Motion in two dimensions 28 03 2015

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

THE PATH TAKEN BY THE BALL
4. 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  2
1
)(222
oxoxx xxavv 
2
2
1
tatvxx xoxo 

3.2 EQUATIONS OF KINEMATICS VERTICALLY (ALONG THE Y)
gtvv oyy 
2
2
1
gttvyy oyo 
 tvvyy yoyo  2
1
)(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 𝒚
are 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 𝒗 𝒙 = 𝒗 𝒐𝒙
because the acceleration 𝒂 𝒙 = 𝟎.
The magnitude of the final velocity is given by 𝒗 = (𝒗 𝒙) 𝟐+(𝒗 𝒚) 𝟐 and the
direction is given by 𝜽 = 𝒕𝒂𝒏−𝟏
(
𝒗 𝒚
𝒗 𝒙
)
The magnitude of velocity along the y changes because of the acceleration 𝒂 =
− 𝒈 𝐢𝐟 𝐮𝐩 > 𝟎

Note:
The magnitude of velocity along the 𝒙
doesn’t change and is given by
𝒗 𝒐𝒙 = 𝑣 𝑜 𝑐𝑜𝑠𝜃0 because the acceleration
𝒂 𝒙 = 𝟎.
Hence:
𝒗 𝒙 = 𝒗 𝒐𝒙 + 𝒂 𝒙 𝒕 = 𝑣 𝑜 𝑐𝑜𝑠𝜃0
Note:
The velocity along the y
changes due to the gravitational
acceleration g
𝒗 𝒐𝒚 = 𝑣 𝑜 𝑠𝑖𝑛𝜃0
because the acceleration 𝒂 𝒚 =
𝒈 = −𝟗. 𝟖
𝐦
𝒔 𝟐 𝐢𝐟 𝐮𝐩 > 𝟎
Hence the final velocity along
y is
𝒗 𝒚 = 𝒗 𝒐𝒚 + 𝒈𝒕 = 𝒗 𝒐 𝒔𝒊𝒏𝜽 𝟎 + 𝒈𝒕

THE MAXIMUM HEIGHT H AND THE RANGE R
(1) Maximum Height H:
Data: set 𝑎𝑛𝑑 𝒚 𝟎 = 𝟎 𝒎 𝑎𝑛𝑑 𝑦 = 𝐻
Then 𝒗 𝒐𝒙 = 𝑣 𝑜 𝑐𝑜𝑠𝜃0and 𝒗 𝒐𝒚 = 𝑣 𝑜 𝑠𝑖𝑛𝜃0
From the diagram , at maximum height
𝑦 = 𝐻 𝒗 𝒚 = 𝟎 𝒎/𝒔
Hence the maximum height can be
found using the equation:
𝑣 𝑦
2 = 𝑣 𝑜𝑦
2 + 2𝑔(𝑦 − 𝑦0)
𝑦 − 𝑦0 =
𝑣 𝑦
2−𝑣 𝑜𝑦
2
2𝑔
Substituting:
𝐻 − 0 =
02−(𝑣 𝑜 𝑠𝑖𝑛𝜃0)2
2(−𝑔)
=
𝑣0
2 𝑠𝑖𝑛2 𝜃0
2𝑔
The maximum height H 𝐻 =
𝑣0
2 𝑠𝑖𝑛2 𝜃0
2𝑔
(2)The time to reach the maximum
height 𝑣 𝑦 = 𝑣 𝑜𝑦 + 𝑔𝑡
𝑡 =
−𝑣 𝑜𝑦
𝑔
=
−𝑣 𝑜 𝑠𝑖𝑛𝜃0
−𝑔
=
𝑣 𝑜 𝑠𝑖𝑛𝜃0
𝑔
(3):Range R: Horizontal distance
Data set: 𝒙 𝟎 = 𝟎𝒎 and 𝒙 = 𝑹
The range R can be given by
:𝑥 − 𝑥0 = 𝑣 𝑜𝑥 𝑡 +
1
2
𝒂 𝒙 𝑡2 = 𝑣 𝑜𝑥 𝑡 since 𝒂 𝒙 = 𝟎
Substituting
R−0 = 𝑣 𝑜𝑥 𝑡, hence R= 𝑣 𝑜𝑥 𝑡 = 𝑣 𝑜 𝑐𝑜𝑠𝜃0t
But due to symmetry 𝒕𝒊𝒎𝒆 𝒖𝒑 =
𝒕𝒊𝒎𝒆 𝒅𝒐𝒘𝒏
Hence the total time is 𝑡 =
2𝑣 𝑜 𝑠𝑖𝑛𝜃0
𝑔
The range R= 𝑣 𝑜 𝑐𝑜𝑠𝜃0
2𝑣 𝑜 𝑠𝑖𝑛𝜃0
𝑔
=
2𝑣0
2 𝑐𝑜𝑠𝜃0 𝑠𝑖𝑛𝜃0
𝑔
Or 𝑹 =
𝒗 𝟎
𝟐 𝒔𝒊𝒏𝟐𝜽 𝟎
𝒈

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 𝜽 𝟎 = 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, 𝑣 𝑦 = 0 𝑚/𝑠 and 𝑣 𝑥 = 𝑣0𝑥 = 20𝑠𝑖𝑛37° = 12 𝑚/𝑠
(a) For maximum height: 𝑣 𝑦
2
= 𝑣 𝑜𝑦
2
+ 2𝑔(𝑦 − 𝑦0) hence 𝑦 − 𝑦0 =
𝑣 𝑦
2−𝑣 𝑜𝑦
2
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 = 𝑣0𝑦 𝑡 + 1
2
𝑔𝑡2
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 = 𝑣0𝑥 𝑡
𝑥 − 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

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

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 sThe 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.
 21
0 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

Motion in two dimensions 28 03 2015

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Motion in two dimensions 28 03 2015