note chapter2 sf017
TRANSCRIPT
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)
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
PHYSICS CHAPTER 2
2
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)
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
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
17
Solution :Solution :
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
Solution :Solution :
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
19
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)
Figure 2.2Figure 2.2
40 45 50
PHYSICS CHAPTER 2
20
Solution :Solution :
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
Solution :Solution :
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
Solution :Solution :
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
Solution :Solution :
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
Figure 2.3Figure 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 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)
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
atuv 2
2
1atuts
asuv 222
PHYSICS CHAPTER 2
27
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
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
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
31
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 2.16t
?v
tvus 2
1
1s m 148 v 2.1602
1 1200 v
0u
m 1200s
?a
PHYSICS CHAPTER 2
32
Solution :Solution :
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
33
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.
Solution :Solution :
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
35
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 :
ttts 23 243
m 12.0s
ttts 23 243 2.0022.0042.003 23 s
PHYSICS CHAPTER 2
36
Solution :Solution :
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
Solution :Solution :
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)
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
PHYSICS CHAPTER 2
41
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
Figure 2.5Figure 2.5
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,
Figure 2.6Figure 2.6
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
Figure 2.7Figure 2.7
PHYSICS CHAPTER 2
47
Solution :Solution :
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
Solution :Solution :
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
Solution :Solution :
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)
Solution :Solution :
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
Solution :Solution :
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
Figure 2.8Figure 2.8
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)
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
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
Figure 2.10Figure 2.10
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)
Figure 2.11Figure 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)
Solution :Solution :
The component of Initial velocity :1s m 1000.60cos200
xu1s m 1730.60sin200
yu
PHYSICS CHAPTER 2
66
Solution :Solution :
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
Solution :Solution :
a. ii. magnitude and direction of ball’s velocity at t = 2.0 s ,
Horizontal component :
Vertical component :
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
Solution :Solution :
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
Solution :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:
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
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 :
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
Solution :Solution :
Vertical component :
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 ,
Horizontal component :
Vertical component :
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
Solution :Solution :
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)
Solution :Solution :
Example 10 :
300 m
d
1s m 50 u
X
PHYSICS CHAPTER 2
74
Solution :Solution :
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
Solution :Solution :
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
Solution :Solution :
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
Figure 2.13Figure 2.13
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