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2.4 Momentum

Here are the solutions to questions 1-4 from the Mastery Practice 2.4 worksheet on page 37: 1. A bullet of mass 5 g moving at 500 m/s penetrates a block of wood of mass 2 kg. What is the common velocity of the bullet and block after impact? 25 m/s 2. A rocket of mass 1000 kg ejects exhaust gases at a rate of 100 kg/s with a velocity of 500 m/s. What is the thrust on the rocket? 50,000 N 3. A ball of mass 0.5 kg moving at 4 m/s collides elastically with another ball of mass 0.2 kg initially at rest. Find

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105 views

2.4 Momentum

Here are the solutions to questions 1-4 from the Mastery Practice 2.4 worksheet on page 37: 1. A bullet of mass 5 g moving at 500 m/s penetrates a block of wood of mass 2 kg. What is the common velocity of the bullet and block after impact? 25 m/s 2. A rocket of mass 1000 kg ejects exhaust gases at a rate of 100 kg/s with a velocity of 500 m/s. What is the thrust on the rocket? 50,000 N 3. A ball of mass 0.5 kg moving at 4 m/s collides elastically with another ball of mass 0.2 kg initially at rest. Find

Uploaded by

khodijahamin
Copyright
© Attribution Non-Commercial (BY-NC)
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Download as PPT, PDF, TXT or read online on Scribd
You are on page 1/ 28

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

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