð‘“

0 = 122 âˆ’ 2 9.8 âˆ†ð‘¦

Öœ âˆ†ð‘¦ = 7.35 m

(b) ð‘£ð‘¦ = ð‘£ð‘¦0 âˆ’ ð‘”ð‘¡

âˆ’12 = 12 âˆ’ 9.8ð‘¡ Öœ ð‘¡ = 12+12

9.8 = 2.45 ð‘

(c) ð‘¥ = ð‘£ð‘¥ð‘œ ð‘¡ = 16 2.45 = 39.2 ð‘š

Physics Group – Basic Sciences Department

ð‘£ð‘¥ð‘œ = ð‘£ð‘œ ð‘ð‘œð‘ ðœƒ = 20 ð‘ð‘œð‘ 37Â° = 16 ð‘š/ð‘

ð‘£ð‘¦ð‘œ = ð‘£ð‘œ ð‘ ð‘–ð‘›ðœƒ = 20 ð‘ ð‘–ð‘›37Â° = 12 ð‘š/ð‘

6

A) What is the vertical component of the ballâ€™s velocity just before it hits the floor?

B) What is the horizontal component of the ballâ€™s velocity just before it hits the floor?

C) How high above the floor is the tabletop?

D) At what horizontal distance from the table the ball land?â€™

Answers: A) 6 m/s downward

, B) 5 m/s Right

, C) 1.8 m

D) 3 m

Physics Group – Basic Sciences Department

A ball rolls off a table with horizontal velocity of 5m/s if it takes 0.6 seconds for it to

reach the floor.

Tuesday, August 2, 2022

Active Learning: inside the Class

7

Tuesday, August 2, 2022

Solution :

Physics Group – Basic Sciences Department

Example:

8

Ex. 3: A ball is thrown at an original speed of 8.0 m/s at an angle of 35Â°

above the horizontal. If there is no air resistance, what is the speed of the

ball when it returns to the same horizontal level?

A) 4.0 m/s

C) 16 m/s

B

D) 9.8 m/s

Ex. 4: A boy jumps with a velocity of magnitude 20.0 m/s at an angle of

25.0Â° above the horizontal. What is the horizontal component of the

boy’s velocity?

A) 18.1 m/s

B) 15.6 m/s

C) 8.45 m/s

D) 12.6 m/s

Physics Group – Basic Sciences Department

B) 8.0 m/s

Tuesday, August 2, 2022

Exercises, and Synthesis Problems

A

9

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