Physics Malaysian Matriculation Semester 1 Notes Complete
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Nota Padat Fizik Matrikulasi Malaysia Sesi 2013/2014...
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PHYSICS
CHAPTER 2
CHAPTER 2: Kinematics of linear motion
1
PHYSICS
CHAPTER 2
Kinematics of linear motion 2.1 2.2 2.3 2.4
Linear Motion Uniformly Accelerated Motion Free Falling Body Projectile Motion
2
PHYSICS CHAPTER 2 Learning Outcome: 2.1 Linear Motion At the end of this chapter, students should be able to: Define and distinguish between i. distance and displacement ii. speed and velocity iii. instantaneous velocity, average velocity and uniform velocity. iv. instantaneous acceleration, average acceleration and uniform acceleration. Sketch graphs of displacement-time, velocity-time and acceleration-time. Determine the distance travelled, displacement, velocity and acceleration from appropriate graphs. 3
PHYSICS
CHAPTER 2
2.1. Linear motion (1-D) 2.1.1. Distance,
d
scalar quantity. is defined as the length of actual path between two points. points For example : Q
P
The length of the path from P to Q is 25 cm.
4
PHYSICS 2.1.2
CHAPTER 2
Displacement,s
vector quantity is defined as the distance between initial point and final point in a straight line. 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 N relative to the original position. Solution : O
W
20 m
θ P
θ 10 m
S
E 10 m
20 m 5
PHYSICS
CHAPTER 2
The magnitude of the displacement is given by
and its direction is
2.1.3 Speed, v
is defined the rate of change of distance. distance scalar quantity. Equation:
change of distance speed = time interval
Δd v= Δt 6
PHYSICS 2.1.4
CHAPTER 2
Velocity, v
is a vector quantity. The S.I. unit for velocity is m s-1.
Average velocity, vav
is defined as the rate of change of displacement. displacement Equation: change of displacement
vav =
time interval
s2 − s1 vav = t 2 − t1 Δs vav = Δt
Its direction is in the same direction of the change in displacement. displacement
7
PHYSICS
CHAPTER 2
Instantaneous velocity, v is defined as the instantaneous rate of change of displacement. displacement Equation:
limit ∆ s v= ∆t→ 0∆t
ds v= dt
An object is moving in uniform velocity if
ds = constant dt 8
PHYSICS
CHAPTER 2
s
s1
0
Q
The gradient of the tangent to the curve at point Q = the instantaneous velocity at time, t = t1
t
t1 Therefore
Gradient of s-t graph = velocity 9
PHYSICS 2.1.5
CHAPTER 2
Acceleration, a
vector quantity The S.I. unit for acceleration is m s-2.
Average acceleration, aav
is defined as the rate of change of velocity. velocity Equation: a = change of velocity av
time interval
v2 − v1 aav = t 2 − t1
Δv aav = Δt
Its direction is in the same direction of motion. motion The acceleration of an object is uniform when the magnitude of velocity changes at a constant rate and along fixed 10
PHYSICS
CHAPTER 2
Instantaneous acceleration, a is defined as the instantaneous rate of change of velocity. velocity Equation: limit
∆v a= ∆t→ 0∆t 2
dv d s a= = 2 dt dt
An object is moving in uniform acceleration if
dv = constant dt 11
PHYSICS
v
CHAPTER 2
Deceleration, a is a negative acceleration. acceleration The object is slowing down meaning the speed of the object decreases with time. time
Q
v1
The gradient of the tangent to the curve at point Q = the instantaneous acceleration at time, t = t1
0
Therefore
t1
t
Gradient of v-t graph = acceleration 12
PHYSICS 2.1.6
CHAPTER 2 Graphical methods
Displacement against time graph (s-t)
s
s
Gradient increases with time
Gradient = constant 0
s (a) Uniform velocity
t
t
0
(b) The velocity increases with time
Q
(c) P
R
Gradient at point R is negative.
Gradient at point Q is zero.
0
t
The direction of velocity is changing.
The velocity is zero.
13
PHYSICS
CHAPTER 2
Velocity versus time graph (v-t)
v
v Uniform velocity
v
Uniform acceleration
B
C
A 0
t1 (a) t2
t
0
t1
(b) t2
t
0
t1
t
t2 (c)
Area under the v-t graph = displacement
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) 14
PHYSICS
CHAPTER 2 From the equation of instantaneous velocity,
ds v= dt
∫ ds = ∫ vdt Therefore
s=
∫
t2 t1
vdt
s = shaded area under the v − t graph Simulation 2.1
Simulation 2.2
Simulation 2.3 15
PHYSICS
CHAPTER 2
Example 2 : A toy train moves slowly along a straight track according to the displacement, s against time, t graph in figure 2.1.
s (cm)
10 8 6 4 2 Figure 2.1 0
2
4
6
8
10 12 14
t (s)
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.
16
PHYSICS
CHAPTER 2
Solution : 0 to 6 s
:
6 to 10 s : 10 to 14 s : b.
v (cm s−1) 1.50
0.68
0
2
4
6
8
10 12 14
t (s) 17
PHYSICS
CHAPTER 2
Solution : c. v
d.
av
s2 − s1 = t 2 − t1
v = average velocity from 10 s to 14 s s2 − s1 v= t 2 − t1
18
PHYSICS
CHAPTER 2
Example 3 : A velocity-time (v-t) graph in figure 2.2 shows the motion of a lift.
v (m s −1) 4 2 0 -2 Figure 2.2
5
10 15
20 25 30 35 40 45
50
t (s)
-4
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. 19
PHYSICS
CHAPTER 2
Solution : a. 0 to 5 s 5 to 15 s 15 to 20 s 20 to 25 s 25 to 30 s 30 to 35 s
: Lift moves upward from rest with acceleration of 0.4 m s−2. : The velocity of the lift from 2 m s−1 to 4 m s−1 but the acceleration to 0.2 m s−2. : Lift : Lift : Lift : Lift moves
35 to 40 s :
Lift moving
40 to 50 s :
20
PHYSICS
CHAPTER 2
Solution : −2 b. a (m s ) 0.8 0.6 0.4 0.2 0 -0.2
5
10 15
20 25 30 35 40 45
50
t (s)
-0.4 -0.6 -0.8
21
PHYSICS
CHAPTER 2
Solution : −1 v (m s ) c. i. 4 2 0 -2
A1 5
A2 10 15
A3 20 25 30 A35 40 45 4 A5
50
t (s)
-4
Total distance = area under the graph of v-t = A1 + A 2 + A 3 + A 4 + A 5
1 1 1 1 1 Total distance = ( 2 )( 5) + ( 2 + 4 )(10 ) + ( 5 + 10 )( 4 ) + ( 5)( 4 ) + (15 + 5)( 4 ) 2 2 2 2 2
22
PHYSICS
CHAPTER 2
Solution : c. ii. Displacement
= area under the graph of v-t = A1 + A 2 + A 3 + A 4 + A 5
1 1 1 1 1 Displacement = ( 2 )( 5) + ( 2 + 4 )(10 ) + ( 5 + 10 )( 4 ) + ( 5)( − 4 ) + (15 + 5)( − 4 ) 2 2 2 2 2
d.
v2 − v1 aav = t 2 − t1
23
PHYSICS
CHAPTER 2
Exercise 2.1 : 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.
Figure 2.3
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. 24 ANS. : 6 m
PHYSICS
CHAPTER 2
Exercise 2.1 : 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−2,0 m s−2,-0.267 m s−2, 80 m, 800 m, 120 m; 6.67 m s−1.
25
PHYSICS CHAPTER 2 Learning Outcome: 2.2 Uniformly accelerated motion At the end of this chapter, students should be able to: Derive and apply equations of motion with uniform acceleration:
v = u + at 1 2 s = ut + at 2 2 2 v = u + 2as
26
PHYSICS
CHAPTER 2
2.2. Uniformly accelerated motion
From the definition of average acceleration, uniform (constant) constant acceleration is given by
v− u a= t
v = u + at where
v u a t
(1)
: final velocity : initial velocity : uniform (constant) acceleration : time
27
PHYSICS
CHAPTER 2 From equation (1), the velocity-time graph is shown in figure velocity 2.4:
v
u Figure 2.4
t
0
time
From the graph,
The displacement after time, s = shaded area under the graph = the area of trapezium Hence,
1 s = ( u + v)t 2
(2) 28
PHYSICS
CHAPTER 2 By substituting eq. (1) into eq. (2) thus
1 s = [ u + ( u + at ) ]t 2
1 2 s = ut + at 2
From eq. (1), From eq. (2),
( v − u ) = at 2s (v + u) =
( v + u )( v − u ) =
(3)
multiply
t 2s ( at ) t
v 2 = u 2 + 2as
(4) 29
PHYSICS
CHAPTER 2 Notes: equations (1) – (4) can be used if the motion in a straight line with constant acceleration.
For a body moving at constant velocity, ( a = 0) the equations (1) and (4) become
v= u
Therefore the equations (2) and (3) can be written as
s = vt
constant velocity
30
PHYSICS
CHAPTER 2
Example 4 : 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 : a= ?
u= 0
a. Use
v= ?
s = 1200 m t = 16.2 s 1 s = ( u + v )t 2 31
PHYSICS
CHAPTER 2
Solution : b. By using the equation of linear motion,
v 2 = u 2 + 2as
OR
1 2 s = ut + at 2
32
PHYSICS
CHAPTER 2
Example 5 : 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. −1 −2 Solution : vb = 30 m s = constant ; u c = 0; ac = 2 ms −1 v = v = 30 m s a. Given c b Use vc = u c + ac t c
33
PHYSICS
CHAPTER 2
b.
b c
vb = 30 m s − 1
b
b
ac = 2 m s − 2
uc = 0
tb = 0 s
vb
c
tb = 5 s
tb = t
s c = sb
From the diagram,
tb = t ; t c = t − 5 s c = sb
1 2 uc tc + ac tc = vbtb 2
vb
Therefore
sc = vb t
34
PHYSICS
CHAPTER 2
Example 6 : A particle moves along horizontal line according to the equation
s = 3t 3 − 4t 2 + 2t
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 : a. t =2.00 s ;
3
2
s = 3t − 4t + 2t
35
PHYSICS
CHAPTER 2
Solution : b. Instantaneous velocity at t = 2.00 s, Use
ds v= dt
(
d 3 v= 3t − 4t 2 + 2t dt Thus
)
v = 9( 2.00 ) − 8( 2.00 ) + 2 2
36
PHYSICS
CHAPTER 2
Solution : c. Instantaneous acceleration at t = 2.00 s, Use
dv a= dt
Hence
a = 18( 2.00 ) − 8
37
PHYSICS
CHAPTER 2
Exercise 2.2 : 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−1
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. 38 ANS. : 14.4 s
PHYSICS
CHAPTER 2
Exercise 2.2 : 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. ANS. : 1.73 m 39
PHYSICS CHAPTER 2 Learning Outcome: 2.3 Free falling body At the end of this chapter, students should be able to: Describe and use equations for free falling body. For upward and downward motion, use
a = −g = −9.81 m s−2
40
PHYSICS
CHAPTER 2
2.3. Free falling body
is defined as the vertical motion of a body at constant acceleration, g under gravitational field without air resistance. In the earth’s gravitational field, the constant acceleration known as acceleration due to gravity or free-fall acceleration or gravitational acceleration. acceleration the value is g = 9.81 m s−2 the direction is towards the centre of the earth (downward). Note: In solving any problem involves freely falling bodies or free fall motion, the assumption made is ignore the air resistance. resistance
41
PHYSICS
CHAPTER 2 Sign convention:
+ From the sign convention thus, +
-
a= −g
Table 2.1 shows the equations of linear motion and freely falling bodies. Linear motion
v = u + at 2
Table 2.1
2
v = u + 2 as 1 2 s = ut + at 2
Freely falling bodies
v = u − gt
v 2 = u 2 − 2 gs 1 2 s = ut − gt 2 42
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. velocity = 0
v= u
H
Figure 2.5
u v
Assuming air resistance is negligible, the acceleration of the ball, a = −g when the ball moves upward and its velocity decreases to zero when the ball reaches the maximum height, H. 43
PHYSICS
CHAPTER 2
The graphs in figure 2.6 show the motion of the ball moves up and down. Derivation of equations At the maximum height or displacement, H where t = t1, its velocity,
v= 0 hence
v = u − gt
0 = u − gt1
therefore the time taken for the ball reaches H, Simulation 2.4
u t1 = g
Figure 2.6
s
v =0
H
0 v u
t1
2t1
0
t1
2t1
t1
2t1
t
t
−u a 0
−g
t 44
PHYSICS
CHAPTER 2 To calculate the maximum height or displacement, H: 1 2 use either
s = ut1 −
2
gt1
Where s
OR 2
=H
2
v = u − 2 gs
0 = u 2 − 2 gH
maximum height,
2
u H= 2g
Another form of freely falling bodies expressions are
v = u − gt v 2 = u 2 − 2 gs
1 2 s = ut − gt 2
v y = u y − gt 2 2 v y = u y − 2 gs y 1 2 s y = u y t − gt 2
45
PHYSICS
CHAPTER 2
Example 7 : 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 B figure 2.7. Calculate a. the maximum height of the stone from point A. b. the time taken from point A to C. u =10.0 m s−1 c. the time taken from point A to D. A d. the velocity of the stone when it reaches point D.
C
(Given g = 9.81 m s−2)
30.0 m
Figure 2.7
46
D
PHYSICS
CHAPTER 2
Solution : a. At the maximum height, H, vy = 0 and u
B
v 2y = u 2y − 2 gs y
= uy = 10.0 m s−1 thus
u C
A
b. From point A to C, the vertical displacement,
1 2 s y = u y t − gt 2
sy= 0 m thus
30.0 m
D 47
PHYSICS
CHAPTER 2
Solution : c. From point A to D, the vertical displacement, sy= −30.0 m thus
B
1 2 s y = u y t − gt 2
u C
A
a 30.0 m
By using
b
t=
− b±
c 2
b − 4ac 2a OR
D
Time don’t have negative value. 48
PHYSICS
CHAPTER 2
Solution : d. Time taken from A to D is t = 3.69 s thus
B
v y = u y − gt v y = (10.0 ) − ( 9.81)( 3.69 )
u C
A
OR From A to D, sy = −30.0 m 2
30.0 m
D
2
v y = u y − 2 gs y
v y = (10.0 ) − 2( 9.81)( − 30.0 ) 2
2
Therefore the ball’s velocity at D is 49
PHYSICS
CHAPTER 2
Example 8 : 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 :
uy = 0 m s
−1
a. The vertical displacement is
sy = −150 m Hence
s y = − 150 m
150 m
1 2 s y = u y t − gt 2
50
PHYSICS
CHAPTER 2
Solution : b. The book’s velocity is given by
uy = 0
v y = u y − gt
OR
s y = − 150 m
2
2
v y = u y − 2 gs y vy = ?
Therefore the book’s velocity is
51
PHYSICS
CHAPTER 2
Exercise 2.3 : 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−1
2. A falling stone takes 0.30 s to travel past a window 2.2 m tall as shown in figure 2.8.
2.2 m
to travel this distance took 0.30 s
Figure 2.8
From what height above the top of the windows did the stone fall? ANS. : 1.75 m
52
PHYSICS CHAPTER 2 Learning Outcome: 2.4 Projectile motion At the end of this chapter, students should be able to: Describe and use equations for projectile,
ux = uy = ax = ay =
u cos θ u sin θ 0 −g
Calculate time of flight, maximum height, range, maximum range, instantaneous position and velocity.
53
PHYSICS CHAPTER 2 2.4. Projectile motion
A projectile motion consists of two components: vertical component (y-comp.)
horizontal component (x-comp.)
motion under constant acceleration, ay= −g motion with constant velocity thus ax= 0
The path followed by a projectile is called trajectory is shown in y figure 2.9.
v1y P
Simulation 2.5 Figure 2.9
uy A
v1
θ1 v1x
u
θ ux
B
v Q
sy=H
v2y
v2x θ2 v2 C
t1
sx= R
t2
x
54
PHYSICS
CHAPTER 2 From figure 2.9, The x-component of velocity along AC (horizontal) at any point is constant,
u x = u cos θ
The y-component (vertical) of velocity varies from one point to another point along AC. but the y-component of the initial velocity is given by
u y = u sin θ
55
PHYSICS
CHAPTER 2 Table 2.2 shows the x and y-components, magnitude and direction of velocities at points P and Q.
Velocity
Point P
x-comp.
v1 x = u x = u cos θ
v2 x = u x = u cos θ
y-comp.
v1 y = u y − gt1
v2 y = u y − gt 2
magnitude
( v1x )
direction
v1 =
2
Point Q
( )
+ v1 y
2
− 1
v1 y θ1 = tan v1 x
v2 =
( v2 x )
2
( )
+ v2 y
2
− 1
v2 y θ 2 = tan v2 x
Table 2.2
56
PHYSICS
CHAPTER 2
2.4.1 Maximum height, H
The ball reaches the highest point at point B at velocity, v where x-component of the velocity, v = v = u = u cos θ x x y-component of the velocity, v = 0 y y-component of the displacement, s = H y
Use
v y2 = u y2 − 2 gs y
0 = ( u sin θ
)
2
− 2 gH
u sin θ H= 2g 2
2
57
PHYSICS
CHAPTER 2
2.4.2 Time taken to reach maximum height, ∆t’ At maximum height, H
Time, t
Use
= ∆t’ and vy= 0
v y = u y − gt
0 = ( u sin θ ) − g∆ t '
u sin θ ∆ t' = g
2.4.3 Flight time, ∆t (from point A to point C)
∆ t = 2∆ t ' 2u sin θ ∆t= g 58
PHYSICS
CHAPTER 2
2.4.4 Horizontal range, R and value of R maximum
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
u x = v x = u cos θ
s x = u x t and s x = R R = ( u cos θ )( ∆ t ) 2u sin θ R = ( u cos θ ) g 2 u ( 2 sin θ cos θ ) R= g
59
PHYSICS
CHAPTER 2 From the trigonometry identity, thus
sin 2θ = 2 sin θ cos θ 2
u R= sin 2θ g
The value of R maximum when θ = 45° and sin 2θ = 1 therefore 2
Rmax
u = g
Simulation 2.6 60
PHYSICS
CHAPTER 2
2.4.5 Horizontal projectile
Figure 2.10 shows a ball bearing rolling off the end of a table with an initial velocity, u in the horizontal direction.
u
u
vx vy
h Figure 2.10
A
Horizontal component along path AB.
Vertical component along path AB.
v
B
x
velocity, u x = u = v x = constant displacement, s x = x initial velocity, u y = 0
displacement, s y = − h
Simulation 2.7 61
PHYSICS
CHAPTER 2
Time taken for the ball to reach the floor (point B), t By using the equation of freely falling bodies,
1 2 s y = u y t − gt 2 1 2 − h = 0 − gt 2
t=
2h g
Horizontal displacement, x Use condition below : The time taken for the ball free fall to point A
Figure 2.11
=
The time taken for the ball to reach point B
(Refer to figure 2.11) 62
PHYSICS
CHAPTER 2 Since the x-component of velocity along AB is constant, thus the horizontal displacement, x
sx = u xt
and
x = u
2h g
sx = x
Note : In solving any calculation problem about projectile motion, the air resistance is negligible. negligible
63
PHYSICS
CHAPTER 2
Example 9 : y
u Figure 2.12 O
H
θ = 60.0°
P
R
v1y 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.
v1x v1 Q
v2y
x v2x v2 64
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) Solution : The component of Initial velocity :
65
PHYSICS
CHAPTER 2
Solution : a. i. position of the ball when t = 2.0 s , Horizontal component :
sx = u xt
Vertical component :
1 2 s y = u y t − gt 2
therefore the position of the ball is 66
PHYSICS
CHAPTER 2
Solution : a. ii. magnitude and direction of ball’s velocity at t = 2.0 s , Horizontal component :
v x = u x = 100 m s − 1
Vertical component :
v y = u y − gt
Magnitude,
v=
v x2 + v 2y =
(100) 2 + (153) 2
vy − 1 153 θ = tan = tan 100 vx
Direction,
− 1
67
PHYSICS
CHAPTER 2
Solution : b. i. At the maximum height, H :
vy = 0
Thus the time taken to reach maximum height is given by
v y = u y − gt
ii. Apply
1 s y = u y t − gt 2
68
PHYSICS
CHAPTER 2
Solution : c.
Flight time = 2×(the time taken to reach the maximum height)
t = 2(17.6 )
Hence the horizontal range, R is
s x = u xt
d.
When the ball reaches point P thus s y = 0 The velocity of the ball at point P, −1 Horizontal component: v1 x = u x = 100 m s Vertical component: v1 y = u y − gt
69
PHYSICS
CHAPTER 2
Solution : Magnitude, v1
Direction,
=
v +v = 2 1x
2 1y
(100)
2
+ ( − 172)
v1 y − 1 − 172 = tan θ = tan 100 v1x −1
therefore the direction of ball’s velocity is θ = 300 from positive x-axis anticlockwise 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 :
sx = u xt
70
2
PHYSICS
CHAPTER 2
Solution : Vertical component :
1 2 s y = u y t − gt 2
therefore the position of the ball is (4500 m, −2148 m) e. ii. magnitude and direction of ball’s velocity at t = 45.0 s , Horizontal component :
v2 x = u x = 100 m s
−1
Vertical component :
v2 y = u y − gt 71
PHYSICS
CHAPTER 2
Solution : Magnitude, v2
=
v2 =
v +v 2 2x
(100)
2
2 2y
+ ( − 269 )
2
v2 y θ = tan v2 x
Direction,
−1
therefore the direction of ball’s velocity is
72
PHYSICS
CHAPTER 2
Example 10 : 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 :
u = 50 m s − 1 300 m X
d
73
PHYSICS
CHAPTER 2
Solution : The parcel’s velocity = plane’s velocity thus a.
u = 50 m s − 1 u x = u = 50 m s − 1 and u y = 0 m s − 1
The vertical displacement is given by Thus the flight time of the parcel is
1 2 s y = u y t − gt 2
74
PHYSICS
CHAPTER 2
Solution : b. The components of velocity of impact of the parcel: −1 Horizontal component: v x = u x = 50 m s Vertical component: v y = u y − gt
v y = 0 − ( 9.81)( 7.82 )
Magnitude, v
=
v +v = 2 x
2 y
( 50)
2
+ ( − 76.7 )
2
vy − 1 − 76.7 θ = tan = tan 50 vx
Direction,
−1
therefore the direction of parcel’s velocity is 75
PHYSICS
CHAPTER 2
Solution : c.
Let the distance from X to the point of impact is d. Thus the distance, d is given by
sx = u xt
76
PHYSICS
CHAPTER 2
Exercise 2.4 : 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.
Figure 2.13 ANS. : 10.7 m s−1
77
PHYSICS
CHAPTER 2
Exercise 2.4 : 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.
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PHYSICS
CHAPTER 2
THE END… Next Chapter…
CHAPTER 3 : Force, Momentum and Impulse
79
PHYSICS
CHAPTER 3
CHAPTER 3 MOMENTUM AND IMPULSE
1
PHYSICS 3.0
CHAPTER 3
MOMENTUM AND IMPULSE
3.1 Momentum and impulse
3.2
Conservation of linear momentum
2
PHYSICS
CHAPTER 3
Learning Outcome: 3.1 Momentum and impulse At the end of this chapter, students should be able to:
Define momentum.
Define impulse determine impulse
Use
and use F-t graph to
3
PHYSICS 3.1.1 Linear momentum,
p
CHAPTER 3
is defined as the product between mass and velocity. velocity is a vector quantity. Equation :
p = mv
The S.I. unit of linear momentum is kg m s-1. The direction of the momentum is the same as the direction of the velocity. velocity It can be resolve into vertical (y) component and horizontal (x) component.
p
py θ
p x = p cos θ = mv cos θ
p y = p sin θ = mv sin θ
px
4
PHYSICS
3.1.2 Impulse, J
CHAPTER 3
Let a single constant force, F acts on an object in a short time interval (collision), thus the Newton’s 2nd law can be written as
∑ where
dp F= F= = constant dt
J = Fdt = dp = p2 − p1
p2 : final momentum p1 : initial momentum
F : impulsive force
is defined as the product of a force, F and the time, t OR the change of momentum. momentum is a vector quantity whose direction is the same as the constant force on the object. 5
PHYSICS
CHAPTER 3
The S.I. unit of impulse is N s or kg m s−1. If the force acts on the object is not constant then
J=
∫
t2 t1
Fdt = Fav dt
Fav : average impulsive force
where Since impulse and momentum are both vector quantities, then it is often easiest to use them in component form :
consider 2-D collision only
J x = ( Fav ) x dt = p2 x − p1 x = m( v x − u x )
(
J y = ( Fav ) y dt = p2 y − p1 y = m v y − u y
)
J z = ( Fav ) z dt = p2 z − p1 z = m( v z − u z ) 6
PHYSICS
CHAPTER 3
When two objects in collision, the impulsive force, F against time, t graph is given by the Figure 3.20.
F
Figure 3.20
0 t1
t2
t
Shaded area under the F−t graph = impulse Picture 3.1
Picture 3.3
Picture 3.2 7
PHYSICS
CHAPTER 3
Example 3.1 : 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. Solution : m1 = 0.20 kg
u1 = 100 m s − 1 1 Wall (2)
v1 = 70 m s
−1
1
v2 = u 2 = 0 8
PHYSICS
CHAPTER 3
Solution : a. From the equation of impulse that the force is constant,
J = dp = p2 − p1
J = m1 ( v1 − u1 )
Therefore the magnitude of the impulse is 34 N s. s b. Given the contact time,
J = Fav dt
9
PHYSICS
CHAPTER 3
Example 3.2 : F ( kN )
18
0 0.2
1.0
1.8
t ( ms )
Figure 3.21
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 being served so it is nearly at rest initially. 10
PHYSICS
CHAPTER 3 −3
Solution : m = 60.0 × 10 kg a. From the force-time graph,
J = area under the F − t graph
= 0 J = dp = m( v − u )
b. Given the ball’s initial speed, u
11
PHYSICS
CHAPTER 3
Exercise 3.1 : 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 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.) ANS. : 293 N 12
PHYSICS
CHAPTER 3
Exercise 3.1 : 3.
Figure 3.22
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 −2.4 N s 13
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 of that system is constant.” constant OR “When the net external force on a system is zero, the total momentum of that system is constant.” constant In a Closed system,
∑
F= 0
From the Newton’s second law, thus
∑
dp F= = 0 dt
dp = 0 14
PHYSICS Therefore
CHAPTER 3 p = constant
∑ ∑
then
p x = constant p y = constant
According to the principle of conservation of linear momentum, we obtain The total of initial momentum = the total of final momentum OR
∑
pi =
∑
pf
15
PHYSICS
CHAPTER 3
Linear momentum in one dimension collision
Example 3.3 :
u A = 6 m s− 1
uB = 3 m s− 1 B
A
Figure 3.14
Figure 3.14 shows an object A of mass 200 g collides head-on with object B of mass 100 g. After the collision, B moves at a speed .of 2 m s-1 to the left. Determine the velocity of A after Collision :Solution m = 0.200 kg; m = 0.100 kg; u = − 6 m s − 1 A
B
−1
u B = 3 m s ; vB = − 2 m s
∑
pi =
∑
−1
A
pf
16
PHYSICS
CHAPTER 3
Linear momentum in two dimension collision
Example 3.4 : m1
u1
m2
50 m1
Before collision
v1
After collision
Figure 3.15
A tennis ball of mass m1 moving with initial velocity u1 collides with a soccer ball of mass m2 initially at rest. After the collision, the tennis ball is deflected 50° from its initial direction with a velocity v1 as shown in figure 3.15. Suppose that m1 = 250 g, m2 = 900 g, u1 = 20 m s−1 and v1 = 4 m s−1. Calculate the magnitude and direction of soccer ball after the collision. Simulation 3.3 17
PHYSICS
CHAPTER 3 m1 = 0.250 kg; m2 = 0.900 kg; u1 = 20 m s − 1 ;
Solution :
u 2 = 0; v1 = 4 m s − 1 ; θ1 = 50
From the principle of conservation of linear momentum,
∑
pi =
∑
pf
The x-component of linear momentum,
pix = p fx m1u1 x + m2 u 2 x = m1v1 x + m2 v2 x
∑
∑
18
PHYSICS
CHAPTER 3
Solution : The y-component oflinear momentum,
∑
piy =
∑
p fy
0 = m1v1 y + m2 v2 y
Magnitude of the soccer ball,
v2 =
( v2 x ) 2 +
(v )
2
2y
v2 y − 1 0.851 = tan θ 2 = tan 4.84 v2 x
Direction of the soccer ball,
− 1
19
PHYSICS
CHAPTER 3
Exercise 3.2 : 1. An object P of mass 4 kg moving with a velocity 4 m s−1 collides elastically with another object Q of mass 2 kg moving with a velocity 3 m s−1 towards it. a. Determine the total momentum before collision. b. If P immediately stop after the collision, calculate the final velocity of Q. c. If the two objects stick together after the collision, calculate the final velocity of both objects. ANS. : 10 kg m s−1; 5 m s−1 to the right; 1.7 m s−1 to the right 2. A marksman holds a rifle of mass mr = 3.00 kg loosely in his hands, so as to let it recoil freely when fired. He fires a bullet of mass mb = 5.00 g horizontally with a velocity 300 m s-1. Determine a. the recoil velocity of the rifle, b. the final momentum of the system. ANS. : −0.5 m s−1; U think.
20
PHYSICS
CHAPTER 3
3. 1.20 kg
1.80 kg
Before 0.630 m s-1
After
1.40 m s-1
Figure 3.16
In Figure 3.16 show a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless tabletop. The bullet passes through the first block, with mass 1.20 kg, and embeds itself in the second block, with mass 1.80 kg. Speeds of 0.630 m s−1 and 1.40 m s-1, respectively, are thereby given to the blocks. Neglecting the mass removed from the first block by the bullet, determine a. the speed of the bullet immediately after it emerges from the first block and .b. the initial speed of the bullet ANS. : 721 m s−1; 937.4 m s−1
21
PHYSICS
CHAPTER 3
Exercise 3.2 : 4. A ball moving with a speed of 17 m s−1 strikes an identical ball that is initially at rest. After the collision, the incoming ball has been deviated by 45° from its original direction, and the struck ball moves off at 30° from the original direction as shown in Figure 3.17. Calculate the speed of each ball after the collision.
Figure 3.17
ANS. : 8.80 m s− 1; 12.4 m s−1 22
PHYSICS
CHAPTER 3
3.2.2 Collision
is defined as an isolated event in which two or more bodies (the colliding bodies) exert relatively strong forces on each other for a relatively short time. time
Two types of collisions :
Elastic collision
Inelastic (non-elastic) collision
23
PHYSICS
CHAPTER 3
Elastic collision is defined as one in which the total kinetic energy (as well as total momentum) of the system is the same before and after the collision. collision Figure 3.18 shows the head-on collision of two billiard balls. 1
Before collision
At collision
After collision
m1u1 m2 u 2
1
m1v1
2
1
2
Simulation 3.4
2 Figure 3.18
m2 v 2 24
PHYSICS
CHAPTER 3
The properties of elastic collision are a. The total momentum is conserved. conserved
∑
pi =
∑
pf
b. The total kinetic energy is conserved. conserved
∑
Ki =
∑
Kf
OR
1 1 1 1 2 2 2 m1u1 + m2 u 2 = m1v1 + m2 v22 2 2 2 2 25
PHYSICS
CHAPTER 3
Inelastic (non-elastic) collision is defined as one in which the total kinetic energy of the system is not the same before and after the collision (even though the total momentum of the system is conserved). conserved) Figure 3.19 shows the model of a completely inelastic collision of two billiard balls. u = 0 Before collision
At collision
After collision (stick together)
1
m1u1
2
2
m2 1
1
Simulation 3.5
2
2
v
Figure 3.19
26
PHYSICS
CHAPTER 3
Caution: Not all the inelastic collision is stick together. together In fact, inelastic collisions include many situations in which the bodies do not stick
The properties of inelastic collision are a. The total momentum is conserved. conserved
∑
pi =
∑
pf
b. The total kinetic energy is not conserved because some of the energy is converted to internal energy and some of it is transferred away by means of sound or heat. heat But the total energy is conserved. conserved
∑
Ei =
∑
Ef OR
∑
Ki =
∑
K f + losses energy 27
PHYSICS
CHAPTER 3
Example 3.5 : Ball A of mass 400 g and velocity 4 m s-1 collides with ball B of mass 600 g and velocity 10 m s-1. After collision, A and B will move together. Determine the final velocity of both balls if A and B moves in the opposite direction initially Solution : mA = 0.4 kg, uA = 4 m s-1 , mB = 0.6 kg, uB = -10 m s-1, inelastic collision Before collision
A
A B
After collision
∑
uA
pi =
∑
uB
B
v= ?
pf
By using the principle of conservation of linear momentum, thus A A B B A B
m u + m u = ( m + m )v
28
PHYSICS
CHAPTER 3
Solution :
m Au A + mB u B v= m A + mB
Final velocity of both balls is - 4.4 m s
-1
29
PHYSICS
CHAPTER 3
Example 3.6 : A ball A of mass 1 kg moving at a velocity of 4 m s-1 collides with ball B of mass 2 kg which at rest. Calculate the velocity of both balls after collision if the collision is an elastic collision. Solution : Given mA = 1 kg, uA = 4 m s-1, mB = 2 kg, uB = 0 m s-1, elastic collision Before collision
After collision
A
u A u B = 0ms − 1
vA = ?
B
vB = ? A
B
30
PHYSICS
CHAPTER 3
Solution : Apply principle of conservation of momentum,
m Au A + m B u B = m A v A + m B v B 1(4) + 2(0) = 1(v A ) + 2(vB ) vA = 4 - 2 v B m s − 1 ……..(1) Apply principle of conservation of kinetic energy,
1 1 1 2 2 2 m A (u A ) = m A (v A ) + mB (vB ) 2 2 2 m A (u A ) = m A (v A ) + mB (vB ) 2
2
2
31
PHYSICS
CHAPTER 3
Solution :
1(4) 2 = 1(v A ) 2 + 2(vB ) 2 2
16 = v A + vB
2
………..(2)
Substitute equation (1) into equation (2) 2
16 = (4 − 2vB ) + 2vB 2 2 16 = 4(2 − vB ) + 2vB 2
8 = 2(2 − vB ) + vB 2
2
32
PHYSICS
CHAPTER 3
Solution :
8vB = 3vB
2
Substitute vB = 2.67 ms
−1
into equation (1),
vA = 4 - 2 (2.67)
33
PHYSICS
CHAPTER 3
THE END… Next Chapter… CHAPTER 4 :
Forces
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PHYSICS
CHAPTER 4
CHAPTER 4: FORCES
1
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PHYSICS
CHAPTER 4
4. FORCES 4.1 Basic of Forces and Free Body Diagram 4.2 Newton’s Laws of Motion
2
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PHYSICS CHAPTER 4 Learning Outcome: 4.1 Basic of Forces and Free Body Diagram At the end of this chapter, students should be able to: Identify the forces acting on a body in different situations. Weight Tension Normal force Friction Determine weight, static friction and kinetic friction Draw free body diagram Determine the resultant force
3
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PHYSICS
CHAPTER 4
4.1 Basic of Forces and Free Body Diagram Weight, is defined as the force exerted on a body under gravitational field. It is a vector quantity.
It is dependant on where it is measured, because the value of g varies at different localities on the earth’s surface. It always directed toward the centre of the earth or in the same direction of acceleration due to gravity, g. The S.I. unit is kg m s-2 or Newton (N). Equation:
W mg
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PHYSICS
CHAPTER 4
Tension, T The tension force is the force that is transmitted through a string, rope, cable or wire when it is pulled tight by forces acting from opposite ends. The tension force is directed along the length of the wire and pulls equally on the objects on the opposite ends of the wire.
Figure 4.1
5
PHYSICS
CHAPTER 4
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Normal (reaction) force,
Figure 4.2
N or R
is defined as a reaction force that exerted by the surface to an object interact with it and the direction always perpendicular to the surface. An object lies at rest on a flat horizontal surface as shown in Figure 4.2. Action: weight of an object is exerted on the N horizontal surface Reaction: surface is exerted a force, N on the object
W mg
F
y
N mg 0
Therefore
A free body diagram is defined as a diagram showing the chosen body by itself, with vectors drawn to show the magnitude and directions of all the forces applied to the body by the other bodies that interact with it.
N mg 6
PHYSICS
CHAPTER 4
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Friction
is defined as a force that resists the motion of one surface relative to another with which it is in contact. is independent of the area of contact between the two surfaces.. is directly proportional to the reaction force. OR
f N
f N
where
f : frictional force
μ : coefficient of friction
N : reaction force
Coefficient of friction, is defined as the ratio between frictional force to reaction force. OR
f N
is dimensionless and depends on the nature of the surfaces. 7
PHYSICS www.kms.matrik.edu.my/physics
CHAPTER 4
There are three types of frictional force :
Static, fs (frictional force act on the object before its move)
Kinetic, fk (frictional force act on the object when its move)
Rolling, fr (frictional force act on the object when its rolling)
f s s N
f k k N
f r r N
Can be ignored where thus
fr fk fs r k s
Caution: The direction of the frictional force exerted by a surface on an object is always in the opposite direction of the motion. The frictional and the reaction forces are always perpendicular. Simulation 4.1 8
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PHYSICS
CHAPTER 4
Example 4.1: A mass is resting on a flat surface which has a normal force of 98N, with a coefficient of static friction of 0.35. What force would it take to move the object? Solution:
N = 98N, μs = 0.35
f s s N
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PHYSICS
CHAPTER 4
Example 4.2: A 15 kg piece of wood is placed on top of another piece of wood. There is 35N of static friction measured between them. Determine the coefficient of static friction between the two pieces of wood. Solution: N = mg = 15(9.81) = 147.15 N, Fs = 35 N
fs s N
10
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PHYSICS
CHAPTER 4
Example 4.3 A dock worker loading crates on a ship finds that a 15 kg crate, initially at rest on a horizontal surface, requires a 50 N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 30 N is required to keep it moving with a constant speed. The acceleration of gravity is 9.8 ms-2. Find the coefficient of kinetic friction. Solution: Mass of crate = m = 15 kg Force required to set the crate in motion = F1 = 50 N Force required to keep the crate in moving at constant speed = fk = 30 N Acceleration of gravity = g = 9.81 ms-2 Normal force, N = mg = =
11
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PHYSICS Solution: From
CHAPTER 4 fk k N
12
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PHYSICS Resultant force
CHAPTER 4
Is defined as a single force that represents the combined effect of two or more forcesy
Example 4.4:
F1 (10 N) 30o
x
O 30o
F3 (40 N)
F2 (30 N)
The figure above shows three forces F1, F2 and F3 acted on a particle O. Calculate the magnitude and direction of the 13 resultant force on particle O.
PHYSICS
CHAPTER 4 y
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Solution :
F2
F3 x
F2 x
30o 60o
30o
x O
F3 y
F3
Fr F F1 F2 F3 Fr Fx Fy Fx F1x F2 x F3x Fy F1 y F2 y F3 y
F2 y F1
14
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PHYSICS Vector
CHAPTER 4 x-component
y-component
F1
F1x 0 N
F1 y F1 F1 y 10 N
F2
F2 x 30 cos 60 F2 x 15 N
F3
F3 x 40 cos 30 F3 x 34.6 N
F2 y 30 sin 60 F2 y 26 N
F3 y 40 sin 30 F3 y 20 N
Vector sum
15
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PHYSICS
CHAPTER 4
Solution : The magnitude of the resultant force is
Fr
F F 2
2
x
y
y
1 θ tan
and
Fy Fx
Fr
Fy
162 18
Fx
O
x
Its direction is 162 from positive x-axis OR 18 above negative xaxis. 16
PHYSICS
CHAPTER 4
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Exercise 4.1: 1. Given three vectors P, Q and R as shown in Figure 4.3.
Q 24 m s 2
R 10 m s 2
P 35 m s 2
y
50 0
x
Figure 4.3 Calculate the resultant vector of P, Q and R. ANS. : 49.4 m s2; 70.1 above + x-axis
17
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PHYSICS CHAPTER 4 Learning Outcome: 4.2 Newton’s laws of motion At the end of this chapter, students should be able to: State Newton’s First Law Define mass as a measure of inertia. Define the equilibrium of a particle. Apply Newton’s First Law in equilibrium of forces State and apply Newton’s Second Law
dp d dv dm F mv v m dt dt dt dt
State and apply Newton’s Third Law.
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PHYSICS
CHAPTER 4
4.2 Newton’s laws of motion 4.2.1 Newton’s first law of motion
states “an object at rest will remain at rest, or continues to move with uniform velocity in a straight line unless it is acted upon by a external forces” OR
Fnett
F 0
The first law gives the idea of inertia.
19
PHYSICS
CHAPTER 4
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4.2.2 Inertia Inertia is defined as the tendency of an object to resist any change in its state of rest or motion. is a scalar quantity.
Mass, m is defined as a measure of a body’s inertia. is a scalar quantity. The S.I. unit of mass is kilogram (kg). The value of mass is independent of location. If the mass of a body increases then its inertia will increase.
20
PHYSICS www.kms.matrik.edu.my/physics
CHAPTER 4
Figures 4.4a and 4.4b show the examples of real experience of inertia.
Figure 4.4
21
PHYSICS
CHAPTER 4
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4.2.3 Equilibrium of a particle
is defined as the vector sum of all forces acting on a particle (point) must be zero. The equilibrium of a particle ensures the body in translational equilibrium and its condition is given by
F Fnett 0
Newton’s first law of motion This is equivalent to the three independent scalar equations along the direction of the coordinate axes,
F
x
0,
F
y
0,
F
z
0
There are two types of equilibrium of a particle. It is
Static equilibrium (v=0) body remains at rest (stationary).
Dynamic equilibrium (a=0) body moving at a uniform (constant) velocity. 22
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PHYSICS
CHAPTER 4
Problem solving strategies for equilibrium of particle
a
The following procedure is recommended when dealing with problems involving the equilibrium of a particle: Sketch a simple diagram of the system to help conceptualize the problem. Sketch a separate free body diagram for each body. Choose a convenient coordinate axes for each body and construct a table to resolve the forces into their components. Apply the condition for equilibrium of a particle in component form : Fx 0 and Fy 0
Solve the component equations for the unknowns.
23
PHYSICS
CHAPTER 4
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Example 4.5: A load of 250 kg is hung by a crane’s cable. The load is pulled by a horizontal force such that the cable makes a 30 angle to the vertical plane. If the load is in the equilibrium, calculate a. the magnitude of the tension in the cable, b. the magnitude of the horizontal force. (Given g =9.81 m s2) Solution : m 250 kg Free body diagram of the load :
30
T Ty 30 F
60
F
Tx
mg 24
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PHYSICS
CHAPTER 4
Solution : m 250 kg 1st method : a. Force x-component (N)
mg F T
y-component (N)
mg 250 9.81 2453 0 T sin 60
0
F T cos 60
Since the load is in the equilibrium, then Thus
F
F 0
Fx 0 y
0
F T cos 60 0
(1)
T sin 60 2453 0 (2)
b. By substituting eq. (2) into eq. (1), therefore
F 2833 cos 60 0
25
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PHYSICS
CHAPTER 4
Solution : m 250 kg 2nd method : a. Since the load is in the equilibrium, then a closed triangle of forces can be sketched as shown below.
30 mg
T
F b.
From the closed triangle of forces, hence
mg cos30 T
F sin 30 T
F sin 30 2833
26
PHYSICS
CHAPTER 4
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Example 4.6:
F1 12 N
F2 20 N
30.0 45.0
55.0
A
F3 30 N Figure 4.5 Calculate the magnitude and direction of a force that balance the three forces acted at point A as shown in Figure 4.5.
27
PHYSICS
CHAPTER 4 F1 12 N; F2 20 N; F3 30 N
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Solution :
Force
x-component (N)
y-component (N)
F1 F2 F3 F
12 cos 55.0 6.88 20 cos 30.0 17.3 30 cos 45.0 21.2 Fx
12 sin 55.0 9.83 20 sin 30.0 10.0 30 sin 45.0 21.2 Fy
To find a force to balance the three forces means the system must be in equilibrium hence
F
x
0
28
PHYSICS
CHAPTER 4
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Solution :
F
0 9.83 10.0 21.2 Fy 0 y
The magnitude of the force,
F and its direction,
Fx 2 Fy 2
31.6 2 1.37 2
1
Fy θ tan Fx 1 1.37
θ tan 31.6 29
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Example 4.7:
F
50.0
Figure 4.6 A window washer pushes his scrub brush up a vertical window at constant speed by applying a force F as shown in Figure 4.6. The brush weighs 10.0 N and the coefficient of kinetic friction is k= 0.125. Calculate
a. the magnitude of the force F , b. the normal force exerted by the window on the brush.
30 30
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Solution : W 10.0 N; μ 0.125 k a. The free body diagram of the brush :
F
constant speed
50.0
N
fk
W
Force
F W N fk
x-component (N)
y-component (N)
F cos 50.0
F sin 50.0
0
10.0
N
0 μk N 0.125 N
0
The brush moves up atconstant speed (a=0) so that
Thus
F
Fx 0 y
0
F ma 0
N F cos 50.0
(1)
F sin 50.0 0.125 N 10.0
(2)
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Solution : a. By substituting eq. (1) into eq. (2), thus
F sin 50.0 0.125 F cos 50.0 10.0 b. Therefore the normal force exerted by the window on the brush is given by
N F cos 50.0
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Exercise 4.2: Use gravitational acceleration, g = 9.81 m s2 1.
Figure 4.7 The system in Figure 5.8 is in equilibrium, with the string at the centre exactly horizontal. Calculate
a. the tensions T1, T2 and T3. b. the angle . ANS. : 49 N, 28 N, 57 N; 29 33
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Exercise 4.2: 2.
Figure 4.8 A 20 kg ball is supported from the ceiling by a rope A. Rope B pulls downward and to the side on the ball. If the angle of A to the vertical is 20 and if B makes an angle of 50 to the vertical as shown in Figure 4.8, Determine the tension in ropes A and B. ANS. : 134 N; 300 N 34
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Exercise 4.2: 3.
Figure 4.9 A block of mass 3.00 kg is pushed up against a wall by a force P that makes a 50.0 angle with the horizontal as show in Figure 4.9. The coefficient of static friction between the block and the wall is 0.250. Determine the possible values for the magnitude of P that allow the block to remain stationary. ANS. : 31.8 N; 48.6 N 35
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Newton’s second law of motion
states “the rate of change of linear momentum of a moving body is proportional to the resultant force and is in the same direction as the force acting on it” OR its can be represented by
dp F dt
where
F : resultant force
dp : change in linear momentum
dt : time interval
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CHAPTER 4
From the Newton’s 2nd law of motion, it also can be written as
dp an p mv F dt d dm d mv dv F v dt m dt F dt
Case 1: Object at rest or in motion with constant velocity but with changing mass. For example : Rocket
dm dv F v m dt dt dm F v dt
and
dv 0 dt
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Case 2: Object at rest or in motion with constant velocity and constant mass.
dm dv dv dm 0 F v m where 0 and dt dt dt dt st law of motion Newton’s 1 F 0 dp Thus F 0 dt
p constant
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Case 3: Object with constant mass but changing velocity.
dm dv dm an F v m 0 dt dt dt d dv dv and a F m dt dt
F ma
where F : resultant force
m : mass of an object a : acceleration
The direction of the resultant force always in the same direction of the motion or acceleration.
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CHAPTER 4
Newton’s 2nd law of motion restates that “The acceleration of an object is directly proportional to the nett force acting on it and inversely proportional to its mass”. OR F
a
m
One newton(1 N) is defined as the amount of nett force that gives an acceleration of one metre per second squared to a body with a mass of one kilogramme. OR 1 N = 1 kg m s-2 Notes: F is a nett force or effective force or resultant force. The force which causes the motion of an object. If the forces act on an object and the object moving at uniform acceleration (not at rest or not in the equilibrium) hence
Fnett
F ma
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Newton’s third law of motion
states “every action force has a reaction force that is equal in magnitude but opposite in direction”. For example : When the student push on the wall it will push back with the same force. (refer to Figure 4.10) B (wall)
FBA
A (hand)
FAB
FAB FBA
Figure 4.10
Where
FAB is a force by the hand on the wall (action) FBA is a force by the wall on the hand (reaction) 41
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When a book is placed on the table. (refer to Figure 4.11) Force by the table on the book (reaction)
Figure 4.11
Force by the book on the table (action)
If a car is accelerating forward, it is because its tyres are pushing backward on the road and the road is pushing forward on the tyres. A rocket moves forward as a result of the push exerted on it by the exhaust gases which the rocket has pushed out.
In all cases when two bodies interact, the action and reaction forces act on different bodies. 42
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Applications of Newton’s 2nd law of motion
From the Newton’s second law of motion, we arrived at equation
F F ma There are five steps in applying the equation above to solve nett
problems in mechanics: Identify the object whose motion is considered. Determine the forces exerted on the object. Draw a free body diagram for each object. is defined as a diagram showing the chosen body by itself, with vectors drawn to show the magnitude and directions of all the forces applied to the body by the other bodies that interact with it. Choose a system of coordinates so that calculations may be simplified. Apply the equation above, Along x-axis: Fx ma
Along y-axis: F
y
x
may
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Example 4.8: Three wooden blocks connected by a rope of negligible mass are being dragged by a horizontal force, F in Figure 4.12.
F
m1
T1
m2
T2
m3
Figure 4.12
Suppose that F = 1000 N, m1 = 3 kg, m2 = 15 kg and m3 = 30 kg. Determine a. the acceleration of blocks system. b. the tension of the rope, T1 and T2. Neglect the friction between the floor and the wooden blocks.
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Solution : a. For the block, m1 = 3 kg
F
a
m1
T1
F F
x
x
F T1 m1a 1000 T1 3a (1)
For the block, m2 = 15 kg
T1
a
m2
T2
F F
x
T1 T2 m2 a
x
T1 T2 15a (2)
For the block, m3 = 30 kg
T2
a
F
x
T2 m3 a
m3
(3) 45
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Solution : a. By substituting eq. (3) into eq. (2) thus
T1 45a 0
(4)
Eq. (1)(4) :
b. By substituting the value of acceleration into equations (4) and (3), therefore
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Example 4.9: Two objects of masses m1 = 10 kg and m2 = 15 kg are connected by a light string which passes over a smooth pulley as shown in Figure 4.13. Calculate a. the acceleration of the object of mass 10 kg. b. the tension in the each string. (Given g = 9.81 m s2) Solution : a. For the object m1= 10 kg,
T1
a
Simulation 4.2
F
T1 m1 g m1a
where
T1 T2 T
y
T 10 g 10a W1 m1 g
m1 m2 Figure 4.13
(1) 47
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Solution : a. For the object m2= 15 kg,
T2
a
F F
m2 g T2 m2 a y 15 g T 15a T 15 g 15a y
(2)
Eq. (1) + (2) :
W2 m2 g b. Substitute the value of acceleration into equation (1) thus
T 109.81 101.96
Therefore 48
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Example 4.10: Two blocks, A of mass 10 kg and B of mass 30 kg, are side by side and in contact with each another. They are pushed along a smooth floor under the action of a constant force F of magnitude 200 N applied to A as shown in Figure 4.14. Determine a. the acceleration of the blocks, B A F b. the force exerted by A on B. Simulation 4.3 Solution :
Figure 4.14
mA 10 kg; mB 30 kg; F 200 N
a. Let the acceleration of the blocks is a. Therefore
F m
mB a F mA mB a x
A
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Solution : b. For the object A,
a
F
A
FBA
F
x
F FBA mAa
200 FBA 105.0
From the Newton’s 3rd law, thus OR For the object B,
a
F
x
FAB
FAB mB a
B 50
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Exercise 4.3: 1. A block is dragged by forces, F1 and F2 of the magnitude 20 N and 30 N respectively as shown in Figure 4.15. The frictional force f exerted on the block is 5 N. If the weight of the block is 200 N and it is move horizontally, determine the acceleration of the block. (Given g = 9.81 m
s2)
F1
a
50
f
20
F2
Figure 4.15
ANS. : 1.77 m s2
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Exercise 4.3: 2. One 3.5 kg paint bucket is hanging by a massless cord from another 3.5 kg paint bucket, also hanging by a massless cord as shown in Figure 4.16. If the two buckets are pulled upward with an acceleration of 1.60 m s2 by the upper cord, calculate the tension in each cord. (Given g = 9.81 m s2)
ANS. : 39.9 N; 79.8 N
Figure 4.16
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THE END… Next Chapter… CHAPTER 5 : Work, Energy and Power
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CHAPTER 5
WORK, ENERGY AND POWER
CHAPTER 5: Work, Energy and Power (3 Hours)
1
CHAPTER 5
WORK, ENERGY AND POWER
Learning Outcome: 5.1 Work (1 hour)
At the end of this chapter, students should be able to: (a) Define and use work done by a force.
W F s (b) Determine work done from the forcedisplacement graph. 2
CHAPTER 5 5.1
WORK, ENERGY AND POWER
Work, W
Work done by a constant force
is defined as the product of the component of the force parallel to the displacement times the displacement of a body. OR is defined as the scalar (dot) product between force and displacement of a body. 3
CHAPTER 5 Mathematically :
WORK, ENERGY AND POWER
W F s W F cos θ s Fs cos θ
Where,
F : magnitude of force
s : displaceme nt of the body
θ : the angle between F and s
4
CHAPTER 5
WORK, ENERGY AND POWER
It is a scalar quantity.
Dimension :
W F s W ML2T 2
The S.I. unit of work is kg m2 s2 or joule (J).
The joule (1 J) is defined as the work done by a force of 1 N which results in a displacement of 1 m in the direction of the force.
1 J 1 N m 1 kg m 2 s 2 5
CHAPTER 5
WORK, ENERGY AND POWER
Work done by a variable force Figure 5.1 shows a force, F whose magnitude changes with the displacement, s. For a small displacement, s1 the force remains almost constant at F1 and work done therefore becomes W1=F1 s1 .
6
CHAPTER 5
WORK, ENERGY AND POWER F/N FN
F4 F1 0
s1
Figure 5.1
W1 s1
s4
s2
ssN
To find the total work done by a variable force, W when the displacement changes from s=s1 to s=s2, we can divide the displacement into N small successive displacements :
s1 , s2 , s3 , …, sN Thus
W F1s1 F2 s2 ... FN s N
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CHAPTER 5
WORK, ENERGY AND POWER
When N , s 0, therefore s2
W Fds s1
W the area under the force - displaceme nt graph F/N
Work = Area
0 s1
s2 s/m
8
CHAPTER 5
WORK, ENERGY AND POWER
Applications of work’s equation Case 1 :
Work done by a horizontal force, F on an object (Figure 4.2).
F
Figure 5.2
W Fs cos θ and θ 0 W Fs s
Case 2 :
Work done by a vertical force, F on an object (Figure 4.3).
F
Figure 5.3
W Fs cos θ and θ 90 W 0J s 9
CHAPTER 5
WORK, ENERGY AND POWER
Case 3 :
Work done by a horizontal forces, F1 and F2 on an object (Figure 5.4). F
W F s cos 0 1 1 1 W F s cos 0 2 2 s W W1 W2 F1s F2 s W F1 F2 s and Fnett F1 F2
F2
Figure 5.4
W W
nett
Case 4 :
Fnett s
Work done by a force, F and frictional force, f on an object (Figure 5.5). F
f
Figure 5.5
s
Wnett Fnett s and Fnett F cos θ f ma Wnett F cos f s OR Wnett mas
10
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WORK, ENERGY AND POWER
Caution :
Work done on an object is zero when F = 0 or s = 0 and = 90.
11
CHAPTER 5
Sign for work.
WORK, ENERGY AND POWER
W Fs cos
If 0 0 (positive) work done on the system ( by the external force) where energy is transferred to the system.
If 90
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