Note Chapter2 SF017

PHYSICS CHAPTER 2

1

xs

ys

xv

yv

xa

ya

g

CHAPTER 2:CHAPTER 2:Kinematics of linear motionKinematics of linear motion

(5 hours)(5 hours)

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PHYSICS CHAPTER 2

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2.0 Kinematics of Linear motion is defined as the studies of motion of an objects without studies of motion of an objects without

considering the effects that produce the motionconsidering the effects that produce the motion.. There are two types of motion:

Linear or straight line motion (1-D) with constant (uniform) velocity with constant (uniform) acceleration, e.g. free fall motion

Projectile motion (2-D) x-component (horizontal) y-component (vertical)

PHYSICS CHAPTER 2

3

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DefineDefine distance, displacement, speed, velocity, distance, displacement, speed, velocity,

acceleration and related parameters: uniform velocity, acceleration and related parameters: uniform velocity, average velocity, instantaneous velocity, uniform average velocity, instantaneous velocity, uniform acceleration, average acceleration and instantaneous acceleration, average acceleration and instantaneous acceleration.acceleration.

SketchSketch graphs of displacement-time, velocity-time and graphs of displacement-time, velocity-time and acceleration-time.acceleration-time.

Learning Outcome:

2.1 Linear Motion (1 hour)

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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

Q

PHYSICS CHAPTER 2

5

vector quantity is defined as the distance between initial point and final the distance between initial point and final

point in a straight linepoint 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

EW

S

O

P

20 m

10 m

10 m 20 m

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:

west-south theor to 4510

10tan

1θ

interval time

distance of changespeed

m 14.11010 22 OP

Δt

Δdv

PHYSICS CHAPTER 2

7

is a vector quantity. The S.I. unit for velocity is m s-1.

Average velocity, Average velocity, vvavav

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

Its direction is in the same direction of the change in same direction of the change in displacementdisplacement.

2.1.4 Velocity,v

interval time

ntdisplaceme of changeavv

Δt

Δsvav

12

12av tt

ssv

PHYSICS CHAPTER 2

8

Instantaneous velocity, Instantaneous velocity, vv is defined as the instantaneous rate of change of the instantaneous rate of change of

displacementdisplacement. Equation:

An object is moving in uniform velocitymoving in uniform velocity if

t

s

0tv

limit

constantdt

ds

dt

dsv

PHYSICS CHAPTER 2

9

Therefore

Q

s

t0

s1

t1

The gradient of the tangent to the curve at point Q

= the instantaneous velocity at time, t = t1

Gradient of s-t graph = velocity

PHYSICS CHAPTER 2

10

vector quantity The S.I. unit for acceleration is m s-2.

Average acceleration, Average acceleration, aaavav

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

velocityof changeava

12

12av tt

vva

Δt

Δvaav

PHYSICS CHAPTER 2

11

Instantaneous acceleration, Instantaneous acceleration, aa 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

0ta

limit

constantdt

dv

2

2

dt

sd

dt

dva

PHYSICS CHAPTER 2

12

Deceleration,Deceleration, aa is a negative accelerationnegative acceleration. The object is slowing downslowing down meaning the speed of the object speed of the object

decreases with timedecreases with time.

Therefore

v

t

Q

0

v1

t1

The gradient of the tangent to the curve at point Q

= the instantaneous acceleration at time, t = t1

Gradient of v-t graph = acceleration

PHYSICS CHAPTER 2

13

Displacement against time graph (Displacement against time graph (s-ts-t))2.1.6 Graphical methods

s

t0

s

t0(a) Uniform velocity (b) The velocity increases with time

Gradient = constant

Gradient increases with time

(c)

s

t0

Q

RP

The direction of velocity is changing.

Gradient at point R is negative.

Gradient at point Q is zero.

The velocity is zero.

PHYSICS CHAPTER 2

14

Velocity versus time graph (Velocity versus time graph (v-tv-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

t0 (a) t2t1

v

t0(b) t1 t2

v

t0(c)

Uniform velocity

Uniform acceleration

Area under the v-t graph = displacement

BC

A

PHYSICS CHAPTER 2

15

From the equation of instantaneous velocity,

Therefore

dt

dsv

vdtds

2

1

t

tvdts

graph under the area dedsha tvs

Simulation 2.1 Simulation 2.2 Simulation 2.3

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 2.1Figure 2.1

PHYSICS CHAPTER 2

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a. 0 to 6 s : The train moves at a constant velocity of 0.68 cm s1.

6 to 10 s : The train stops.

10 to 14 s : The train moves in the same direction at a constant velocity of 1.50 cm s1.

b.

0 2 4 6 8 10 12 14 t (s)

0.68

1.50

v (cm s1)

PHYSICS CHAPTER 2

18


c.

d.

12

12

tt

ssvav

014

010

avv

1s cm 714.0 avv

s 14 tos 10 from velocity averagev

12

12

tt

ssv

1014

410

v

1s cm 50.1 v

PHYSICS CHAPTER 2

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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

displacement.

d. Calculate the average acceleration between 20 s to 40 s.

Example 3 :

05 10 15 20 25 30 35 t (s)

-4

-2

2

4

v (m s1)


40 45 50

PHYSICS CHAPTER 2

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a. 0 to 5 s : Lift moves upward from rest with a constant acceleration of 0.4 m s2.

5 to 15 s : The velocity of the lift increases from 2 m s1 to 4 m s1 but the acceleration decreasing to

0.2 m s2.

15 to 20 s : Lift moving with constant velocity of 4 m s1.

20 to 25 s : Lift decelerates at a constant rate of 0.8 m s2.

25 to 30 s : Lift at rest or stationary.

30 to 35 s : Lift moves downward with a constant acceleration of 0.8 m s2.

35 to 40 s : Lift moving downward with constant velocity

of 4 m s1.

40 to 50 s : Lift decelerates at a constant rate of 0.4 m s2 and comes to rest.

PHYSICS CHAPTER 2

21


b.

t (s)5 10 15 20 25 30 35 40 45 500

-0.4

-0.2

0.2

0.6

a (m s2)

-0.6

-0.8

0.8

0.4

PHYSICS CHAPTER 2

22


c. i.

05 10 15 20 25 30 35 t (s)

-4

-2

2

4

v (m s1)

40 45 50

A1

A2 A3

A4 A5

v-t ofgraph under the area distance Total 54321 AAAAA

45152

145

2

14105

2

11042

2

152

2

1distance Total

m 115distance Total

PHYSICS CHAPTER 2

23


c. ii.

d.

v-t ofgraph under the areant Displaceme

54321 AAAAA

45152

145

2

14105

2

11042

2

152

2

1ntDisplaceme

m 15ntDisplaceme

12

12

tt

vvaav

2040

44

ava

2s m 4.0 ava

PHYSICS CHAPTER 2

24


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 mANS. : 6 m

Exercise 2.1 :

PHYSICS CHAPTER 2

25

2. A train pulls out of a station and accelerates steadily for 20 s until its velocity reaches 8 m s1. 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 sANS. : 0.4 m s22,0 m s,0 m s22,-0.267 m s,-0.267 m s22, 80 m, 800 m, 120 m; , 80 m, 800 m, 120 m;

6.67 m s6.67 m s11..

Exercise 2.1 :

PHYSICS CHAPTER 2

26

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Derive and applyDerive and apply equations of motion with uniform equations of motion with uniform

acceleration:acceleration:

Learning Outcome:

2.2 Uniformly accelerated motion (1 hour)

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atuv 2

2

1atuts

asuv 222

PHYSICS CHAPTER 2

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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

atuv (1)

t

uva

PHYSICS CHAPTER 2

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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

timetFigure 2.4Figure 2.4

tvu2

1s (2)

PHYSICS CHAPTER 2

29

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

From eq. (1),

From eq. (2),

tatuus 2

1

(3)2

2

1atuts

atuv

t

suv

2

multiply

att

suvuv

2

asuv 222 (4)

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, ( ( aa = 0) = 0) the

equations (1) and (4) become

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

uv

vts constant velocityconstant velocity

PHYSICS CHAPTER 2

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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.


a. Use

Example 4 :

s 2.16t

?v

tvus 2

1

1s m 148 v 2.1602

1 1200 v

0u

m 1200s

?a

PHYSICS CHAPTER 2

32


b. By using the equation of linear motion,

asuv 222

2s m 13.9 a

120020 148 2 a

OROR

2

2

1atuts

2s m 14.9 a

22.162

101200 a

PHYSICS CHAPTER 2

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A bus travelling steadily at 30 m s1 along a straight road passes a stationary car which, 5 s later, begins to move with a uniform acceleration of 2 m s2 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.


a. Given

Use

Example 5 :

21 ms 2 0; ;constant s m 30 ccb auv

s 15ct

cccc tauv 1s m 30 bc vv

ct2030

PHYSICS CHAPTER 2

34

b.

From the diagram,

c

b

1s m 30 bv

0cu

s 0t s 5t

2s m 2 cab

bvb

c

bv

tt bc ss

bc ss ttt bc

m 900cs

tvtatu bcc 2

2

1

s 30t

tt0 2 3022

1

Therefore

tvs bc 3030cs

PHYSICS CHAPTER 2

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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.


a. t =2.00 s ;

Example 6 :

ttts 23 243

m 12.0s

ttts 23 243 2.0022.0042.003 23 s

PHYSICS CHAPTER 2

36


b. Instantaneous velocity at t = 2.00 s,

Use

Thus

dt

dsv

tttdt

dv 243 23

289 2 ttv

1s m 22.0 v

22.0082.009 2 v

PHYSICS CHAPTER 2

37


c. Instantaneous acceleration at t = 2.00 s,

Use

Hence

dt

dva

289 2 ttdt

da

818 ta

2s m 28.0 a

82.0018 a

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 sANS. : 4.53 s; 14.1 m s11

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 sANS. : 14.4 s

Exercise 2.2 :

PHYSICS CHAPTER 2

39

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 sANS. : 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.

ANS. : 1.73 mANS. : 1.73 m

Exercise 2.2 :

PHYSICS CHAPTER 2

40

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

For For upward and downwardupward and downward motion, use motion, use

aa = = gg = = 9.81 m s9.81 m s22

Learning Outcome:

2.3 Freely falling bodies (1 hour)

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PHYSICS CHAPTER 2

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2.3. Freely falling bodies is defined as the vertical motion of a body at constant the vertical motion of a body at constant

acceleration, acceleration, gg under gravitational field under gravitational field without air without air resistanceresistance..

In the earth’s gravitational field, the constant acceleration known as acceleration due to gravityacceleration due to gravity or free-fall free-fall

accelerationacceleration or gravitational accelerationgravitational acceleration. the value is gg = 9.81 m s= 9.81 m s22

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

fall motion, the assumption made is ignore the air assumption made is ignore the air resistanceresistance.

PHYSICS CHAPTER 2

42

Sign convention:

Table 2.1 shows the equations of linear motion and freely falling bodies.

Table 2.1Table 2.1

Linear motion Freely falling bodies

atuv gtuv

as2uv 22 gs2uv 22

2at2

1uts 2gt

2

1uts

+

- +

-

From the sign convention thus,

ga

PHYSICS CHAPTER 2

43

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 decreases to zerodecreases to zero when the ball reaches the maximum maximum

height, height, HH.

H

u

v

velocity = 0


uv

PHYSICS CHAPTER 2

44

The graphs in figure 2.6 show the motion of the ball moves up and down.

Derivation of equationsDerivation of equations At the maximum height or

displacement, H where t = t1,

its velocity,

hence

therefore the time taken for

the ball reaches H,


t0

vu

u

t1 2t1

t0

a

g

t1 2t1

t

s

0

H

t1 2t1

v =0

gtuv 1gtu 0

0v

g

ut1

Simulation 2.4

PHYSICS CHAPTER 2

45

To calculate the maximum height or displacement, H:

use either

maximum height,

Another form of freely falling bodies expressions are

211 gtuts

2

1

gsuv 22 2Where s = H

gHu 20 2

OROR

g

uH

2

2

gtuv gsuv 222

2

2

1gtuts

gtuv yy

yyy gsuv 222 2

2

1gttus yy

PHYSICS CHAPTER 2

46

A ball is thrown from the top of a building is given an initial velocity of 10.0 m s1 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 s2)

Example 7 :

A

B

C

D

u =10.0 m s1

30.0 m


PHYSICS CHAPTER 2

47


a. At the maximum height, H, vy = 0 and u = uy = 10.0 m s1 thus

b. From point A to C, the vertical displacement, sy= 0 m thus

y2y

2y gsuv 2

H9.81210.00 2 m 5.10H

2yy gttus

2

1

s 2.03t

2tt 9.812

110.00

A

B

C

D

u

30.0 m

PHYSICS CHAPTER 2

48


c. From point A to D, the vertical displacement, sy= 30.0 m thus

By using

2yy gttus

2

1

s 3.69t

2tt 9.812

110.030.0

A

B

C

D

u

30.0 m

030.010.04.91 tt 2

2a

4acbb 2 t

OR s 1.66Time don’t Time don’t have have negative negative value.value.

a b c

PHYSICS CHAPTER 2

49


d. Time taken from A to D is t = 3.69 s thus

From A to D, sy = 30.0 m

Therefore the ball’s velocity at D is

A

B

C

D

u

30.0 m

gtuv yy

1s m 26.2 yv

3.699.8110.0 yv

OR

y2

y2

y gsuv 2

1s m 26.2 yv 30.09.81210.0 22

yv

1s m 26.2 yv

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

sy = 150 m

Hence

Example 8 :

uy = 0 m s1

150 mm 150ys

2yy gttus

2

1

s 5.53t

2t9.812

10150

PHYSICS CHAPTER 2

51


b. The book’s velocity is given by

Therefore the book’s velocity is

gtuv yy

1s m 54.2 yv

5.539.810 yv

OR

y2

y2

y gsuv 2

1s m 54.2 yv 1509.8120 2

yv

1s m 54.2 yv

m 150ys

0yu

?yv

PHYSICS CHAPTER 2

52

1. A ball is thrown directly downward, with an initial speed of 8.00 m s1, 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 sANS. : 1.79 s; 25.6 m s11

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 mANS. : 1.75 m

Exercise 2.3 :

m 2.2


to travel this distance took 0.30 s

PHYSICS CHAPTER 2

53

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

CalculateCalculate: time of flight, maximum height, range and : time of flight, maximum height, range and maximum range, instantaneous position and velocity.maximum range, instantaneous position and velocity.

Learning Outcome:

2.4 Projectile motion (2 hours)

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θuux cosθuu y sin

0xagay

PHYSICS CHAPTER 2

54

2.4. Projectile motion A projectile motion consists of two components:

vertical component (y-comp.)

motion under constant acceleration, ay= g horizontal component (x-comp.)

motion with constant velocity thus ax= 0

The path followed by a projectile is called trajectory is shown in figure 2.9.

v

u

sx= R

sy=H

ux

v2y

uy

v1x

v1y

v2x

v1

1

v2

2

t1 t2

B

A

P Q

C

y

xFigure 2.9Figure 2.9

Simulation 2.5

PHYSICS CHAPTER 2

55

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

θuux cos

θuu y sin

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

11 gtuv yy

θuuv xx1 cos

22 gtuv yy θuuv xx2 cos

2y12

x11 vvv

x1

y111 v

vθ tan

2y22

x22 vvv

x2

y212 v

vθ tan

Table 2.2Table 2.2

PHYSICS CHAPTER 2

57

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

θuuvv xx cos0yv

yyy gsuv 222

gHu 2sin0 2

g

uH

2

sin 22

Hsy

PHYSICS CHAPTER 2

58

At maximum height, H

Time, t = t’ and vy= 0

Use

2.4.2 Time taken to reach maximum height, t’

gtuv yy 'sin0 tgu

g

ut

sin'

2.4.3 Flight time, t (from point A to point C)

'2 tt

g

θut

sin2

PHYSICS CHAPTER 2

59

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

tus xx

cosuvu xx

tuR cos

g

uuR

sin2cos

cossin22

g

uR

and Rsx

PHYSICS CHAPTER 2

60

From the trigonometry identity,

thus

The value of R maximum when = = 4545 and sin 2sin 2 = = 11 therefore

cossin22sin

2sin2

g

uR

g

uR

2

max

Simulation 2.6

PHYSICS CHAPTER 2

61

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.

Vertical component along path AB.

2.4.5 Horizontal projectile

h

xA B

u u

v

xv

yv


constant velocity, xx vuuxsx nt,displaceme

0u y velocity,initialhsy nt,displaceme

Simulation 2.7

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), tt By using the equation of freely falling bodies,

Horizontal displacement, Horizontal displacement, xx Use condition below :

2yy gttus

2

1

2gt0h2

1

g

ht

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)


PHYSICS CHAPTER 2

63

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.

tus xx

g

hux

2

and xsx

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, = 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 2.12Figure 2.12 xO

u

= 60.0

y

R

H

v2y

v1x

v1y v2xQ

v1

P

v2

PHYSICS CHAPTER 2

65

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

calculate the value of H.

c. the horizontal range, Rd. 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)


The component of Initial velocity :1s m 1000.60cos200

xu1s m 1730.60sin200

yu

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 (200 m, 326 m)(200 m, 326 m)

2yy gttus

2

1

ground theabove m 326ys

22.009.812

12.00173 ys

tus xx

Opoint from m 200xs 2.00100xs

PHYSICS CHAPTER 2

67


a. ii. magnitude and direction of ball’s velocity at t = 2.0 s ,



Magnitude,

Direction,

gtuv yy

1yv s m 153

1xx uv s m 100

2.009.81173 yv

1v s m 183

2222 153100 yx vvv

56.8θ

100

153tantan 11

x

y

v

vθ

from positive x-axis anticlockwisefrom positive x-axis anticlockwise

PHYSICS CHAPTER 2

68


b. i. At the maximum height, H :

Thus the time taken to reach maximum height is given by

ii. Apply

t9.811730 s 17.6t

gtuv yy

0yv

217.69.812

117.6173 H

m 1525H

gttus yy 2

1

PHYSICS CHAPTER 2

69


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:

s 35.2t 17.62t

35.2100Rm 3520R

tus xx

11 s m 100 xx uv

0ys

gtuv yy 1

35.29.811731 yv1

1 s m 172 yv

PHYSICS CHAPTER 2

70


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,


11 s m 200 v

2221

211 172100 yx vvv

60.0θ

100

172tantan 1

1

11

x

y

v

vθ

300θ from positive x-axis anticlockwisefrom positive x-axis anticlockwise

tus xx

Opoint from m 4500xs 45.0100xs

PHYSICS CHAPTER 2

71



therefore the position of the ball is (4500 m, (4500 m, 268 m)268 m)

e. ii. magnitude and direction of ball’s velocity at t = 45.0 s ,



2yy gttus

2

1

ground thebelow m 268ys

245.09.812

145.0173 ys

gtuv yy 2

12 s m 269 yv

12 s m 100 xx uv

45.09.811732 yv

PHYSICS CHAPTER 2

72


Magnitude,

Direction,

therefore the direction of ball’s velocity is

12 s m 287 v

222 269100 v

69.6θ

100

269tan 1θ

from positive x-axis anticlockwisefrom positive x-axis anticlockwise 290θ

22

222 yx vvv

x

y

v

vθ

2

21tan

PHYSICS CHAPTER 2

73

A transport plane travelling at a constant velocity of 50 m s1 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)


Example 10 :

300 m

d

1s m 50 u

X

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

1s m 50 uux

s 82.7t

29.812

10300 t

1s m 50 u

m 300ys

and1s m 0 yu

2

2

1gttus yy

PHYSICS CHAPTER 2

75


b. The components of velocity of impact of the parcel:

Horizontal component:

Vertical component:

Magnitude,

Direction,

therefore the direction of parcel’s velocity is

1s m 50 xx uv

7.829.810 yvgtuv yy

1s m 6.77 yv

1s m 6.19 v

56.9θ

50

6.77tantan 11

x

y

v

vθ

2222 6.7750 yx vvv

from positive x-axis anticlockwisefrom positive x-axis anticlockwise 033 θ

PHYSICS CHAPTER 2

76


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

Thus the distance, d is given by

tus xx

7.8250d

m 391d

PHYSICS CHAPTER 2

77


Use gravitational acceleration, g = 9.81 m s2

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.

ANS. : 10.7 m sANS. : 10.7 m s11

Exercise 2.4 :

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 s1. Calculatea. 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 mANS. : 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 s1 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.ANS. : 10.3 m below the original level.

Exercise 2.4 :

PHYSICS CHAPTER 2

79

THE END…Next Chapter…

CHAPTER 3 :Force, Momentum and Impulse