2.
4
ANALYSING
MOMENTUM
Lesson 1
INTRODUCTION
A softball keeper catching a hard
and fast moving softball ......
Ouch .
Will it hurt that much if
(a) the ball is moving slow ?
(b) the ball is a ping-pong ball ?
Hands-on activity 2.7 pg 22
Aim : To gain an idea of momentum
Discussion :
1. The amount of linear momentum of
the object depends on its mass and
velocity.
2. Momentum increases when
(i) mass increases
(ii) velocity increases
Any object which is moving in a
straight line has linear momentum .
Linear momentum is the
product of mass and velocity.
Momentum = Mass x Velocity
p = mxv
momentum is a vector quantity.
SI unit for momentum is kg m s-1 or N s
Example 2 m s-1 At rest
25 m s-1
Mass = 200 kg Mass = 23 kg Mass = 217 kg
Momentum Momentum Momentum
=200(25) = 5000 N s =23(-2)= - 46 N s =217(0) = 0 N s
2000 m s-1 120m s-1 0.5m s-1
Mass = 1500 kg Mass = 4 kg
Mass = 3000 kg Momentum
Momentum =1500(-120) Momentum
=2000(3000)=6x106 N s = -180000 N s =4(-0.5) = -2.0 N s
Motion to the right / upward Motion to the left / downward
has positive momentum has negative momentum
Conservation of Momentum
u1 u2 v1 v2
Elastic collision
Total momentum before and after
the collision are equal.
m1u1 +m2u2 = m1v1 + m2v2
Conservation of Momentum
u
v
0 v
Inelastic collision
Total momentum before and after
the collision are equal.
m1u +0 = m1v + m2v m1u = ( m1 + m2 )v
Conservation of Momentum
v1
0
0 v2
explosion
Total momentum before and after
the explosion are equal.
0 +0 = m1v1 + m2v2 m1v1 + m2v2 = 0
Principle of conservation of
momentum
The total momentum in a closed
system remains unchanged ,if
no external force acts on the
system.
Situation involving
conservation of
momentum
The gun will recoil when a bullet is fired
The hose pipe will
move backward
when water is
spurting from the
fire hose .
• Jets of hot gases,
expelled downwards at
a very high speed.
• A very large downward
momentum is
produced.
View : SC CHIA • An equal but opposite
1. canon momentum on the
2. Shuttle rocket propelling
3. Challenger the rocket upwards.
Applications in daily life
Hammering a nail Piling
Applications in daily life
Cuttlefish move
Snooker player
forward by spurting
use it instinctively
out liquid to the
with every shot
back
Applications in daily life
Boat using fan is
Boat using
more suitable in
propeller to move
swampy area
forwards
Lesson 2
Experiment 2.1 pg 23
Aim : To verify the principle of conservation
of momentum
Group 1 & 2 : Experiment A
(Elastic collision)
Group 3 & 4 : Experiment B
(Inelastic collision)
Group 5 & 6 : Experiment C
Lesson 3
Creativity Time 2.1 pg 28
Aim : To build a water rocket
Problem involving
Linear Momentum
Example pg 36
Steel ball 1 Steel ball 2
m1= 0.04 kg m2= 0.02 kg
u1= 8 m s-1 u2= - 10 m s-1
v1= - 4 m s-1 v2= ?
Total momentum = Total momentum
before collision after collision
m1u1 + m2u2 = m1v1 + m2v2
0.04(8) + 0.02(-10) = 0.04(-4) + 0.02 v2
0.12 = - 0.16 + 0.02 v2
-1
Example (Exploring pg 84)
A bullet of mass 20 g and travelling at a velocity
of 390 ms -1 penetrates a block of wood of mass
0.5 kg. What is the velocity of both the wooden
block and bullet immediately after the bullet hit
the block ?
v
390 m s -1
0.5 kg
20 g
Before collision After collision
bullet Wooden block
m1= 20g = 0.02 kg m2= 0.5 kg
u1= 390 m s-1 u2= 0
v1 = v v2= v
bullet Wooden block
m1= 20g = 0.02 kg m2= 0.5 kg
u1= 390 m s-1 u2= 0
v1= v v2= v
Total momentum = Total momentum
before collision after collision
m1u1 + m2u2 = m1v1 + m2v2
0.02(390) + 0.5(0) = 0.02 v + 0.5 v
7.8 = 0.52 v
in the original
v = 7.8 / 0.52
direction
= 15 m s-1
Example(Exploring pg 84)
An instructor fires a pistol which has a mass of
1.50 kg. If the bullet weighs 10 g and it reaches
a velocity of 300 m s-1 after shooting, what is the
recoil velocity of the pistol ?
v 300 m s-1
10 g
1.50 kg
pistol bullet
m1= 1.50 kg m2= 10g = 0.01 kg
u1= 0 u2= 0
v1= -v v2= 300 m s-1
pistol bullet
m1= 1.50 kg m2= 10 g=0.01 kg
u1= 0 u2= 0
v1= -v v2= 300 m s-1
Total momentum = Total momentum
before explosion after explosion
m1u1 + m2u2 = m1v1 + m2v2
1.50 (0) + 0.01(0) = 1.50(- v )+ 0.01 (300)
0 = - 1.50 v + 3
in the original
1.50 V = 3
direction
v = 2 m s-1
Summary
1. Momentum = mass x velocity
2. Principle of conservation of momentum states that
the total momentum in a closed system remains
unchanged, if no external force acts on the system
3. The above principle can be applied to
(a) Elastic collision
(b) Inelastic collision
(c) Explosion
Mastery practice 2.4 pg 37
Question 1, 2, 3, 4
