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Tutorial 1_Physics 132_Summer 2022
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’—ðŸŽ
ð’—ðŸŽð’š
Î¸
ð’—ðŸŽð’™
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 ðœƒ = 100Ã—ð‘ð‘œð‘ 30Â° = 87 m/s
ð‘£!# = ð‘£’ sin ðœƒ = 100Ã—ð‘ ð‘–ð‘›30Â° = 50 m/s
ð’—ðŸŽ
ð’—ðŸŽð’š = ðŸ“ðŸŽ ð’Ž/ð’”
Î˜= 30 Â°
ð’—ðŸŽð’™ = ðŸ–ðŸ•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’—ðŸŽ
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 ðœƒ = 100Ã—ð‘ð‘œð‘ 30Â° = 87 m/s
ð‘£’# = ð‘£’ sin ðœƒ = 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(2ðœƒ, ) = 1 ð‘†ð‘œ 2ðœƒ, = 90 Â°
ðœƒ, = 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(2ðœƒ, )
ð‘…=
g
ð‘…g (320)(9.8)
sin(2ðœƒ, ) = * =
= 0.871
*
(60)
ð‘£,
2ðœƒ, = sin-. 0.871 = 60.6Â° ð‘œð‘Ÿ 119.4Â°
ðœƒ, = 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 ðš¤Ì‚ ð‘˜ð‘š/â„Ž
velocity of the boat relative to the water is ð’—1/ = 10 ðš¥Ì‚ ð‘˜ð‘š/â„Ž
velocity of the boat relative to Earth is ð’—10 = ? ?
ð’—10 = ð’—/0 + ð’—1/ = ( 5 ðš¤Ì‚ + 10 ðš¥Ì‚ ) ð‘˜ð‘š/â„Ž
ð‘£10 = ð‘£/0 * + ð‘£1/ * =
= 11.2 ð‘˜ð‘š/â„Ž
(5)* +(10)*
3
ðœƒ = 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 ðœƒ0
Â§ Initial velocity
20
Friday, March 18, 2022
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— = 4ðš¤Ì‚ + ðš¥Ì‚ , ð‘ = âˆ’12ðš¤Ì‚ + 5ðš¥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 ðš¤Ì‚ + 5 + 1 ðš¥Ì‚ = âˆ’8ðš¤Ì‚ + 6ðš¥Ì‚
b) |ð‘Ž + ð‘|! = (âˆ’8)!+(6)!
ð‘Ž + ð‘ = 64 + 36 = 10
\$
“#
c) ðœƒ = 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 = ðŸðš¥Ì‚ , where ð‘¨ = ðš¤Ì‚ âˆ’ 2ð‘˜ and ð‘© = âˆ’ðš¥Ì‚ + ! ð‘˜.
Solution
2ð‘¨ âˆ’ 6ð‘© + 3ð‘ª = ðŸðš¥Ì‚
3ð‘ª = ðŸðš¥Ì‚ âˆ’ 2ð‘¨ + 6ð‘©
!
!
ð‘ª = & ðš¥Ì‚ âˆ’ & ð‘¨ + 2ð‘©
!
!
> + 2(âˆ’ðš¥Ì‚ + # ð‘˜)
>
ð‘ª = ðš¥Ì‚ âˆ’ (ðš¤Ì‚ âˆ’ 2ð‘˜)
&
&
!
!
‘>
ð‘ª = & ðš¥Ì‚ âˆ’ & ðš¤Ì‚ + & ð‘˜ âˆ’ 2ðš¥Ì‚ + ð‘˜>
!

(>
ð‘ª = âˆ’ & ðš¤Ì‚ + & ðš¥ 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 ðµ has x and y components of 14 cm and -6 cm respectively .
If ð‘¨ âˆ’ ð‘© + ðŸ‘ð‘ª = 0, what are the components of ð‘ª ?
A)ð¶* = 2,
ð¶+ = âˆ’7
ðµ)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:
ðš¤Ì‚
ðš¥Ì‚ ð‘˜*
ð‘Žð‘¥ð‘
âƒ— = ð‘âƒ— = 3 âˆ’4 0
âˆ’2 0 3
(“#”)
=ðš¤(âˆ’1)
Ì‚
(-12-0) +
(“#%)
ðš¥(âˆ’1)
Ì‚
(9-0)+
ð‘˜( âˆ’1 “#& 0 âˆ’ 8
=ðš¤Ì‚ (-12)-ðš¥Ì‚ (9)+ð‘˜( (-8)= -12ðš¤-9
Ì‚ ðš¥-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‘ ðœƒ = 30 ð‘ð‘œð‘ 45Â° = 21.2 ð‘š/ð‘
ð‘£ð‘¦0 = ð‘£0 ð‘ ð‘–ð‘›ðœƒ = 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 =
ðœƒð‘£ = 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 ðœ½ðŸŽ = ðŸ‘ðŸ•. ðŸŽÂ° 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‘ ðœƒ = 20 ð‘ð‘œð‘ 37Â° = 16 ð‘š/ð‘
ð‘£ð‘¦ð‘œ = ð‘£ð‘œ ð‘ ð‘–ð‘›ðœƒ = 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€™
, 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