Module_SP015

TOPIC 1
PHYSICAL QUANTITIES
AND MEASUREMENTS

1.1 Dimensions of Physical Quantities
1.2 Scalars and Vectors
1.3 Significant figures and uncertainties analysis

(Laboratory works)

MODE Face to face Non Face to face
SLT SLT
Lecture 0.5
0.5
Tutorial 3
3

1

Learning outcomes

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

1.1 Dimensions of physical quantities b) Resolve vector into two perpendicular
a) Define dimension.
components (x and y axes)
(Lecture : C2,PLO1, MQF LOD1)
(Tutorial : C4, PLO4, CTPS3, MQF LOD6)
b) Determine the dimensions of
derived quantities. c) Illustrate unit vectors (iˆ, ˆj, kˆ) in Cartesian
coordinate
(Tutorial : C4, PLO4, CTPS3, MQF LOD6)
(Tutorial : C4, PLO4, CTPS3, MQF LOD6)
c) Verify the homogeneity of
equations using dimensional d) State the physical meaning of dot (scalar)
analysis product:
A  B  AB cos
(Tutorial : C4, PLO4, CTPS3, MQF LOD6) (Lecture : C2,PLO1, MQF LOD1)

1.2 Scalars and vectors e) (Svteactteotrh) eprpohdyuscict:alAmeBaninAgBosifncronˆss
a) Define scalar and vector
quantities. Note : Direction of cross product is
determined by corkscrew method
(Lecture : C2,PLO1, MQF LOD1) or right hand rule.

(Lecture : C2,PLO1, MQF LOD1)

1.1 Dimensions of Table 1 shows the dimension of basic quantities.
Physical
Quantities Basic Quantity SI unit / symbol Dimension Symbol
of
Dimension is defined as a Length meter / m [l]
method used to describe a dimension
physical quantity in terms of
its basic quantity regardless of L
the system of units used.
Mass kilogram / kg [m] M
example:
100 cm, 1m, 2 mi and 3 light years Time second / s [t] T
all have the dimension of length
but are expressed in different Electric current ampere / A [I] A
units.
Temperature kelvin / K [T] θ
Can be written as
Amount of mole / mol [n] N
[physical quantity or its symbol] substance, n J
candela / cd []
Luminous
intensity Table 1

Physical Quantities Equation Dimension SI unit Use
Area L × L = m2 dimension
Density = ×
= kg m‒3 analysis to
Momentum determine
Velocity = MLT kg m s‒1 the SI unit
= = LT m s‒1
Acceleration of a physical
= LT m s‒2 quantity
Force =
Pressure MLT kg m s‒2 (N)
=ML T kg m‒1 s‒2 (Pa)
Work =
= M L T (L) kg m2 s‒2 (J)
=ML T
kg m2 s‒3 (W)
= =ML T
= kg m2
ML kg m2 s‒2
Power M L T (L)
= =ML T s‒1
Moment of inertia =
Torque = Force × distance 1
T=T
Frequency 1
= period

o Pure numerical factors/number like ½ , ¾,  have no dimensions.
o Quantities can be added or subtracted only if they have the same dimensions.

o Dimension analysis can be used to derive or check equations by treating

dimensions as algebraic quantities.

o Dimensional homogeneity : the dimensions in each equation on both sides

equal.

Dimension of LHS = Dimension of RHS

Let consider equation, =   where is velocity, is acceleration and is height.

Dimension of LHS = [ ] = L T‒1 = = L T‒1 Try Tutorial 1 Q1-10
Dimension of RHS
[ ]

[  ] =   × =  

Dimension of LHS = Dimension of RHS ∴ equation =   is dimensionally homogenous.

1.2 Scalars and vectors  Vectors are typically
illustrated by drawing an
Physical ARROW above the symbol
Quantities e.g ⃗ , ⃗ , ⃗ , ⃗

 Vector can be represented graphically
using arrow.

Scalar quantities Vector quantities

o Is defined as quantity o Is defined as quantity  When doing any mathematical
with magnitude only. with both magnitude operation on a vector quantity (like
and direction. adding, subtracting, multiplying ..)
o Example: we have to consider both the
mass, distance, o Example: magnitude and the direction.
speed, work, displacement, velocity,
pressure, current, force, momentum,
temperature. impulse, torque, electric
field, magnetic field.

Unit vectors = =

A unit vector is a vector that has a The hypotenuse in The LEGS of the right triangle
magnitude of 1 with no units. Physics is called the are called the COMPONENTS
RESULTANT or
It only purpose is to point − that is, to VECTOR SUM. Vertical
describe a direction in space. component
( − comp)
& is used to represent unit

vectors pointing in the positive ,
y
& directions.

iˆ  ˆj  kˆ  1 ˆj x NOTE : When
kˆ iˆ drawing a right
Horizontal component ( − comp) triangle, you must
of vector ⃗ draw components
HEAD TO TAIL.
z

Resolving a vector into 2 perpendicular components

A vector may be expressed in terms For the comp which is
of its components.
with the aid of trigonometry: adjacent to angle θ, we
y+ use cos

– component of cos   adj  Ax  Ax  A cos
vector A hyp A

Vertical sin   opp  Ay  Ay  A sin 
hyp A

hyp A oApyp Magnitude of vector A : For the comp which is
opposite to angle θ,
| A | Ax2  Ay2 we use sin

θ Direction of vector A :

adj Ax Horizonxtal   tan 1 Ay * θ is measured relative
Ax to x axis.
– component of vector A

VECTOR ANALYSIS : Determine resultant vector

Flow map : Problem solving strategy (using components to add
vectors and determine Resultant vector)

❶Resolve each ❷Add comp. x and ❸Visualize the ❹ Calculate the ❺Calculate the
vector(s) into its x magnitude of direction of
y separately total component resultant vector :
D&enyocteosmpone+nt resultant vector :
for x and y, | R | Rx 2  Ry 2
STRICTLY DO  tan 1 Ry
resultant vector Rx
= 55 N with + − + NOT add 
or − components to and the
⃗ = 60 N 55 30 components direction. θ is measured
signs. − relative to x axis.

−60 45+60 45❶ ⃗ Comp Comp ❸ ❹ ❺
55 30
−60 45 60 45 =   + =
= 90.06−42.43 N42.43 N
❷ =   (−14.93) +(90.06) 90.06
55 30 55 30 = −14.93 = −14.93
27.5 47.63 N = . N
= . °
− . N . N
above −

Multiplication of vectors

DOT (scalar) Cross (vector)
product product

• = scalar =
quantity

Result in a Result in a
scalar vector


Dot (scalar) Product ( A  B)

o Is defined as : • = o The angle  ranges from 0 to 180 .
 0
 ⃗ θ  90 A  B  AB cos 90  0
A
scalar product is zero

  1
B θ  0 A  B  AB cos 0  AB

scalar product is maximum
⃗

where : magnitude of vector o Example of dot product is work done
by a constant force :
: magnitude of vector
: angle between the two = •= )=
vectors

o Dot product results a scalar quantity scalar ⃗ vectors
and AB  B A

⃗

Cross (vector) Product (   )
A B

o Cross product A × B is defined as o Cross product result in a vector
a vector that is perpendicular quantity.
(orthogonal) to both A and B, with
a direction given by the right- o its magnitude is given by
hand rule.
: angle between the two vectors
=
o The angle  ranges from 0 to 180 
⃗ × results in a so the cross product always positive
new vector which
are perpendicular value.  1
┴ to vectors ⃗ and
⃗ θ  90 A B  AB sin 90  AB

cross product is maximum
0

θ 0 A B  AB sin 0  0


cross product is zero
⃗

Cross (vector) Product (   )
A B

o The direction of cross product A  B Example of cross (vector) product :

is determined by (1) Magnetic force on a moving charge in

magnetic field where the expression is

given by

   q v  
B
F

vector vectors

(2) Torque (turning effect) produced by a

force where the expression is given by

CORKSCREW OR RIGHT-HAND   r  
METHOD RULE F

vector vectors

o How to use right hand rule :

1) Point the 4 fingers to the direction of the 1st vector (a).

2) Swept the 4 fingers from the 1st vector towards the 2nd vector (b).

c3) The thumb shows the direction of the vector product ( ) .

considering  and  : 
two vectors A B A

 
B

Physical meaning of • Physical meaning of

 
A •= A | |=

  
B B

   
 B is the magnitude of B | AB | is equals the magnitude of B
A component of A PARALLEL multiplied by by the component of A
to B . multiplied
the PERPENDICULAR to B .

Subtopic : MODE Face Non 1
2.1 Linear motion to face Face
2.2 Uniformly accelerated motion Lecture to face
2.3 Projectile motions Tutorial SLT SLT

1.5 1.5

7 7

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

2.1 Linear motion 2.2 Uniformly accelerated motion

a) Define a) Apply equations of motion with uniform
acucelearattions.
i. instantaneous velocity, average velocity v  ut  1 at 2 v2  u2  2as
and uniform velocity. 2
(Tutorial : C4, PLO4, CTPS3, MQF LOD6)
ii. instantaneous acceleration, average
acceleration and uniform acceleration. 2.3 Projectile Motion

b) Discuss the physical meaning of a) Describe projectile motions launched
displacement-time, velocity-time and at an angle , θ as well as special case
acceleration-time graphs. when θ = 0º and θ = 90º (free fall)

(Lecture : C2,PLO1, MQF LOD1) (Lecture : C2,PLO1, MQF LOD1)

c) Determine the distance travelled, b) Solve problems related to projectile
displacement, velocity and acceleration motion.
from appropriate graphs.

(Tutorial : C4, PLO4, CTPS3, MQF LOD6)

(Tutorial : C4, PLO4, CTPS3, MQF LOD6)

Kinematics

Description of the motion of objects without consideration of what
causes the motion (mass or force).

1 dimension (1D) 2 dimension (2D) Projectile
(linear/straight line) motion

Horizontal Vertical uniform Launched Launched at
uniform accelerated horizontally angle
motion
accelerated e.g Free Fall
motion

2.1 Linear motion

Distance, Displacement,

o Is defined as o Is defined as
total path shortest
length distance
traversed in
moving from (straight line)
one location to between initial
another.
and final point.
o scalar o Vector quantity.
quantity. o can be positive,

o always negative @
positive. zero

SI unit : meter (m) ; basic quantity

Speed, velocity,

o Is defined as o Speed in a
distance particular
traveled per direction
unit time
interval. o Is defined as
time rate of
change of
displacement

o scalar o Vector quantity.
quantity.

SI unit : m s‒1 ; derived quantity

Instantaneous velocity Average velocity Uniform velocity
The rate of change of
velocity at a specified displacement over a Velocity always constant
position or instant of time finite interval of time
along the path of motion. ( for object moves with uniform
= velocity, its instantaneous

velocity equals to the average
velocity at any time)

1 : initial point ; 2 : final point

9.4 m If the body speeds
up or slows down
11.7 m during the
displacement, then
4.3 s 3.6 s the average velocity
is not the same as
the velocity at a
given instant of time
during the motion

Average velocity Instantaneous velocity
defined by the change between two points Velocity at a point - the slope of the tangent line

= . .
= . = . = = . = .

Acceleration, Acceleration Average acceleration,
(increasing speed) and
o Is defined as the rate of deceleration o Is defined as change in velocity
change of velocity (decreasing speed) divided by the time taken to make
should not be the change.
o vector quantity confused with the
o S.I. unit is m s−2. directions of velocity o Direction of :
and acceleration
 Same direction as direction of
o Velocity is vector quantity,
 a change in velocity motion if an object accelerates
may thus involve either (increase in velocity)
or both magnitude &  Opposite direction to direction of
direction. motion if the object decelerates
(decrease in velocity)
o An acceleration may due to
change in:
1) speed (magnitude),
2) direction or
3) both speed and direction.

Instantaneous acceleration, Uniform acceleration

o Is defined as the acceleration at a o An object moves in a uniform acceleration
particular instant of time or position. when

The gradient of magnitude of velocity changes at a
constant rate and along fixed direction.
v the tangent line at o For object moves with uniform
acceleration its instantaneous acceleration
point Q = the equals to the average acceleration at any
instantaneous time.
acceleration at
time, t = t1 t

v1 Q

0 t1

Graphical representation of motion

Displacement−time Displacement, Uniform Stationary or stop Uniform
(fast) ( = 0) returning to
Gradient = velocity, starting point
(Slope)

o Horizontal line − zero Uniform (slow) moving time,
velocity /at rest away from starting point
/stationary/stop increasing
Decreasing returning to start
o Straight line slope −
uniform (constant) Displacement, Increasing Displacement,
velocity
time,
o Curve line − changing time, decreasing returning to start
velocity

 Steeper slope  fast
 Genter slope  slow

Graphical representation of motion

velocity−time ( ) Velocity, (m s−1) Forward direction (+ ) opposite direction (− )
Gradient = acceleration,
Uniform ( = 0)
(Slope)

Area under graph = Accelerates quickly Slow down - rapid deceleration
distance, OR Accelerates slowly
displacement, Turn back
(change
stopped direction)

Time, (s)

Speed up

Speeding up, (accelerate) Slow down
accelerating in opposite (decelerate)
direction

Distance travelled, = 50 + 8 = 58 m Positive slope implies + in opposite
Displacement, = 50 − 8 = 42 m Negative slope implies − Uniform in direction
Horizontal line (zero slope) implies = 0 opposite direction

Steeper slope  greater

Graphical representation of motion

acceleration−time (
Area under graph = change of velocity,

Positive Positive
acceleration acceleration

acceleration, a (m s−2) Zero Zero
acceleration acceleration

Negative acceleration Time, (s)

Example (TLO 2.1(c)) (ii) Determine the instantaneous velocity and
instantaneous acceleration at time 8 s.
The velocity-time graph of a car which starts from
rest and travels along a highway is shown in figure = m s‒1 Read from − graph at
below. time 8 s

9 = gradient of the − graph

(i) Determine the displacement of the car within 0−4
10 s. = 8 − 6

= area under the − graph = − m s‒2
111
(iii) Explain qualitatively the motion of the car.
= 2 6 4 + 2 2 4 − 2 (2)(2)
= m 0 ‒ 6 s: car moves forwards with
increasing velocity (accelerates).

6 ‒ 8 s: velocity decreases with time
(decelerates).

At 8 s: car turns back (change direction).
8 ‒ 9 s: car accelerates in opposite

direction.
9 ‒ 10 s: car decelerates and finally stop.

2.2 Uniformly accelerated motion

For uniform (constant) accelerated 4 kinematics equations
motion  velocity changes at a
uniform rate.

Consider the motion for an object
under uniform accelerated motion:

① − + ②
(initial) (final)
⃗

= 0 ⃗ = are vector quantities, must taking into
After time, account direction of these quantities when do
substitution into the kinematics equations.
Let
= initial velocity Implicit data in problem(s) :
= final velocity
= acceleration Initially at rest  =
= displacement Decelerate or brake  − (* has negative sign)
= time taken Finally stop  =
uniform velocity  =

Derivation of the 4 kinematics equations

velocity Displacement after time, s = shaded area under the graph
v
= the area of trapezium

s  1 u  v t (2)
2

u   By substituting eq. (1) into eq. (2): 1
0 t time s 2 u u  at t

s  ut  1 at 2 (3)
2
Gradient of − graph = acceleration,

a  v  u s  1 (v  u )( v u )
t 2 a

v  u  at (1) Substituting
into eq. (2)
v u v2  u2  2as
t  a (4)

Example (TLO 2.2(a)) (ii) How hong does it takes for the car to stop?

A car is moving with velocity of 90 km h‒1 along a Find: =? How long refer to time taken
straight road. Upon seeing the trafic light turns red, = +
the driver applies the brakes and slows with an 0 = 25 + (−16.5)
acceleration of 16.5 m s‒2. = . s
(i) How far does the car moves before it stops?

Given : {List out all the data given and convert

into SI unit (if any)}

= 90 km h‒1 = × = 25 m s

×

= −16.5 m s‒2 Negative because
applies brake and slows

= 0 Car
stopped

Find: =? How far refer to displacement

= + 2
0 = 25 + 2(−16.5)

= . m

2.3 Projectile motions

Definition of a projectile

Projectile motion is a form of motion

experienced by an object or particle (a
projectile) that is thrown near the Earth's

surface and moves along a curved
path under the action of gravity
only (in particular, the effects of air

resistance are assumed to be negligible).

The path of the motion is a parabolic arc.

o 3 cases in projectile motions:

General case Special case

Launch vertically or

Launch at angle θ Launch horizontally Dropped

θ = 0º θ = 90º

(1D projectile/Free fall)

u u u u
 u   90
  90
u   90
 u Note :
θ is measured
u from horizontal
axis


17

o As the object moves upwards or
downward it also moving horizontally.

o There are two components of the
projectile’s motion:

Horizontal motion Vertical motion
(x direction) ( direction)

Forces Yes
(present? – Yes or No) (if No Gravity
present, what direction?)
Acceleration (downward)
(present? – Yes or No) (if
present, what direction?) Yes
Velocity No ( is downward at
(constant or changing?)
9.81 m s−2)

constant Changing
(by 9.81 m s−1
each second)

o Horizontal and vertical components of
motion (velocity , displacement ,

acceleration ) MUST be discussed

separately.

uy = u sin θ u ax = 0 m s–2

Resolve into Vertical
& components Displacement,
Sy @ Height
ux = u cos θ ay = – g At any
Horizontal Displacement,Sx @ Range vx position, the
velocity is
θ always
composed
of and
components

Most important concept in projectile motion is that horizontal and vertical vy V
motions are independent, meaning that they don’t influence one another.
19
We can analyze them separately, along perpendicular axes. To do this, we separate
projectile motion into the two components of its motion, one along the horizontal
axis ( ) and the other along the vertical ( ).

General case: launch at angle θ

①Notice that y B v = vx = − ②At every location
is constant. during vertical motion,
The projectile vy v H Qvx the acceleration is
never speeds ALWAYS, ALWAYS
up or slows P sy vy v −9.81 m s−2 (near Earth)
down in the x no matter whether the
direction! vx C (final position) instantaneous velocity
uy u is positive (upward),
t  vx negative (downward),
(Initial position) A  ux t sx x or zero (at the highest
R point).
T vy v
③Due to gravity which
Terms used to u = initial velocity , θ = launch angle acts downwards,.
describe the ux = initial component-x (horizontal) velocity changes . It slows
motion uy = initial component-y (vertical) velocity down as it rises, then
v = velocity at time t speeds up as it falls.
vX = velocity component-x (horizontal) at time t
vy = velocity component-y (vertical) at time t 20
sx = Horizontal displacement ; sy = vertical displacement
R = range ; H = maximum height ; T = time of flight

o Maximum height, is a characteristic of the vertical part of motion.
o When a projectile reaches maximum height, the vertical component of its velocity is

momentarily zero ( m s−1). However, the horizontal component of its
velocity is not zero.

y v = vx = −

vy v

uy u  vx

vx H sy vy v

 ux t sx R t  vx x

vy v

o Range, is the horizontal distance traveled between launching and landing, 21
assuming the projectile returns to the same vertical level at which it was fired.

o Range depends on the angle at which the projectile is fired above the
horizontal. The maximum range results when

o Because projectiles Equations of projectile motion
move differently in
the and TREAT AND MOTION SEPARATELY
directions, there are
two separate sets of Quantity x component (Horizontal) y component
equations for (Vertical)
modeling projectile Acceleration (a)
motion: one set for ax  0 ay  g
the axis one set Initial velocity (u)
for the axis ux  u cos θ uy  u sin θ
Displacement (s)
o x & y don’t talk to sx  uxt  1 0 sx  uxt sy  uyt  1 gt 2
each other. The subscript “ ” or 2 2
“ ” tells you that the axt 2 
o Only variable that go quantity relates to
into both is time, t. motion in the or 0 vy  uy  gt
direction.
o Always include +/– vx  ux  axt  vx  ux v 2  u 2  2 gs y
sign to indicate the Velocity at any y y
direction for , , , time (v)
Magnitude :  v  vx 2  vy 2
+
Direction : θ  tan 1  vy  * is measured from
+ vx horizontal axis

Special case : launch at angle θ = 0º

uu sx

h sy vx
vy v

Ax B

 Horizontal component along path AB.

velocity,ux  u  vx  constant

 Vertical component along path AB.

initial velocity,uy  0
displacement, sy  h

23

Special case : launch at angle θ = 90º Implicit data in
projectile motion
At max height : vy=0  Horizontal component does not exist. problem(s) solving:

 Vertical component for the motion is under Drop  =
the sole influence of gravity.
Thrown vertically
sy vy Quantity y component (vertical) downwards  −
u
Acceleration (a) ay  g Thrown vertically
θ=90º Initial velocity (u) uy  u upwards  +
Launch horizontally
vy Displacement (s) sy  uyt  1 gt2  = ; =
2
vy Velocity at any Reach maximum
time (v) vy  uy  gt height  =

v 2  u 2  2 gs y Misconception :
y y hit the ground 

=0

 This motion also known as free fall. Hit ground object 24
has velocity,

Example 1 (Launch at y component (Vertical)

A student drops a ball from the top of a tall s
building, it takes 2.8 s for the ball to reach the m s−2
ground. m s−1 (Dropped)
(a) What was the ball’s speed just before hitting

the ground ?
(b) What is the height of the building ?

+ (a)

+ m s−1

(b) (–) indicates v is downward )






m

(–) indicates displacement is downward )

25

Example 2 (Launch at x component y component (Vertical)
(Horizontal)
FIGURE 1 shows an airplane moving horizontally with a Acceleration vector for gravity points
= downward in negative direction
constant velocity of 115 m s−1 at an altitude of 1050 m. = m s−1

The plane releases a “care package” that falls to the = ?
=?
ground along a curved trajectory. Package has = − = − . m s−2
Package travelling = m s−1
plane’s horizontally in direction
horizontal at instant of release not in = − m
velocity at direction
instant of =?
release. Negative, since
upward is positive
and package falls = +
downward (below
initial point) −

= + (− . )

= . s

FIGURE 1 Tips : Always start solve the
problem from the
Ignoring air resistance, determine the time required for the components that you have
package to hit the ground. most information given.

Strategy 26
Breaking this two-dimensional motion into two independent one-
dimensional motions

Example 3 (Launch at x component y component (Vertical)
(Horizontal)
A baseball player hits a home run, and the
ball lands in the left-field seats, 7.5 m above = = − = − . m s−2
the point at which it was hit. It lands with a =? =?
velocity of 36 m s−1 at an angle of 28 below = ?
the horizontal (see Figure above). Ignoring air = = +7.5 m
resistance, find the magnitude and direction = . m s−1 = −
of the initial velocity with which the ball =− . m s−1
leaves the bat. = +
. = + = +
Strategy = . m s−1 (− . )
Breaking this two-dimensional motion into two independent = + − . .
one-dimensional motions
= . m s−1

Magnitude of initial velocity
=   + =   31.79 + 20.80 = . m s−1

Direction of initial velocity 20.80 = . ° Above x
31.79

= =

Non Face-to-face SLT

LEARN MORE THROUGH VIDEOS: SLIDE 85−110

(Scan QR codes given, WATCH the videos before attending tutorial class)

Video 1 : Video 4 :
Introduction to Problem solving
Projectile motion strategy for kinematics
problem

Video 2 : Video 5 :
Demonstrating the Solving Projectile
Components of motion_Part1
Projectile Motion

Video 3 : Video 6 :
Projectile motion Solving Projectile
motion_Part2

28

TOPIC 3
MOMENTUM & IMPULSE

Subtopics: MODE Face to Non Face
3.1 Momentum and Impulse face to face
3.2 Conservation of Linear Momentum Lecture SLT
Tutorial SLT
0.5 0.5

3 3

3.1 MOMENTUM AND IMPULSE

Momentum (

Definition

a) Linear momentum (or “momentum” for short) of an object is defined as

the product of its mass and velocity.

Equation

b) SI unit of momentum : kg m s−1 or N s.

c) Momentum is vector quantity that has the same direction as
the velocity.

d) A particle’s momentum vector can be resolve into and component.

Note:
Momentum vector
can be resolved into
perpendicular
components ( and )
and individual
components can be
then be added to find
the components of
the total momentum
just as we do with any
vector components in
Topic 1.

e) The more momentum the object has, the harder it is to stop it (in
other words more force needed to stop it), and the greater effect it

will have on another object if it is brought to rest by striking the object.

Both are hard to stop. Charging elephant has great mass, single
bullet has high velocity.

Impulse (

Definition

a) Impulse is define as the product of force, and the time interval
( ).

Equation We say that the
bat delivers an
b) SI unit of impulse is N s or kg m s-1 impulse to the
ball

c) Vector quantity whose direction is the same as the constant force

on the object

f) Impulsive force against time is
given by figure below:

d) When 2 objects – such as baseball A force acting of time-
and a bat, a hammer and a nail or the object may averaged
even two cars – collide, they vary in time force

exert large forces on one
another for a short period of
time, . Impulse, is equal to the shaded area
e) Forces of this type, which exist under the graph.
only over a very short time
during the collision or

impact are called impulsive
force.

Impulse-momentum theorem

From newton second law:
“Net force is directly proportional to the rate of change of momentum”

Since impulse is equal to , impulse-momentum theorem states that

an impulse changes a particle’s momentum.
where

Impulse = change of momentum