Physics Malaysian Matriculation Semester 1 Notes Complete

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PHYSICS

CHAPTER 2 CHAPTER 2:

CHAPTER 2:

Kinematics of linear motion

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PHYSICS

CHAPTER 2 Kinematics of linear motion

Kinematics of linear motion

2.1 Linear Motion

2.2 Uniformly Accelerated Motion

2.3 Free Falling Body

2.4 Projectile Motion

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PHYSICS

CHAPTER 2

At the end of this chapter, students should be able to:

At the end of this chapter, students should be able to:  Define_Define and distinguish between_{and distinguish between}

i.

i. distance and displacement distance and displacement

ii.

ii. speed and velocityspeed and velocity

iii.

iii. instantaneous velocity, average velocity and uniform instantaneous velocity, average velocity and uniform

velocity. velocity.

iv.

iv. instantaneous acceleration, average acceleration and instantaneous acceleration, average acceleration and

uniform acceleration. uniform acceleration.

 Sketch_Sketch graphs of displacement-time, velocity-time and _{graphs of displacement-time, velocity-time and} acceleration-time.

acceleration-time.

 Determine_Determine the distance travelled, displacement, velocity _{the distance travelled, displacement, velocity}

Learning Outcome:

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PHYSICS

CHAPTER 2

4

2.1. Linear motion (1-D)

2.1.1. Distance,

d

 scalar quantity.

 is defined as the length of actual path between two pointslength of actual path between two points.  For example :

 The length of the path from P to Q is 25 cm.

P

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PHYSICS

CHAPTER 2

 vector quantity

 is defined as the distance between initial point and final the distance between initial point and final point in a straight line

point in a straight line.

 The S.I. unit of displacement is metre (m).

Example 1:

An object P moves 20 m to the east after that 10 m to the south and finally moves 30 m to west. Determine the displacement of P relative to the original position.

Solution : Solution :

2.1.2 Displacement,

s



N E W O P θ θ 20 m 10 m 10 m 20 m

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PHYSICS

CHAPTER 2

6

The magnitude of the displacement is given by

and its direction is

2.1.3 Speed,

v

 is defined the rate of change of distancerate of change of distance.  scalar quantity.  Equation:

interval

time

distance

of

change

speed

=

Δt

Δd

v

=

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PHYSICS

CHAPTER 2

 is a vector quantity.

 The S.I. unit for velocity is m s-1_.

Average velocity, Average velocity,

v

_av_av

 is defined as the rate of change of displacementthe rate of change of displacement.  Equation:

2.1.4 Velocity,

v



interval

time

nt

displaceme

of

change

=

av

v

Δt

Δs

v

_av

=

1 2 1 2 av

t

s

v

−

=

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PHYSICS

CHAPTER 2

8

Instantaneous velocity, Instantaneous velocity,

v

 is defined as the instantaneous rate of change of the instantaneous rate of change of displacement

displacement.

 Equation:

 An object is moving in uniform velocitymoving in uniform velocity if

t

s

0 t

v

∆

→

∆

=

limit

constant

=

dt

ds

dt

ds

v

=

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PHYSICS

CHAPTER 2

 Therefore Q

s

t

0

s

₁

t

₁

The gradient of the tangent to the curve at point Q = the instantaneous velocity at time, t = t₁

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PHYSICS

CHAPTER 2

10

 vector quantity

 The S.I. unit for acceleration is m s-2_.

Average acceleration, Average acceleration,

a

_av_av

 is defined as the rate of change of velocitythe rate of change of velocity.  Equation:

 Its direction is in the same direction of motionsame direction of motion.

 The accelerationacceleration of an object is uniformuniform when the magnitude magnitude of velocity changes at a constant rate and along fixed

of velocity changes at a constant rate and along fixed

direction. direction.

2.1.5 Acceleration,

a



interval

time

velocity

of

change

=

av

a

1 2 1 2 av

t

v

a

−

=

Δt

Δv

a

_av

=

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PHYSICS

CHAPTER 2

Instantaneous acceleration,

a

 is defined as the instantaneous rate of change of velocityinstantaneous rate of change of velocity.  Equation:

 An object is moving in uniform acceleration moving in uniform acceleration if

t

v

0 t

a

∆

→

∆

=

limit

constant

=

dt

dv

2 2

dt

s

d

dt

dv

a

=

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PHYSICS

CHAPTER 2

12

Deceleration, Deceleration,

a

 is a negative accelerationnegative acceleration.

 The object is slowing downslowing down meaning the speed of the object speed of the object decreases with time

decreases with time.

 Therefore

v

t

Q 0

v

₁

t

₁

The gradient of the tangent to the curve at point Q = the instantaneous acceleration at time, t = t₁

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PHYSICS

CHAPTER 2

Displacement against time graph ( Displacement against time graph (

s-t

))

2.1.6 Graphical methods

s

t

0

s

t

0

(a) Uniform velocity _{(b) The velocity increases with time} Gradient = constant Gradient increases with time (c)

s

Q R P The direction of Gradient at point R is negative.

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PHYSICS

CHAPTER 2

14

Velocity versus time graph ( Velocity versus time graph (

v-t

))

 The gradient at point A is positive – a > 0(speeding up)  The gradient at point B is zero – a= 0

 The gradient at point C is negative – a < 0(slowing down)

t₁ t₂

v

t

0 _(a) _t 2 t₁

v

t

0 (b) t1 t₂

v

t

0 (c) Uniform velocity Uniform acceleration

Area under the v-t graph = displacement

B

C

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PHYSICS

CHAPTER 2

 From the equation of instantaneous velocity,

Therefore

dt

ds

v

=

∫

ds

=

vdt

∫

=

2 1

t

vdt

s

graph

under the

area

ded

sha

v

t

s

=

−

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PHYSICS

CHAPTER 2

16

A toy train moves slowly along a straight track according to the displacement,

s

against time,

t

graph in figure 2.1.

a. Explain qualitatively the motion of the toy train. b. Sketch a velocity (cm s-1) against time (s) graph.

c. Determine the average velocity for the whole journey. d. Calculate the instantaneous velocity at

t

= 12 s.

Example 2 :

0 ₂ ₄ ₆ ₈ _{10 12 14}

_{t (s)}

2 4 6 8 10

s (cm)

Figure 2.1 Figure 2.1

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PHYSICS

CHAPTER 2

Solution : Solution : 0 to 6 s : 6 to 10 s : 10 to 14 s : b. 0.68 1.50

v (cm s

−1

₎

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PHYSICS

CHAPTER 2

18 Solution : Solution : c. d. 1 2 1 2

t

s

v

_av

−

=

s

14 to

s

10 from

velocity

average

=

v

1 2 1 2

t

s

v

−

=

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PHYSICS

CHAPTER 2

A velocity-time (

v

-

t

) graph in figure 2.2 shows the motion of a lift.

a. Describe qualitatively the motion of the lift.

b. Sketch a graph of acceleration (m s-1_{) against time (s).}

c. Determine the total distance travelled by the lift and its

Example 3 :

0 5 10 15 20 25 30 35

t (s)

-4 -2 2 4

v (m s

−1

)

Figure 2.2 Figure 2.2 40 45 50

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PHYSICS

CHAPTER 2

20

a. 0 to 5 s : Lift moves upward from rest with acceleration of 0.4 m s−2_.

5 to 15 s : The velocity of the lift from 2 m s−1 to

4 m s−1_{but the acceleration} _{to 0.2 m s}−2_.

15 to 20 s : Lift 20 to 25 s : Lift 25 to 30 s : Lift 30 to 35 s : Lift moves 35 to 40 s : Lift moving 40 to 50 s :

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PHYSICS

CHAPTER 2

Solution : Solution : b.

t (s)

5 10 15 20 25 30 35 40 45 50 0 -0.4 -0.2 0.2 0.6

a (m s

−2

₎

-0.6 -0.8 0.8 0.4

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PHYSICS

CHAPTER 2

22 Solution : Solution : c. i. 0 5 10 15 20 25 30 35

t (s)

-4 -2 2 4

v (m s

−1

₎

40 45 50 A₁ A2 A3 A₄ A₅

v-t

of

graph

under the

area

distance

Total

=

5 4 3 2 1

A

+

=

( )( )

(

)( )

(

)( )

( )( )

(

15

5 )( )

4

2

1

4

5

2

1

4

10

5

2

1

10

4

2

1

5

2

1 distance

Total

=

+

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PHYSICS

CHAPTER 2

Solution : Solution : c. ii. d.

v-t

of

graph

under the

area

nt

Displaceme

=

5 4 3 2 1

A

+

=

( )( )

(

)( )

(

)( )

( )( )

(

15 5

)( )

4 2 1 4 5 2 1 4 10 5 2 1 10 4 2 2 1 5 2 2 1 nt Displaceme = + + + + + − + + − 1 2 1 2

t

v

a

_av

−

=

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PHYSICS

CHAPTER 2

24

Figure 2.3

1. Figure 2.3 shows a velocity versus time graph for an object constrained to move along a line. The positive direction is to the right.

a. Describe the motion of the object in 10 s.

b. Sketch a graph of acceleration (m s-2_{) against time (s) for}

the whole journey.

c. Calculate the displacement of the object in 10 s. ANS. : 6 m

ANS. : 6 m

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PHYSICS

CHAPTER 2

2. A train pulls out of a station and accelerates steadily for 20 s until its velocity reaches 8 m s−1_{. It then travels at a constant}

velocity for 100 s, then it decelerates steadily to rest in a further time of 30 s.

a. Sketch a velocity-time graph for the journey.

b. Calculate the acceleration and the distance travelled in each part of the journey.

c. Calculate the average velocity for the journey.

Physics For Advanced Level, 4th_{edition, Jim Breithaupt, Nelson}

Thornes, pg.15, no. 1.11 ANS. : 0.4 m s ANS. : 0.4 m s−−22,0 m s_{,0 m s}−₋22,-0.267 m s_{,-0.267 m s}−₋22, 80 m, 800 m, 120 m; _{, 80 m, 800 m, 120 m;} 6.67 m s6.67 m s−−11_._.

Exercise 2.1 :

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PHYSICS

CHAPTER 2

26

 Derive and applyDerive and apply equations of motion with uniform equations of motion with uniform acceleration:

acceleration:

Learning Outcome:

2.2 Uniformly accelerated motion

at

u

v

=

+

2

1 at

ut

s

=

+

as

u

v

2

=

2

+

2

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PHYSICS

CHAPTER 2 2.2. Uniformly accelerated motion

 From the definition of average acceleration, uniform (constantconstant)

acceleration is given by

where

v

: final velocity

u

: initial velocity

a

: uniform (constant) acceleration

t

: time

at

u

v

=

+

(1)

t

u

v

a

=

−

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PHYSICS

CHAPTER 2

28

 From equation (1), the velocity-time graph is shown in figure

2.4:

 From the graph,

The displacement after time,

s

= shaded area under the graph

= the area of trapezium

 Hence,

velocity

0 v

u

time

t

(

u

v

)

t

2

1 s

=

+

(2)

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PHYSICS

CHAPTER 2

 By substituting eq. (1) into eq. (2) thus

 From eq. (1),  From eq. (2),

(

)

[

u

at

]

t

s

=

+

2

1

(3) 2

2

1 at

ut

s

=

+

(

v

−

u

)

=

at

(

)

t

s

u

v

+

=

2

multiply

(

)(

)

( )

at

t

s

u

v

u

v

_











=

−

+

2 as

u

v

2

=

2

+

2

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PHYSICS

CHAPTER 2

30

 Notes:

 equations (1) – (4) can be used if the motion in a straight motion in a straight

line with constant acceleration. line with constant acceleration.

 _{For a body moving at constant velocity, (}₍

a

_{= 0)}_{= 0)}_the

equations (1) and (4) become

Therefore the equations (2) and (3) can be written as

u

v

=

vt

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PHYSICS

CHAPTER 2

A plane on a runway takes 16.2 s over a distance of 1200 m to take off from rest. Assuming constant acceleration during take off, calculate

a. the speed on leaving the ground, b. the acceleration during take off.

Solution : Solution : a. Use

Example 4 :

s

2 .

16 =

t

?

=

v

(

u

v

)

t

s

=

+

2

1

0 =

u

m

1200

=

s

?

=

a



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PHYSICS

CHAPTER 2

32

b. By using the equation of linear motion,

as

u

v

2

=

2

+

2

OR OR 2

2

1 at

ut

s

=

+

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PHYSICS

CHAPTER 2

A bus travelling steadily at 30 m s−1_{along a straight road passes a}

stationary car which, 5 s later, begins to move with a uniform

acceleration of 2 m s−2_{in the same direction as the bus. Determine}

a. the time taken for the car to acquire the same velocity as the bus,

b. the distance travelled by the car when it is level with the bus. Solution : Solution : a. Given Use

Example 5 :

2 1

_constant

_;

_0;

₂

_ms

s

m

30

−

=

−

=

_c _c b

u

a

v

c c c c

u

a

t

v

=

+

1

s

m

30

−

=

_b c

v

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PHYSICS

CHAPTER 2

34

b.

From the diagram,

c b 1

s

m

30

−

=

b

v

0 =

c

u

s

0 =

b

t

_b

=

5 s

2

s

m

2

−

=

c

a

b

v

b b c b

v

t

_b

=

b c

s

=

b c

s

=

b b c c c c

t

a

t

v

t

u

+

2

=

2

1

Therefore

t

v

s

_c

=

_b

;

t

_b

=

t

_c

=

t

−

5

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PHYSICS

CHAPTER 2

A particle moves along horizontal line according to the equation Where

s

is displacement in meters and

t

is time in seconds.

At time,

t

=2.00 s, determine

a. the displacement of the particle, b. Its velocity, and

c. Its acceleration. Solution : Solution : a.

t

=2.00 s ;

Example 6 :

t

s

=

3

−

4

2

+

2 t

t

s

=

3

−

4

2

+

2

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PHYSICS

CHAPTER 2

36 Solution : Solution : b. Instantaneous velocity at

t

= 2.00 s, Use Thus

dt

ds

v

=

(

t

)

dt

d

v

=

3

−

4

2

+

2 (

2.00 )

8 (

2.00 )

2

9

2

−

+

=

v

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PHYSICS

CHAPTER 2

Solution : Solution : c. Instantaneous acceleration at

t

= 2.00 s, Use Hence

dt

dv

a

=

(

2.00 )

8

18 −

=

a

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PHYSICS

CHAPTER 2

38

1. A speedboat moving at 30.0 m s-1_{approaches stationary}

buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of -3.50 m s-2_{by reducing the throttle.}

a. How long does it take the boat to reach the buoy?

b. What is the velocity of the boat when it reaches the buoy?

No. 23,pg. 51,Physics for scientists and engineers with modern physics, Serway & Jewett,6th_edition.

ANS. : 4.53 s; 14.1 m s

ANS. : 4.53 s; 14.1 m s−−11

2. An unmarked police car travelling a constant 95 km h-1 _is

passed by a speeder traveling 140 km h-1_{. Precisely 1.00 s}

after the speeder passes, the policemen steps on the accelerator; if the police car’s acceleration is 2.00 m s-2_{, how}

much time passes before the police car overtakes the speeder (assumed moving at constant speed)?

No. 44, pg. 41,Physics for scientists and engineers with modern physics, Douglas C. Giancoli,3rd_edition.

ANS. : 14.4 s

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PHYSICS

CHAPTER 2

3. A car traveling 90 km h-1_{is 100 m behind a truck traveling}

75 km h-1_{. Assuming both vehicles moving at constant}

velocity, calculate the time taken for the car to reach the truck.

No. 15, pg. 39,Physics for scientists and engineers with modern physics, Douglas C. Giancoli,3rd_edition.

ANS. : 24 s

4. A car driver, travelling in his car at a constant velocity of 8 m s-1_{, sees a dog walking across the road 30 m ahead. The}

driver’s reaction time is 0.2 s, and the brakes are capable of producing a deceleration of 1.2 m s-2_{. Calculate the distance}

from where the car stops to where the dog is crossing, assuming the driver reacts and brakes as quickly as possible.

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PHYSICS

CHAPTER 2

40

At the end of this chapter, students should be able to:  Describe and useDescribe and use equations for free falling body. equations for free falling body.

 For _Forupward and downward_{upward and downward} motion, use_{motion, use}

a

= = −₋

g

= = −₋9.81 m s9.81 m s−−22

Learning Outcome:

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PHYSICS

CHAPTER 2 2.3. Free falling body

 is defined as the vertical motion of a body at constant the vertical motion of a body at constant acceleration,

acceleration,

g

under gravitational field under gravitational field without air without air resistance

resistance..

 In the earth’s gravitational field, the constant acceleration

 known as acceleration due to gravityacceleration due to gravity or free-fall free-fall

acceleration

acceleration or gravitational accelerationgravitational acceleration.

 _{the value is}

g

_{= 9.81 m s}_{= 9.81 m s}−−22

 the direction is towards the centre of the earth towards the centre of the earth

(downward). (downward).  Note:

 In solving any problem involves freely falling bodies or free

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PHYSICS

CHAPTER 2

42

 Sign convention:

 Table 2.1 shows the equations of linear motion and freely

falling bodies.

Table 2.1

Linear motion Freely falling bodies

at

u

v

=

+

v

=

u

−

gt

as

2 u

v

2

=

2

+

v

2

=

u

2

−

2 gs

2

at

2

1 ut

s

=

+

gt

2

1 ut

s

=

−

+ - +

-From the sign convention thus,

g

a

=

−

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PHYSICS

CHAPTER 2

 An example of freely falling body is the motion of a ball thrown

vertically upwards with initial velocity,

u

as shown in figure 2.5.

 Assuming air resistance is negligible, the acceleration of the

ball,

a

= −

g

when the ball moves upward and its velocity velocity

H

u

v

velocity = 0 Figure 2.5 Figure 2.5

u

v

=

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PHYSICS

CHAPTER 2

44

 The graphs in figure 2.6 show

the motion of the ball moves up and down.

Derivation of equations Derivation of equations

 At the maximum height or

displacement,

H

where

t = t

₁, its velocity,

hence

therefore the time taken for the ball reaches

H

,

t

0 v

u

−

u

t

₁

2t

₁

t

0 a

−

g

t

₁

2t

₁

t

s

0 H

t

₁

_2t

₁

v =0

gt

u

v

=

−

1

gt

u

−

=

0

0 =

v

g

u

t

₁

=

Simulation 2.4

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PHYSICS

CHAPTER 2

 _{To calculate the maximum height or displacement,}

H

_:

use either

maximum height,

 Another form of freely falling bodies expressions are

2 1 1

gt

ut

s

2

1 −

=

gs

u

v

2

=

2

−

2

Where

s = H

gH

u

2

0 =

2

−

OR OR

g

u

H

2

=

gt

u

v

=

−

gs

u

v

2

=

2

−

2 gt

u

v

_y

=

_y

−

gs

u

v

2

=

2

−

2

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PHYSICS

CHAPTER 2

46

A ball is thrown from the top of a building is given an initial velocity of 10.0 m s−1_{straight upward. The building is 30.0 m high and the}

ball just misses the edge of the roof on its way down, as shown in figure 2.7. Calculate

a. the maximum height of the stone from point A. b. the time taken from point A to C.

c. the time taken from point A to D.

d. the velocity of the stone when it reaches point D. (Given

g

= 9.81 m s−2)

Example 7 :

A B C D

u

=10.0 m s−1 30.0 m Figure 2.7 Figure 2.7

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PHYSICS

CHAPTER 2

Solution :

a. At the maximum height,

H

,

v

_y= 0 and

u = u

_y= 10.0 m s−1_thus

b. From point A to C, the vertical displacement,

s

_y= 0 m thus

y 2 y 2 y

u

gs

v

=

−

2

2 y y

u

t

gt

s

2

1 −

=

A B C

u

30.0 m

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PHYSICS

CHAPTER 2

48

c. From point A to D, the vertical displacement,

s

_y= −30.0 m thus

By using 2 y y

u

t

gt

s

2

1 −

=

A B C D

u

30.0 m

2a

4ac

b

±

2

−

=

t

OR Time don’t Time don’t have have negative negative value. value.

a

b

c

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PHYSICS

CHAPTER 2

Solution :

d. Time taken from A to D is

t

= 3.69 s thus

From A to D,

s

_y = −30.0 m A B C

u

30.0 m

gt

u

v

_y

=

_y

−

(

10.0 ) (

−

9.81 )(

3.69 )

=

y

v

OR y 2 y 2 y

u

gs

v

=

−

2 (

10.0 )

−

2 (

9.81 )(

−

30.0 )

=

2 2 y

v

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PHYSICS

CHAPTER 2

50

A book is dropped 150 m from the ground. Determine a. the time taken for the book reaches the ground.

b. the velocity of the book when it reaches the ground. (given

g

= 9.81 m s-2)

a. The vertical displacement is

s

_y = −150 m Hence

Example 8 :

u

_y = 0 m s−1 150 m

m

150 −

=

y

s

2 y y

u

t

gt

s

2

1 −

=

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PHYSICS

CHAPTER 2

Solution :

b. The book’s velocity is given by

Therefore the book’s velocity is

gt

u

v

_y

=

_y

−

OR y 2 y 2 y

u

gs

v

=

−

2 m

150 −

=

y

s

0 =

y

u

?

=

y

v

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PHYSICS

CHAPTER 2

52

1. A ball is thrown directly downward, with an initial speed of 8.00 m s−1_{, from a height of 30.0 m. Calculate}

a. the time taken for the ball to strike the ground, b. the ball’s speed when it reaches the ground.

ANS. : 1.79 s; 25.6 m s

ANS. : 1.79 s; 25.6 m s−−11

2. A falling stone takes 0.30 s to travel past a window 2.2 m tall as shown in figure 2.8.

From what height above the top of the windows did the stone fall? ANS. : 1.75 m ANS. : 1.75 m

Exercise 2.3 :

m 2.2 Figure 2.8 Figure 2.8 to travel this distance took 0.30 s

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PHYSICS

CHAPTER 2

At the end of this chapter, students should be able to:  Describe and useDescribe and use equations for projectile, equations for projectile,

 Calculate_Calculate time of flight, maximum height, range, _{time of flight, maximum height, range,}

maximum range, instantaneous position and velocity. maximum range, instantaneous position and velocity.

Learning Outcome:

2.4 Projectile motion

θ

u

_x

=

cos

θ

u

_y

=

sin

0 =

x

a

g

a

_y

=

−

(54)

PHYSICS

CHAPTER 2

54

2.4. Projectile motion

 A projectile motion consists of two components:

 vertical component (y-comp.)

 motion under constant acceleration,

a

y= −

g

 horizontal component (x-comp.)

 motion with constant velocity thus

a

x= 0

 The path followed by a projectile is called trajectory is shown in

figure 2.9.

v

u

θ

s

_x

= R

s

_y

=H

u

_x

v

_2y

u

_y

v

_1x

v

_1y

v

_2x

v

₁

θ

1

v

₂

θ

2 t₁ t₂ B A P _Q C

y

x

Figure 2.9 Figure 2.9 Simulation 2.5

(55)

PHYSICS

CHAPTER 2

 From figure 2.9,

 The x-component of velocityx-component of velocity along AC (horizontal) at any

point is constant,constant,

 The y-component (vertical) of velocity variesy-component (vertical) of velocity varies from one

point to another point along AC.

but the y-component of the initial velocity is given by

θ

u

_x

=

cos

θ

u

_y

=

sin

(56)

PHYSICS

CHAPTER 2

56

 Table 2.2 shows the x and y-components, magnitude and

direction of velocities at points P and Q.

Velocity Point P Point Q

x-comp. y-comp. magnitude direction 1 1

u

gt

v

_y

=

_y

−

θ

u

v

₁_x

=

_x

=

cos

2 2

u

gt

v

_y

=

_y

−

θ

u

v

₂_x

=

_x

=

cos

( )

2 y 1 2 x 1 1

v

=

+









=

− x 1 y 1 1 1

v

θ

tan

( )

2 y 2 2 x 2 2

v

=

+









=

− x 2 y 2 1 2

v

θ

tan

Table 2.2 Table 2.2

(57)

PHYSICS

CHAPTER 2

 _{The ball reaches the highest point at point B at velocity,}

v

where

 x-component of the velocity,  y-component of the velocity,

 y-component of the displacement,

 Use

2.4.1 Maximum height,

H

θ

u

v

_x

=

_x

=

cos

0 =

y

v

y y y

u

gs

v

2

=

2

−

2 (

u

sin

)

2 gH

0 =

θ

2

−

g

u

H

2 sin

2 2

θ

=

H

s

_y

=

(58)

PHYSICS

CHAPTER 2

58  _{At maximum height,}

H

 _Time,

t =

∆

t’

_and

v

y

= 0

 Use

2.4.2 Time taken to reach maximum height,

∆

t’

gt

u

v

_y

=

_y

−

(

sin

)

'

0 =

u

θ

−

g

∆

t

g

u

t

'

=

sin

θ

∆

2.4.3 Flight time,

∆

t

(from point A to point C)

'

2 t

t

=

∆

g

θ

u

t

=

2 sin

∆

(59)

PHYSICS

CHAPTER 2

 Since the x-component for velocity along AC is constant hence

 From the displacement formula with uniform velocity, thus the

x-component of displacement along AC is

2.4.4 Horizontal range,

R

and value of

R

maximum

t

u

s

_x

=

_x

θ

cos

u

v

u

_x

=

_x

=

(

u

)( )

t

R

=

cos

θ

∆

(

)

_



_



=

g

u

R

cos

θ

2 sin

θ

(

2 sin

θ

cos

θ

)

2

g

u

R

=

and

s

_x

=

R

(60)

PHYSICS

CHAPTER 2

60

 From the trigonometry identity,

thus

 _{The value of}

R

_{maximum when}

θ

₌₌₄₅₄₅°_°_and

sin 2

θ

=

1

therefore

θ

2 sin

cos

2 sin

=

θ

2 sin

2

g

u

R

=

g

u

R

2 max

=

Simulation 2.6

(61)

PHYSICS

CHAPTER 2

 Figure 2.10 shows a ball bearing rolling off the end of a table

with an initial velocity,

u

in the horizontal direction.

 Horizontal component along path AB.

2.4.5 Horizontal projectile

h

x

A B

u

_u

v

x

v

y

v

Figure 2.10 Figure 2.10

constant

velocity,

u

_x

=

u

=

v

_x

=

x

s

_x

=

nt,

displaceme

(62)

PHYSICS

CHAPTER 2

62

Time taken for the ball to reach the floor (point B), Time taken for the ball to reach the floor (point B),

t

 By using the equation of freely falling bodies,

Horizontal displacement, Horizontal displacement,

x

 Use condition below :

2 y y

u

t

gt

s

2

1 −

=

2

gt

0 h

2

1 −

=

−

g

h

t

=

2

The time taken for the ball free fall to point A

The time taken for the ball to reach point B

=

(Refer to figure 2.11)

Figure 2.11

(63)

PHYSICS

CHAPTER 2

 Since the x-component of velocity along AB is constant, thus

the horizontal displacement,

x

 Note :

 In solving any calculation problem about projectile motion,

the air resistance is negligibleair resistance is negligible.

t

u

s

_x

=

_x













=

g

h

u

x

2

and

s

_x

=

x

(64)

PHYSICS

CHAPTER 2

64

Figure 2.12 shows a ball thrown by superman with an initial speed,

u

= 200 m s-1_{and makes an}

angle,

θ

= 60.0° to the horizontal. Determine

a. the position of the ball, and the magnitude and

direction of its velocity, when t = 2.0 s.

Example 9 :

Figure 2.12 Figure 2.12 O

x

u

θ

= 60.0

°

y

R

H

v

_2y

v

_1x

v

_1y

_v

2x Q

v

₁ P

v

₂

(65)

PHYSICS

CHAPTER 2

b. the time taken for the ball reaches the maximum height,

H

and calculate the value of

H

.

c. the horizontal range,

R

d. the magnitude and direction of its velocity when the ball reaches the ground (point P).

e. the position of the ball, and the magnitude and direction of its velocity at point Q if the ball was hit from a flat-topped hill with the time at point Q is 45.0 s.

(given

g

= 9.81 m s-2)

(66)

PHYSICS

CHAPTER 2

66

a. i. position of the ball when t = 2.0 s , Horizontal component :

Vertical component :

therefore the position of the ball is 2 y y

u

t

gt

s

2

1 −

=

t

u

s

_x

=

_x

(67)

PHYSICS

CHAPTER 2

Solution :

a. ii. magnitude and direction of ball’s velocity at t = 2.0 s , Horizontal component : Vertical component : Magnitude, Direction,

gt

u

v

_y

=

_y

−

1 x x

u

v

=

100 m

s

−

( ) ( )

2 2 2 2

₊

₌

₁₀₀

₊

₁₅₃

=

v

_x

v

_y

v









=





=

tan

−1

v

y

tan

−1

153 θ

(68)

PHYSICS

CHAPTER 2

68

b. i. At the maximum height,

H

:

Thus the time taken to reach maximum height is given by

ii. Apply

gt

u

v

_y

=

_y

−

0 =

y

v

gt

t

u

s

_y _y

2

1 −

=

(69)

PHYSICS

CHAPTER 2

Solution :

c. Flight time

=

2×(the time taken to reach the maximum height)

Hence the horizontal range,

R

is

d. When the ball reaches point P thus The velocity of the ball at point P, Horizontal component: Vertical component:

(

17.6 )

2 =

t

u

s

_x

=

_x 1 1x

=

u

x

=

100 m

s

−

v

0 =

y

s

gt

u

v

₁_y

=

_y

−

(70)

PHYSICS

CHAPTER 2

70 Solution : Solution : Magnitude, Direction,

therefore the direction of ball’s velocity is e. The time taken from point O to Q is 45.0 s. i. position of the ball when t = 45.0 s,

Horizontal component :

( ) (

2

)

2 2 1 2 1 1

=

v

x

+

v

y

=

100 +

−

172 v











 −

=









=

− −

100

172 tan

tan

1 1 1 1 x y

v

θ

300



=

θ

from positive x-axis anticlockwisefrom positive x-axis anticlockwise

t

u

s

_x

=

_x

(71)

PHYSICS

CHAPTER 2

Solution :

Vertical component :

therefore the position of the ball is (4500 m, (4500 m, −₋2148 m)2148 m) e. ii. magnitude and direction of ball’s velocity at t = 45.0 s ,

Horizontal component : Vertical component : 2 y y

u

t

gt

s

2

1 −

=

gt

u

v

₂_y

=

_y

−

1 2x

=

u

x

=

100 m

s

−

v

(72)

PHYSICS

CHAPTER 2

72 Solution : Solution : Magnitude, Direction,

therefore the direction of ball’s velocity is

( ) (

2

)

2 2

=

100 +

−

269 v

2 2 2 2 2

v

x

v

y

v

=

+









=

− x y

v

θ

2 2 1

tan

(73)

PHYSICS

CHAPTER 2

A transport plane travelling at a constant velocity of 50 m s−1_{at an}

altitude of 300 m releases a parcel when directly above a point X on level ground. Calculate

a. the flight time of the parcel,

b. the velocity of impact of the parcel,

c. the distance from X to the point of impact. (given

g

= 9.81 m s-2₎ Solution : Solution :

Example 10 :

300 m 1

s

m

50

−

=

u

(74)

PHYSICS

CHAPTER 2

74

The parcel’s velocity = plane’s velocity thus

a. The vertical displacement is given by Thus the flight time of the parcel is

1

s

m

50

−

=

u

_x 1

s

m

50

−

=

u

and

u

_y

=

0 m

s

− 1 2

2

1 gt

t

u

s

_y

=

_y

−

(75)

PHYSICS

CHAPTER 2

Solution :

b. The components of velocity of impact of the parcel: Horizontal component: Vertical component: Magnitude, Direction, 1

s

m

50

−

=

_x x

u

v

(

9.81 )(

7.82 )

0 −

=

y

v

gt

u

v

_y

=

_y

−











 −

=









=

− −

50

6.7

7 tan

tan

1 1 x y

v

θ

( ) (

2

)

2 2 2

₊

₌

₅₀

₊

₋

₇

_6.7

=

v

_x

v

_y

v

(76)

PHYSICS

CHAPTER 2

76

c. Let the distance from X to the point of impact is

d

. Thus the distance,

d

is given by

t

u

s

_x

=

_x

(77)

PHYSICS

CHAPTER 2

Use gravitational acceleration, g = 9.81 m s−2

1. A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket, as in figure 2.13. If he shoots the ball at a 40.0° angle above the horizontal, at what initial

speed must he throw so that it goes through the hoop without striking the backboard? The basket height is 3.05 m.

(78)

PHYSICS

CHAPTER 2

78

2. An apple is thrown at an angle of 30° above the horizontal from the top of a building 20 m high. Its initial speed is 40 m s−1_{. Calculate}

a. the time taken for the apple to strikes the ground,

b. the distance from the foot of the building will it strikes the ground,

c. the maximum height reached by the apple from the ground.

ANS. : 4.90 s; 170 m; 40.4 m

3. A stone is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m s−1

at 40° above the horizontal. How far above or below its original level will the stone strike the opposite wall?

ANS. : 10.3 m below the original level.

Exercise 2.4 :

(79)

PHYSICS

CHAPTER 2

THE END…

Next Chapter…

CHAPTER 3 :

(80)

PHYSICS

CHAPTER 3

1

MOMENTUM AND

IMPULSE

(81)

PHYSICS

CHAPTER 3 

3.0 MOMENTUM AND IMPULSE



3.1 Momentum and impulse

(82)

PHYSICS

CHAPTER 3

3

At the end of this chapter, students should be able to:  Define_Define momentum.

 Define_Define impulse and use F-t graph to  determine impulse

 Use _Use

Learning Outcome:

(83)

PHYSICS

CHAPTER 3 3.1.1 Linear momentum,

 is defined as the product between mass and velocitythe product between mass and velocity.  is a vector quantity.

 Equation :

 The S.I. unit of linear momentum is kg m skg m s-1-1_.

 The direction of the momentumdirection of the momentum is the samesame as the direction direction of the velocity

of the velocity.

 _{It can be resolve into vertical (}

y

_{) component and horizontal (}

x

₎

component.

p



v

m

p



=



p



y

p

x

=

p

cos

θ

=

mv

cos

θ

mv

θ

p

_y

=

sin

=

sin

(84)

PHYSICS

CHAPTER 3

5

3.1.2 Impulse,

 Let a single constant force, constant force, F_F acts on an object in a short time

interval (collision), thus the Newton’s 2nd law can be written as

 _{is defined as the product of a force,}_{the product of a force,}

F

_{and the time,}_{and the time,}

t

OR the change of momentumthe change of momentum.

 is a vector quantityvector quantity whose directiondirection is the samesame as the constant force

constant force on the object.

J



constant

=

∑

F

d

_dt

p



1 2

p

d

dt

F

J



=



=



=



−



momentum

final

:

2

p



where

momentum

initial

:

1

p



force

impulsive

:

F



(85)

PHYSICS

CHAPTER 3

 The S.I. unit of impulse is N sN s or kg m skg m s−−11_.

 If the forceforce acts on the object is not constantnot constant then

 Since impulse and momentum are both vector quantities, then it

is often easiest to use them in component form :

dt

F

dt

F

J

t _av t



=

∫

2 1

where

F



_av

:

average

impulsive

force

( )

av _x 2x 1x

(

x x

)

x

F

dt

p

m

v

u

J

=

−

=

−

( )

av _y 2 y 1y

(

y y

)

y

F

dt

p

m

v

u

J

=

−

=

−

( )

F

dt

p

m

(

v

u

)

J

=

−

=

−

consider 2-D consider 2-D collision only collision only

(86)

PHYSICS

CHAPTER 3

7

 _{When two objects in collision, the impulsive force,}

F

_against

time,

t

graph is given by the Figure 3.20.

1

t

₂ Figure 3.20 Figure 3.20

t

0 F

Shaded area under the F

−

t graph = impulse

Picture 3.1

Picture 3.2

(87)

PHYSICS

CHAPTER 3

A 0.20 kg tennis ball strikes the wall horizontally with a speed of 100 m s−1_{and it bounces off with a speed of 70 m s}−1_{in the}

opposite direction.

a. Calculate the magnitude of impulse delivered to the ball by the wall,

b. If the ball is in contact with the wall for 10 ms, determine the magnitude of average force exerted by the wall on the ball.

Example 3.1 :

Wall (2) 1 1 1

s

m

100

−

=

1

u

1 1

v

=

u

=

0 kg

0.20

1

=

m

(88)

PHYSICS

CHAPTER 3

9

a. From the equation of impulse that the force is constant,

Therefore the magnitude of the impulse is 34 N s34 N s. b. Given the contact time,

1 2

p

dp

J

=

−

(

1 1

)

1

v

u

m

J

=

−

dt

F

J

=

_av

(89)

PHYSICS

CHAPTER 3

An estimated force-time curve for a tennis ball of mass 60.0 g struck by a racket is shown in Figure 3.21. Determine

a. the impulse delivered to the ball,

b. the speed of the ball after being struck, assuming the ball is

Example 3.2 :

0.2

1.8 t

( )

ms

0 ( )

kN

F

1.0

18

Figure 3.21 Figure 3.21

(90)

PHYSICS

CHAPTER 3

11

a. From the force-time graph,

b. Given the ball’s initial speed,

graph

under the

area

F

t

J

=

−

0 =

u

(

v

u

)

m

dp

J

=

−

kg

10

60.0 ×

− 3

=

m

(91)

PHYSICS

CHAPTER 3

1. A steel ball with mass 40.0 g is dropped from a height of 2.00 m onto a horizontal steel slab. The ball rebounds to a height of 1.60 m.

a. Calculate the impulse delivered to the ball during impact. b. If the ball is in contact with the slab for 2.00 ms, determine

the average force on the ball during impact. ANS. : 0.47 N s; 237. 1 N

ANS. : 0.47 N s; 237. 1 N

2. A golf ball (m = 46.0 g) is struck with a force that makes an angle of 45° with the horizontal. The ball lands 200 m away on a flat fairway. If the golf club and ball are in contact for 7.00 ms, calculate the average force of impact. (neglect the air resistance.)

(92)

PHYSICS

CHAPTER 3

13

Figure 3.22

3.

A tennis ball of mass, m = 0.060 kg and a speed, v = 28 m s−1 strikes a wall at a 45° angle and rebounds with

the same speed at 45° as shown in Figure 3.22. Calculate the impulse given by the wall.

ANS. : 2.4 N s to the left or

ANS. : 2.4 N s to the left or −₋2.4 N s2.4 N s

Exercise 3.1 :

(93)

PHYSICS

CHAPTER 3 3.2 Conservation of linear momentum

3.2.1 Principle of conservation of linear momentum

 states “In an isolated (closed) system, the total momentum In an isolated (closed) system, the total momentum of that system is constant

of that system is constant.”

OR

“When the net external force on a system is zero, the total When the net external force on a system is zero, the total momentum of that system is constant