PRE-U STPM PHYSICS TERM 1

STPM Text

Summary of Key Quantities and Units





QUANTITY BASE QUANTITY SYMBOL UNIT
KUANTITI KUANTITI ASAS SIMBOL UNIT
Mass Jisim m kg
Length Panjang l m
Time Masa t s
Electric current Arus elektrik I A
Thermodynamic temperature Suhu termodinamik T K
Amount of substance Amaun bahan n mol

Other Quantities Kuantiti Lain
Distance Jarak d m
Displacement Sesaran s, x m
Area Luas A m 2
Volume Isi padu V m 3
Density Ketumpatan ρ kg m –3
Speed Laju u, v m s –1
Velocity Halaju u, v m s –1
Acceleration Pecutan a m s –2
Acceleration of free fall Pecutan jatuh bebas g m s –2
Force Daya F N
Weight Berat W N
Momentum Momentum p N s
Work Kerja W J
Energy Tenaga E, U J
Potential energy Tenaga keupayaan U J
Heat Haba Q J
Power Kuasa P W
Pressure Tekanan p Pa
Torque Tork  N m
–2
Gravitational constant Pemalar graviti G N kg m 2
Gravitational field strength Kekuatan medan graviti g N kg –1
Gravitational potential Keupayaan graviti V J kg –1
Moment of inertia Momen inersia I kg m 2
Angular displacement Sesaran sudut θ °, rad
Angular speed Laju sudut ω, θ • rad s –1
Angular velocity Halaju sudut ω, θ • rad s –1
••
Angular acceleration Pecutan sudut α, θ rad s –2
2
Angular momentum Momentum sudut L kg m rad s –1
Period Tempoh T s
Frequency Frekuensi f, v Hz
Angular frequency Frekuensi sudut ω rad s –1
Wavelength Panjang gelombang λ m
Electric charge Cas elektrik Q, q C
Current density Ketumpatan arus J A m –2
Electric potential Keupayaan elektrik V V
Electric potential difference Beza keupayaan elektrik V V
Electromotive force Daya gerak elektrik ε, E V

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Physics Term 1 STPM Summary of Key Quantities and Units

QUANTITY BASE QUANTITY SYMBOL UNIT
KUANTITI KUANTITI ASAS SIMBOL UNIT
Resistance Rintangan R Ω
Resistivity Kerintangan ρ Ω m
Conductance Konduktans G S = Ω –1
–1
Conductivity Kekonduksian σ S m = Ω m –1
–1
Electric field strength Kekuatan medan elektrik E N C –1
Permittivity Ketelusan ε F m –1
Permittivity of free space Ketelusan ruang bebas ε F m –1
0
Relative permittivity Ketelusan relatif ε –
r
Capacitance Kapasitans C F
Time constant Pemalar masa τ s
Magnetic flux Fluks magnet φ Wb
Magnetic flux density Ketumpatan fluks magnet B T
Self inductance Swainduktans L H
Mutual inductance Induktans saling M H
Reactance Reaktans X Ω
Impedance Impedans Z Ω
Permeability Ketelapan μ H m –1
Permeability of free space Ketelapan ruang bebas μ H m –1
0
Relative permeability Ketelapan relatif μ r –
Force constant Pemalar daya k N m –1
Modulus Young Modulus Young E, Y Pa
Tension Tegangan T N
Stress Tegasan σ Pa
Refractive index Indeks biasan n –
Critical angle Sudut genting θ °
c
Temperature Suhu T, θ K, °C
Heat capacity Muatan haba C J K –1
Specific heat capacity Muatan haba tentu c J K kg –1
–1
Latent heat Haba pendam L J
Specific latent heat Haba pendam tentu l J kg –1
Molar heat capacity Muatan haba molar C J K mol –1
–1
m
Principle molar heat Muatan haba molar utama C , C J K mol –1
–1
V.m p.m
capacity
Molar gas constant Pemalar gas molar R J K mol –1
–1
Boltzmann constant Pemalar Boltzmann k J K –1
Avogadro constant Pemalar Avogadro L, N A mol –1
Thermal conductivity Kekonduksian terma k, λ W m K –1
–1
Planck constant Pemalar Planck h J s
Activity of radioactive source Keaktifan sumber radioaktif A s , Bq
–1
Decay constant Pemalar reputan λ s –1
Half-life Setengah hayat t s
1/2
Atomic mass Jisim atom m kg
a
Relative atomic mass Jisim atom relatif A r –
Mass of electron Jisim elektron m e kg, u
Mass of neutron Jisim neutron m kg, u
n
Mass of proton Jisim proton m p kg, u
Molar mass Jisim molar M kg mol –1
Atomic mass unit Unit jisim atom u kg
Relative molecular mass Jisim molekul relatif M –
r
Proton number Nombor proton Z –
Nucleon number Nombor nukleon A –
Neutron number Nombor neutron N –
330






SUMMARY STPM PHY T1.indd 330 4/9/18 8:33 AM

CHAPTER
2 KINEMATICS














Concept Map




Kinematics






Linear Motion Projectile


u
Motion with Constant
Acceleration u sin

• v = u + at u cos

• s = 1 (u + v)t
2
• s = ut + 1 at 2
2
• v = u + 2as Effect of Air Resistance
2
2
v
Graphical Method

s v
t
0
a
0 t 0 t



t
0
Bilingual Keywords

Acceleration: Pecutan Resistance: Rintangan
Displacement: Sesaran Speed: Laju
Kinematics: Kinematik Velocity: Halaju
Projectile: Luncuran


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Physics Term 1 STPM Chapter 2 Kinematics

2.1 Linear Motion

2010/P1/Q3, 2016/P1/Q2,Q18
Learning Outcomes

Students should be able to:
• derive and use equations of motion with constant acceleration
• sketch and use the graphs of displacement-time, velocity-time and acceleration-time for the motion of a body with constant
acceleration 2

1. Kinematics is the study of motion without considering the causes of the motion. Linear motion
is motion along a straight line.

2. The displacement s of a body from a point O to another point P is the distance moved by the body
along the straight line OP. Displacement s has both magnitude and direction, hence it is a vector.
3. The velocity v of a body is its rate of change of displacement. v

ds O ds P
Velocity, v = Figure 2.1
dt
Velocity v is also a vector, its direction is along the direction of the change in displacement ds. (Figure 2.1).
ds
4. The expression v = gives the instantaneous velocity. The instantaneous velocity of a car
dt
travelling along a straight road is shown by the speedometer.
5. If the time taken by a body to move through a displacement s is t, then
Final displacement, s
average velocity =
Time taken, t














Figure 2.2 The speedometer shows
the instantaneous
velocity

6. Acceleration a is the rate of change of velocity.
dv
Acceleration, a =
dt
Acceleration is also a vector. Its direction follows the direction of the change in velocity dv. When
the velocity of a body changes either in magnitude, or in direction, or both, the body accelerates.


Motion With Constant Acceleration

1. The acceleration of a body is uniform if the magnitude of the acceleration is constant and its direction
remains unchanged.


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Physics Term 1 STPM Chapter 2 Kinematics

2. For a body travelling along a straight line, if u = initial velocity, v = final Exam Tips
velocity after a time t, and s = displacement in time t, There are five quantities,
Change in velocity
then uniform acceleration, a = u, v, t, s, a, in the four
Time taken equations for motion
= v – u under constant
acceleration. Each of the
t
equations involves only
v = u + at .............................. a four of the quantities.
2 In a question, usually
Displacement, s = Average velocity × Time values for three of the
quantities are provided,
u + v
s = ( ) t .............................. b and you are asked to
2
calculate the value of
another of the quantities.
( u + u + at Use the equation
2 )
s = t that contains the four
quantities mentioned in
the question.
2 2
1
s = ut + at .............................. c
Info Physics
2s
From ①, v – u = at From ➁, v + u = t A380 Airbus take-off speed
(v – u)(v + u) = at × 2s A wide-body aircraft such as
the Airbus A380 has a take-
t
–1
off speed of 280 km h and
v – u = 2as .............................. d ➃ requires a runway of length
2
2
3000 m. What then is its ac-
celeration?
Example 1
A motorist travelling at 72 km h on a straight road approaches a traffic light which turns red
–1
when he is 55.0 m away from the stop line. The reaction time is 0.7 s. With the brakes applied
fully, the vehicle decelerates at 5.0 m s . How far from the stop line will he stop and on which
–2
side of the stop line?
Solution:

–1
72 km h = 72 000 m = 20 m s –1
3 600 s
–1
The vehicle travels at constant speed of 20 m s during the reaction time of 0.7 s.
Distance travelled during the reaction time = 20 × 0.70 = 14.0 m
When the vehicle stops, final velocity v = 0
–2
Acceleration = –5.0 m s (negative because of deceleration)
2
2
Using v = u + 2as
2
Distance travelled when the brakes are applied, s = v – u 2
2a
= 0 – 20 2
2 (–5)
= 40.0 m
Total distance travelled = (14.0 + 40.0) m = 54.0 m
Hence, the vehicle stops (55.0 – 54.0) m = 1.0 m before the stop line.




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Physics Term 1 STPM Chapter 2 Kinematics

Example 2

The table below is from the handbook of a car. On a dry road, the car driven by an alert driver will
stop in distances as shown below.
Speed/ Thinking Braking Overall stopping
m s –1 distance/m distance/m distance/m
5.0 3.0 1.9 4.9 2
10.0 6.0 7.5 13.5
15.0 9.0 17.0 26.0
20.0 12.0 30.0 42.0
25.0 15.0 47.0 62.0
30.0 18.0 68.0 86.0
35.0 21.0 92.0 113.00

The thinking distance is the distance travelled by the car during the driver’s reaction time. The
braking distance is the distance travelled by the car before the car is stopped when the brakes are
applied.
(a) Explain why the thinking distance is directly proportional to the speed, whereas the braking
distance is not. State the relationship between the braking distance and the speed.
(b) What is the value of deceleration used in calculating the braking distance?
(c) Calculate the overall stopping distance for a car travelling at 40 m s .
–1
Solution:

(a) During the reaction time, t of the driver, the car travels at a constant speed, u.
Hence, thinking distance, s 1 = ut
s 1 ∝ u
The reaction time t of a driver is constant,
The final speed after the brakes are applied = 0
2
If deceleration = a, using v = u + 2as
2
0 = u – 2as 2
2
u 2
Braking distance, s 2 = ∝ u 2
2a
u 2
(b) Using braking distance, s 2 =
2a
When u = 10 m s , s 2 = 7.5 m
–1
u 2
Deceleration, a =
2s
10 2
=
2 × 7.5
= 6.67 m s –2
(c) When u = 40 m s –1 2
u
Thinking distance, s 1 = 40 × Reaction time Braking distance, s 2 = 2a
2
40
3.0
= 40 × (5.0) = 2 × 6.67
= 24 m = 120 m
Overall distance travelled = (24 + 120) m = 144 m

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Physics Term 1 STPM Chapter 2 Kinematics

Quick Check 1


1. A car starts from rest and accelerates at 2. A car decelerates uniformly from 30 m s to
–1
2.0 m s . 15 m s in a distance of 75 m. Calculate the
–2
–1
(a) Find its speed at (i) 1.0 s, (ii) 2.0 s and further distance travelled before the car comes
(iii) 5.0 s. to a stop.
2 (b) What is the average speed in (i) the first
second, (ii) 2.0 s and (iii) 5.0 s? 3. A car accelerates through three gear changes
(c) Calculate the distance travelled in with the following speed:
–1
(i) 1.0 s, (ii) 2.0 s and (iii) 5.0 s. 20 m s for 2.0 s
40 m s for 2.0 s
–1
–1
60 m s for 6.0 s
What is the overall average speed of the car?
Motion Under Gravity 2009/P1/Q2, 2011/P2/Q1, 2017/P1/Q2


1. When a body is released from rest, it falls towards the earth with an acceleration.
2. In free space where there is no air resistance, all objects irrespective of their mass, fall with the same
acceleration, known as the acceleration of free fall or acceleration due to gravity, g.

3. Close to the earth, the value of g can be as assumed constant. In this book, the value of g is assumed
–2
to be 9.81 m s unless otherwise stated.
4. Since the value of g is constant, equations of motion under uniform acceleration such as:
1
2
v = u + at, s = ut + at and v = u + 2as
2
2
2
can be used for motion under gravity by substituting the acceleration a as +g, or as –g where applicable.
5. When the body is only falling downwards, it may be easier to use a = +g, since the direction of
motion is in the direction of g, that is downwards.
6. When the body is projected vertically and upwards, if it is assumed that the velocity of the body
is positive, it implies that the upwards direction is assumed positive. Since the direction of g is
downwards, hence acceleration a = –g.
7. When the body moves in both directions, upwards then falls downwards or vice versa, it would be
better to assume the upwards direction as positive. Hence, the acceleration a = –g.
Upwards journey Downwards journey
Displacement : + E E Displacement : +
Velocity : + F Velocity : –
Acceleration : a = –g D Acceleration : a = –g
G
C

B H

A Below the level AI
I
Displacement : –
Velocity : –
J Acceleration : a = –g
Free Fall
INFO Figure 2.3

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Physics Term 1 STPM Chapter 2 Kinematics

8. Figure 2.3 shows a particle being projected upwards from the point A. It reaches the highest point
E and then falls down to a point below A.


Example 3


A bullet is fired vertically upwards from the ground with an initial velocity of u and takes a time 2
of t 1 to reach a point P at a height h. From then the bullet takes a further time of t 2 to move to the
highest point and back to the ground. Find in terms of t 1 , t 2 and g
(a) the initial velocity u,
(b) height h, and
(c) the greatest height H reached.

Solution:
(a) Assume the upwards direction as positive, then a = –g. H
Total time taken for the bullet to move up and back to the ground,
t = (t 1 + t 2 ) P
When the bullet returns to the ground, H v t 1
displacement, s = 0
1 h
Using s = ut + at 2
2 (t + t )
1
2
1
0 = u(t 1 + t 2 ) – g(t 1 + t 2 ) 2
2
1
u = g(t 1 + t 2 )
2
(b) When s = h, t = t 1 (c) At the highest point, s = H, velocity = 0
Using v = u + 2as
2
2
1
Using s = ut + at 2
2 0 = u – 2gH
2
1 1
2 u = g(t 1 + t 2 )
2 2 u 2 1
h = ut 1 – gt 1
H = 2g u = g(t 1 + t 2 )
2
= 1 1 2
2 g(t 1 + t 2 )t 1 – gt 1 1 1 ] 2
2
= g(t 1 + t 2 )
1 2g[ 2
=
2 gt 1 t 2 1
= g (t 1 + t 2 ) 2
8
Quick Check 2
1. A ball is thrown vertically upwards from the 2. A spaceship descends at a constant velocity
top of a building, 9.0 m above the ground with of 10 m s on the moon. When it is 120 m
–1
a speed of 8.0 m s . The ball hits the ground from the moon’s surface, an object falls off the
–1
after a time T. Calculate spaceship. If the acceleration due to gravity on
(a) the time taken to reach the greatest height. the moon is 1.6 m s , what is the velocity of the
–2
(b) the greatest height reached by the ball. object when it reaches the surface of the moon?
(c) the time, T taken by the ball to reach the
ground.

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Physics Term 1 STPM Chapter 2 Kinematics

3. The acceleration of free fall is determined P
by timing the fall of a ball bearing using
photocells. The ball is released from the point Light X
P, and passes the point X and Y at time t 1 and beam
t 2 after being released. Find an expression for h Photo-cells
the acceleration due to gravity in terms of h, t 1 Y
and t 2 . Light
2 beam

Q

Graphical Methods

The motion of a body can be analysed by studying the various graphs for the motion:
(a) Displacement-time graph
(b) Velocity-time graph

Displacement-time Graph 2010/P1/Q2, 2014/P1/Q2

1. Informations that can be deduced from the displacement-time graph:
(a) The instantaneous displacement.
The displacement s at any time t can be obtained off the graph.
(b) The velocity = Gradient of the graph.
(c) The manner that the velocity changes with time from the shape of the graph.

2. The displacement-time (s-t) graphs of a few types of motion are as shown in Figure 2.4.
(a) s (b) s









t t
0 0
Constant velocity Increasing velocity
Gradient = constant Gradient increasing
(c) (d) s
s







0 t
0 t T 2T
Decreasing velocity Motion of a body projected vertically upwards and
(deceleration) then falling back to the ground. When t = T, the body
Gradient decreasing is at the highest point. When t = 2T, the body is back
on the ground.
Figure 2.4 Displacement-time graphs



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Physics Term 1 STPM Chapter 2 Kinematics

Exam Tips
Note the difference between graph (a) and graph (b) when t = 0.
v v



2



0 t 0 t
(a) (b)
Initial velocity u = 0 Initial velocity u ≠ 0
Gradient at t = 0 is zero. Gradient at t = 0 is not zero.


Velocity-time Graph

1. Informations that can be deduced from the velocity-time graph:
(a) The instantaneous velocity.
t 2
(b) Between the time t = t 1 and t = t 2 , displacement = ∫ v dt = shaded area under the graph.
t 1
(c) The acceleration, a = dv = Gradient of graph.
dt
(d) How the acceleration changes with time from the shape of the graph.

2. Figure 2.5 shows the velocity-time (v-t) graphs for various types of motion.
(a) v (b) v (c) v
u




u
0 t t t 0 t t t 0 t t t
1 2 1 2 1 2
Uniform acceleration. Uniform deceleration. Increasing acceleration.
Gradient = constant Constant negative gradient Gradient increasing
(d) v (e) v








0 t 0 t
t t t t
1 2 1 2
Decreasing acceleration. Constant velocity
Gradient decreasing
Figure 2.5 Velocity-time graphs







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Physics Term 1 STPM Chapter 2 Kinematics

Example 4


The graph shows the speeds of two cars A and B Speed / m s –1
which are travelling in the same direction
over a period of 80 s. Car A travelling at a constant 30 B
speed of 20 m s overtakes car B at time t = 0.
–1
2 In order to catch up with car A, car B immediately 20 A
accelerate uniformly for 30 s to reach a constant
speed of 30 m s .
–1
(a) How far does car A travel during the first 30 s?
(b) Calculate the acceleration of car B in the first 30 s. 5
(c) What is the distance travelled by car B in this 0 Time / s
time? 30 80
(d) What additional time will it take for car B to catch up with car A after car A passes car B?
(e) How far would each car have travelled since t = 0?
(f) What is the maximum distance between the cars before car B catches up with car A?

Solution:
(a) Distance travelled by car A during the first 30 s = 30 × 20
= 600 m
(b) Acceleration of car B = Gradient of graph
30 – 5
=
30
= 0.833 m s –2
(c) Distance travelled by car B during the first 30 s
= Area under the graph from t = 0 to t = 30 s
= 1 (5 + 30) × 30
2
= 525 m
(d) After 30 s, car A is ahead of car B by (600 – 525) m = 75 m
For car B to catch up with car A,
Further distance travelled by car B – Further distance travelled by car A = 75 m
30t – 20t = 75
t = 7.5 s
Hence, car B catches up with car A (30 + 7.5) s from when car A passes car B, i.e. 37.5 s.
(e) Distance travelled by each car = 20 × 37.5 m
= 750 m
(f) The two cars are furthest apart when the two graphs intercept at P.
At P, the difference (Area under graph A – Area under graph B) is maximum.
Refering to the figure:
x = 10 x + y = 30 Speed / m s –1
y 15
2 2
x = y y + y = 30 30 B
3 3
3
y = × 30 P
5 20 y A
= 18 s x
15
Hence, maximum separation = Shaded area
1
= × 18 × 15 5
2 Time / s
= 135 m 0 30 80
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Physics Term 1 STPM Chapter 2 Kinematics

Example 5


(a) A body accelerates uniformly from rest along a straight line. Sketch a graph to show the
variation of displacement with time. How can the instantaneous velocity be deduced from the
displacement-time graph?
(b) A cricket player throws a ball vertically upwards and catches it 3.0 s later. Neglecting air 2
resistance, calculate
(i) speed of the ball when it leaves the player’s hand,
(ii) the maximum height reached by the ball.

(c) Sketch a graph to show how the velocity of the ball varies with time. Mark each of the following
instants on the graph:
(i) The ball leaves the player’s hand (t 1 ).
(ii) The ball at the maximum height (t 2 ).
(iii) The ball returns to the player’s hand again (t 3 ).
(d) You were told that the air resistance is negligible. In actual fact the ball experiences a retarding
force during its motion. Without making any calculation, explain how the air resistance affects
(i) the time taken to reach the highest point,
(ii) the value of the maximum height reached.

(e) Discuss, taking into account the air resistance, whether the time t u taken by the ball to reach
the highest point is greater or smaller than t d , the time taken for the ball to drop.
–2
(Assume g = 10 m s ).
Solution:
(a) Instantaneous velocity = Gradient of the graph at time, t
s





0
t
Note that the gradient of the graph = 0 when t = 0
because initial velocity = 0

(b) (i) Suppose u = velocity of ball when it leaves (ii) At the maximum height H,
the player’s hand. velocity v = 0

Acceleration a = –g Using v = u + 2as
2
2
= –10 m s –2 0 = 15 – 2 × 10 × H
2
Time t = 3.0 s H = 11.25 m
Displacement s = 0
1
Using s = ut + at 2
2
1
0 = 3.0 u – × 10 × (3.0) 2
2
u = 15 m s –1




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Physics Term 1 STPM Chapter 2 Kinematics

(c) Velocity, v


+

0 t t t Time, t
1 2 3
2 –




(d) (i) Neglecting air resistance, the time taken to reach the highest point is given by
v = u + at v = 0, a = –g
0 = u – gt
u
Time taken, t =
g
With air resistance, the retardation a 1 > g
using v = u + at
0 = u – a 1 t 1
u u
Time taken, t 1 = < (a 1 > g)
a 1 g
Hence, the time taken is shorter.

(ii) Without air resistance, the greatest height reached = H
At the greatest height, v = 0
2
Using v = u + 2as
2
0 = u – 2gH a = –g
2
Greatest height, H = u 2
2g
With air resistance retardation a 1 > g.
u 2 u 2
Hence, the greatest height, H 1 = < (a 1 > g)
a 1 g
Therefore, the greatest height reached is lower.

(e) When the ball moves up, When the ball falls,
initial velocity = u, initial velocity = 0,
final velocity = 0, final velocity, v < u velocity of projection because of friction
time = t u time = t d
1 1
Using s = (u + v)t using s = (u + v)t
2 2
1 1

H 1 = (u + 0)t u
2 H 1 = (0 + v)t d
2
2H 1 2H 1
t u = t d = > t u (v < u)
u v
Hence, the time taken to come down is greater than the time taken to move up.








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Physics Term 1 STPM Chapter 2 Kinematics

Quick Check 3


1. A body initially at rest moves with uniform 3. The velocity-time (v-t) graph of a vehicle
acceleration. Which graph best represents the travelling along a straight line is shown below.
variation of displacement s of the body with
time t?
A C 2
s s








t 0 t
0 Which of the following is the displacement-
time (s-t) graph of the vehicle?
B D A s C s
s s





0 t 0 t
t 1 t 2 t 3 t 1 t 2 t 3
0 t 0 t B s D s


2. A brick is dislodged from a tall building and
falls vertically under gravity. Which of the
following curves represents the variation of 0 t t t t 0 t t t t
its height h above the ground with time t if 1 2 3 1 2 3
air resistance is negligible?
A C 4. A tennis ball is released, it falls vertically to
h h the floor and bounces back. Taking velocity
upwards as positive. Which of the following
is the velocity-time (v-t) graph of the ball?
A v C v
+ +

0 t 0 t 0 t 0 t
– –
B D
h h
B v D v
+ +
0 t 0 t
– –
0 t 0 t


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Physics Term 1 STPM Chapter 2 Kinematics

5. The figure below represents the motion of 8. A puck slides across an icy surface and travels a
a ball rebounding from the floor after being distance x in time t under uniform retardation.
released from a point above the floor. What is Which of the following when plotted to
the quantity represented on the y-axis? represent the motion of the puck would give
a straight-line graph?
y
A x against t C x t against t 2
2 B x against t D x against t 2
t
9. A car starts from rest and accelerates at
–1
–2
0 Time 2.0 m s when a lorry moving at 16 m s in the
same direction passes it. What is the distance
A Displacement C Acceleration travelled by the car when it overtakes the
B Velocity D Kinetic energy lorry?
A 16 m C 64 m
6. A falling stone strikes a soft ground at speed u B 32 m D 128 m
and decelerates uniformly until it stops. Which
one of the following graphs best represents the 10. The velocity-time graph of an object is as
variation of the stone’s speed v with distance shown in the figure.
s moved into the ground?
A v C v Velocity

u u



s 0 s
0
0 Time
B v D v t 1 t 2
Sketch
u u
(a) the displacement-time graph,
(b) the acceleration-time graph.

11. A particle moves in a straight line with
0 s 0 s
uniform retardation. At time t = 0, its speed is
u and its displacement from the origin is zero.
7. The graph below shows the variation with time Sketch labelled graphs to show how
t of the velocity v of a bouncing ball, released (a) the velocity v of the particle,
from rest. Downwards velocities are taken as (b) the displacement s of the particle from the
positive. At which time does the ball start to origin,
leave the floor on the rebounce. vary with time t.
Explain the relation between the graphs.
v
12. A ball is released from rest from a height of
1.00 m. It falls freely to the floor and rebounces
to a height of 0.80 m. Neglecting the time of
0 Time impact with the floor, sketch the velocity-time
A B C D
graph of the motion, taking upwards velocity
as positive.






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Physics Term 1 STPM Chapter 2 Kinematics

2.2 Projectile
2008/P2/Q1, 2013/P1/Q2, 2014/P1/Q16
Learning Outcomes

Students should be able to:
• solve problems on projectile motion without air resistance
• explain the effects of air resistance on the motion of bodies in air
2
1. A projectile is a body that travels under the action of gravity after being projected at an angle to the
horizontal.
2. The motion of a projectile consists of two components:
(a) a vertical component which is motion with uniform acceleration, g the acceleration due to
gravity,
(b) a horizontal component which is motion with constant velocity.
3. The path of a projectile is a curve known as parabola.

10 m s –1 30 m s –1

20 m s –1 30 m s –1
30 m s –1 –1
30 m s –1 30 m s –1 –10 m s –1 30 m s
–20 m s –1 30 m s –1
–1
40 m s –1 30 m s
–1
50 m s –30 m s –1 –1
53° a = –g 30 m s
30 m s –1 –1 –1
–40 m s 50 m s
Figure 2.6 Projectile
4. Figure 2.6 shows the motion of a tennis ball with an initial velocity of 50 m s at an angle of
–1
projection 53° to the horizontal. An analysis of the motion shows that:
(a) The horizontal component of velocity is constant at 30 m s , because there is no horizontal
–1
force on the ball.
(b) The horizontal displacements are the same for equal time intervals.
(c) The vertical component of velocity continuously changes during the motion.
(d) At the highest point of the trajectory, the vertical component of velocity is zero.
(e) The acceleration a is constant and vertically downwards, a = –g.
5. To study the motion of a projectile, we consider the horizontal component and vertical components
separately.

u sin θ u


a = –g
H
y
θ
O
u cos θ
x
R
Figure 2.7

45







02 STPM PHY T1.indd 45 4/9/18 8:19 AM

Physics Term 1 STPM Chapter 2 Kinematics

6. Figure 2.7 shows a body projected with a velocity u at an angle θ to the horizontal.
7. Consider the vertical component of motion:
(a) Initial velocity = u sin θ
Acceleration = a = –g
Suppose H = maximum height reached
At the maximum height, vertical component of velocity = 0
2
2
2 Using v = u + 2as
2
0 = (u sin θ) – 2gH
2
2
Maximum height, H = u sin θ
2g
(b) If t 1 = Time taken by the body to reach the maximum height,
using v = u + at
0 = u sin θ – gt 1

t 1 = u sin θ
g

1
(c) The instantaneous height y, at any time t is given by s = ut + at 2
2
1
y = (u sin θ) t – gt 2
2
(d) Let T = Total time of flight, the time taken by the body to travel up and fall back to the ground.
When the body lands on the ground, the vertical displacement s = 0.
1
Using s = ut + at 2
2
1
0 = (u sin θ) T – gT 2
2
2u sin θ
Time of flight, T = g = 2t 1

This means that the time taken by the object to go to its maximum height is the same as the
time it takes to move from the maximum height to the ground.

8. Consider the horizontal component of motion:
(a) Horizontal component of velocity = u cos θ = constant
Instantaneous horizontal displacement at any time t, is
x = (u cos θ) t

(b) The range R of the projectile,
R = (u cos θ) T
( g )
= (u cos θ) 2u sin θ


u sin 2θ
2
R = g






46







02 STPM PHY T1.indd 46 4/9/18 8:19 AM

Physics Term 1 STPM Chapter 2 Kinematics

u 2
(c) The maximum range is g and it occurs when sin 2θ = 1
2θ = 90°
θ = 45°
To obtain the maximum range for a particular speed of projection, the body must be projected
at an angle of 45° to the horizontal.

9. Figure 2.8 shows three possible ranges for a body projected with the same speed but at different 2
angles of projection.





Exam Tips
At the highest point
60° 1. velocity of projectile is
u cos θ horizontally and
45° not zero.
30°
Maximum range 2. kinetic energy
1
= m (u cos θ) 2
The range is maximum when the angle of projection is 45° 2
Figure 2.8




Example 6

A motorcycle stunt-rider moving horizontally takes off from a point 5.0 m above the ground with
a speed of 30 m s . How far away does the motorcycle land?
–1



5.0 m


x
Solution:

Consider the vertical component of motion:
Initial velocity = 0

Acceleration a = +g (Consider downwards direction as positive)
Displacement s = 5.0 m
1
Using s = ut + at 2
2
1
5.0 = 0 + (9.81)t 2
2
 2 × 5.0
Time of flight, t = 9.81 = 1.01 s

Consider horizontal component of motion:
Horizontal displacement x = 30 × t = 30 × 1.01 = 30.3 m


47







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Physics Term 1 STPM Chapter 2 Kinematics

Example 7


An aircraft flies at a height h with a constant horizontal velocity u so as to fly over a cannon. When
the aircraft is directly over the cannon, a shot is fired to hit the aircraft. Neglecting air resistance,
find in terms of u, h and g, the acceleration due to gravity the minimum speed of the shell in order
to hit the aircraft.
2
Solution:
v
If v = minimum speed u
θ = angle of projection
With this minimum speed, the shell hits the aircraft at the maximum h
height reached by the shell, and since the shell is fired when the air- θ
craft is above the cannon,
horizontal component of shell velocity = speed of aircraft, u
v cos θ = u .............................. ①
v sin θ
2
2
Using maximum height, h =
2g
2
2
v sin θ = 2gh .............................. ②
2
2
① + ②: v sin θ + v cos θ = 2gh + u 2
2
2
v = u + 2gh
2
Example 8
The diagram shows the path of a bullet fired 20.0 m s –1
horizontally with a velocity of 20.0 m s from a
–1
height of 2.0 m. Calculate
(a) the speed of the bullet v, 2.0 m
(b) the angle θ when the bullet hits the ground.
v
θ
Solution:
The horizontal component of velocity v x = 20.0 m s (constant)
–1
To find the vertical component of velocity when the bullet hit the ground, consider vertical
component of motion:

Initial vertical component of velocity = 0
Acceleration, a = g
Vertical displacement, s = 2.0 m
1
Using s = ut + at 2
2
1
2.0 = 0 + × 9.81 × t 2
2
t = 0.639 s





48







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Physics Term 1 STPM Chapter 2 Kinematics

Vertical component of velocity when bullet hits the ground v y = u + at
= 0 + 9.81 × 0.639
v x = 6.27 m s –1

2 2
Speed of bullet v = v x + v y
θ = 20 + 6.27
2
2
v y v
= 21.0 m s –1 2
v y
tan θ =
v x
6.27
=
20
θ = 17.4°



Quick Check 4



1. A stone is thrown from O and follows a v/m s –1 h/m
parabolic path. The highest point reached is P. A 20 15
Which of the following is zero at the highest B 20 25
point P? C 15 25
A Acceleration D 25 20
B Vertical component of velocity 4. An aeroplane flying in a straight line at
C Kinetic energy constant height of 2 000 m with a speed of
D Momentum 200 m s drops an object. The object takes
–1
2. A projectile is fired with an initial velocity a time t to reach the ground and travels a
–2
u at an angle θ to the horizontal. Neglecting horizontal distance s. Taking g as 10 m s and
air resistance, the horizontal distance x it has ignoring air resistance, which of the
travelled, and the height y reached after a following gives the values of t and s?
time t are t s
1 A 200 s 10 km
A x = ut sin θ, y = ut cos θ – gt 2
2 B 100 s 05 km
1
B x = ut sin θ – gt , y = ut cos θ C 020 s 04 km
2
2 D 010 s 02 km
1
C x = ut cos θ, y = ut sin θ + gt 2
2
1 5. A bullet is fired horizontally from the top
D x = ut cos θ, y = ut sin θ – gt 2
2 of a cliff on the surface of the earth with a
–1
speed of 40 m s . Assuming no air resistance,
3. A projectile is fired with a horizontal velocity what is the speed of the bullet 3 s later?
v from a height h as shown in the figure. (g = 10 m s )
–2
It strikes the ground at an angle of θ to the A 30 m s C 50 m s –1
–1
horizontal. Which of the following values of v B 40 m s D 70 m s –1
–1
and h will give the greatest value of the angle θ?
v 6. A projectile leaves the ground at an angle of
60° to the horizontal. Its initial kinetic energy
is K. Neglecting air resistance, find in terms of
h K its kinetic energy at the highest point of the
motion.
θ
49







02 STPM PHY T1.indd 49 4/9/18 8:19 AM

Physics Term 1 STPM Chapter 2 Kinematics

7. A ball is thrown with a velocity of 8.0 m s at (b) Calculate the maximum range, R.
–1
60° to the horizontal. (c) What is its maximum height above the
(a) Draw on the same axes, a graph to ground when its range is maximum?
represent the variation with time of (Neglect air resistance)
(i) v H , the horizontal component of 9. An aeroplane is flying at a constant horizontal
velocity, –1
(ii) v V , the vertical component of velocity. velocity of 50 m s at a height of 1 000 m.
2 Identity your graphs and show suitable What is its horizontal distance from a target
values of velocity and time. on the ground, so that a parcel released from
(b) Use your graph to find the maximum the plane will hit the target.
height reached by the ball. 10. A coin is pushed off from the smooth
horizontal surface of a table of height 2.0 m.
8.
It falls and strikes the floor at a horizontal
distance of 3.2 m from the edge of the table.
v
Calculate
(a) the time taken by the coin to fall through
θ the air,
R (b) the speed at which the coin leaves the
table,
–1
A missile is fired with a speed of 500 m s from (c) the velocity of the coin when it strikes the
the ground. floor.
(a) What is the angle of projection θ for the
missile to achieve the maximum range?



Effects of Air Resistance

1. When a body moves through the air, the air resistance against the motion of the body is known as
the viscous drag.
2. The viscous drag on a body depends on
(a) shape of the body
Objects which are streamlined experience less drag.
(b) velocity of the body
The viscous drag is proportional to the square of the velocity.
3. When a body is released from rest and falls through the air, its velocity initially increases. As its
velocity increases, the viscous drag increases. The acceleration of the body decreases. Finally when the
acceleration is zero, the velocity is constant. This maximum constant velocity is known as terminal
velocity.
4. Figure 2.9 shows how the velocity of the body varies with time. Figure 2.10 shows the variation of
acceleration with time.
Velocity Acceleration
g

Terminal
velocity

0 Time 0 Time

Figure 2.9 Figure 2.10



50







02 STPM PHY T1.indd 50 4/9/18 8:19 AM

Physics Term 1 STPM Chapter 2 Kinematics










2



Figure 2.11
Info Physics
5. To slow down his fall, a skydiver spreads himself wide Highest free-fall
to increase the viscous drag (Figure 2.11). What happens October 14, 2012, Austria’s Felix Baumgartner
to the skydiver when his parachute opens? jumped off a helium-filled ballon at an altitude of
39,045 m and free-fall for 4 minutes, He reached
–1
a speed of 1342.0 km h (1.24 times the speed
of sound)
Example 9


A ball bearing is released from rest below the surface of a viscous liquid in a tall and wide container.
Sketch a graph to show how the height h of the ball bearing from the base of the container varies
with time. Explain the shape of the graph.
Solution:
z
If z = Instantaneous displacement from the liquid surface.
H = Height of liquid
z = H – h h = height from base of container
H
h = H – z h
dh = – dz dH = 0
dt dt dt
= –v v = velocity of ball bearing

The initial velocity of the ball bearing is zero. Hence, when t = 0, gradient of graph = 0
The velocity of the ball bearing increases until the terminal velocity is achieved.

h
H





0 l


Since the gradient of h - t graph, dh = –v, the gradient becomes more negative until a constant
dt
negative value is obtained. When the terminal velocity is achieved, the h-t graph is straight.
Hence, the h-t graph is as shown.


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Physics Term 1 STPM Chapter 2 Kinematics

Quick Check 5


1. A small metal sphere is held just below the C Acceleration / m s –2
surface of a viscous liquid in a tall vessel. It
is then released and its displacement s with 10
time t is plotted. Which of the following is the
2 s-t graph?
A s C s
0 Time


D Acceleration / m s –2

t t 10
0 0
B s D s


0 Time

4. A ball is thrown vertically upwards in a viscous
t t medium. The time of flight for the upwards
0 0
motion is t u and for the downwards motion is
t d . Which of the following is correct?
2. A steel sphere is held a short distance below
the surface of a liquid in a deep vessel and A t d > t u because the ball moves faster on its
then released. The terminal velocity of the downwards motion and the viscous drag
sphere in the liquid is independent of is greater.
A the height of the liquid in the vessel B t d > t u because the magnitude of the
B the density of the liquid acceleration when the ball is moving
C the diameter of the sphere down is smaller than the magnitude of the
D the temperature of the liquid retardation during the upwards motion.
C t d < t u because the viscous drag is the
3. Which one of the following graphs represents greatest at the moment of projection.
the acceleration of a small rain drop falling D t d < t u because at a given speed the
through the air. Acceleration due to gravity net accelerating force when the ball
–2
g = 10 m s . is moving downwards is greater that the
A Acceleration / m s –2 net retarding force when it is moving
upwards.
10

5. Four spheres are released from the top floor of
a skyscraper and fall through air to the ground.
The spheres attain the terminal velocity before
0 Time reaching the ground. The sphere with the
B Acceleration / m s –2 greatest terminal velocity is
A steel sphere of radius 1 cm
10
B steel sphere of radius 2 cm
C polystyrene sphere of radius 1 cm
D polystyrene sphere of radius 2 cm

0 Time

52







02 STPM PHY T1.indd 52 4/9/18 8:19 AM

Physics Term 1 STPM Chapter 2 Kinematics

Important Formulae



1. Equations for motion with constant accelaration
1 1
v = u + at s = (u + v)t s = ut + at 2 2 2
v = u + 2as
2 2
2. Projectile
2
2
u sin θ 2
Maximum height, H =
2g
2
Range, R = u sin 2θ
g
Range is maximum when = 45°
At time = t, horizontal displacement, x = (u cos θ)t
1
vertical displacement, y = (u sin θ)t – gt 2
2


STPM PRACTICE 2


1. An object is projected vertically upwards from 3. An object starts from rest and its acceleration-
the top of a building of height 12.0 m with a time graph is as shown below.
–1
speed of 6.0 m s . What is the time taken by Acceleration / ms –2
the object to reach the ground?
[Acceleration of free fall is 10.0 m s .] 4
–2
A 1.06 s 2
B 2.00 s
0 Time / s
C 2.26 s 4 8 12 16
D 4.90 s –2

2. The displacement-time graph of a body is What is the maximum velocity of the object?
shown below. A 16 m s –1
B 22 m s –1
Displacement C 24 m s –1
D 26 m s
–1
Q
4. A vehicle decelerates uniformly from a speed
P R u. It takes a time t, and travels a distance
S d before stopping. Its speed, and distance
0 Time t
travelled after a time is given by
2
Which statement about the motion of the body Speed Distance travelled
is true? u d
A At P, the body is accelerating A 2 less than 2
B At Q, the speed of the body is zero B u more than d
C The velocities at P and R are the same 2 u d 2
D At S, the speed is constant C Less than less than
2
2
u d
D More than more than
2 2

53







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Physics Term 1 STPM Chapter 2 Kinematics

5. An object is projected vertically from the top 8. (a) An object is projected with a speed of v at
of a building and it falls to the ground below. an angle θ to the horizontal. Derive the
Neglecting air resistance, which is the graph Cartesian equation for the trajectory of
of displacement-y against time t? the object.
A y C y (b) The figure below shows a boat approaching
a cliff of height 15.0 m at a constant
speed of 12.0 m s . A bullet is fired
-1
2 horizontally with a speed u when the boat
0
t
0 t
is 800 m from the cliff. The bullet strikes
the boat.
B y D y Bullet v
Cliff
15.0 m
0 t 0 t


800 m

6. A stone is thrown vertically upwards with (i) The bullet strikes the boat how many
an intial speed of 24.0 m s from the top of a seconds after been fired?
-1
tower 60.0 m above the ground. (ii) What is the distance of the boat
(a) Determine the speed of the stone when it from the cliff when it is struck by the
strikes the ground. bullet?
(b) Calculate the time taken to reach the (iii) Find the initial speed v of the bullet.
ground from the instant it is thrown. (iv) Determine the magnitude and
direction of the velocity of the bullet
7. A stone is thrown from a cliff which is at when it strikes the boat.
a height h above the water surface with a
velocity v at an angle θ to the horizontal as 9. A ball is kicked horizontally from a cliff of
shown in the figure below. After a time t, the height H with a velocity of 8.0 m s . The
–1
stone is at P(x,y). ball hits the water at an angle of 60° to the
y horizontal as shown in the diagram.
v
P(x, y) 8.0 m s –1
Clif
θ Cliff f
x
Cliff H
h 60º

R (a) Neglecting air resistance, calculate
(i) speed of the ball when it hits the
(a) Write the expressions for the x- and water,
y-coordinates of P. (ii) H, the height of the cliff.
(b) The height h = 10.0 m, v = 12.0 m s and (b) If air resistance is not negligible, discuss
–1
θ = 60 . Calculate the horizontal distance the change in the angle that the ball hits
o
R. the water.





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02 STPM PHY T1.indd 54 4/9/18 8:19 AM

Physics Term 1 STPM Chapter 2 Kinematics

10. (a) An object in linear motion has an initial 13. (a) Define acceleration.
velocity, u. The object has an uniform Explain how it is possible for a body to
acceleration of a. When the displacement undergo acceleration when its speed
is s, the velocity is v. Define uniform remains constant.
acceleration and deduce the equation (b) A ball falls 2.00 m from rest and a further
2
2
v = u + 2as distance of 0.10 m as shown in the figure.
–1
(b) A car is travelling at 54 km h , and when 2
the car is 25 m from the stop line, the
traffic light turns red. The reaction time
of the driver is 0.60 s. 2.00 m
(i) Find the minimum deceleration of
the car for the car to stop behind the
stop line.
(ii) Sketch a graph to show how the
displacement of the car from the stop 0.10 m
line varies with time.
(c) A ball is kicked from the ground and
it follows the path shown in the figure Neglect air resistance, calculate
below. The greatest height reached is (i) the speed of the ball after falling 2.00 m,
2.0 m and the range in 22.0 m. (ii) the time to fall a distance of 0.10 m.

2.0 m
14. (a) Define velocity.
22.0 m
Explain how the displacement and the
Determine acceleration of an object may be deduced
(i) the velocity of projection of the ball, from a velocity-time graph.
(ii) the time of flight. (b) A trolley is placed at the top of a slope
as shown in the figure. A block is fixed
11. A car starts from rest and travels along a rigidly to the lower end of the slope.
straight road. The acceleration-time graph of
the car is as shown below.
Acceleration / m s –2

4

2

0 Time / s At time t = 0, the trolley is released from
2 4 6 8 10
the top of the incline and its velocity v
–2 varies with time t as shown below.
v
(a) What is the maximum speed of the car? A
(b) What is the speed after 10 s?
12. A missile is to be launched from a point P,
500 m from the peak of a hill of height
2000 m into enemy territory. If the missile C
is to just pass over the hill and strike a target O
1200 m from P, calculate
(a) the velocity at which the missile should
be launched, B
(b) the time of flight of the missile.
55







02 STPM PHY T1.indd 55 4/9/18 8:19 AM

Physics Term 1 STPM Chapter 2 Kinematics

(i) Describe qualitatively the motion of The total stopping distance is the sum of the
the trolley during the periods OA, AB distance travelled by the vehicle during the
and BC. reaction time of the driver and the distance
(ii) Calculate the acceleration of the travelled when the brakes are applied.
x
trolley down the incline. Draw a graph of against v. Use the graph to
(iii) Find the length of the incline. find v
(iv) Calculate the distance the trolley (a) the total stopping distance if the velocity
2 moved up the incline on the of the vehicle is 35 m s .
–1
rebounce. (b) the reaction time of the driver.
15. The table below shows the total stopping
distance x for different velocities v of a
vehicle.

v/m s –1 10.0 15.0 20.0 25.0 30.0

x/m 20.0 37.5 60.0 87.5 120.0














ANSWERS



1 2. 22 m s –1
2h
1. (a) Use v = u + at 3. g = 2 2
t 2 – t 1
(i) 2.0 m s –1
(ii) 4.0 m s
–1
(iii) 10.0 m s –1 3
1 1. C 2. B 3. C 4. C 5. A
(b) Use average speed = (u + v)
2 6. C 7. C 8. B 9. C
(i) 1.0 m s –1
(ii) 2.0 m s –1 10. (a) Displacement
(ii) 5.0 m s –1
(c) Use s = average speed × time
(i) 1.0 m (ii) 4.0 m

(iii) 25.0 m
2. 25 m Time
3. 48 m s –1 0 t 1 t 2
(b) Acceleration
2
1. (a) Use v = u + at, t = 0.815 s
2
(b) Use v = u – 2gH
2
0 Time
H = 3.26 m above the building. t 1 t 2
= 12.26 m from the ground
1
(c) Use s = uT – gT , s = –9.0 m
2
2
T = 2.40 s
56





02 STPM PHY T1.indd 56 4/9/18 8:19 AM

Physics Term 1 STPM Chapter 2 Kinematics

11. (a) v (b) s 2. B : Velocity = gradient of graph
u At Q, gradient = 0.
3. C : Increase in velocity = area between graph and
time-axis
1
= (12)(4) m s = 24 m s –1
–1
0 t 0 t 2
4. B : Shaded area under graph > total area under graph
ds Speed 2
v = = gradient of (s-t) graph
dt Area = distance
or s =v dt = area under (v-t) graph. u travelled in time t/2
12. v u/2
0 t/2 t Time

0 t 1 2
5. C : y = ut – gt
2
Gradient of y-t graph decrease until zero at the
4 highest point.
2
2
6. (a) v = u + 2as
1. B 2. D 3. C 4. C 5. C = 24.0 + 2(–9.81)(–60.0)
2
1 v = 41.9 m s –1
6. K
4 1
(b) s = (u + v)t
7. (a) v /m s –1 2
1
–60.0 m = (24.0 + (–41.9)t
6.9 2
t = 6.70 s
4.0 v H
Alternative:
1
0 t /s s = ut + at 2
1.41 2
1
–60.0 m = (24.0)t + (–9.81)t 2
2
Solving t = 6.70 s
–6.9 v V
7. (a) x = (v cos θ)t

1
(b) Maximum height = 2.43 m s = ut + at 2
2
8. (a) 45° 1 2

4
(b) 2.55 × 10 m = (v sin θ)t – gt
2
(c) 6.37 × 10 m (b) Vertical motion,
3
1
9. 714 m –10.0 = (12.0 sin 60 )t – (9.81)t 2
o
1 2
10. (a) 0.639 s (Use s = ut + gt for vertical motion) 2
2
2 9.81t – 20.8t – 20 = 0
(b) 5.01 m s (Use 3.2 m = vt) 20.8 ± 20.8 – 4(9.81)(–20)
–1
2
–1
(c) v x = 5.01 m s , v y = 0 + (9.81)(0.639) t = 2(9.81)
= 6.27 m s –1 = 2.84 s
–1
2
2
V = v x + v y = 8.03 m s Horizontal motion,
R = (12.0 cos 60 )(2.84) m
o
v y
At an angle to the horizontal, tan ( ) = 51.4° = 17.0 m
–1
v x
8. (a)
5
v sin θ v
1. D 2. A 3. C 4. B 5. B P(x, y)
y
STPM Practice 2
θ v cos θ
1 x
1. C : s = ut + at 2 x
2
1
–12.0 m = (6.0)t – (10.0)t 2
2
t = 2.26 s
57
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Physics Term 1 STPM Chapter 2 Kinematics

x 10. (a) Refer to page 34
x = (v cos θ)t, t = v cos θ 3

–1
–1
–1
1 (b) (i) 54 km h = 54 × 10 m s = 15 m s
y = (v sin θ)t – gt 2 Distance travelled under deceleration
60 × 60

2
( x ) 1 ( x ) 2 = (25 – 15 × 0.60) m = 16 m
= (v sin θ) v cos θ – g v cos θ
2
2
v = u + 2as
2
gx 2
= x tan θ – 2 2 2

2 2v cos θ a = 0 – 15 m s = –7.0 m s -2
-2
2(16)

(b) (i) Vertical motion
(ii) s (m)
1
s = ut + at 2
2 25
1
15.0 = 0 + (9.81)t 2 16
2
2(15.0)
t = s = 1.75 s
9.81 0 t
(ii) Distance of boat from the cliff u sin θ
2
2
= 800m – (12.0)(1.75) m (c) (i) H = 2.0 m = 2g
= 779 m u sin2θ
2
R = 22.0 m =
(iii) v(1.75) = 779 m g
Solving, θ = 10.3 and u = 35 m s –1
o
v = 445 m s –1
22.0
(ii) T = = 0.64 s
(iv) V x = 445 m s –1 ucos θ
1
V y = 0 + (9.81)(1.75) m s –1 11. (a) v max = (8)(4) m s = 16 m s –1
–1
= 17.2 m s –1 2
1
(b) Speed = 16 – (2)(2) m s = 14 m s –1
–1
2
V = 445 + 17.2 m s –1 2
2
–1.
= 445.3 m s . 12. (a) 140 m s at 19° to the horizontal
–1
V y (b) 9.1 s
At an angle tan ) to the horizontal = 2.2°
-1
( V x 13. (a) Acceleration is the rate of change of velocity.
Eg. A body accelerates from rest.
9. (a) (i) v x = 8.0 m s –1
(b) (i) 6.26 m s
–1
v = 8.0 ms –1
x (ii) 0.0157 s
60º
14. (a) Refer to page 33
v v y
(b) (i) OA – constant acceleration
AB – constant deceleration

v x BC – constant acceleration
Cos 60° = v (ii) 1.67 m s –2
(iii) 1.20 m
8.0
–1
v = m s = 16.0 m s –1 (iv) 0.65 m
cos 60°
(ii) v y = v x tan 60° = 13.9 m s –1 15. (a) 157.5 m
Vertical motion: (b) 1.0 s
2
2
v y = 0 + 2(9.81)H
13.9 2
H = m = 9.85 m
2(9.81)
(b) If there is air resistance, path of the ball is
shown in the diagram.
New path
Cliff
θ

The angle θ  60°
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